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Example 8.3 Let \( F = \mathbb{Q} \) and \( K = \mathbb{Q}\left( \sqrt[3]{2}\right) \) . We will determine the norm and trace of \( \sqrt[3]{2} \) . An \( F \) -basis for \( K \) is \( \{ 1,\sqrt[3]{2},\sqrt[3]{4}\} \) . We can check that \( {L}_{\sqrt[3]{2}}\left( 1\right) = \sqrt[3]{2},{L}_{\sqrt[3]{2}}\left( \sqrt[3]{2}\right) = \sqrt[3]{4} \), and \( {L}_{\sqrt[3]{2}}\left( \sqrt[3]{4}\right) = 2 \) . Therefore, the matrix representing \( {L}_{\sqrt[3]{2}} \) using this basis is | \[ \left( \begin{array}{lll} 0 & 0 & 2 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right) ,\] so \( {N}_{K/F}\left( \sqrt[3]{2}\right) = 2 \) and \( {T}_{K/F}\left( \sqrt[3]{2}\right) = 0 \) . | Yes |
Lemma 8.5 Let \( K \) be a finite extension of \( F \) with \( n = \left\lbrack {K : F}\right\rbrack \) .\n\n1. If \( a \in K \), then \( {N}_{K/F}\left( a\right) \) and \( {T}_{K/F}\left( a\right) \) lie in \( F \) .\n\n2. The trace map \( {T}_{K/F} \) is an \( F \) -linear transformation.\n\n3. If \( \alpha \in F \), then \( {T}_{K/F}\left( \alpha \right) = {n\alpha } \) .\n\n4. If \( a, b \in K \), then \( {N}_{K/F}\left( {ab}\right) = {N}_{K/F}\left( a\right) \cdot {N}_{K/F}\left( b\right) \) .\n\n5. If \( \alpha \in F \), then \( {N}_{K/F}\left( \alpha \right) = {\alpha }^{n} \) . | Proof. These properties all follow immediately from the definitions and properties of the determinant and trace functions. | No |
Proposition 8.6 Let \( K \) be an extension of \( F \) with \( \left\lbrack {K : F}\right\rbrack = n \) . If \( a \in K \) and \( p\left( x\right) = {x}^{m} + {\alpha }_{m - 1}{x}^{m - 1} + \cdots + {\alpha }_{1}x + {\alpha }_{0} \) is the minimal polynomial of a over \( F \), then \( {N}_{K/F}\left( a\right) = {\left( -1\right) }^{n}{\alpha }_{0}^{n/m} \) and \( {T}_{K/F}\left( a\right) = - \frac{n}{m}{\alpha }_{m - 1} \) . | Proof. Let \( \varphi : K \rightarrow {\operatorname{End}}_{F}\left( K\right) \) be the map \( \varphi \left( a\right) = {L}_{a} \) . It is easy to see that \( {L}_{a + b} = {L}_{a} + {L}_{b} \) and \( {L}_{ab} = {L}_{a} \circ {L}_{b} \), so \( \varphi \) is a ring homomorphism. Also, if \( \alpha \in F \) and \( a \in K \), then \( {L}_{\alpha a} = \alpha {L}_{a} \) . Thus, \( \varphi \) is also an \( F \) -vector space homomorphism. The kernel of \( \varphi \) is necessarily trivial, since \( \varphi \) is not the zero map. Since \( \varphi \) is injective, the minimal polynomials of \( a \) and \( {L}_{a} \) are equal. Let \( \chi \left( x\right) \) be the characteristic polynomial of \( {L}_{a} \), and say \( \chi \left( x\right) = {x}^{n} + {\beta }_{n - 1}{x}^{n - 1} + \cdots + {\beta }_{0} \) . By the Cayley-Hamilton theorem, Theorem 2.1 of Appendix D, the characteristic and minimal polynomials of a linear transformation have the same irreducible factors, and the minimal polynomial divides the characteristic polynomial. Since \( p \) is irreducible, by comparing degrees we see that \( \chi \left( x\right) = p{\left( x\right) }^{n/m} \) . Note that \( m \) divides \( n \) , because \( m = \left\lbrack {F\left( a\right) : F}\right\rbrack \) and\n\n\[ n = \left\lbrack {K : F}\right\rbrack = \left\lbrack {K : F\left( a\right) }\right\rbrack \cdot \left\lbrack {F\left( a\right) : F}\right\rbrack .\n\]\n\nNow, recalling the relation between the determinant and trace of a matrix and its characteristic polynomial, we see that \( {N}_{K/F}\left( a\right) = \det \left( {L}_{a}\right) = \) \( {\left( -1\right) }^{n}{\beta }_{0} \) and \( {T}_{K/F}\left( a\right) = \operatorname{Tr}\left( {L}_{a}\right) = - {\beta }_{n - 1} \) . Multiplying out \( p{\left( x\right) }^{n/m} \) shows that \( {\beta }_{0} = {\alpha }_{0}^{n/m} \) and \( {\beta }_{n - 1} = \frac{n}{m}{\alpha }_{m - 1} \), which proves the proposition. | Yes |
If \( F \) is any field and if \( K = F\left( \sqrt{d}\right) \) for some \( d \in F - {F}^{2} \), then a short calculation shows that the minimal polynomial of \( a + b\sqrt{d} \) is \( {x}^{2} - {2ax} + \left( {{a}^{2} - {b}^{2}d}\right) \). | Proposition 8.6 yields \( {N}_{K/F}\left( {a + b\sqrt{d}}\right) = {a}^{2} - {b}^{2}d \) and \( {T}_{K/F}\left( {a + b\sqrt{d}}\right) = {2a} \), as we had obtained before. | Yes |
Lemma 8.9 Let \( K \) be a finite extension of \( F \), and let \( S \) be the separable closure of \( F \) in \( K \). Then \( \left\lbrack {S : F}\right\rbrack \) is equal to the number of \( F \)-homomorphisms from \( K \) to an algebraic closure of \( F \). | Proof. Let \( M \) be an algebraic closure of \( F \). We may assume that \( K \subseteq M \). If \( S \) is the separable closure of \( F \) in \( K \), then \( S = F\left( a\right) \) for some \( a \) by the primitive element theorem. If \( r = \left\lbrack {S : F}\right\rbrack \), then there are \( r \) distinct roots of \( \min \left( {F, a}\right) \) in \( M \). Suppose that these roots are \( {a}_{1},\ldots ,{a}_{r} \). Then the map \( {\sigma }_{i} : S \rightarrow M \) defined by \( f\left( a\right) = f\left( {a}_{i}\right) \) is a well-defined \( F \)-homomorphism for each \( i \). Moreover, any \( F \)-homomorphism from \( S \) to \( M \) must be of this form since \( a \) must map to a root of \( \min \left( {F, a}\right) \). Therefore, there are \( r \) distinct \( F \)-homomorphisms from \( S \) to \( M \). The field \( K \) is purely inseparable over \( S \); hence, \( K \) is normal over \( S \). Therefore, each \( {\sigma }_{i} \) extends to an \( F \)-homomorphism from \( K \) to \( M \) by Proposition 3.28. We will be done once we show that each \( {\sigma }_{i} \) extends in a unique way to \( K \). To prove this, suppose that \( \tau \) and \( \rho \) are extensions of \( {\sigma }_{i} \) to \( K \). Then \( \tau \left( K\right) = K \) by Proposition 3.28, and so \( {\tau }^{-1}\rho \) is an automorphism of \( K \) that fixes \( S \). However, \( \operatorname{Gal}\left( {K/S}\right) = \{ \mathrm{{id}}\} \), since \( K/S \) is purely inseparable. Therefore, \( {\tau }^{-1}\rho = \mathrm{{id}} \), so \( \tau = \rho \). | Yes |
Lemma 8.10 Let \( K \) be a finite dimensional, purely inseparable extension of \( F \). If \( a \in K \), then \( {a}^{\left\lbrack K : F\right\rbrack } \in F \). More generally, if \( N \) is a finite dimensional, Galois extension of \( F \) and if \( a \in {NK} \), then \( {a}^{\left\lbrack K : F\right\rbrack } \in N \). | Proof. Let \( K \) be purely inseparable over \( F \), and let \( n = \left\lbrack {K : F}\right\rbrack \). If \( a \in K \), then \( {a}^{\left\lbrack F\left( a\right) : F\right\rbrack } \in F \) by Lemma 4.16. Since \( \left\lbrack {F\left( a\right) : F}\right\rbrack \) divides \( n = \left\lbrack {K : F}\right\rbrack \), we also have \( {a}^{n} \in F \). To prove the second statement, let \( N \) be a Galois extension of \( F \). Then \( N \cap K \) is both separable and purely inseparable over \( F \), so \( N \cap K = F \). Therefore, \( \left\lbrack {{NK} : K}\right\rbrack = \left\lbrack {N : F}\right\rbrack \) by the theorem of natural irrationalities, so \( \left\lbrack {{NK} : N}\right\rbrack = \left\lbrack {K : F}\right\rbrack \). The extension \( {NK}/N \) is purely inseparable, so by the first part of the proof, we have \( {a}^{n} \in N \) for all \( a \in {NK} \). This finishes the proof. | Yes |
Lemma 8.11 Suppose that \( F \subseteq L \subseteq K \) are fields with \( \left\lbrack {K : F}\right\rbrack < \infty \) . Then \( {\left\lbrack K : F\right\rbrack }_{i} = {\left\lbrack K : L\right\rbrack }_{i} \cdot {\left\lbrack L : F\right\rbrack }_{i}. | Proof. Let \( {S}_{1} \) be the separable closure of \( F \) in \( L \), let \( {S}_{2} \) be the separable closure of \( L \) in \( K \), and let \( S \) be the separable closure of \( F \) in \( K \) . Since any element of \( K \) that is separable over \( F \) is also separable over \( L \), we see that \( S \subseteq {S}_{2} \) . Moreover, \( {SL} \) is a subfield of \( {S}_{2} \) such that \( {S}_{2}/{SL} \) is both separable and purely inseparable, so \( {S}_{2} = {SL} \) . We claim that this means that \( \left\lbrack {L : {S}_{1}}\right\rbrack = \left\lbrack {{S}_{2} : S}\right\rbrack \) . If this is true, then\n\n\[ \n{\left\lbrack K : F\right\rbrack }_{i} = \left\lbrack {K : S}\right\rbrack \n\]\n\n\[ \n= \left\lbrack {K : {S}_{2}}\right\rbrack \cdot \left\lbrack {{S}_{2} : S}\right\rbrack \n\]\n\n\[ \n= \left\lbrack {K : {S}_{2}}\right\rbrack \cdot \left\lbrack {L : {S}_{1}}\right\rbrack \n\]\n\n\[ \n= {\left\lbrack K : L\right\rbrack }_{i} \cdot {\left\lbrack L : F\right\rbrack }_{i} \n\]\n\nproving the result. We now verify that \( \left\lbrack {L : {S}_{1}}\right\rbrack = \left\lbrack {{S}_{2} : S}\right\rbrack \) . By the primitive element theorem, \( S = {S}_{1}\left( a\right) \) for some \( a \) . Let \( f\left( x\right) = \min \left( {{S}_{1}, a}\right) \), and let \( g\left( x\right) = \min \left( {L, a}\right) \) . Then \( g \) divides \( f \) in \( L\left\lbrack x\right\rbrack \) . However, since \( L \) is purely inseparable over \( {S}_{1} \), some power of \( g \) lies in \( {S}_{1}\left\lbrack x\right\rbrack \) . Consequently, \( f \) divides a power of \( g \) in \( F\left\lbrack x\right\rbrack \) . These two divisibilities force \( f \) to be a power of \( g \) . The polynomial \( f \) has no repeated roots since \( a \) is separable over \( {S}_{1} \), so the only possibility is for \( f = g \) . Thus, \( \left\lbrack {S : {S}_{1}}\right\rbrack = \left\lbrack {L\left( a\right) : L}\right\rbrack \), and since \( L\left( a\right) = {SL} = {S}_{2} \), we see that \( \left\lbrack {S : {S}_{1}}\right\rbrack = \left\lbrack {{S}_{2} : L}\right\rbrack \) . Therefore,\n\n\[ \n\left\lbrack {{S}_{2} : S}\right\rbrack = \frac{\left\lbrack {S}_{2} : {S}_{1}\right\rbrack }{\left\lbrack S : {S}_{1}\right\rbrack } = \frac{\left\lbrack {S}_{2} : {S}_{1}\right\rbrack }{\left\lbrack {S}_{2} : L}\right\rbrack } = \left\lbrack {L : {S}_{1}}\right\rbrack .\n\]\n\nThis finishes the proof. | Yes |
Corollary 8.13 If \( K/F \) is Galois with Galois group \( G \), then for all \( a \in K \) , \[ {N}_{K/F}\left( a\right) = \mathop{\prod }\limits_{{\sigma \in G}}\sigma \left( a\right) \;\text{ and }\;{T}_{K/F}\left( a\right) = \mathop{\sum }\limits_{{\sigma \in G}}\sigma \left( a\right) . \] | Example 8.14 Let \( F \) be a field of characteristic not 2, and let \( K = F\left( \sqrt{d}\right) \) for some \( d \in F - {F}^{2} \) . Then \( \operatorname{Gal}\left( {K/F}\right) = \{ \mathrm{{id}},\sigma \} \), where \( \sigma \left( \sqrt{d}\right) = - \sqrt{d} \) . Therefore, \[ {N}_{K/F}\left( {a + b\sqrt{d}}\right) = \left( {a + b\sqrt{d}}\right) \left( {a - b\sqrt{d}}\right) = {a}^{2} - {b}^{2}d, \] \[ {T}_{K/F}\left( {a + b\sqrt{d}}\right) = \left( {a + b\sqrt{d}}\right) + \left( {a - b\sqrt{d}}\right) = {2a}. \] | No |
Example 8.14 Let \( F \) be a field of characteristic not 2, and let \( K = F\\left( \\sqrt{d}\\right) \) for some \( d \\in F - {F}^{2} \) . Then \( \\operatorname{Gal}\\left( {K/F}\\right) = \\{ \\mathrm{id},\\sigma \\} \), where \( \\sigma \\left( \\sqrt{d}\\right) = - \\sqrt{d} \) . | Therefore,\n\n\\[ \n{N}_{K/F}\\left( {a + b\\sqrt{d}}\\right) = \\left( {a + b\\sqrt{d}}\\right) \\left( {a - b\\sqrt{d}}\\right) = {a}^{2} - {b}^{2}d, \n\\]\n\n\\[ \n{T}_{K/F}\\left( {a + b\\sqrt{d}}\\right) = \\left( {a + b\\sqrt{d}}\\right) + \\left( {a - b\\sqrt{d}}\\right) = {2a}. \n\\] | No |
Example 8.15 Suppose that \( F \) is a field containing a primitive \( n \) th root of unity \( \omega \), and let \( K \) be an extension of \( F \) of degree \( n \) with \( K = F\left( \alpha \right) \) and \( {\alpha }^{n} = a \in F \) . By the isomorphism extension theorem, there is an automorphism of \( K \) with \( \sigma \left( \alpha \right) = {\omega \alpha } \) . From this, we can see that the order of \( \sigma \) is \( n \), so \( \operatorname{Gal}\left( {K/F}\right) = \langle \sigma \rangle \) . Therefore, | \[ {N}_{K/F}\left( \alpha \right) = {\alpha \sigma }\left( \alpha \right) \cdots {\sigma }^{n - 1}\left( \alpha \right) = \alpha \cdot {\omega \alpha }\cdots {\omega }^{n - 1}\alpha \] \[ = {\omega }^{n\left( {n - 1}\right) /2}{\alpha }^{n} = {\left( -1\right) }^{n}a. \] If \( n \) is odd, then \( n\left( {n - 1}\right) /2 \) is a multiple of \( n \), so \( {\omega }^{n\left( {n - 1}\right) /2} = 1 \) . If \( n \) is even, then this exponent is not a multiple of \( n \), so \( {\omega }^{n\left( {n - 1}\right) /2} \neq 1 \) . However, \( {\left( {\omega }^{n\left( {n - 1}\right) /2}\right) }^{2} = 1 \), so \( {\omega }^{n\left( {n - 1}\right) /2} = - 1 \) . This justifies the final equality \( {N}_{K/F}\left( \alpha \right) = {\left( -1\right) }^{n}a \) . As for the trace, \[ {T}_{K/F}\left( \alpha \right) = \alpha + {\omega \alpha } + \cdots + {\omega }^{n - 1}\alpha = \left( {1 + \omega + \cdots + {\omega }^{n - 1}}\right) \alpha \] \[ = 0 \] because \( \omega \) is a root of \( \left( {{x}^{n} - 1}\right) /\left( {x - 1}\right) = 1 + x + \cdots + {x}^{n - 1} \) . These norm and trace calculations could also have been obtained by using the minimal polynomial of \( \alpha \), which is \( {x}^{n} - a \) . | Yes |
Theorem 8.16 If \( F \subseteq L \subseteq K \) are fields with \( \left\lbrack {K : F}\right\rbrack < \infty \), then\n\n\[ \n{N}_{K/F} = {N}_{L/F} \circ {N}_{K/L}\;\text{ and }\;{T}_{K/F} = {T}_{L/F} \circ {T}_{K/L};\n\]\n\nthat is, \( {N}_{K/F}\left( a\right) = {N}_{L/F}\left( {{N}_{K/L}\left( a\right) }\right) \) and \( {T}_{K/F}\left( a\right) = {T}_{L/F}\left( {{T}_{K/L}\left( a\right) }\right) \) for each \( a \in K \) . | Proof. Let \( M \) be an algebraic closure of \( F \), let \( {\sigma }_{1},\ldots ,{\sigma }_{r} \) be the distinct \( F \) -homomorphisms of \( L \) to \( M \), and let \( {\tau }_{1},\ldots ,{\tau }_{s} \) be the distinct \( L \) - homomorphisms of \( K \) to \( M \) . By the isomorphism extension theorem, we can extend each \( {\sigma }_{j} \) and \( {\tau }_{k} \) to automorphisms \( M \rightarrow M \), which we will also call \( {\sigma }_{j} \) and \( {\tau }_{k} \), respectively. Each \( {\sigma }_{j}{\tau }_{k} \) is an \( F \) -homomorphism from \( K \) to \( M \) . In fact, any \( F \) -homomorphism of \( K \) to \( M \) is of this type, as we now prove. If \( \rho : K \rightarrow M \) is an \( F \) -homomorphism, then \( {\left. \rho \right| }_{L} : L \rightarrow M \) is equal to \( {\sigma }_{j} \) for some \( j \) . The map \( {\sigma }_{j}^{-1}\rho \) is then an \( F \) -homomorphism \( K \rightarrow M \) which fixes \( L \) . Thus, \( {\sigma }_{j}^{-1}\rho = {\tau }_{k} \) for some \( k \), so \( \rho = {\sigma }_{j}{\tau }_{k} \) . If \( a \in K \), then by\n\nTheorem 8.12 we have\n\n\[ \n{N}_{K/F}\left( a\right) = {\left( \mathop{\prod }\limits_{{j, k}}{\sigma }_{j}{\tau }_{k}\left( a\right) \right) }^{{\left\lbrack K : F\right\rbrack }_{i}}\;\text{ and }\;{N}_{K/L}\left( a\right) = {\left( \mathop{\prod }\limits_{k}{\tau }_{k}\left( a\right) \right) }^{{\left\lbrack K : L\right\rbrack }_{i}}.\n\]\n\nTherefore,\n\n\[ \n{N}_{L/F}\left( {{N}_{K/L}\left( a\right) }\right) = {\left( \mathop{\prod }\limits_{j}{\sigma }_{j}{\left( \mathop{\prod }\limits_{k}{\tau }_{k}\left( a\right) \right) }^{{\left\lbrack K : L\right\rbrack }_{i}}\right) }^{{\left\lbrack L : F\right\rbrack }_{i}}\n\]\n\n\[ \n= {\left( \mathop{\prod }\limits_{{j, k}}{\sigma }_{j}{\tau }_{k}\left( a\right) \right) }^{{\left\lbrack K : L\right\rbrack }_{i}{\left\lbrack L : F\right\rbrack }_{i}}.\n\]\n\nSince \( {\left\lbrack K : F\right\rbrack }_{i} = {\left\lbrack K : L\right\rbrack }_{i} \cdot {\left\lbrack L : F\right\rbrack }_{i} \) by Lemma 8.11, this proves that \( {N}_{K/F}\left( a\right) = {N}_{L/F}\left( {{N}_{K/L}\left( a\right) }\right) \) . A similar calculation shows that \( {T}_{K/F}\left( a\right) = \) \( {T}_{L/F}\left( {{T}_{K/L}\left( a\right) }\right) \) . | Yes |
Corollary 8.17 A finite extension \( K/F \) is separable if and only if \( {T}_{K/F} \) is not the zero map; that is, \( K/F \) is separable if and only if there is an \( a \in K \) with \( {T}_{K/F}\left( a\right) \neq 0 \) . | Proof. Suppose that \( K/F \) is not separable. Then \( \operatorname{char}\left( F\right) = p > 0 \) . Let \( S \) be the separable closure of \( F \) in \( K \) . Then \( S \neq K \) and \( K/S \) is a purely inseparable extension. Moreover, \( \left\lbrack {K : S}\right\rbrack = {p}^{t} \) for some \( t \geq 1 \) by Lemma 4.17. If \( a \in K \), then by Theorem 8.16 we have \( {T}_{K/F}\left( a\right) = {T}_{S/F}\left( {{T}_{K/S}\left( a\right) }\right) \) . However by Theorem 8.12, if \( {\sigma }_{1},\ldots ,{\sigma }_{r} \) are the distinct \( S \) -homomorphisms from \( K \) to an algebraic closure of \( F \), then\n\n\[ \n{T}_{K/S}\left( a\right) = {\left\lbrack K : S\right\rbrack }_{i}\left( {{\sigma }_{1}\left( a\right) + \cdots + {\sigma }_{r}\left( a\right) }\right) .\n\]\n\nBut \( {\left\lbrack K : S\right\rbrack }_{i} = \left\lbrack {K : S}\right\rbrack = {p}^{t} \), since \( K \) is purely inseparable over \( S \) . Since \( \operatorname{char}\left( F\right) = p \), this forces \( {T}_{K/S}\left( a\right) = 0 \), so \( {T}_{K/F}\left( a\right) = {T}_{S/F}\left( 0\right) = 0 \) . Thus, \( {T}_{K/F} \) is the zero map.\n\nConversely, suppose that \( K \) is separable over \( F \) . Let \( N \) be the normal closure of \( K/F \) . By Theorem 8.16, we see that if \( {T}_{N/F} \) is nonzero, then so is \( \left. {{T}_{K/F}\text{. Say }\operatorname{Gal}\left( {N/F}\right) = \left\{ {{\sigma }_{1},\ldots ,{\sigma }_{n}}\right\} \text{. If }a \in N\text{, then }{T}_{N/F}\left( a\right) = \mathop{\sum }\limits_{j}{\sigma }_{j}\left( a\right) }\right\} \) by the corollary to Theorem 8.12. By Dedekind’s lemma, \( {\sigma }_{1}\left( a\right) + \cdots + {\sigma }_{n}\left( a\right) \) is not zero for all \( a \in N \), so \( {T}_{N/F} \) is not the zero map. Therefore, \( {T}_{K/F} \) is not the zero map. | Yes |
Let \( \omega \) be a primitive fifth root of unity in \( \mathbb{C} \), let \( F = \mathbb{Q}\left( \omega \right) \) , and let \( K = F\left( \sqrt[5]{2}\right) \) . Then \( K \) is the splitting field of \( {x}^{5} - 2 \) over \( F \), so \( K \) is Galois over \( F \) . Also, \( \left\lbrack {F : \mathbb{Q}}\right\rbrack = 4 \) and \( \left\lbrack {\mathbb{Q}\left( \sqrt[5]{2}\right) : \mathbb{Q}}\right\rbrack = 5 \) . The field \( K \) is the composite of these two extensions of \( \mathbb{Q} \) . The degree \( \left\lbrack {K : \mathbb{Q}}\right\rbrack \) is divisible by 4 and 5 ; hence, it is divisible by 20 . Moreover, \( \left\lbrack {K : F}\right\rbrack \leq 5 \), so \( \left\lbrack {K : \mathbb{Q}}\right\rbrack \leq {20} \) . Therefore, \( \left\lbrack {K : \mathbb{Q}}\right\rbrack = {20} \), and so \( \left\lbrack {K : F}\right\rbrack = 5 \) . | Let \( \alpha = \sqrt[5]{2} \) . The roots of \( \min \left( {F,\alpha }\right) \) are \( \alpha ,{\omega \alpha },{\omega }^{2}\alpha ,{\omega }^{3}\alpha \), and \( {\omega }^{4}\alpha \) . By the isomorphism extension theorem, there is a \( \sigma \in \operatorname{Gal}\left( {K/F}\right) \) with \( \sigma \left( \alpha \right) = {\omega \alpha } \) . Then \( {\sigma }^{i}\left( \alpha \right) = {\omega }^{i}\alpha \) . Consequently, \( {\sigma }^{5} = \) id and \( {\sigma }^{i} \neq \) id if \( i < 5 \) . The order of \( \sigma \) is thus equal to 5. This means that \( \operatorname{Gal}\left( {K/F}\right) = \langle \sigma \rangle \), so \( K/F \) is a cyclic extension. | Yes |
Lemma 9.4 Let \( F \) be a field containing a primitive nth root of unity \( \omega \) , let \( K/F \) be a cyclic extension of degree \( n \), and let \( \sigma \) be a generator of \( \operatorname{Gal}\left( {K/F}\right) \) . Then there is an \( a \in K \) with \( \omega = \sigma \left( a\right) /a \) . | Proof. The automorphism \( \sigma \) is an \( F \) -linear transformation of \( K \) . We wish to find an \( a \in K \) with \( \sigma \left( a\right) = {\omega a} \) ; that is, we want to show that \( \omega \) is an eigenvalue for \( \sigma \) . To do this, we show that \( \omega \) is a root of the characteristic polynomial of \( \sigma \) . Now, since \( \sigma \) has order \( n \) in \( \operatorname{Gal}\left( {K/F}\right) \), we have \( {\sigma }^{n} = \mathrm{{id}} \) . Therefore, \( \sigma \) satisfies the polynomial \( {x}^{n} - 1 \) . Moreover, if there is a polynomial \( g\left( x\right) \in F\left\lbrack x\right\rbrack \) of degree \( m < n \) satisfied by \( \sigma \), then the automorphisms id, \( \sigma ,\ldots ,{\sigma }^{m - 1} \) are linearly dependent over \( F \), a contradiction to the Dedekind independence lemma. Thus, \( {x}^{n} - 1 \) is the minimal polynomial of \( \sigma \) over \( F \) . However, the characteristic polynomial of \( \sigma \) has degree \( n = \left\lbrack {K : F}\right\rbrack \) and is divisible by \( {x}^{n} - 1 \), so \( {x}^{n} - 1 \) is the characteristic polynomial of \( \sigma \) . Since \( \omega \) is a root of this polynomial, \( \omega \) is an eigenvalue for \( \sigma \) . Thus, there is an \( a \in K \) with \( \sigma \left( a\right) = {\omega a} \) . | Yes |
Theorem 9.5 Let \( F \) be a field containing a primitive nth root of unity, and let \( K/F \) be a cyclic Galois extension of degree \( n \) . Then there is an \( a \in K \) with \( K = F\left( a\right) \) and \( {a}^{n} = b \in F \) ; that is, \( K = F\left( \sqrt[n]{b}\right) \) . | Proof. By the lemma, there is an \( a \) with \( \sigma \left( a\right) = {\omega a} \) . Therefore, \( {\sigma }^{i}\left( a\right) = {\omega }^{i}a \) , so \( a \) is fixed by \( {\sigma }^{i} \) only when \( n \) divides \( i \) . Since the order of \( \sigma \) is \( n \), we see that \( a \) is fixed only by id, so \( \operatorname{Gal}\left( {K/F\left( a\right) }\right) = \langle \mathrm{{id}}\rangle \) . Thus, \( K = F\left( a\right) \) by the fundamental theorem. We see that \( \sigma \left( {a}^{n}\right) = {\left( \omega a\right) }^{n} = {a}^{n} \), so \( {a}^{n} \) is fixed by \( \sigma \) . Hence, \( b = {a}^{n} \in F \), so \( K = \left( \sqrt[n]{b}\right) \) . | Yes |
Corollary 9.7 Let \( K/F \) be a cyclic extension of degree \( n \), and suppose that \( F \) contains a primitive nth root of unity. If \( K = F\left( \sqrt[n]{a}\right) \) with \( a \in F \), then any intermediate field of \( K/F \) is of the form \( F\left( \sqrt[m]{a}\right) \) for some divisor \( m \) of \( n \). | Proof. Let \( \sigma \) be a generator for \( \operatorname{Gal}\left( {K/F}\right) \). Then any subgroup of \( \operatorname{Gal}\left( {K/F}\right) \) is of the form \( \left\langle {\sigma }^{t}\right\rangle \) for some divisor \( t \) of \( n \). By the fundamental theorem, the intermediate fields are the fixed fields of the \( {\sigma }^{t} \). If \( t \) is a divisor of \( n \), write \( n = {tm} \), and let \( \alpha = \sqrt[n]{a} \). Then \( {\sigma }^{t}\left( {\alpha }^{m}\right) = {\left( {\omega }^{t}\alpha \right) }^{m} = {\alpha }^{m} \), so \( {\alpha }^{m} \) is fixed by \( {\sigma }^{t} \). However, the order of \( {a}^{t}{F}^{*n} \) in \( {F}^{ * }/{F}^{*n} \) is \( m \), so \( F\left( \sqrt[m]{a}\right) \) has degree \( m \) over \( F \) by Proposition 9.6. By the fundamental theorem, the fixed field of \( {\sigma }^{t} \) has degree \( m \) over \( F \), which forces \( F\left( \sqrt[m]{a}\right) \) to be the fixed field of \( {\sigma }^{t} \). This shows that any intermediate field of \( K/F \) is of the form \( F\left( \sqrt[m]{a}\right) \) for some divisor \( m \) of \( n \). | Yes |
Theorem 9.8 Let \( \operatorname{char}\left( F\right) = p \), and let \( K/F \) be a cyclic Galois extension of degree \( p \) . Then \( K = F\left( \alpha \right) \) with \( {\alpha }^{p} - \alpha - a = 0 \) for some \( a \in F \) ; that is, \( K = F\left( {{\wp }^{-1}\left( a\right) }\right) \) . | Proof. Let \( \sigma \) be a generator of \( \operatorname{Gal}\left( {K/F}\right) \), and let \( T \) be the linear transformation \( T = \sigma - \mathrm{{id}} \) . The kernel of \( T \) is\n\n\[ \ker \left( T\right) = \{ b \in K : \sigma \left( b\right) = b\} \]\n\n\[ = F\text{.} \]\n\nAlso, \( {T}^{p} = {\left( \sigma - \mathrm{{id}}\right) }^{p} = {\sigma }^{p} - \mathrm{{id}} = 0 \), since the order of \( \sigma \) is \( p \) and \( \operatorname{char}\left( F\right) = p \) . Thus, \( \operatorname{im}\left( {T}^{p - 1}\right) \subseteq \ker \left( T\right) \) . Because \( \ker \left( T\right) = F \) and \( \operatorname{im}\left( {T}^{p - 1}\right) \) is an \( F \) - subspace of \( K \), we get \( \operatorname{im}\left( {T}^{p - 1}\right) = \ker \left( T\right) \) . Therefore, \( 1 = {T}^{p - 1}\left( c\right) \) for some \( c \in K \) . Let \( \alpha = {T}^{p - 2}\left( c\right) \) . Then \( T\left( \alpha \right) = 1 \), so \( \sigma \left( \alpha \right) - \alpha = 1 \) or \( \sigma \left( \alpha \right) = \alpha + 1 \) . Since \( \alpha \) is not fixed by \( \sigma \), we see that \( \alpha \notin F \), so \( F\left( \alpha \right) = K \) because \( \left\lbrack {K : F}\right\rbrack = p \) is prime. Now,\n\n\[ \sigma \left( {{\alpha }^{p} - \alpha }\right) = \sigma {\left( \alpha \right) }^{p} - \sigma \left( \alpha \right) = {\left( \alpha + 1\right) }^{p} - \left( {\alpha + 1}\right) \]\n\n\[ = {\alpha }^{p} + 1 - \alpha - 1 = {\alpha }^{p} - \alpha . \]\n\nIf \( a = {\alpha }^{p} - \alpha \), then \( \wp \left( \alpha \right) = a \in F \), so \( {\alpha }^{p} - \alpha - a = 0 \) . | Yes |
Theorem 9.9 Let \( F \) be a field of characteristic \( p \), and let \( a \in F - {\wp }^{-1}\left( F\right) \) . Then \( f\left( x\right) = {x}^{p} - x - a \) is irreducible over \( F \), and the splitting field of \( f \) over \( F \) is a cyclic Galois extension of \( F \) of degree \( p \) . | Proof. Let \( K \) be the splitting field of \( f \) over \( F \) . If \( \alpha \) is a root of \( f \), it is easy to check that \( \alpha + 1 \) is also a root of \( f \) . Hence, the \( p \) roots of \( f \) are \( \alpha ,\alpha + 1,\ldots ,\alpha + p - 1 \) . Therefore, \( K = F\left( \alpha \right) \) . The assumption on \( a \) assures us that \( \alpha \notin F \) . Assume for now that \( f \) is irreducible over \( F \) . Then \( \left\lbrack {K : F}\right\rbrack = \deg \left( f\right) = p \) . By the isomorphism extension theorem, there is a \( \sigma \in \operatorname{Gal}\left( {K/F}\right) \) with \( \sigma \left( \alpha \right) = \alpha + 1 \) . From this, it follows that the order of \( \sigma \) is \( p \), so \( \operatorname{Gal}\left( {K/F}\right) = \langle \sigma \rangle \) . This proves that \( K/F \) is a cyclic Galois extension.\n\nIt remains for us to prove that \( f\left( x\right) \) is irreducible over \( F \) . If not, then \( f \) factors over \( F \) as \( f\left( x\right) = {g}_{1}\left( x\right) \cdots {g}_{r}\left( x\right) \), with each \( {g}_{i} \) irreducible over \( F \) . If \( \beta \) is a root of \( {g}_{i} \) for some \( i \), then the paragraph above shows that \( K = F\left( \beta \right) \) , so \( \left\lbrack {K : F}\right\rbrack = \deg \left( {g}_{i}\right) \) . This forces all degrees of the \( {g}_{i} \) to be the same, so \( \deg \left( f\right) = r\deg \left( {g}_{1}\right) \) . Since \( \deg \left( f\right) \) is prime and \( f \) does not split over \( F \), we see that \( r = 1 \) ; hence, \( f \) is irreducible over \( F \) . | Yes |
Example 9.10 Let \( F = {\mathbb{F}}_{p}\left( x\right) \) be the rational function field in one variable over \( {\mathbb{F}}_{p} \). We claim that \( x \notin {\wp }^{-1}\left( F\right) \), so the extension \( F\left( {{\wp }^{-1}\left( x\right) }\right) \) is a cyclic extension of \( F \) of degree \( p \). To prove this, suppose instead that \( x \in {\wp }^{-1}\left( F\right) \), so \( x = {a}^{p} - a \) for some \( a \in F \). We can write \( a = f/g \) with \( f, g \in {\mathbb{F}}_{p}\left\lbrack x\right\rbrack \) relatively prime. Then \( x = {f}^{p}/{g}^{p} - f/g \), or \( {g}^{p}x = {f}^{p} - f{g}^{p - 1} \). Solving for \( {f}^{p} \) gives \( {f}^{p} = {g}^{p - 1}\left( {{gx} - f}\right) \), so \( g \) divides \( {f}^{p} \). This is impossible; thus, \( x \notin {\wp }^{-1}\left( F\right) \), and then \( F\left( {{\wp }^{-1}\left( F\right) }\right) \) is a cyclic extension of \( F \) of degree \( p \) as we claimed. | To prove this, suppose instead that \( x \in {\wp }^{-1}\left( F\right) \), so \( x = {a}^{p} - a \) for some \( a \in F \). We can write \( a = f/g \) with \( f, g \in {\mathbb{F}}_{p}\left\lbrack x\right\rbrack \) relatively prime. Then \( x = {f}^{p}/{g}^{p} - f/g \), or \( {g}^{p}x = {f}^{p} - f{g}^{p - 1} \). Solving for \( {f}^{p} \) gives \( {f}^{p} = {g}^{p - 1}\left( {{gx} - f}\right) \), so \( g \) divides \( {f}^{p} \). This is impossible; thus, \( x \notin {\wp }^{-1}\left( F\right) \), and then \( F\left( {{\wp }^{-1}\left( F\right) }\right) \) is a cyclic extension of \( F \) of degree \( p \) as we claimed. | Yes |
Proposition 10.1 Let \( K \) be a Galois extension of \( F \) with Galois group \( G \) , and let \( f : G \rightarrow {K}^{ * } \) be a crossed homomorphism. Then there is an \( a \in K \) with \( f\left( \tau \right) = \tau \left( a\right) /a \) for all \( \sigma \in G \) . | \( \\textbf{Proof. The Dedekind independence lemma shows that }\\mathop{\\sum }\\limits_{{\\sigma \\in G}}f\\left( \\sigma \\right) \\sigma \\left( c\\right) \\neq 0 \\) for some \( c \\in K \\), since each \( f\\left( \\sigma \\right) \\neq 0 \\) . Let \( b = \\mathop{\\sum }\\limits_{{\\sigma \\in G}}f\\left( \\sigma \\right) \\sigma \\left( c\\right) \\) . Then \( \\tau \\left( b\\right) = \\mathop{\\sum }\\limits_{{\\sigma \\in G}}\\tau \\left( {f\\left( \\sigma \\right) }\\right) \\left( {\\tau \\sigma }\\right) \\left( c\\right) , \\) so\n\n\\[ \nf\\left( \\tau \\right) \\tau \\left( b\\right) = \\mathop{\\sum }\\limits_{{\\sigma \\in G}}f\\left( \\tau \\right) \\tau \\left( {f\\left( \\sigma \\right) }\\right) \\cdot \\left( {\\tau \\sigma }\\right) \\left( c\\right) \n\\]\n\n\\[ \n= \\mathop{\\sum }\\limits_{{\\sigma \\in G}}f\\left( {\\tau \\sigma }\\right) \\cdot \\left( {\\tau \\sigma }\\right) \\left( c\\right) = b. \n\\]\n\nThus, \( f\\left( \\tau \\right) = b/\\tau \\left( b\\right) \\) . Setting \( a = {b}^{-1} \\) proves the result. | Yes |
Theorem 10.2 (Hilbert Theorem 90) Let \( K/F \) be a cyclic Galois extension, and let \( \sigma \) be a generator of \( \operatorname{Gal}\left( {K/F}\right) \). If \( u \in K \), then \( {N}_{K/F}\left( u\right) = 1 \) if and only if \( u = \sigma \left( a\right) /a \) for some \( a \in K \). | Proof. One direction is easy. If \( u = \sigma \left( a\right) /a \), then \( {N}_{K/F}\left( {\sigma \left( a\right) }\right) = {N}_{K/F}\left( a\right) \), so \( N\left( u\right) = 1 \). Conversely, if \( {N}_{K/F}\left( u\right) = 1 \), then define \( f : G \rightarrow {K}^{ * } \) by \( f\left( \mathrm{{id}}\right) = 1, f\left( \sigma \right) = u \), and \( f\left( {\sigma }^{i}\right) = {u\sigma }\left( u\right) \cdots {\sigma }^{i - 1}\left( u\right) \) for \( i < n \). To show that \( f \) is a crossed homomorphism, let \( 0 \leq i, j < n \). If \( i + j < n \), then\n\n\[ f\left( {{\sigma }^{i}{\sigma }^{j}}\right) = f\left( {\sigma }^{i + j}\right) = {u\sigma }\left( u\right) \cdots {\sigma }^{i + j - 1}\left( u\right) \]\n\n\[ = \left( {{u\sigma }\left( u\right) \cdots {\sigma }^{i - 1}\left( u\right) }\right) \cdot {\sigma }^{i}\left( {{u\sigma }\left( u\right) \cdots {\sigma }^{i - 1}\left( u\right) }\right) \]\n\n\[ = f\left( {\sigma }^{i}\right) \cdot {\sigma }^{i}\left( {f\left( {\sigma }^{j}\right) }\right) \]\n\nIf \( i + j \geq n \), then \( 0 \leq i + j - n < n \), so\n\n\[ f\left( {{\sigma }^{i}{\sigma }^{j}}\right) = f\left( {\sigma }^{i + j}\right) = f\left( {\sigma }^{i + j - n}\right) = {u\sigma }\left( u\right) \cdots {\sigma }^{i + j - n - 1}\left( u\right) .\n\]\n\nHowever,\n\n\[ f\left( {\sigma }^{i}\right) {\sigma }^{i}\left( {f\left( {\sigma }^{j}\right) }\right) = \left( {{u\sigma }\left( u\right) \cdots {\sigma }^{i - 1}\left( u\right) }\right) \cdot {\sigma }^{i}\left( {{u\sigma }\left( u\right) \cdots {\sigma }^{j - 1}\left( u\right) }\right) \]\n\n\[ = \left( {{u\sigma }\left( u\right) \cdots {\sigma }^{i + j - n - 1}\left( u\right) }\right) \cdot {\sigma }^{i + j - n}\left( {{u\sigma }\left( u\right) \cdots {\sigma }^{n - 1}\left( u\right) }\right) \]\n\n\[ = f\left( {{\sigma }^{i}{\sigma }^{j}}\right) \cdot {N}_{K/F}\left( u\right) \]\n\n\[ = f\left( {{\sigma }^{i}{\sigma }^{j}}\right) \text{.} \]\n\nTherefore, \( f \) is a crossed homomorphism. By Proposition 10.1, there is an \( a \in K \) with \( f\left( {\sigma }^{i}\right) = {\sigma }^{i}\left( a\right) /a \) for all \( i \). Thus, \( u = f\left( \sigma \right) = \sigma \left( a\right) /a \). | Yes |
Proposition 10.3 Let \( K/F \) be a Galois extension with Galois group \( G \) , and let \( g : G \rightarrow K \) be a 1-cocycle. Then there is an \( a \in K \) with \( g\left( \tau \right) = \) \( \tau \left( a\right) - a \) for all \( \tau \in G \) . | Proof. Since \( K/F \) is separable, the trace map \( {T}_{K/F} \) is not the zero map. Thus, there is a \( c \in K \) with \( {T}_{K/F}\left( c\right) \neq 0 \) . If \( \alpha = {T}_{K/F}\left( c\right) \), then \( \alpha \in {F}^{ * } \) and \( {T}_{K/F}\left( {{\alpha }^{-1}c}\right) = 1 \) . By replacing \( c \) with \( {\alpha }^{-1}c \), we may assume that \( {T}_{K/F}\left( c\right) = \) 1. Recall that \( {T}_{K/F}\left( x\right) = \mathop{\sum }\limits_{{\sigma \in G}}\sigma \left( x\right) \) for all \( x \in K \) . Let \( b = \mathop{\sum }\limits_{{\sigma \in G}}g\left( \sigma \right) \sigma \left( c\right) \) . Then \( \tau \left( b\right) = \mathop{\sum }\limits_{{\sigma \in G}}\tau \left( {g\left( \sigma \right) }\right) \left( {\tau \sigma }\right) \left( c\right) \) . Since \( g\left( {\tau \sigma }\right) = g\left( \tau \right) + \tau \left( {g\left( \sigma \right) }\right) \) ,\n\n\[ \tau \left( b\right) = \mathop{\sum }\limits_{{\sigma \in G}}\left( {g\left( {\tau \sigma }\right) - g\left( \tau \right) }\right) \left( {\tau \sigma }\right) \left( c\right) \]\n\n\[ = \mathop{\sum }\limits_{{\sigma \in G}}g\left( {\tau \sigma }\right) \left( {\tau \sigma }\right) \left( c\right) - \mathop{\sum }\limits_{{\sigma \in G}}g\left( \tau \right) \left( {\tau \sigma }\right) \left( c\right) \]\n\n\[ = b - g\left( \tau \right) \cdot \tau \left( {\mathop{\sum }\limits_{{\sigma \in G}}\sigma \left( c\right) }\right) \]\n\n\[ = b - g\left( \tau \right) \text{.} \]\n\nTherefore, \( g\left( \tau \right) = b - \tau \left( b\right) \) . Setting \( a = - b \) gives \( g\left( \tau \right) = \tau \left( a\right) - a \) for all \( \tau \in G \) . | Yes |
Theorem 10.5 (Additive Hilbert Theorem 90) Let \( K \) be a cyclic Galois extension of \( F \), and let \( \sigma \) be a generator of \( \operatorname{Gal}\left( {K/F}\right) \) . If \( u \in K \) , then \( {T}_{K/F}\left( u\right) = 0 \) if and only if \( u = \sigma \left( a\right) - a \) for some \( a \in K \) . | Proof. If \( u = \sigma \left( a\right) - a \), then \( {T}_{K/F}\left( u\right) = 0 \) . Conversely, suppose that \( {T}_{K/F}\left( u\right) = 0 \) . Let \( n = \left\lbrack {K : F}\right\rbrack \), and define \( g : G \rightarrow K \) by \( g\left( \mathrm{{id}}\right) = 0, g\left( \sigma \right) = \) \( u \), and for \( i < n \) by\n\n\[ g\left( {\sigma }^{i}\right) = u + \sigma \left( u\right) + \cdots + {\sigma }^{i - 1}\left( u\right) . \]\n\nIf \( 0 \leq i, j < n \), then as \( 0 = {T}_{K/F}\left( u\right) = \mathop{\sum }\limits_{{i = 1}}^{n}{\sigma }^{i}\left( u\right) \), we see that regardless of whether \( i + j < n \) or \( i + j \geq n \), we have\n\n\[ g\left( {{\sigma }^{i}{\sigma }^{j}}\right) = u + \sigma \left( u\right) + \cdots + {\sigma }^{i + j - 1}\left( u\right) \]\n\n\[ = \left( {u + \sigma \left( u\right) + \cdots + {\sigma }^{i - 1}\left( u\right) }\right) + {\sigma }^{i}\left( {u + \sigma \left( u\right) + \cdots + {\sigma }^{j - 1}\left( u\right) }\right) \]\n\n\[ = g\left( {\sigma }^{i}\right) + {\sigma }^{i}\left( {g\left( {\sigma }^{j}\right) }\right) . \]\n\nTherefore, \( g \) is a cocycle. By Proposition 10.3, there is an \( a \in K \) with \( g\left( {\sigma }^{i}\right) = {\sigma }^{i}\left( a\right) - a \) for all \( i \) . Hence, \( u = g\left( \sigma \right) = \sigma \left( a\right) - a \) . | Yes |
Theorem 10.5 (Additive Hilbert Theorem 90) Let \( K \) be a cyclic Galois extension of \( F \), and let \( \sigma \) be a generator of \( \operatorname{Gal}\left( {K/F}\right) \) . If \( u \in K \) , then \( {T}_{K/F}\left( u\right) = 0 \) if and only if \( u = \sigma \left( a\right) - a \) for some \( a \in K \) . | Proof. If \( u = \sigma \left( a\right) - a \), then \( {T}_{K/F}\left( u\right) = 0 \) . Conversely, suppose that \( {T}_{K/F}\left( u\right) = 0 \) . Let \( n = \left\lbrack {K : F}\right\rbrack \), and define \( g : G \rightarrow K \) by \( g\left( \mathrm{{id}}\right) = 0, g\left( \sigma \right) = \) \( u \), and for \( i < n \) by\n\n\[ g\left( {\sigma }^{i}\right) = u + \sigma \left( u\right) + \cdots + {\sigma }^{i - 1}\left( u\right) . \]\n\nIf \( 0 \leq i, j < n \), then as \( 0 = {T}_{K/F}\left( u\right) = \mathop{\sum }\limits_{{i = 1}}^{n}{\sigma }^{i}\left( u\right) \), we see that regardless of whether \( i + j < n \) or \( i + j \geq n \), we have\n\n\[ g\left( {{\sigma }^{i}{\sigma }^{j}}\right) = u + \sigma \left( u\right) + \cdots + {\sigma }^{i + j - 1}\left( u\right) \]\n\n\[ = \left( {u + \sigma \left( u\right) + \cdots + {\sigma }^{i - 1}\left( u\right) }\right) + {\sigma }^{i}\left( {u + \sigma \left( u\right) + \cdots + {\sigma }^{j - 1}\left( u\right) }\right) \]\n\n\[ = g\left( {\sigma }^{i}\right) + {\sigma }^{i}\left( {g\left( {\sigma }^{j}\right) }\right) . \]\n\nTherefore, \( g \) is a cocycle. By Proposition 10.3, there is an \( a \in K \) with \( g\left( {\sigma }^{i}\right) = {\sigma }^{i}\left( a\right) - a \) for all \( i \) . Hence, \( u = g\left( \sigma \right) = \sigma \left( a\right) - a \) . | Yes |
Let \( {Q}_{8} \) be the quaternion group. Then \( {Q}_{8} = \) \( \{ \pm 1, \pm i, \pm j, \pm k\} \), and the operation on \( {Q}_{8} \) is given by the relations \( {i}^{2} = \) \( {j}^{2} = {k}^{2} = - 1 \) and \( {ij} = k = - {ji} \) . We show that \( {Q}_{8} \) is a group extension of \( M = \langle i\rangle \) by \( \mathbb{Z}/2\mathbb{Z} \), and we determine the cocycle for this extension. | First note that \( M \) is an Abelian normal subgroup of \( {Q}_{8} \) and that \( {Q}_{8}/M \cong \mathbb{Z}/2\mathbb{Z} \) . Therefore, \( {Q}_{8} \) is a group extension of \( M \) by \( \mathbb{Z}/2\mathbb{Z} \) . We use 1 and \( j \) as coset representatives of \( M \) in \( {Q}_{8} \) . Our cocycle \( f \) that represents this group extension is then given by\n\n\[ f\left( {1,1}\right) = f\left( {1, j}\right) = f\left( {j,1}\right) = 1, \]\n\n\[ f\left( {j, j}\right) = {j}^{2} = - 1\text{.} \]\n\nThis cocycle is not trivial, so \( {Q}_{8} \) is not the semidirect product of \( M \) and \( \mathbb{Z}/2\mathbb{Z} \) . In fact, \( {Q}_{8} \) is not the semidirect product of any two subgroups, because one can show that there do not exist two subgroups of \( {Q}_{8} \) whose intersection is \( \langle 1\rangle \) . | Yes |
Example 10.11 Let \( \mathbb{H} \) be Hamilton’s quaternions. The ring \( \mathbb{H} \) consists of all symbols \( a + {bi} + {cj} + {dk} \) with \( a, b, c, d \in \mathbb{R} \), and multiplication is given by the relations \( {i}^{2} = {j}^{2} = {k}^{2} = - 1 \) and \( {ij} = k = - {ji} \) . This was the first example of a noncommutative division ring. The field of complex numbers \( \mathbb{C} \) can be viewed as the subring of \( \mathbb{H} \) consisting of all elements of the form \( a + {bi} \), and \( \mathbb{H} = \mathbb{C} \oplus \mathbb{C}j \) . The extension \( \mathbb{C}/\mathbb{R} \) is Galois with Galois group \( \{ \mathrm{{id}},\sigma \} \), where \( \sigma \) is complex conjugation. Let \( {x}_{\mathrm{{id}}} = 1 \) and \( {x}_{\sigma } = j \) . Then | \[ {x}_{\sigma }\left( {a + {bi}}\right) {x}_{\sigma }^{-1} = j\left( {a + {bi}}\right) {j}^{-1} = a - {bi} = \sigma \left( {a + {bi}}\right) . \] The cocycle \( f \) associated to this algebra is given by \[ f\left( {\mathrm{{id}},\mathrm{{id}}}\right) = {x}_{\mathrm{{id}}}{x}_{\mathrm{{id}}}{x}_{\mathrm{{id}}}^{-1} = 1 \] \[ f\left( {\mathrm{{id}},\sigma }\right) = {x}_{\mathrm{{id}}}{x}_{\sigma }{x}_{\sigma }^{-1} = 1 \] \[ f\left( {\sigma ,\mathrm{{id}}}\right) = {x}_{\sigma }{x}_{\mathrm{{id}}}{x}_{\sigma }^{-1} = 1 \] \[ f\left( {\sigma ,\sigma }\right) = {x}_{\sigma }{x}_{\sigma }{x}_{\mathrm{{id}}}^{-1} = {j}^{2} = - 1. \] On the other hand, if we start with this cocycle and construct the crossed product \( A = \left( {\mathbb{C}/\mathbb{R},\operatorname{Gal}\left( {\mathbb{C}/\mathbb{R}}\right), f}\right) \), then \( A = \mathbb{C}{x}_{\mathrm{{id}}} \oplus \mathbb{C}{x}_{\sigma } \), and the map \( A \rightarrow \mathbb{H} \) given by \( c{x}_{\mathrm{{id}}} + d{x}_{\sigma } \mapsto c + {dj} \) is an isomorphism of \( \mathbb{R} \) -algebras. | Yes |
Example 10.12 Let \( K/F \) be a Galois extension of degree \( n \) with Galois group \( G \), and consider the crossed product \( A = \left( {K/F, G,1}\right) \), where 1 represents the trivial cocycle. We will show that \( A \cong {M}_{n}\left( F\right) \), the ring of \( n \times n \) matrices over \( F \) . | First, note that \( A = { \oplus }_{\sigma \in G}K{x}_{\sigma } \), where multiplication on \( A \) is determined by the relations \( {x}_{\sigma }{x}_{\tau } = {x}_{\sigma \tau } \) and \( {x}_{\sigma }a = \sigma \left( a\right) {x}_{\sigma } \) for \( a \in K \) . If \( f = \sum {a}_{\sigma }{x}_{\sigma } \in A \), then \( f \) induces a map \( {\varphi }_{f} : K \rightarrow K \) given by \( {\varphi }_{f}\left( k\right) = \sum {a}_{\sigma }\sigma \left( k\right) \) . In other words, \( {\varphi }_{f} \) is the linear combination \( \sum {a}_{\sigma }\sigma \) . Each \( \sigma \) is an \( F \) -linear transformation of \( K \), so \( {\varphi }_{f} \in {\operatorname{End}}_{F}\left( K\right) \) . The relations governing multiplication in \( A \) show that the map \( \varphi : A \rightarrow {\operatorname{End}}_{F}\left( K\right) \) given by \( \varphi \left( f\right) = {\varphi }_{f} \) is an \( F \) -algebra homomorphism. Moreover, \( \varphi \) is injective since if \( \sum {a}_{\sigma }\sigma \) is the zero transformation, then each \( {a}_{\sigma } = 0 \) by the Dedekind independence lemma. Both \( A \) and \( {\operatorname{End}}_{F}\left( K\right) \) have dimension \( {n}^{2} \) over \( F \), so \( \varphi \) is automatically surjective. This proves that \( A \cong {\operatorname{End}}_{F}\left( K\right) \) , and so \( A \cong {M}_{n}\left( F\right) \) . | Yes |
Let \( K = \mathbb{Q}\left( {\sqrt{2},\sqrt{3}}\right) \). The field \( K \) is the splitting field of \( \left( {{x}^{2} - 2}\right) \left( {{x}^{2} - 3}\right) \) over \( \mathbb{Q} \), so \( K \) is a Galois extension of \( \mathbb{Q} \). A short calculation shows that \( \left\lbrack {K : \mathbb{Q}}\right\rbrack = 4 \), and the Galois group of \( K/\mathbb{Q} \) consists of the four automorphisms | \[
\text{id} : \sqrt{2} \rightarrow \sqrt{2},\;\sqrt{3} \rightarrow \sqrt{3}\text{,}
\]
\[
\sigma : \sqrt{2} \rightarrow - \sqrt{2},\;\sqrt{3} \rightarrow \sqrt{3}
\]
\[
\tau : \sqrt{2} \rightarrow \sqrt{2},\;\sqrt{3} \rightarrow - \sqrt{3}
\]
\[
{\sigma \tau } : \sqrt{2} \rightarrow - \sqrt{2},\;\sqrt{3} \rightarrow - \sqrt{3}.
\]
The Galois group \( \operatorname{Gal}\left( {K/\mathbb{Q}}\right) \) is isomorphic to \( \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \), an Abelian group of exponent 2. Since \( \mathbb{Q} \) contains the primitive second root of unity, -1, the extension \( K/\mathbb{Q} \) is a 2-Kummer extension. | Yes |
Lemma 11.8 Let \( B : G \times H \rightarrow C \) be a bilinear pairing. If \( h \in H \), let \( {B}_{h} : G \rightarrow C \) be defined by \( {B}_{h}\left( g\right) = B\left( {g, h}\right) \) . Then the map \( \varphi : h \mapsto {B}_{h} \) is a group homomorphism from \( H \) to \( \hom \left( {G, C}\right) \) . If \( B \) is nondegenerate, then \( \exp \left( G\right) \) divides \( \left| C\right| \), the map \( \varphi \) is injective, and \( \varphi \) induces an isomorphism \( G \cong H \) . | Proof. The property \( B\left( {g,{h}_{1}{h}_{2}}\right) = B\left( {g,{h}_{1}}\right) B\left( {g,{h}_{2}}\right) \) translates to \( {B}_{{h}_{1}{h}_{2}} = \) \( {B}_{{h}_{1}}{B}_{{h}_{2}} \) . Thus, \( \varphi \left( {{h}_{1}{h}_{2}}\right) = \varphi \left( {h}_{1}\right) \varphi \left( {h}_{2}\right) \), so \( \varphi \) is a homomorphism. The kernel of \( \varphi \) is\n\n\[ \ker \left( \varphi \right) = \left\{ {h \in H : {B}_{h} = 0}\right\} \]\n\n\[ = \{ h \in H : B\left( {g, h}\right) = e\text{ for all }h \in H\} . \]\n\nIf \( \varphi \) is nondegenerate, then \( \ker \left( \varphi \right) = \langle e\rangle \), so \( \varphi \) is injective. Suppose that \( m = \left| C\right| \) . Then\n\n\[ e = B\left( {e, h}\right) = B{\left( g, h\right) }^{m} = B\left( {{g}^{m}, h}\right) . \]\n\nNondegeneracy of \( B \) forces \( {g}^{m} = e \), so \( \exp \left( G\right) \) divides \( \left| G\right| \) . By a group theory exercise (see Problems 4 and 5), \( \hom \left( {G, C}\right) \) is isomorphic to the character group \( \hom \left( {G,{\mathbb{C}}^{ * }}\right) \), which is isomorphic to \( G \) . Therefore, there are group isomorphisms\n\n\[ H \cong \operatorname{im}\left( \varphi \right) = \hom \left( {G, C}\right) \cong G. \] | Yes |
Proposition 11.9 Let \( K \) be an \( n \) -Kummer extension of \( F \), and let \( B \) : \( \operatorname{Gal}\left( {K/F}\right) \times \operatorname{kum}\left( {K/F}\right) \rightarrow \mu \left( F\right) \) be the associated Kummer pairing. Then \( B \) is nondegenerate. Consequently, \( \operatorname{kum}\left( {K/F}\right) \cong \operatorname{Gal}\left( {K/F}\right) \) . | Proof. First, we show that \( B \) is a bilinear pairing. Let \( \sigma ,\tau \in \operatorname{Gal}\left( {K/F}\right) \) and \( \alpha {F}^{ * } \in \operatorname{kum}\left( {K/F}\right) \) . Then\n\n\[ B\left( {{\sigma \tau },\alpha {F}^{ * }}\right) = \frac{{\sigma \tau }\left( \alpha \right) }{\alpha } = \frac{\sigma \left( {\tau \left( \alpha \right) }\right) }{\tau \left( \alpha \right) } \cdot \frac{\tau \left( \alpha \right) }{\alpha } \]\n\n\[ = \tau \left( \frac{\sigma \left( \alpha \right) }{\alpha }\right) \cdot \frac{\tau \left( \alpha \right) }{\alpha } \]\n\nthe final equality is true because \( \operatorname{Gal}\left( {K/F}\right) \) is Abelian. But \( \sigma {\left( \alpha \right) }^{n} = {\alpha }^{n} \) , since \( {\alpha }^{n} \in F \) . Therefore, \( \sigma \left( \alpha \right) /\alpha \) is an \( n \) th root of unity, so \( \sigma \left( \alpha \right) /\alpha \in F \) . The automorphism \( \tau \) then fixes \( \sigma \left( \alpha \right) /\alpha \), so\n\n\[ B\left( {{\sigma \tau },\alpha {F}^{ * }}\right) = \frac{\sigma \left( \alpha \right) }{\alpha } \cdot \frac{\tau \left( \alpha \right) }{\alpha }. \]\n\nThe pairing \( B \) is thus linear in the first component. For the second component, if \( \alpha ,\beta \in \operatorname{KUM}\left( {K/F}\right) \), then\n\n\[ B\left( {\sigma ,\alpha {F}^{ * }\beta {F}^{ * }}\right) = \frac{\sigma \left( {\alpha \beta }\right) }{\alpha \beta } = \frac{\sigma \left( \alpha \right) \sigma \left( \beta \right) }{\alpha \beta } = \frac{\sigma \left( \alpha \right) }{\alpha } \cdot \frac{\sigma \left( \beta \right) }{\beta }. \]\n\nTherefore, \( B \) is a bilinear pairing.\n\nFor nondegeneracy, suppose that \( \sigma \in \operatorname{Gal}\left( {K/F}\right) \) with \( B\left( {\sigma ,\alpha {F}^{ * }}\right) = 1 \) for all \( a{F}^{ * } \in \operatorname{kum}\left( {K/F}\right) \) . Then \( \sigma \left( \alpha \right) = \alpha \) for all \( \alpha \in \operatorname{KUM}\left( {K/F}\right) \) . However, the elements in \( \operatorname{KUM}\left( {K/F}\right) \) generate \( K \) as a field extension of \( F \), and so automorphisms of \( K \) are determined by their action on this set. Therefore, \( \sigma = \mathrm{{id}} \) . Also, if \( B\left( {\sigma ,\alpha {F}^{ * }}\right) = 1 \) for all \( \sigma \in \operatorname{Gal}\left( {K/F}\right) \), then \( \sigma \left( \alpha \right) = \alpha \) for all \( \sigma \) . But then \( \alpha \in \mathcal{F}\left( {\operatorname{Gal}\left( {K/F}\right) }\right) \), and this fixed field is \( F \) by the fundamental theorem. Therefore, \( \alpha {F}^{ * } = {F}^{ * } \), so \( B \) is nondegenerate. The isomorphism \( \operatorname{kum}\left( {K/F}\right) \cong \operatorname{Gal}\left( {K/F}\right) \) then follows from Lemma 11.8. | Yes |
Proposition 11.10 Let \( K/F \) be an \( n \) -Kummer extension. Then there is an injective group homomorphism \( f : \operatorname{kum}\left( {K/F}\right) \rightarrow {F}^{ * }/{F}^{*n} \), given by \( f\left( {\alpha {F}^{ * }}\right) = {\alpha }^{n}{F}^{*n} \) . The image of \( f \) is then a finite subgroup of \( {F}^{ * }/{F}^{*n} \) of order equal to \( \left\lbrack {K : F}\right\rbrack \) . | Proof. It is easy to see that \( f \) is well defined and that \( f \) preserves multiplication. For injectivity, let \( \alpha {F}^{ * } \in \ker \left( f\right) \) . Then \( {\alpha }^{n} \in {F}^{*n} \), so \( {\alpha }^{n} = {a}^{n} \) for some \( a \in F \) . Hence, \( \alpha /a \) is an \( n \) th root of unity, and so \( \alpha /a \in F \) . Therefore, \( \alpha \in F \), so \( \alpha {F}^{ * } = {F}^{ * } \) is the identity. The group \( \operatorname{kum}\left( {K/F}\right) \) is then isomorphic to the image of \( f \) . The final statement of the proposition follows immediately from Proposition 11.9. | Yes |
Example 11.11 Let \( F = \mathbb{C}\left( {x, y, z}\right) \) be the rational function field in three variables over \( \mathbb{C} \), and let \( K = F\left( {\sqrt[4]{xyz},\sqrt[4]{{y}^{2}z},\sqrt[4]{x{z}^{2}}}\right) \) . Then \( K/F \) is a 4- Kummer extension. The image of \( \operatorname{kum}\left( {K/F}\right) \) in \( {F}^{ * }/{F}^{*4} \) is generated by the cosets of \( {xyz},{yz} \), and \( x{z}^{2} \) . For simplicity we will call these three cosets \( a, b, c \) respectively. We claim that the subgroup of \( {F}^{ * }/{F}^{*4} \) generated by \( a, b, c \) has order 32, which shows that \( \left\lbrack {K : F}\right\rbrack = {32} \) by Proposition 11.10. | The subgroup \( \langle a, b\rangle \) of \( {F}^{ * }/{F}^{*4} \) generated by \( a \) and \( b \) has order 16, since the 16 elements \( {a}^{i}{b}^{j} \) with \( 1 \leq i, j \leq 4 \) are all distinct. To see this, suppose that \( {a}^{i}{b}^{j} = {a}^{k}{b}^{l} \) . Then there is an \( h \in {F}^{ * } \) with\n\n\[{\left( xyz\right) }^{i}{\left( {y}^{2}z\right) }^{j} = {\left( xyz\right) }^{k}{\left( {y}^{2}z\right) }^{l}{h}^{4}\]\n\nWriting \( h = f/g \) with \( f, g \in \mathbb{C}\left\lbrack {x, y, z}\right\rbrack \) relatively prime gives\n\n\[{\left( xyz\right) }^{i}{\left( {y}^{2}z\right) }^{j}f\left( {x, y, z}\right) = {\left( xyz\right) }^{k}{\left( {y}^{2}z\right) }^{l}g{\left( x, y, z}\right) }^{4}.\]\n\nBy unique factorization, comparing powers of \( x \) and \( z \) on both sides of this equation, we obtain\n\n\[i \equiv k\left( {\;\operatorname{mod}\;4}\right)\]\n\n\[i + j \equiv k + l\left( {\;\operatorname{mod}\;4}\right) .\]\n\nThese equations force \( i \equiv k\left( {\;\operatorname{mod}\;4}\right) \) and \( j \equiv l\left( {\;\operatorname{mod}\;4}\right) \), so the elements \( {a}^{i}{b}^{j} \) for \( 1 \leq i, j \leq 4 \) are indeed distinct. Note that \( {abc} = {x}^{2}{y}^{2}{z}^{4}{F}^{*4} \), so \( {\left( abc\right) }^{2} = {x}^{4}{y}^{4}{z}^{8}{F}^{*4} = {F}^{*4} \) . Therefore, \( {c}^{2} = {\left( ab\right) }^{2} \), so either the subgroup \( \langle a, b, c\rangle \) of \( {F}^{ * }/{F}^{*4} \) generated by \( a, b, c \) is equal to \( \langle a, b\rangle \), or \( \langle a, b\rangle \) has index 2 in \( \langle a, b, c\rangle \) . For the first to happen, we must have \( c = {a}^{i}{b}^{j} \) for some \( i, j \) . This leads to an equation\n\n\[x{z}^{2}f{\left( x, y, z\right) }^{4} = {\left( xyz\right) }^{i}{\left( {y}^{2}z\right) }^{j}g{\left( x, y, z\right) }^{4}\]\n\nfor some polynomials \( f, g \) . Again applying unique factorization and equating powers of \( x \) and \( y \) gives \( 1 \equiv i\left( {\;\operatorname{mod}\;4}\right) \) and \( 0 \equiv i + {2j}\left( {\;\operatorname{mod}\;4}\right) \) . A simultaneous solution of these equations does not exist, so \( c \) is not in the group \( \langle a, b\rangle \), so \( \langle a, b\rangle \) has index 2 in \( \langle a, b, c\rangle \) . This proves that \( \langle a, b, c\rangle \) has order 32 , as we wanted to show. | Yes |
Lemma 12.3 Let \( F \) be a field with \( \operatorname{char}\left( F\right) \neq 2 \), let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) be an irreducible, separable polynomial, and let \( K \) be the splitting field of \( f\left( x\right) \) over \( F \) . If \( \Delta \) is defined as in Definition 12.2, then \( \sigma \in \operatorname{Gal}\left( {K/F}\right) \) is an even permutation if and only if \( \sigma \left( \Delta \right) = \Delta \), and \( \sigma \) is odd if and only if \( \sigma \left( \Delta \right) = - \Delta \) . Furthermore, \( \operatorname{disc}\left( f\right) \in F \) . | Proof. Before we prove this, we note that the proof we give is the same as the typical proof that every permutation of \( {S}_{n} \) is either even or odd. In fact, the proof of this result about \( {S}_{n} \) is really about discriminants. It is easy to see that each \( \sigma \in G = \operatorname{Gal}\left( {K/F}\right) \) fixes \( \operatorname{disc}\left( f\right) \), so \( \operatorname{disc}\left( f\right) \in F \) . For the proof of the first statement, if \( n = \deg \left( f\right) \), let \( M = F\left( {{x}_{1},\ldots ,{x}_{n}}\right) \) . We saw in Example 2.22 that \( {S}_{n} \) acts as field automorphisms on \( M \) by permuting the variables. Let \( h\left( x\right) = \mathop{\prod }\limits_{{i < j}}\left( {{x}_{i} - {x}_{j}}\right) \) . Suppose that \( \sigma \in {S}_{n} \) is a transposition, say \( \sigma = \left( {ij}\right) \) with \( i < j \) . Then \( \sigma \) affects only those factors of \( h \) that involve \( i \) or \( j \) . We break up these factors into four groups:\n\n\[ \n{x}_{i} - {x}_{j} \n\] \n\n\[ \n{x}_{k} - {x}_{i},{x}_{k} - {x}_{j}\text{ for }k < i \n\] \n\n\[ \n{x}_{i} - {x}_{l},{x}_{j} - {x}_{l}\;\text{ for }\;j < l, \n\] \n\n\[ \n{x}_{i} - {x}_{m},{x}_{m} - {x}_{j}\text{ for }i < m < j. \n\] \n\nFor \( k < i \), the permutation \( \sigma = \left( {ij}\right) \) maps \( {x}_{k} - {x}_{i} \) to \( {x}_{k} - {x}_{j} \) and vice versa, and \( \sigma \) maps \( {x}_{i} - {x}_{l} \) to \( {x}_{j} - {x}_{l} \) and vice versa for \( j < l \) . If \( i < m < j \), then \n\n\[ \n\sigma \left( {{x}_{i} - {x}_{m}}\right) = {x}_{j} - {x}_{m} = - \left( {{x}_{m} - {x}_{j}}\right) \n\] \n\nand \n\n\[ \n\sigma \left( {{x}_{m} - {x}_{j}}\right) = {x}_{m} - {x}_{i} = - \left( {{x}_{i} - {x}_{m}}\right) . \n\] \n\nFinally, \n\n\[ \n\sigma \left( {{x}_{i} - {x}_{j}}\right) = {x}_{j} - {x}_{i} = - \left( {{x}_{i} - {x}_{j}}\right) . \n\] \n\nMultiplying all the terms together gives \( \sigma \left( h\right) = - h \) . Thus, we see for an arbitrary \( \sigma \in {S}_{n} \) that \( \sigma \left( h\right) = h \) if and only if \( \sigma \) is a product of an even number of permutations, and \( \sigma \left( h\right) = - h \) if and only if \( \sigma \) is a product of an odd number of permutations. By substituting the roots \( {\alpha }_{i} \) of \( f \) for the \( {x}_{i} \) , we obtain the desired conclusion. | Yes |
Corollary 12.4 Let \( F, K \), and \( f \) be as in Lemma 12.3, and let \( G = \) \( \operatorname{Gal}\left( {K/F}\right) \) . Then \( G \subseteq {A}_{n} \) if and only if \( \operatorname{disc}\left( f\right) \in {F}^{2} \) . Under the correspondence of the fundamental theorem, the field \( F\left( \Delta \right) \subseteq K \) corresponds to the subgroup \( G \cap {A}_{n} \) of \( G \) . | Proof. This follows from the lemma, since \( G \subseteq {A}_{n} \) if and only if each \( \sigma \in G \) is even, and this occurs if and only if \( \sigma \left( \Delta \right) = \Delta \) . Therefore, \( G \subseteq {A}_{n} \) if and only if \( \operatorname{disc}\left( f\right) \in {F}^{2} \) . | Yes |
If \( K \) is a field and \( {\alpha }_{1},\ldots ,{\alpha }_{n} \in K \), then the determinant of the Vandermonde matrix \( V\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) is \( \mathop{\prod }\limits_{{i < j}}\left( {{\alpha }_{j} - {\alpha }_{i}}\right) \). | Let \( A = V\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) . That \( \det \left( A\right) = \mathop{\prod }\limits_{{i < j}}\left( {{\alpha }_{j} - {\alpha }_{i}}\right) \) is a moderately standard fact from linear algebra. For those who have not seen this, we give a proof. Note that if \( {\alpha }_{i} = {\alpha }_{j} \) with \( i \neq j \), then \( \det \left( A\right) = 0 \), since two rows of \( A \) are the same, so the determinant formula is true in this case. We therefore assume that the \( {\alpha }_{i} \) are distinct, and we prove the result using induction on \( n \) . If \( n = 1 \), this is clear, so suppose that \( n > 1 \) . Let \( h\left( x\right) = \det \left( {V\left( {{\alpha }_{1},{\alpha }_{2},\ldots ,{\alpha }_{n - 1}, x}\right) }\right) \) . Then \( h\left( x\right) \) is a polynomial of degree less than \( n \) . By expanding the determinant about the last row, we see that the leading coefficient of \( h \) is \( \det \left( {V\left( {{\alpha }_{1},{\alpha }_{2},\ldots ,{\alpha }_{n - 1}}\right) }\right) \) . Moreover, \( h\left( {\alpha }_{i}\right) = \det \left( {V\left( {{\alpha }_{1},\ldots ,{\alpha }_{n - 1},{\alpha }_{i}}\right) }\right) \), so \( h\left( {\alpha }_{i}\right) = 0 \) if \( 1 \leq i \leq n - 1 \) . Therefore, \( h\left( x\right) \) is divisible by each \( x - {\alpha }_{i} \) . Since \( \deg \left( h\right) < n \) and \( h \) has \( n - 1 \) distinct factors, \( h\left( x\right) = c\left( {x - {\alpha }_{1}}\right) \cdots \left( {x - {\alpha }_{n - 1}}\right) \), where \( c = \det \left( {V\left( {{\alpha }_{1},{\alpha }_{2},\ldots ,{\alpha }_{n - 1}}\right) }\right) \) . By evaluating \( h \) at \( {\alpha }_{n} \) and using induction, we get\n\n\[ h\left( {\alpha }_{n}\right) = \det \left( {V\left( {{\alpha }_{1},{\alpha }_{2},\ldots ,{\alpha }_{n}}\right) }\right) \]\n\n\[ = \mathop{\prod }\limits_{{i < j \leq n - 1}}\left( {{\alpha }_{j} - {\alpha }_{i}}\right) \mathop{\prod }\limits_{{i < n}}\left( {{\alpha }_{n} - {\alpha }_{i}}\right) \]\n\n\[ = \mathop{\prod }\limits_{{i < j}}\left( {{\alpha }_{j} - {\alpha }_{i}}\right) \]\n\nThis finishes the proof that \( \det \left( {V\left( {{\alpha }_{1},{\alpha }_{2},\ldots ,{\alpha }_{n}}\right) }\right) = \mathop{\prod }\limits_{{i < j}}\left( {{\alpha }_{j} - {\alpha }_{i}}\right) \) . The last statement of the lemma is an immediate consequence of this formula and the definition of discriminant. | Yes |
Proposition 12.6 (Newton’s Identities) Let \( f\left( x\right) = {a}_{0} + {a}_{1}x + \cdots + \) \( {a}_{n - 1}{x}^{n - 1} + {x}^{n} \) be a monic polynomial over \( F \) with roots \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) . If \( {t}_{i} = \mathop{\sum }\limits_{j}{\alpha }_{j}^{i}, \) then\n\n\[ \left. \begin{matrix} {t}_{m} + {a}_{n - 1} & {t}_{m - 1} + \cdots + {a}_{n - m + 1} \\ {t}_{1} + m & {a}_{n - m} = 0 \end{matrix}\right\} \;\text{ for }m \leq n, \]\n\n\[ {t}_{m} + {a}_{n - 1}{t}_{m - 1} + \cdots + {a}_{0}{t}_{m - n} = 0\;\text{ for }m > n. \] | Proof. An alternative way of stating Newton's identities is to use the elementary symmetric functions \( {s}_{i} \) in the \( {a}_{i} \), instead of the \( {a}_{i} \) . Since \( {s}_{i} = \) \( {\left( -1\right) }^{i}{a}_{n - i} \), Newton’s identities can also be written as\n\n\[ {t}_{m} - {s}_{1}{t}_{m - 1} + {s}_{2}{t}_{m - 2} - + \cdots {\left( -1\right) }^{m}m{s}_{m} = 0\;\text{ for }m \leq n \]\n\n\[ {t}_{m} - {a}_{1}{t}_{m - 1} - + \cdots + {\left( -1\right) }^{n}{s}_{n}{t}_{m - n} = 0\;\text{ for }m > n. \]\n\nThe proof we give here is from Mead [21]. The key is arranging the terms in the identities in a useful manner. We start with a bit of notation. If \( \left( {{a}_{1},{a}_{2},\ldots ,{a}_{r}}\right) \) is a sequence of nonincreasing, nonnegative integers, let\n\n\[ {f}_{\left( {a}_{1},{a}_{2},\ldots ,{a}_{r}\right) } = \sum {\alpha }_{\sigma \left( 1\right) }^{{a}_{1}}\cdots {\alpha }_{\sigma \left( n\right) }^{{a}_{r}}, \]\n\nwhere the sum is over all permutations \( \sigma \) of \( \{ 1,2,\ldots, n\} \) that give distinct terms. Then \( {s}_{i} = {f}_{\left( 1,1,\ldots ,1\right) } \) ( \( i \) ones) and \( {t}_{i} = {f}_{\left( i\right) } \) . To simplify the notation a little, the sequence of \( i \) ones will be denoted \( \left( {1}_{i}\right) \), and the sequence\n\n\( \left( {a,1,\ldots ,1}\right) \) of length \( i + 1 \) will be denoted \( \left( {a,{1}_{i}}\right) \) . It is then straightforward to see that\n\n\[ {f}_{\left( m - 1\right) }{f}_{\left( 1\right) } = {f}_{\left( m\right) } + {f}_{\left( m - 1,1\right) } \]\n\n\[ {f}_{\left( m - 2\right) }{f}_{\left( 1,1\right) } = {f}_{\left( m - 1,1\right) } + {f}_{\left( m - 2,1,1\right) } \]\n\n\[ {f}_{\left( m - 3\right) }{f}_{\left( 1,1,1\right) } = {f}_{\left( m - 2,1,1\right) } + {f}_{\left( m - 3,1,1,1\right) } \]\n\nand, in general,\n\n\[ {f}_{\left( m - i\right) }{f}_{\left( {1}_{i}\right) } = {f}_{\left( m - i + 1,{1}_{i}\right) } + {f}_{\left( m - i,{1}_{i}\right) }\;\text{ for }1 \leq i < \min \{ m - 1, n\} . \]\n\n(12.1)\n\nMoreover, if \( m \leq n \) and \( i = m - 1 \), then\n\n\[ {f}_{\left( 1\right) }{f}_{\left( {1}_{m - 1}\right) } = {f}_{\left( 2,{1}_{m - 2}\right) } + m{f}_{\left( {1}_{m}\right) }. \]\n\nIf \( m > n = i \), then\n\n\[ {f}_{\left( m - n\right) }{f}_{\left( {1}_{n}\right) } = {f}_{\left( m - n + 1,{1}_{n - 1}\right) }. \]\n\nNewton’s identities then follow from these equations by multiplying the \( i \) th equation in (12.1) by \( {\left( -1\right) }^{i - 1} \) and summing over \( i \) . | Yes |
Example 12.7 Let \( f\left( x\right) = {x}^{2} + {bx} + c \) . Then \( {t}_{0} = 2 \) . Also, Newton’s identities yield \( {t}_{1} + b = 0 \), so \( {t}_{1} = - b \) . For \( {t}_{2} \), we have \( {t}_{2} + b{t}_{1} + {2c} = 0 \), so \( {t}_{2} = - b{t}_{1} - {2c} = {b}^{2} - {2c} \) . | Therefore,\n\n\[\n\operatorname{disc}\left( f\right) = \left| \begin{matrix} 2 & - b \\ - b & {b}^{2} - {2c} \end{matrix}\right| = 2\left( {{b}^{2} - {2c}}\right) - {b}^{2} = {b}^{2} - {4c}\n\]\n\nthe usual discriminant of a monic quadratic. | Yes |
Proposition 12.9 Let \( L = F\left( \alpha \right) \) be a field extension of \( F \) . If \( f\left( x\right) = \) \( \min \left( {F,\alpha }\right) \), then \( \operatorname{disc}\left( f\right) = {\left( -1\right) }^{n\left( {n - 1}\right) /2}{N}_{L/F}\left( {{f}^{\prime }\left( \alpha \right) }\right) \), where \( {f}^{\prime }\left( x\right) \) is the formal derivative of \( f \) . | Proof. Let \( K \) be a splitting field for \( f \) over \( F \), and write \( f\left( x\right) = (x - \) \( \left. {\alpha }_{1}\right) \cdots \left( {x - {\alpha }_{n}}\right) \in K\left\lbrack x\right\rbrack \) . Set \( \alpha = {\alpha }_{1} \) . Then a short calculation shows that \( {f}^{\prime }\left( {\alpha }_{j}\right) = \mathop{\prod }\limits_{{i = 1, i \neq j}}^{n}\left( {{\alpha }_{j} - {\alpha }_{i}}\right) \) . If \( {\sigma }_{1},\ldots ,{\sigma }_{n} \) are the \( F \) -homomorphisms of \( L \) to \( K \) that satisfy \( {\sigma }_{i}\left( \alpha \right) = {\alpha }_{i} \), then by Proposition 8.12,\n\n\[ \n{N}_{L/F}\left( {{f}^{\prime }\left( \alpha \right) }\right) = \mathop{\prod }\limits_{j}{\sigma }_{j}\left( {{f}^{\prime }\left( \alpha \right) }\right) = \mathop{\prod }\limits_{j}{f}^{\prime }\left( {\alpha }_{j}\right) . \n\]\n\nUsing the formula above for \( {f}^{\prime }\left( {\alpha }_{j}\right) \), we see by checking signs carefully that\n\n\[ \n{N}_{L/F}\left( {{f}^{\prime }\left( \alpha \right) }\right) = \mathop{\prod }\limits_{j}{f}^{\prime }\left( {\alpha }_{j}\right) = \mathop{\prod }\limits_{j}\mathop{\prod }\limits_{\substack{{i = 1} \\ {i \neq j} }}^{n}\left( {{\alpha }_{j} - {\alpha }_{i}}\right) = {\left( -1\right) }^{n\left( {n - 1}\right) /2}\operatorname{disc}\left( f\right) . \n\] | Yes |
Example 12.10 Let \( p \) be an odd prime, and let \( \omega \) be a primitive \( p \) th root of unity in \( \mathbb{C} \) . We use the previous result to determine \( \operatorname{disc}\left( \omega \right) \) . Let \( K = \) \( \mathbb{Q}\left( \omega \right) \), the \( p \) th cyclotomic extension of \( \mathbb{Q} \) . If \( f\left( x\right) = \min \left( {\mathbb{Q},\omega }\right) \), then \( f\left( x\right) = \) \( 1 + x + \cdots + {x}^{p - 1} = \left( {{x}^{p} - 1}\right) /\left( {x - 1}\right) \) . We need to calculate \( {N}_{K/\mathbb{Q}}\left( {{f}^{\prime }\left( \omega \right) }\right) \) . | First,\n\n\[ \n{f}^{\prime }\left( x\right) = \frac{p{x}^{p - 1}\left( {x - 1}\right) - \left( {{x}^{p} - 1}\right) }{{\left( x - 1\right) }^{2}} \n\]\n\nso \( {f}^{\prime }\left( \omega \right) = p{\omega }^{p - 1}/\left( {\omega - 1}\right) \) . We claim that \( {N}_{K/\mathbb{Q}}\left( \omega \right) = 1 \) and \( {N}_{K/\mathbb{Q}}\left( {\omega - 1}\right) = \) \( p \) . To prove the first equality, by the description of \( \operatorname{Gal}\left( {K/\mathbb{Q}}\right) \) given in\n\nCorollary 7.8, we have\n\n\[ \n{N}_{K/\mathbb{Q}}\left( \omega \right) = \mathop{\prod }\limits_{{i = 1}}^{{p - 1}}{\omega }^{i} = {\omega }^{p\left( {p - 1}\right) /2} = 1 \n\]\n\nsince \( p \) is odd. For the second equality, note that\n\n\[ \n1 + x + \cdots + {x}^{p - 1} = \mathop{\prod }\limits_{{i = 1}}^{{p - 1}}\left( {x - {\omega }^{i}}\right) \n\]\n\nso \( p = \mathop{\prod }\limits_{{i = 1}}^{{p - 1}}\left( {1 - {\omega }^{i}}\right) \) . However,\n\n\[ \n{N}_{K/\mathbb{Q}}\left( {\omega - 1}\right) = \mathop{\prod }\limits_{{i = 1}}^{{p - 1}}\left( {{\omega }^{i} - 1}\right) \n\]\n\nso \( {N}_{K/\mathbb{Q}}\left( {\omega - 1}\right) = p \), where again we use \( p \) odd. From this, we see that\n\n\[ \n{N}_{K/\mathbb{Q}}\left( {{f}^{\prime }\left( \omega \right) }\right) = {N}_{K/\mathbb{Q}}\left( \frac{p{\omega }^{p - 1}}{\omega - 1}\right) = \frac{{N}_{K/\mathbb{Q}}\left( p\right) {N}_{K/\mathbb{Q}}{\left( \omega \right) }^{p - 1}}{{N}_{K/\mathbb{Q}}\left( {\omega - 1}\right) } \n\]\n\n\[ \n= \frac{{p}^{p - 1} \cdot 1}{p} = {p}^{p - 2}. \n\] | Yes |
Lemma 12.12 Let \( K \) be a separable field extension of \( F \) of degree \( n \), and let \( {\alpha }_{1},\ldots ,{\alpha }_{n} \in K \) . Then \( \operatorname{disc}\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) = \det \left( {{\operatorname{Tr}}_{K/F}\left( {{\alpha }_{i}{\alpha }_{j}}\right) }\right) \) . Consequently, \( \operatorname{disc}\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \in F \) . | Proof. Let \( {\sigma }_{1},\ldots ,{\sigma }_{n} \) be the distinct \( F \) -homomorphisms from \( K \) to an algebraic closure of \( F \) . If \( A = \left( {{\sigma }_{i}\left( {\alpha }_{j}\right) }\right) \), then the discriminant of the \( n \) - tuple \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) is the determinant of the matrix \( {A}^{t}A \), whose \( {ij} \) entry is\n\n\[ \mathop{\sum }\limits_{k}{\sigma }_{k}\left( {\alpha }_{i}\right) {\sigma }_{k}\left( {\alpha }_{j}\right) = \mathop{\sum }\limits_{k}{\sigma }_{k}\left( {{\alpha }_{i}{\alpha }_{j}}\right) \]\n\n\[ = {\operatorname{Tr}}_{K/F}\left( {{\alpha }_{i}{\alpha }_{j}}\right) \]\n\nTherefore, \( \operatorname{disc}\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) = \det \left( {{\operatorname{Tr}}_{K/F}\left( {{\alpha }_{i}{\alpha }_{j}}\right) }\right) \) . | Yes |
Proposition 12.13 Let \( K \) be a separable field extension of \( F \) of degree \( n \) , and let \( {\alpha }_{1},\ldots ,{\alpha }_{n} \in K \) . Then \( \operatorname{disc}\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) = 0 \) if and only if \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) are linearly dependent over \( F \) . Thus, \( \left\{ {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right\} \) is an \( F \) -basis for \( K \) if and only if \( \operatorname{disc}\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \neq 0 \) . | Proof. Suppose that the \( {\alpha }_{i} \) are linearly dependent over \( F \) . Then one of the \( {\alpha }_{i} \) is an \( F \) -linear combination of the others. If \( {\alpha }_{i} = \mathop{\sum }\limits_{{k \neq i}}{a}_{k}{\alpha }_{k} \) with \( {a}_{j} \in F \), then\n\n\[ \n{\operatorname{Tr}}_{K/F}\left( {{\alpha }_{i}{\alpha }_{j}}\right) = \mathop{\sum }\limits_{k}{a}_{k}{\operatorname{Tr}}_{K/F}\left( {{\alpha }_{k}{\alpha }_{j}}\right) \n\]\n\nTherefore, the columns of the matrix \( \left( {{\operatorname{Tr}}_{K/F}\left( {{\alpha }_{i}{\alpha }_{j}}\right) }\right) \) are linearly dependent over \( F \), so \( \det \left( {{\operatorname{Tr}}_{K/F}\left( {{\alpha }_{i}{\alpha }_{j}}\right) }\right) = 0 \) .\n\nConversely, suppose that \( \det \left( {{\operatorname{Tr}}_{K/F}\left( {{\alpha }_{i}{\alpha }_{j}}\right) }\right) = 0 \) . Then the rows \( {R}_{1},\ldots ,{R}_{n} \) of the matrix \( \left( {{\operatorname{Tr}}_{K/F}\left( {{\alpha }_{i}{\alpha }_{j}}\right) }\right) \) are dependent over \( F \), so there are \( {a}_{i} \in F \), not all zero, with \( \mathop{\sum }\limits_{i}{a}_{i}{R}_{i} = 0 \) . The vector equation \( \mathop{\sum }\limits_{i}{a}_{i}{R}_{i} = 0 \) means that \( \mathop{\sum }\limits_{i}{a}_{i}{\operatorname{Tr}}_{K/F}\left( {{\alpha }_{i}{\alpha }_{j}}\right) = 0 \) for each \( j \) . Let \( x = \mathop{\sum }\limits_{i}{a}_{i}{\alpha }_{i} \) . By linearity of the trace, we see that \( {\operatorname{Tr}}_{K/F}\left( {x{\alpha }_{j}}\right) = 0 \) for each \( j \) . If the \( {\alpha }_{i} \) are independent over \( F \), then they form a basis for \( K \) . Consequently, linearity of the trace then implies that \( {\operatorname{Tr}}_{K/F}\left( {xy}\right) = 0 \) for all \( y \in K \) . This means that the trace map is identically zero, which is false by the Dedekind independence lemma. Thus, the \( {\alpha }_{i} \) are dependent over \( F \) . | Yes |
Proposition 12.14 Let \( \left\{ {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right\} \) and \( \left\{ {{\beta }_{1},\ldots ,{\beta }_{n}}\right\} \) be two \( F \) -bases for \( K \) . Let \( A = \left( {a}_{ij}\right) \) be the \( n \times n \) transition matrix between the two bases; that \( {is},\;{\beta }_{j} = \mathop{\sum }\limits_{i}{a}_{ij}{\alpha }_{i}.\; \) Then disc \( \left( {{\beta }_{1},\ldots ,{\beta }_{n}}\right) = \det {\left( A\right) }^{2}\; \) disc \( \left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) .\; \) Consequently, the coset of \( \operatorname{disc}\left( {K/F}\right) \) in \( {F}^{ * }/{F}^{*2} \) is well defined, independent of the basis chosen. | Proof. Since \( {\beta }_{j} = \mathop{\sum }\limits_{k}{a}_{kj}{\alpha }_{k} \), we have \( {\sigma }_{i}\left( {\beta }_{j}\right) = \mathop{\sum }\limits_{k}{a}_{kj}{\sigma }_{i}\left( {\alpha }_{k}\right) \) . In terms of matrices, this says that\n\n\[ \left( {{\sigma }_{i}\left( {\beta }_{j}\right) }\right) = {\left( {a}_{ij}\right) }^{t}\left( {{\sigma }_{i}\left( {\alpha }_{j}\right) }\right) = {A}^{t}\left( {{\sigma }_{i}\left( {\alpha }_{j}\right) }\right) . \]\n\nTherefore, by taking determinants, we obtain\n\n\[ \operatorname{disc}\left( {{\beta }_{1},\ldots ,{\beta }_{n}}\right) = \det {\left( A\right) }^{2}\operatorname{disc}\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) . \]\n\nThe final statement of the proposition follows immediately from this relation, together with the fact that the discriminant of a basis is nonzero, by Proposition 12.13. | Yes |
In this example, we show that the discriminant of a polynomial is equal to the discriminant of an appropriate field extension. Suppose that \( K = F\left( \alpha \right) \) is an extension of \( F \) of degree \( n \) . Then \( 1,\alpha \) , \( {\alpha }^{2},\ldots ,{\alpha }^{n - 1} \) is a basis for \( K \) . We calculate \( \operatorname{disc}\left( {K/F}\right) \) relative to this basis. | We have \( \operatorname{disc}\left( {K/F}\right) = \det {\left( {\sigma }_{i}\left( {\alpha }^{j - 1}\right) \right) }^{2} \) . Consequently, if \( {\alpha }_{i} = {\sigma }_{i}\left( \alpha \right) \) , then\n\n\[ \operatorname{disc}\left( {K/F}\right) = \det {\left( \begin{matrix} 1 & {\sigma }_{1}\left( \alpha \right) & \cdots & {\sigma }_{1}\left( {\alpha }^{n - 1}\right) \\ 1 & {\sigma }_{2}\left( \alpha \right) & \cdots & {\sigma }_{2}\left( {\alpha }^{n - 1}\right) \\ \vdots & \vdots & \ddots & \vdots \\ 1 & {\sigma }_{n}\left( \alpha \right) & \cdots & {\sigma }_{n}\left( {\alpha }^{n - 1}\right) \end{matrix}\right) }^{2} \]\n\n\[ = \det {\left( V\left( {\alpha }_{1},{\alpha }_{2},\ldots ,{\alpha }_{n}\right) \right) }^{2}. \]\n\nTherefore, \( \operatorname{disc}\left( {K/F}\right) = \operatorname{disc}\left( \alpha \right) = \operatorname{disc}\left( {\min \left( {F,\alpha }\right) }\right) \) . | Yes |
Example 12.16 Let \( K = \mathbb{Q}\left( \sqrt{-1}\right) \). If \( i = \sqrt{-1} \), then using the basis \( 1, i \) of \( K/\mathbb{Q} \), we get | \[ \operatorname{disc}\left( {\mathbb{Q}\left( i\right) /\mathbb{Q}}\right) = \det {\left( \begin{matrix} 1 & i \\ 1 & - i \end{matrix}\right) }^{2} = {\left( -2i\right) }^{2} = - 4. \] | Yes |
We now show that the discriminant of a field extension is the discriminant of the trace form. Let \( K \) be a finite separable extension of \( F \) . Let \( B : K \times K \rightarrow F \) be defined by \( B\left( {a, b}\right) = {T}_{K/F}\left( {ab}\right) \) . Then \( B \) is a bilinear form because the trace is linear. The discriminant of \( B \) relative to a basis \( \mathcal{V} = \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) is \( \det \left( {{T}_{K/F}\left( {{v}_{i}{v}_{j}}\right) }\right) \) . | But, by Lemma 12.12, this is the discriminant of \( K/F \) . Therefore, the previous notions of discriminant are special cases of the notion of discriminant of a bilinear form. | Yes |
Theorem 13.1 Let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) be an irreducible, separable polynomial of degree 3 over \( F \), and let \( K \) be the splitting field of \( f \) over \( F \) . If \( D \) is the discriminant of \( f \), then \( \operatorname{Gal}\left( {K/F}\right) \cong {S}_{3} \) if and only if \( D \notin {F}^{2} \), and \( \operatorname{Gal}\left( {K/F}\right) \cong {A}_{3} \) if and only if \( D \in {F}^{2} \) . | Proof. Let \( G = \operatorname{Gal}\left( {K/F}\right) \) . By Corollary \( {12.4}, G \subseteq {A}_{3} \) if and only if \( D \in {F}^{2} \) . But \( G \cong {S}_{3} \) or \( G \cong {A}_{3} \), so \( G \cong {S}_{3} \) if and only if \( D \) is a square in \( F \) . | Yes |
Consider \( {x}^{3} - {3x} + 1 \) . Then \( \Gamma = - D/{108} = - {81}/{108} = \) \( - 3/4 \) . We have \( p = - 3 \) and \( q = 1 \) . Then \( A = - 1/2 + i\sqrt{3}/2 \) and \( B = - 1/2 - i\sqrt{3}/2 \), so \( A = \exp \left( {{2\pi i}/3}\right) \) and \( B = \exp \left( {-{2\pi i}/3}\right) \) . We can then set \( u = \exp \left( {{2\pi i}/9}\right) \) and \( v = \exp \left( {-{2\pi i}/9}\right) \) . Also, \( \omega = \exp \left( {{2\pi i}/3}\right) \) . By simplifying the formulas for the roots of \( f \), we see that the three roots are \( 2\cos \left( {{2\pi }/9}\right) ,2\cos \left( {{8\pi }/9}\right) \), and \( 2\cos \left( {{14\pi }/9}\right) \) . | Suppose that the polynomial \( f\left( x\right) = {x}^{3} + {px} + q \) has real coefficients. If \( \Gamma > 0 \), then \( D < 0 \), so \( D \) is not a square in \( F \) . We can then take the real cube roots of \( A \) and \( B \) for \( u \) and \( v \) . Furthermore, if \( \omega = \left( {-1 + i\sqrt{3}}\right) /2 \), we see that the three roots of \( f \) are\n\n\[ \n{\alpha }_{1} = \sqrt[3]{A} + \sqrt[3]{B} \in \mathbb{R} \n\] \n\n\[ \n{\alpha }_{2} = - \left( \frac{\sqrt[3]{A} + \sqrt[3]{B}}{2}\right) + i\sqrt{3}\left( \frac{\sqrt[3]{A} - \sqrt[3]{B}}{2}\right) \n\] \n\nand \n\n\[ \n{\alpha }_{3} = - \left( \frac{\sqrt[3]{A} + \sqrt[3]{B}}{2}\right) - i\sqrt{3}\left( \frac{\sqrt[3]{A} - \sqrt[3]{B}}{2}\right) . \n\] \n\nOn the other hand, if \( \Gamma < 0 \), then \( A = - q/2 + i\sqrt{-\Gamma } \) and \( B = - q/2 - i\sqrt{-\Gamma } \) . If we choose \( \sqrt[3]{A} = a + {bi} \) to satisfy \( \sqrt[3]{A}\sqrt[3]{B} = - p/3 \), we must then have \( \sqrt[3]{B} = a - {bi} \) . The roots of \( f \) are then \( {\alpha }_{1} = {2a},{\alpha }_{2} = - a - b\sqrt{3} \), and \( {\alpha }_{3} = - a + b\sqrt{3} \), and all three are real numbers. | Yes |
Theorem 13.4 With the notation above, let \( m = \left\lbrack {L : F}\right\rbrack \) . 1. \( G \cong {S}_{4} \) if and only if \( r\left( x\right) \) is irreducible over \( F \) and \( D \notin {F}^{2} \), if and only if \( m = 6 \) . | Proof. We first point out a couple of things. First, \( \left\lbrack {K : L}\right\rbrack \leq 4 \), since \( K = L\left( {\alpha }_{1}\right) \) . This equality follows from the fundamental theorem, since only the identity automorphism fixes \( L\left( {\alpha }_{1}\right) \) . Second, \( r\left( x\right) \) is irreducible over \( F \) if and only if \( m = 3 \) or \( m = 6 \) . Also, \( r\left( x\right) \) has a unique root in \( F \) if and only if \( m = 2 \) . Finally, if \( \sigma \) is a 4-cycle, then \( {\sigma }^{2} \in V \) . Suppose that \( r\left( x\right) \) is irreducible over \( F \) . Then \( m \) is either 3 or 6, so 3 divides \( \left| G\right| \) . This forces \( G \) to be isomorphic to either \( {S}_{4} \) or \( {A}_{4} \) . In either case, \( V \subseteq G \), so \( L = \mathcal{F}\left( V\right) \) by the fundamental theorem. Thus, \( \left\lbrack {K : L}\right\rbrack = 4 \) , so \( G = {S}_{4} \) if \( m = 6 \), and \( G = {A}_{4} \) if \( m = 3 \) . Alternatively, \( G = {S}_{4} \) if and only if \( D \notin {F}^{2} \), and \( G = {A}_{4} \) if and only if \( D \in {F}^{2} \) . Conversely, if \( G = {S}_{4} \) , then \( m = \left| {{S}_{4} : V}\right| = 6 \), and if \( G = {A}_{4} \), then \( m = \left| {{A}_{4} : V}\right| = 3 \) . In either case,3 divides \( \left| G\right| \), so \( r\left( x\right) \) is irreducible over \( F \) . | Yes |
Example 13.5 Let \( f\left( x\right) = {x}^{4} + {x}^{3} + {x}^{2} + x + 1 \) . Then \( a = b = c = d = 1 \) , so \( {s}_{1} = {s}_{3} = - 1 \) and \( {s}_{2} = {s}_{4} = 1 \) . Also,\n\n\[ r\left( x\right) = {x}_{3} - {x}_{2} - {3x} + 2 = \left( {x - 2}\right) \left( {{x}^{2} + x - 1}\right) . \] | Set \( {\beta }_{1} = 2 \) . Then \( u = \sqrt{5} \) . Also,\n\n\[ {v}^{2} = \frac{1}{4}{\left( -1 + u\right) }^{2} - 2\left( {2 + {u}^{-1}\left( {-2 + 2}\right) }\right) \]\n\n\[ = \frac{1}{4}\left( {{u}^{2} - {2u} + 1}\right) - 4 = - \frac{5 + u}{2}. \]\n\nThus, \( v = \frac{i}{2}\sqrt{{10} - 2\sqrt{5}} \) . In addition, we see that \( {v}^{\prime } = \frac{i}{2}\sqrt{{10} - 2\sqrt{5}} \) . The roots of \( f \) are then\n\n\[ \frac{1}{2}\left( {\frac{i}{2}\sqrt{{10} + 2\sqrt{5}} + \frac{1}{2}\left( {-1 + \sqrt{5}}\right) }\right) = \frac{1}{4}\left( {-1 + \sqrt{5}}\right) + \frac{i}{4}\sqrt{{10} + 2\sqrt{5}} \]\n\n\[ \frac{1}{2}\left( {-\frac{i}{2}\sqrt{{10} + 2\sqrt{5}} + \frac{1}{2}\left( {-1 + \sqrt{5}}\right) }\right) = \frac{1}{4}\left( {-1 + \sqrt{5}}\right) - \frac{i}{4}\sqrt{{10} + 2\sqrt{5}} \]\n\n\[ \frac{1}{2}\left( {\frac{i}{2}\sqrt{{10} - 2\sqrt{5}} + \frac{1}{2}\left( {-1 - \sqrt{5}}\right) }\right) = \frac{1}{4}\left( {-1 - \sqrt{5}}\right) + \frac{i}{4}\sqrt{{10} - 2\sqrt{5}} \]\n\n\[ \frac{1}{2}\left( {-\frac{i}{2}\sqrt{{10} - 2\sqrt{5}} + \frac{1}{2}\left( {-1 - \sqrt{5}}\right) }\right) = \frac{1}{4}\left( {-1 - \sqrt{5}}\right) - \frac{i}{4}\sqrt{{10} - 2\sqrt{5}}. \]\n\nThe polynomial \( h\left( x\right) = \left( {{x}^{2} - {2x} + 1}\right) \left( {{x}^{2} + x - 1}\right) \) splits over \( L \), so by Theorem 13.4 the Galois group of \( f \) is isomorphic to \( {C}_{4} \) . Alternatively, \( f\left( x\right) \) is the fifth cyclotomic polynomial \( {\Psi }_{5}\left( x\right) \), so Section 7 tells us that the Galois group of \( f \) is cyclic. | Yes |
Theorem 14.1 (Lindemann-Weierstrauss) Let \( {\alpha }_{1},\ldots ,{\alpha }_{m} \) be distinct algebraic numbers. Then the exponentials \( {e}^{{\alpha }_{1}},\ldots ,{e}^{{\alpha }_{m}} \) are linearly independent over \( \mathbb{Q} \) . | Proof of the theorem. Suppose that there are \( {a}_{j} \in \mathbb{Q} \) with\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{m}{a}_{j}{e}^{{\alpha }_{j}} = 0 \]\n\nBy multiplying by a suitable integer, we may assume that each \( {a}_{j} \in \mathbb{Z} \) . Moreover, by eliminating terms if necessary, we may also assume that each\n\n\( {a}_{j} \neq 0 \) . Let \( K \) be the normal closure of \( \mathbb{Q}\left( {{\alpha }_{1},\ldots ,{\alpha }_{m}}\right) /\mathbb{Q} \) . Then \( K \) is a Galois extension of \( \mathbb{Q} \) . Suppose that \( \operatorname{Gal}\left( {K/\mathbb{Q}}\right) = \left\{ {{\sigma }_{1},\ldots ,{\sigma }_{n}}\right\} \) . Since \( \mathop{\sum }\limits_{{j = 1}}^{m}{a}_{j}{e}^{{\alpha }_{j}} = 0 \), we have\n\n\[ 0 = \mathop{\prod }\limits_{{k = 1}}^{n}\left( {\mathop{\sum }\limits_{{j = 1}}^{n}{a}_{j}{e}^{{\sigma }_{k}\left( {\alpha }_{j}\right) }}\right) = \mathop{\sum }\limits_{{j = 0}}^{r}{c}_{j}{e}^{{\beta }_{j}}, \]\n\nwhere the \( {c}_{j} \in \mathbb{Z} \) and the \( {\beta }_{j} \) can be chosen to be distinct elements of \( K \) by gathering together terms with the same exponent. Moreover, some \( {c}_{j} \neq 0 \) (see Problem 4); without loss of generality, say \( {c}_{0} \neq 0 \) . If \( \sigma \in \) \( \operatorname{Gal}\left( {K/\mathbb{Q}}\right) \), then the \( n \) terms \( \mathop{\sum }\limits_{{j = 1}}^{n}{a}_{j}{e}^{\sigma {\sigma }_{k}\left( {\alpha }_{j}\right) } \) for \( 1 \leq k \leq n \) are the terms \( \mathop{\sum }\limits_{{j = 1}}^{n}{a}_{j}{e}^{{\sigma }_{k}\left( {\alpha }_{j}\right) } \) in some order, so the product is unchanged when replacing \( {\sigma }_{k}\left( {\alpha }_{j}\right) \) by \( \sigma {\sigma }_{k}\left( {\alpha }_{j}\right) \) . Since each \( {\beta }_{j} \) is a sum of terms of the form \( {\sigma }_{k}\left( {\alpha }_{l}\right) \) , the exponents in the expansion of \( \mathop{\prod }\limits_{{k = 1}}^{n}\left( {\mathop{\sum }\limits_{{j = 1}}^{n}{a}_{j}{e}^{\sigma {\sigma }_{k}\left( {\alpha }_{j}\right) }}\right) \) are the various \( \sigma \left( {\beta }_{j}\right) \) . Thus, we obtain equations\n\n\[ 0 = \mathop{\sum }\limits_{{j = 0}}^{r}{c}_{j}{e}^{{\sigma }_{i}\left( {\beta }_{j}\right) } \]\n\nfor each \( i \) . Multiplying the \( i \) th equation by \( {e}^{{\sigma }_{i}\left( {\beta }_{0}\right) } \), we get\n\n\[ 0 = {c}_{0} + \mathop{\sum }\limits_{{j = 1}}^{r}{c}_{j}{e}^{{\sigma }_{i}\left( {\gamma }_{j}\right) } \]\n\n(14.1)\n\nwhere \( {\gamma }_{j} = {\beta }_{j} - {\beta }_{0} \) . Note that \( {\gamma }_{j} \neq 0 \) since the \( {\beta }_{j} \) are all distinct. Each \( {\gamma }_{j} \in K \) ; hence, each \( {\gamma }_{j} \) is algebraic over \( \mathbb{Q} \) . Thus, for a fixed \( j \), the elements \( {\sigma }_{i}\left( {\gamma }_{j}\right) \) are roots of a polynomial \( {g}_{j}\left( x\right) \in \mathbb{Q}\left\lbrack x\right\rbrack \), where the leading coefficie | Yes |
Corollary 14.2 The numbers \( \pi \) and \( e \) are transcendental over \( \mathbb{Q} \) . | Proof of the corollary. Suppose that \( e \) is algebraic over \( \mathbb{Q} \) . Then there are rationals \( {r}_{i} \) with \( \mathop{\sum }\limits_{{i = 0}}^{n}{r}_{i}{e}^{i} = 0 \) . This means that the numbers \( {e}^{0} \) , \( {e}^{1},\ldots ,{e}^{n - 1} \) are linearly dependent over \( \mathbb{Q} \) . By choosing \( m = n + 1 \) and \( {\alpha }_{i} = i - 1 \), this dependence is false by the theorem. Thus, \( e \) is transcendental over \( \mathbb{Q} \) . For \( \pi \), we note that if \( \pi \) is algebraic over \( \mathbb{Q} \), then so is \( {\pi i} \) ; hence, \( {e}^{0},{e}^{\pi i} \) are linearly independent over \( \mathbb{Q} \), which is false since \( {e}^{\pi i} = - 1 \) . Thus, \( \pi \) is transcendental over \( \mathbb{Q} \) . | Yes |
Lemma 15.1 Let \( K \) be a subfield of \( \mathbb{R} \) .\n\n1. The intersection of two lines in \( K \) is either empty or is a point in the plane of \( K \) .\n\n2. The intersection of a line and a circle in \( K \) is either empty or consists of one or two points in the plane of \( K\left( \sqrt{u}\right) \) for some \( u \in K \) with \( u \geq 0 \) .\n\n3. The intersection of two circles in \( K \) is either empty or consists of one or two points in the plane of \( K\left( \sqrt{u}\right) \) for some \( u \in K \) with \( u \geq 0 \) . | Proof. The first statement is an easy calculation. For the remaining two statements, it suffices to prove statement 2, since if \( {x}^{2} + {y}^{2} + {ax} + {by} + c = 0 \) and \( {x}^{2} + {y}^{2} + {a}^{\prime }x + {b}^{\prime }y + {c}^{\prime } = 0 \) are the equations of circles \( C \) and \( {C}^{\prime } \) , respectively, then their intersection is the intersection of \( C \) with the line \( \left( {a - {a}^{\prime }}\right) x + \left( {b - {b}^{\prime }}\right) y + \left( {c - {c}^{\prime }}\right) = 0 \) . So, to prove statement 2, suppose that our line \( L \) in \( K \) has the equation \( {dx} + {ey} + f = 0 \) . We assume that \( d \neq 0 \) , since if \( d = 0 \), then \( e \neq 0 \) . By dividing by \( d \), we may then assume that \( d = 1 \) . Plugging \( - x = {ey} + f \) into the equation of \( C \), we obtain\n\n\[ \left( {{e}^{2} + 1}\right) {y}^{2} + \left( {{2ef} - {ae} + b}\right) y + \left( {{f}^{2} - {af} + c}\right) = 0. \]\n\nWriting this equation in the form \( \alpha {y}^{2} + {\beta y} + \gamma = 0 \), if \( \alpha = 0 \), then \( y \in K \) . If \( \alpha \neq 0 \), then completing the square shows that either \( L \cap C = \varnothing \) or \( y \in K\left( \sqrt{{\beta }^{2} - {4\alpha \gamma }}\right) \) with \( {\beta }^{2} - {4\alpha \gamma } \geq 0 \) . | Yes |
Theorem 15.2 A real number \( c \) is constructible if and only if there is a tower of fields \( \mathbb{Q} = {K}_{0} \subseteq {K}_{1} \subseteq \cdots \subseteq {K}_{r} \) such that \( c \in {K}_{r} \) and \( \left\lbrack {{K}_{i + 1} : }\right. \) \( \left. {K}_{i}\right\rbrack \leq 2 \) for each \( i \) . Therefore, if \( c \) is constructible, then \( c \) is algebraic over \( \mathbb{Q} \), and \( \left\lbrack {\mathbb{Q}\left( c\right) : \mathbb{Q}}\right\rbrack \) is a power of 2 . | Proof. If \( c \) is constructible, then the point \( \left( {c,0}\right) \) can be obtained from a finite sequence of constructions starting from the plane of \( \mathbb{Q} \) . We then obtain a finite sequence of points, each an intersection of constructible lines and circles, ending at \( \left( {c,0}\right) \) . By Lemma 15.1, the first point either lies in \( \mathbb{Q} \) or in \( \mathbb{Q}\left( \sqrt{u}\right) \) for some \( u \) . This extension has degree either 1 or 2 . Each time we construct a new point, we obtain a field extension whose degree over the previous field is either 1 or 2 by the lemma. Thus, we obtain a sequence of fields\n\n\[ \mathbb{Q} = {K}_{0} \subseteq {K}_{1} \subseteq {K}_{2} \subseteq \cdots \subseteq {K}_{r} \]\n\nwith \( \left\lbrack {{K}_{i + 1} : {K}_{i}}\right\rbrack \leq 2 \) and \( c \in {K}_{r} \) . Therefore, \( \left\lbrack {{K}_{r} : \mathbb{Q}}\right\rbrack = {2}^{n} \) for some \( n \) . However, \( \left\lbrack {\mathbb{Q}\left( c\right) : \mathbb{Q}}\right\rbrack \) divides \( \left\lbrack {{K}_{r} : \mathbb{Q}}\right\rbrack \), so \( \left\lbrack {\mathbb{Q}\left( c\right) : \mathbb{Q}}\right\rbrack \) is also a power of 2 .\n\nFor the converse, suppose that we have a tower \( \mathbb{Q} = {K}_{0} \subseteq {K}_{1} \subseteq \cdots \subseteq {K}_{r} \) with \( c \in {K}_{r} \) and \( \left\lbrack {{K}_{i + 1} : {K}_{i}}\right\rbrack \leq 2 \) for each \( i \) . We show that \( c \) is constructible by induction on \( r \) . If \( r = 0 \), then \( c \in \mathbb{Q} \), so \( c \) is constructible. Assume then that \( r > 0 \) and that elements of \( {K}_{r - 1} \) are constructible. Since \( \left\lbrack {{K}_{r} : {K}_{r - 1}}\right\rbrack \leq 2 \), the quadratic formula shows that we may write \( {K}_{r} = {K}_{r - 1}\left( \sqrt{a}\right) \) for some \( a \in {K}_{r - 1} \) . Since \( a \) is constructible by assumption, so is \( \sqrt{a} \) . Therefore, \( {K}_{r} = {K}_{r - 1}\left( \sqrt{a}\right) \) lies in the field of constructible numbers; hence, \( c \) is constructible. | Yes |
Theorem 15.3 It is impossible to trisect a \( {60}^{ \circ } \) angle by ruler and compass construction. | Proof. As noted above, a \( {60}^{ \circ } \) angle can be constructed. If a \( {60}^{ \circ } \) angle can be trisected, then it is possible to construct the number \( \alpha = \cos {20}^{ \circ } \) . However, the triple angle formula \( \cos {3\theta } = 4{\cos }^{3}\theta - 3\cos \theta \) gives \( 4{\alpha }^{3} - {3\alpha } = \cos {60}^{ \circ } = \) \( 1/2 \) . Thus, \( \alpha \) is algebraic over \( \mathbb{Q} \) . The polynomial \( 8{x}^{3} - {6x} - 1 \) is irreducible over \( \mathbb{Q} \) because it has no rational roots. Therefore, \( \left\lbrack {\mathbb{Q}\left( \alpha \right) : \mathbb{Q}}\right\rbrack = 3 \), so \( \alpha \) is not constructible. A \( {20}^{ \circ } \) angle cannot then be constructed, so a \( {60}^{ \circ } \) degree angle cannot be trisected. | Yes |
Theorem 15.4 It is impossible to double a cube of length 1 by ruler and compass construction. | Proof. The length of a side of a cube of volume 2 is \( \sqrt[3]{2} \) . The minimal polynomial of \( \sqrt[3]{2} \) over \( \mathbb{Q} \) is \( {x}^{3} - 2 \) . Thus, \( \left\lbrack {\mathbb{Q}\left( \sqrt[3]{2}\right) : \mathbb{Q}}\right\rbrack = 3 \) is not a power of 2, so \( \sqrt[3]{2} \) is not constructible. | Yes |
Theorem 15.5 It is impossible to square a circle of radius 1. | Proof. We are asking whether we can construct a square of area \( \pi \) . To do so requires us to construct a line segment of length \( \sqrt{\pi } \), which is impossible since \( \sqrt{\pi } \) is transcendental over \( \mathbb{Q} \) by the Lindemann-Weierstrauss theorem; hence, \( \sqrt{\pi } \) is not algebraic of degree a power of 2 . | Yes |
Theorem 15.6 A regular \( n \) -gon is constructible if and only if \( \phi \left( n\right) \) is a power of \( 2 \) . | Proof. We point out that a regular \( n \) -gon is constructible if and only if the central angles \( {2\pi }/n \) are constructible, and this occurs if and only if \( \cos \left( {{2\pi }/n}\right) \) is a constructible number. Let \( \omega = {e}^{{2\pi i}/n} = \cos \left( {{2\pi }/n}\right) + \) \( i\sin \left( {{2\pi }/n}\right) \), a primitive \( n \) th root of unity. Then \( \cos \left( {{2\pi }/n}\right) = \frac{1}{2}\left( {\omega + {\omega }^{-1}}\right) \) , since \( {\omega }^{-1} = \cos \left( {{2\pi }/n}\right) - i\sin \left( {{2\pi }/n}\right) \) . Thus, \( \cos \left( {{2\pi }/n}\right) \in \mathbb{Q}\left( \omega \right) \) . However, \( \cos \left( {{2\pi }/n}\right) \in \mathbb{R} \) and \( \omega \notin \mathbb{R} \), so \( \mathbb{Q}\left( \omega \right) \neq \mathbb{Q}\left( {\cos \left( {{2\pi }/n}\right) }\right) \) . But \( \omega \) is a root of \( {x}^{2} - \) \( 2\cos \left( {{2\pi }/n}\right) x + 1 \), as an easy calculation shows, so \( \left\lbrack {\mathbb{Q}\left( \omega \right) : \mathbb{Q}\left( {\cos \left( {{2\pi }/n}\right) }\right) }\right\rbrack = \) 2. Therefore, if \( \cos \left( {{2\pi }/n}\right) \) is constructible, then \( \left\lbrack {\mathbb{Q}\left( {\cos \left( {{2\pi }/n}\right) }\right) : \mathbb{Q}}\right\rbrack \) is a power of 2 . Hence, \( \phi \left( n\right) = \left\lbrack {\mathbb{Q}\left( \omega \right) : \mathbb{Q}}\right\rbrack \) is also a power of 2 .\n\nConversely, suppose that \( \phi \left( n\right) \) is a power of 2 . The field \( \mathbb{Q}\left( \omega \right) \) is a Galois extension of \( \mathbb{Q} \) with Abelian Galois group by Proposition 7.2. If \( H = \operatorname{Gal}\left( {\mathbb{Q}\left( \omega \right) /\mathbb{Q}\left( {\cos \left( {{2\pi }/n}\right) }\right) }\right) \), by the theory of finite Abelian groups there is a chain of subgroups\n\n\[ \n{H}_{0} \subseteq {H}_{1} \subseteq \cdots \subseteq {H}_{r} = H \n\]\n\nwith \( \left| {{H}_{i + 1} : {H}_{i}}\right| = 2 \) . If \( {L}_{i} = \mathcal{F}\left( {H}_{i}\right) \), then \( \left\lbrack {{L}_{i} : {L}_{i + 1}}\right\rbrack = 2 \) ; thus, \( {L}_{i} = \) \( {L}_{i + 1}\left( \sqrt{{u}_{i}}\right) \) for some \( {u}_{i} \) . Since \( {L}_{i} \subseteq \mathbb{Q}\left( {\cos \left( {{2\pi }/n}\right) }\right) \subseteq \mathbb{R} \), each of the \( {u}_{i} \geq \) 0 . Since the square root of a constructible number is constructible, we see that everything in \( \mathbb{Q}\left( {\cos \left( {{2\pi }/n}\right) }\right) \) is constructible. Thus, \( \cos \left( {{2\pi }/n}\right) \) is constructible, so a regular \( n \) -gon is constructible. | Yes |
Lemma 16.6 Let \( K \) be an \( n \)-radical extension of \( F \), and let \( N \) be the normal closure of \( K/F \). Then \( N \) is an \( n \)-radical extension of \( F \). | Proof. Let \( K = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right) \) with \( {\alpha }_{i}^{n} \in F\left( {{\alpha }_{1},\ldots ,{\alpha }_{i - 1}}\right) \). We argue by induction on \( r \). If \( r = 1 \), then \( K = F\left( \alpha \right) \) with \( {\alpha }^{n} = a \in F \). Then \( N = F\left( {{\beta }_{1},\ldots ,{\beta }_{m}}\right) \), where the \( {\beta }_{i} \) are the roots of \( \min \left( {F,\alpha }\right) \). However, this minimal polynomial divides \( {x}^{n} - a \), so \( {\beta }_{i}^{n} = a \). Thus, \( N \) is an \( n \)-radical extension of \( F \). Now suppose that \( r > 1 \). Let \( {N}_{0} \) be the normal closure of \( F\left( {{\alpha }_{1},\ldots ,{\alpha }_{r - 1}}\right) \) over \( F \). By induction, \( {N}_{0} \) is an \( n \)-radical extension of \( F \). Since \( {N}_{0} \) is the splitting field over \( F \) of \( \left\{ {\min \left( {F,{\alpha }_{i}}\right) : 1 \leq i \leq r - 1}\right\} \), and \( N \) is the splitting field of all \( \min \left( {F,{\alpha }_{i}}\right) \), we have \( N = {N}_{0}\left( {{\gamma }_{1},\ldots ,{\gamma }_{m}}\right) \), where the \( {\gamma }_{i} \) are roots of \( \min \left( {F,{\alpha }_{r}}\right) \). Also, \( {\alpha }_{r}^{n} = b \) for some \( b \in F\left( {{\alpha }_{1},\ldots ,{\alpha }_{r - 1}}\right) \subseteq \) \( {N}_{0} \). By the isomorphism extension theorem, for each \( i \) there is a \( {\sigma }_{i} \in \) \( \operatorname{Gal}\left( {N/F}\right) \) with \( {\sigma }_{i}\left( {\alpha }_{r}\right) = {\gamma }_{i} \). Therefore, \( {\gamma }_{i}^{n} = {\sigma }_{i}\left( b\right) \) by Proposition 3.28. However, \( {N}_{0} \) is normal over \( F \), and \( b \in {N}_{0} \), so \( {\sigma }_{i}\left( b\right) \in {N}_{0} \). Thus, each \( {\gamma }_{i} \) is an \( n \) th power of some element of \( {N}_{0} \), so \( N \) is an \( n \)-radical extension of \( {N}_{0} \). Since \( {N}_{0} \) is an \( n \)-radical extension of \( F \), \( N \) is an \( n \)-radical extension of \( F \). | Yes |
Example 16.4 If \( K = \mathbb{Q}\left( \sqrt[4]{2}\right) \), then \( K \) is both a 4-radical extension and a 2-radical extension of \( \mathbb{Q} \). | The second statement is true by considering the tower\n\n\[\n\mathbb{Q} \subseteq \mathbb{Q}\left( \sqrt{2}\right) \subseteq \mathbb{Q}\left( \sqrt{2}\right) \left( \sqrt{\sqrt{2}}\right) = \mathbb{Q}\left( \sqrt[4]{2}\right)\n\] | Yes |
Lemma 16.6 Let \( K \) be an \( n \) -radical extension of \( F \), and let \( N \) be the normal closure of \( K/F \) . Then \( N \) is an \( n \) -radical extension of \( F \) . | Proof. Let \( K = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right) \) with \( {\alpha }_{i}^{n} \in F\left( {{\alpha }_{1},\ldots ,{\alpha }_{i - 1}}\right) \) . We argue by induction on \( r \) . If \( r = 1 \), then \( K = F\left( \alpha \right) \) with \( {\alpha }^{n} = a \in F \) . Then \( N = F\left( {{\beta }_{1},\ldots ,{\beta }_{m}}\right) \), where the \( {\beta }_{i} \) are the roots of \( \min \left( {F,\alpha }\right) \) . However, this minimal polynomial divides \( {x}^{n} - a \), so \( {\beta }_{i}^{n} = a \) . Thus, \( N \) is an \( n \) -radical extension of \( F \) . Now suppose that \( r > 1 \) . Let \( {N}_{0} \) be the normal closure of \( F\left( {{\alpha }_{1},\ldots ,{\alpha }_{r - 1}}\right) \) over \( F \) . By induction, \( {N}_{0} \) is an \( n \) -radical extension of \( F \) . Since \( {N}_{0} \) is the splitting field over \( F \) of \( \left\{ {\min \left( {F,{\alpha }_{i}}\right) : 1 \leq i \leq r - 1}\right\} \), and \( N \) is the splitting field of all \( \min \left( {F,{\alpha }_{i}}\right) \), we have \( N = {N}_{0}\left( {{\gamma }_{1},\ldots ,{\gamma }_{m}}\right) \), where the \( {\gamma }_{i} \) are roots of \( \min \left( {F,{\alpha }_{r}}\right) \) . Also, \( {\alpha }_{r}^{n} = b \) for some \( b \in F\left( {{\alpha }_{1},\ldots ,{\alpha }_{r - 1}}\right) \subseteq \) \( {N}_{0} \) . By the isomorphism extension theorem, for each \( i \) there is a \( {\sigma }_{i} \in \) \( \operatorname{Gal}\left( {N/F}\right) \) with \( {\sigma }_{i}\left( {\alpha }_{r}\right) = {\gamma }_{i} \) . Therefore, \( {\gamma }_{i}^{n} = {\sigma }_{i}\left( b\right) \) by Proposition 3.28. However, \( {N}_{0} \) is normal over \( F \), and \( b \in {N}_{0} \), so \( {\sigma }_{i}\left( b\right) \in {N}_{0} \) . Thus, each \( {\gamma }_{i} \) is an \( n \) th power of some element of \( {N}_{0} \), so \( N \) is an \( n \) -radical extension of \( {N}_{0} \) . Since \( {N}_{0} \) is an \( n \) -radical extension of \( F \), we see that \( N \) is an \( n \) -radical extension of \( F \) . | Yes |
Lemma 16.6 Let \( K \) be an \( n \) -radical extension of \( F \), and let \( N \) be the normal closure of \( K/F \) . Then \( N \) is an \( n \) -radical extension of \( F \) . | Proof. Let \( K = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right) \) with \( {\alpha }_{i}^{n} \in F\left( {{\alpha }_{1},\ldots ,{\alpha }_{i - 1}}\right) \) . We argue by induction on \( r \) . If \( r = 1 \), then \( K = F\left( \alpha \right) \) with \( {\alpha }^{n} = a \in F \) . Then \( N = F\left( {{\beta }_{1},\ldots ,{\beta }_{m}}\right) \), where the \( {\beta }_{i} \) are the roots of \( \min \left( {F,\alpha }\right) \) . However, this minimal polynomial divides \( {x}^{n} - a \), so \( {\beta }_{i}^{n} = a \) . Thus, \( N \) is an \( n \) -radical extension of \( F \) . Now suppose that \( r > 1 \) . Let \( {N}_{0} \) be the normal closure of \( F\left( {{\alpha }_{1},\ldots ,{\alpha }_{r - 1}}\right) \) over \( F \) . By induction, \( {N}_{0} \) is an \( n \) -radical extension of \( F \) . Since \( {N}_{0} \) is the splitting field over \( F \) of \( \left\{ {\min \left( {F,{\alpha }_{i}}\right) : 1 \leq i \leq r - 1}\right\} \), and \( N \) is the splitting field of all \( \min \left( {F,{\alpha }_{i}}\right) \), we have \( N = {N}_{0}\left( {{\gamma }_{1},\ldots ,{\gamma }_{m}}\right) \), where the \( {\gamma }_{i} \) are roots of \( \min \left( {F,{\alpha }_{r}}\right) \) . Also, \( {\alpha }_{r}^{n} = b \) for some \( b \in F\left( {{\alpha }_{1},\ldots ,{\alpha }_{r - 1}}\right) \subseteq \) \( {N}_{0} \) . By the isomorphism extension theorem, for each \( i \) there is a \( {\sigma }_{i} \in \) \( \operatorname{Gal}\left( {N/F}\right) \) with \( {\sigma }_{i}\left( {\alpha }_{r}\right) = {\gamma }_{i} \) . Therefore, \( {\gamma }_{i}^{n} = {\sigma }_{i}\left( b\right) \) by Proposition 3.28. However, \( {N}_{0} \) is normal over \( F \), and \( b \in {N}_{0} \), so \( {\sigma }_{i}\left( b\right) \in {N}_{0} \) . Thus, each \( {\gamma }_{i} \) is an \( n \) th power of some element of \( {N}_{0} \), so \( N \) is an \( n \) -radical extension of \( {N}_{0} \) . Since \( {N}_{0} \) is an \( n \) -radical extension of \( F \), we see that \( N \) is an \( n \) -radical extension of \( F \) . | Yes |
Corollary 16.11 Let \( f\left( x\right) \) be the general \( n \) th degree polynomial over a field of characteristic 0 . If \( n \geq 5 \), then \( f \) is not solvable by radicals. | Example 16.12 Let \( f\left( x\right) = {x}^{5} - {4x} + 2 \in \mathbb{Q}\left\lbrack x\right\rbrack \) . By graphing techniques of calculus, we see that this polynomial has exactly two nonreal roots, as indicated in the graph below.\n\n\n\nFurthermore, \( f \) is irreducible over \( \mathbb{Q} \) by the Eisenstein criterion. Let \( K \) be the splitting field of \( f \) over \( \mathbb{Q} \) . Then \( \left\lbrack {K : \mathbb{Q}}\right\rbrack \) is a multiple of 5, since any root of \( f \) generates a field of dimension 5 over \( \mathbb{Q} \) . Let \( G = \operatorname{Gal}\left( {K/\mathbb{Q}}\right) \) . We can view \( G \subseteq {S}_{5} \) . There is an element of \( G \) of order 5 by Cayley’s theorem, since 5 divides \( \left| G\right| \) . Any element of \( {S}_{5} \) of order 5 is a 5-cycle. Also, if \( \sigma \) is complex conjugation restricted to \( K \), then \( \sigma \) permutes the two nonreal roots of \( f \) and fixes the three others, so \( \sigma \) is a transposition. The subgroup of \( {S}_{5} \) generated by a transposition and a 5-cycle is all of \( {S}_{5} \), so \( G = {S}_{5} \) is not solvable. Thus, \( f \) is not solvable by radicals. | Yes |
Example 16.12 Let \( f\left( x\right) = {x}^{5} - {4x} + 2 \in \mathbb{Q}\left\lbrack x\right\rbrack \) . By graphing techniques of calculus, we see that this polynomial has exactly two nonreal roots, as indicated in the graph below. | Furthermore, \( f \) is irreducible over \( \mathbb{Q} \) by the Eisenstein criterion. Let \( K \) be the splitting field of \( f \) over \( \mathbb{Q} \) . Then \( \left\lbrack {K : \mathbb{Q}}\right\rbrack \) is a multiple of 5, since any root of \( f \) generates a field of dimension 5 over \( \mathbb{Q} \) . Let \( G = \operatorname{Gal}\left( {K/\mathbb{Q}}\right) \) . We can view \( G \subseteq {S}_{5} \) . There is an element of \( G \) of order 5 by Cayley’s theorem, since 5 divides \( \left| G\right| \) . Any element of \( {S}_{5} \) of order 5 is a 5-cycle. Also, if \( \sigma \) is complex conjugation restricted to \( K \), then \( \sigma \) permutes the two nonreal roots of \( f \) and fixes the three others, so \( \sigma \) is a transposition. The subgroup of \( {S}_{5} \) generated by a transposition and a 5-cycle is all of \( {S}_{5} \), so \( G = {S}_{5} \) is not solvable. Thus, \( f \) is not solvable by radicals. | Yes |
Let \( f\left( x\right) = {x}^{3} - {3x} + 1 \in \mathbb{Q}\left\lbrack x\right\rbrack \), and let \( K \) be the splitting field of \( f \) over \( \mathbb{Q} \). We show that \( f \) is solvable by radicals but that \( K \) is not a radical extension of \( \mathbb{Q} \). | Since \( f \) has no roots in \( \mathbb{Q} \) and \( \deg \left( f\right) = 3 \), the polynomial \( f \) is irreducible over \( \mathbb{Q} \). The discriminant of \( f \) is \( {81} = {9}^{2} \), so the Galois group of \( K/\mathbb{Q} \) is \( {A}_{3} \) and \( \left\lbrack {K : \mathbb{Q}}\right\rbrack = 3 \), by Corollary 12.4. Therefore, \( \operatorname{Gal}\left( {K/F}\right) \) is solvable, so \( f \) is solvable by radicals by Galois’ theorem. If \( K \) is a radical extension of \( \mathbb{Q} \), then there is a chain of fields\n\n\[ \mathbb{Q} \subseteq {F}_{1} \subseteq \cdots \subseteq {F}_{r} = K \]\n\nwith \( {F}_{i} = {F}_{i - 1}\left( {\alpha }_{i}\right) \) and \( {\alpha }_{i}^{n} \in {F}_{i - 1} \) for some \( n \). Since \( \left\lbrack {K : \mathbb{Q}}\right\rbrack \) is prime, we see that there is only one proper inclusion in this chain. Thus, \( K = \mathbb{Q}\left( b\right) \) with \( {b}^{n} = u \in \mathbb{Q} \) for some \( n \). The minimal polynomial \( p\left( x\right) \) of \( b \) over \( \mathbb{Q} \) splits in \( K \), since \( K/\mathbb{Q} \) is normal. Let \( {b}^{\prime } \) be another root of \( p\left( x\right) \). Then \( {b}^{n} = {\left( {b}^{\prime }\right) }^{n} = u \), so \( {b}^{\prime }/b \) is an \( n \) th root of unity. Suppose that \( \mu = {b}^{\prime }/b \) is a primitive \( m \) th root of unity, where \( m \) divides \( n \). Then \( \mathbb{Q}\left( \mu \right) \subseteq K \), so \( \left\lbrack {\mathbb{Q}\left( \mu \right) : \mathbb{Q}}\right\rbrack = \phi \left( m\right) \) is either 1 or 3. An easy calculation shows that \( \phi \left( m\right) \neq 3 \) for all \( m \). Thus, \( \left\lbrack {\mathbb{Q}\left( \mu \right) : \mathbb{Q}}\right\rbrack = 1 \), so \( \mu \in \mathbb{Q} \). However, the only roots of unity in \( \mathbb{Q} \) are \( \pm 1 \), so \( \mu = \pm 1 \). Therefore \( {b}^{\prime } = \pm b \). This proves that \( p\left( x\right) \) has at most two roots, so \( \left\lbrack {\mathbb{Q}\left( b\right) : \mathbb{Q}}\right\rbrack \leq 2 < \left\lbrack {K : \mathbb{Q}}\right\rbrack \), a contradiction to the equality \( \mathbb{Q}\left( b\right) = K \). Thus, \( K \) is not a radical extension of \( \mathbb{Q} \). | Yes |
Lemma 17.1 If \( {\alpha }_{1},\ldots ,{\alpha }_{n} \in K \), then there is an \( E \in \mathcal{I} \) with \( {\alpha }_{i} \in E \) for all \( i \) . | Proof. Let \( E \subseteq K \) be the splitting field of the minimal polynomials of the \( {\alpha }_{i} \) over \( F \) . Then, as each \( {\alpha }_{i} \) is separable over \( F \), the field \( E \) is normal and separable over \( F \) ; hence, \( E \) is Galois over \( F \) . Since there are finitely many \( {\alpha }_{i} \), we have \( \left\lbrack {E : F}\right\rbrack < \infty \), so \( E \in \mathcal{I} \) . | Yes |
Lemma 17.2 Let \( N \in \mathcal{N} \), and set \( N = \operatorname{Gal}\left( {K/E}\right) \) with \( E \in \mathcal{I} \) . Then \( E = \mathcal{F}\left( N\right) \) and \( N \) is normal in \( G \) . Moreover, \( G/N \cong \operatorname{Gal}\left( {E/F}\right) \) . Thus, \( \left| {G/N}\right| = \left| {\operatorname{Gal}\left( {E/F}\right) }\right| = \left\lbrack {E : F}\right\rbrack < \infty \). | Proof. Since \( K \) is normal and separable over \( F \), the field \( K \) is also normal and separable over \( E \), so \( K \) is Galois over \( E \) . Therefore, \( E = \mathcal{F}\left( N\right) \) . As in the proof of the fundamental theorem, the map \( \theta : G \rightarrow \operatorname{Gal}\left( {E/F}\right) \) given by \( \sigma \mapsto {\left. \sigma \right| }_{E} \) is a group homomorphism with kernel \( \operatorname{Gal}\left( {K/E}\right) = N \) . Proposition 3.28 shows that \( \theta \) is surjective. The remaining statements then follow. | Yes |
Lemma 17.3 We have \( \mathop{\bigcap }\limits_{{N \in \mathcal{N}}}N = \{ \mathrm{{id}}\} \) . Furthermore, \( \mathop{\bigcap }\limits_{{N \in \mathcal{N}}}{\sigma N} = \{ \sigma \} \) for all \( \sigma \in G \) . | Proof. Let \( \tau \in \mathop{\bigcap }\limits_{{N \in \mathcal{N}}}N \) and let \( a \in K \) . By Lemma 17.1, there is an \( E \in \mathcal{I} \) with \( a \in E \) . Set \( N = \operatorname{Gal}\left( {K/E}\right) \in \mathcal{N} \) . The automorphism \( \tau \) fixes \( E \) since \( \tau \in N \), so \( \tau \left( a\right) = a \) . Thus, \( \tau = \mathrm{{id}} \), so \( \mathop{\bigcap }\limits_{{N \in \mathcal{N}}}N = \{ \mathrm{{id}}\} \) . For the second statement, if \( \tau \in {\sigma N} \) for all \( N \), then \( {\sigma }^{-1}\tau \in N \) for all \( N \) ; thus, \( {\sigma }^{-1}\tau = \mathrm{{id}} \) by the first part. This yields \( \tau = \sigma \), so \( \mathop{\bigcap }\limits_{{N \in \mathcal{N}}}{\sigma N} = \{ \sigma \} \) . | Yes |
Lemma 17.4 Let \( {N}_{1},{N}_{2} \in \mathcal{N} \) . Then \( {N}_{1} \cap {N}_{2} \in \mathcal{N} \) . | Proof. Let \( {N}_{i} = \operatorname{Gal}\left( {K/{E}_{i}}\right) \) with \( {E}_{i} \in \mathcal{I} \) . Each \( {E}_{i} \) is finite Galois over \( F \) ; hence, \( {E}_{1}{E}_{2} \) is also finite Galois over \( F \), so \( {E}_{1}{E}_{2} \in \mathcal{I} \) . However, \( \operatorname{Gal}\left( {K/{E}_{1}{E}_{2}}\right) = {N}_{1} \cap {N}_{2} \) ; to see this, we note that \( \sigma \in {N}_{1} \cap {N}_{2} \) if and only if \( {\left. \sigma \right| }_{{E}_{1}} = \) id and \( {\left. \sigma \right| }_{{E}_{2}} = \) id, if and only if \( {E}_{1} \subseteq \mathcal{F}\left( \sigma \right) \) and \( {E}_{2} \subseteq \mathcal{F}\left( \sigma \right) \) , and if and only if \( {E}_{1}{E}_{2} \subseteq \mathcal{F}\left( \sigma \right) \) . This last condition is true if and only if \( \sigma \in \operatorname{Gal}\left( {K/{E}_{1}{E}_{2}}\right) \) . Thus, \( {N}_{1} \cap {N}_{2} = \operatorname{Gal}\left( {K/{E}_{1}{E}_{2}}\right) \in \mathcal{N} \) . | Yes |
Theorem 17.7 Let \( H \) be a subgroup of \( G \), and let \( {H}^{\prime } = \operatorname{Gal}\left( {K/\mathcal{F}\left( H\right) }\right) \) . Then \( {H}^{\prime } = \bar{H} \), the closure of \( H \) in the topology of \( G \) . | Proof. It is clear that \( H \subseteq {H}^{\prime } \), so it suffices to show that \( {H}^{\prime } \) is closed and that \( {H}^{\prime } \subseteq \bar{H} \) . To show that \( {H}^{\prime } \) is closed, take any \( \sigma \in G - {H}^{\prime } \) . Then there is an \( \alpha \in \mathcal{F}\left( H\right) \) with \( \sigma \left( \alpha \right) \neq \alpha \) . Take \( E \in \mathcal{I} \) with \( \alpha \in E \), and let \( N = \operatorname{Gal}\left( {K/E}\right) \in \mathcal{N} \) . Then, for any \( \tau \in N \), we have \( \tau \left( \alpha \right) = \alpha \), so \( {\sigma \tau }\left( \alpha \right) = \sigma \left( \alpha \right) \neq \alpha \) . Hence, \( {\sigma N} \) is an open neighborhood of \( \sigma \) disjoint from \( {H}^{\prime } \) . Therefore, \( G - {H}^{\prime } \) is open, so \( {H}^{\prime } \) is closed. To prove the inclusion \( {H}^{\prime } \subseteq \) \( \bar{H} \), we first set \( L = \mathcal{F}\left( H\right) \) . Let \( \sigma \in {H}^{\prime } \) and \( N \in \mathcal{N} \) . Set \( E = \mathcal{F}\left( N\right) \in \mathcal{I} \) , and let \( {H}_{0} = \left\{ {{\left. \rho \right| }_{E} : \rho \in H}\right\} \), a subgroup of the finite group \( \operatorname{Gal}\left( {E/F}\right) \) . Since \( \mathcal{F}\left( {H}_{0}\right) = \mathcal{F}\left( H\right) \cap E = L \cap E \), the fundamental theorem for finite Galois extensions shows that \( {H}_{0} = \operatorname{Gal}\left( {E/\left( {E \cap L}\right) }\right) \) . Since \( \sigma \in {H}^{\prime } \), we have \( {\left. \sigma \right| }_{L} = \) id, so \( {\left. \sigma \right| }_{E} \in {H}_{0} \) . Therefore, there is a \( \rho \in H \) with \( {\left. \rho \right| }_{E} = {\left. \sigma \right| }_{E} \) . Thus, \( {\sigma }^{-1}\rho \in \operatorname{Gal}\left( {K/E}\right) = N \), so \( \rho \in {\sigma N} \cap H \) . This shows that every basic open neighborhood \( {\sigma N} \) of \( \sigma \in {H}^{\prime } \) meets \( H \), so \( \sigma \in \bar{H} \) . This proves the inclusion \( {H}^{\prime } \subseteq \bar{H} \) and finishes the proof. | Yes |
Theorem 17.8 (Fundamental Theorem of Infinite Galois Theory) Let \( K \) be a Galois extension of \( F \), and let \( G = \operatorname{Gal}\left( {K/F}\right) \) . With the Krull topology on \( G \), the maps \( L \mapsto \operatorname{Gal}\left( {K/L}\right) \) and \( H \mapsto \mathcal{F}\left( H\right) \) give an inclusion reversing correspondence between the fields \( L \) with \( F \subseteq L \subseteq K \) and the closed subgroups \( H \) of \( G \) . Furthermore, if \( L \leftrightarrow H \), then \( \left| {G : H}\right| < \infty \) if and only if \( \left\lbrack {L : F}\right\rbrack < \infty \), if and only if \( H \) is open. When this occurs, \( \left| {G : H}\right| = \left\lbrack {L : F}\right\rbrack \) . Also, \( H \) is normal in \( G \) if and only if \( L \) is Galois over \( F \), and when this occurs, there is a group isomorphism \( \operatorname{Gal}\left( {L/F}\right) \cong G/N \) . If \( G/N \) is given the quotient topology, this isomorphism is also a homeomorphism. | Proof. If \( L \) is a subfield of \( K \) containing \( F \), then \( K \) is normal and separable over \( L \), so \( K \) is Galois over \( L \) . Thus, \( L = \mathcal{F}\left( {\operatorname{Gal}\left( {K/L}\right) }\right) \) . If \( H \) is a subgroup of \( G \), then Theorem 17.7 shows that \( H = \operatorname{Gal}\left( {K/\mathcal{F}\left( H\right) }\right) \) if and only if \( H \) is closed. The two maps \( L \mapsto \operatorname{Gal}\left( {K/L}\right) \) and \( H \mapsto \mathcal{F}\left( H\right) \) then give an inclusion reversing correspondence between the set of intermediate fields of \( K/F \) and the set of closed subgroups of \( G \) .\n\nLet \( L \) be an intermediate field of \( K/F \), and let \( H = \operatorname{Gal}\left( {K/L}\right) \) . Suppose that \( \left| {G : H}\right| < \infty \) . Then \( G - H \) is a finite union of cosets of \( H \), each of which is closed, since \( H \) is closed. Thus, \( G - H \) is closed, so \( H \) is open. Conversely, if \( H \) is open, then \( H \) contains some basic neighborhood of id, so \( N \subseteq H \) for some \( N \in \mathcal{N} \) . If \( E = \mathcal{F}\left( N\right) \), then \( L \subseteq E \), so \( \left\lbrack {L : F}\right\rbrack < \infty \) . Finally, if \( \left\lbrack {L : F}\right\rbrack < \infty \), then choose an \( E \in \mathcal{I} \) with \( L \subseteq E \), possible by Lemma 17.1. Let \( N = \operatorname{Gal}\left( {K/E}\right) \) . Then \( N \subseteq H \), since \( L \subseteq E \), so \( \left| {G : H}\right| \leq \left| {G : N}\right| < \infty \) . By Lemma 17.2, we have \( G/N \cong \operatorname{Gal}\left( {E/F}\right) \) via the map \( {\sigma N} \mapsto {\left. \sigma \right| }_{E} \) . Thus, \( H/N \) maps to \( \left\{ {{\left. \rho \right| }_{E} : \rho \in H}\right\} \), a subgroup of \( \operatorname{Gal}\left( {E/F}\right) \) with fixed field \( L \cap E = L \) . By the fundamental theorem for finite extensions, the order of this group is \( \left\lbrack {E : L}\right\rbrack \) . Therefore,\n\n\[ \left| {G : H}\right| = \left| {G/N : H/N}\right| = \frac{\left| G/N\right| }{\left| H/N\right| } = \frac{\left\lbrack E : F\right\rbrack }{\left\lbrack E : L\right\rbrack } = \left\lbrack {L : F}\right\rbrack \]\n\nFor the statement about normality, we continue to assume that \( H = \) \( \operatorname{Gal}\left( {K/L}\right) \) . Suppose that \( H \) is a normal subgroup of \( G \) . Let \( a \in L \), and let \( f\left( x\right) = \min \left( {F, a}\right) \) . If \( b \in K \) is any root of \( f \), by the isomorphism extension theorem there is a \( \sigma \in G \) with \( \sigma \left( a\right) = b \) . To see that \( b \in L \), take \( \tau \in H \) . Then\n\n\[ \tau \left( b\right) = {\sigma }^{-1}\left( {{\sigma \tau }{\sigma }^{-1}\left( a\right) }\right) = {\sigma }^{-1}\left( a\right) = b \] since \( {\sigma \tau }{\sigma }^{-1} \in H \), as \( H \) is normal in \( G \) . Thus, \( b \in \mathcal{F}\left( H\right) | Yes |
Let \( K = \mathbb{Q}\left( \left\{ {{e}^{{2\pi ik}/n} : k, n \in \mathbb{N}}\right\} \right) \) be the field generated over \( \mathbb{Q} \) by all roots of unity in \( \mathbb{C} \) . Then \( K \) is the splitting field over \( \mathbb{Q} \) of the set \( \left\{ {{x}^{n} - 1 : n \in \mathbb{N}}\right\} \), so \( K/\mathbb{Q} \) is Galois. If \( L \) is a finite Galois extension of \( \mathbb{Q} \) contained in \( K \), then \( L \) is contained in a cyclotomic extension of \( \mathbb{Q} \) . The Galois group of a cyclotomic extension is Abelian. Consequently, \( \operatorname{Gal}\left( {L/F}\right) \) is Abelian. To see that \( \operatorname{Gal}\left( {K/F}\right) \) is Abelian, by the proof of Theorem 17.8 the Galois group \( \operatorname{Gal}\left( {K/F}\right) \) is isomorphic to a subgroup of the direct product of the \( \operatorname{Gal}\left( {L/F}\right) \) as \( L \) ranges over finite Galois subextensions of \( \mathbb{Q} \) , so \( \operatorname{Gal}\left( {K/F}\right) \) is Abelian. As a consequence of this fact, any subextension of \( K/\mathbb{Q} \) is a Galois extension of \( \mathbb{Q} \). | We give an alternate proof that \( \operatorname{Gal}\left( {K/F}\right) \) is Abelian that does not use the proof of Theorem 17.8. Take \( \sigma ,\tau \in \operatorname{Gal}\left( {K/\mathbb{Q}}\right) \) . If \( a \in K \), then there is an intermediate field \( L \) of \( K/\mathbb{Q} \) that is Galois over \( \mathbb{Q} \) and that \( a \in L \) . The restrictions \( {\left. \sigma \right| }_{L},{\left. \tau \right| }_{L} \) are elements of \( \operatorname{Gal}\left( {L/\mathbb{Q}}\right) \), and this group is Abelian by the previous paragraph. Thus,\n\n\[ \sigma \left( {\tau \left( a\right) }\right) = {\left. \sigma \right| }_{L}\left( {{\left. \tau \right| }_{L}\left( a\right) }\right) = {\left. \tau \right| }_{L}\left( {{\left. \sigma \right| }_{L}\left( a\right) }\right) = \tau \left( {\sigma \left( a\right) }\right) . \]\n\nConsequently, \( {\sigma \tau } = {\tau \sigma } \), so \( \operatorname{Gal}\left( {K/\mathbb{Q}}\right) \) is Abelian. | Yes |
Example 17.11 Let \( K \) be an algebraic closure of \( {\mathbb{F}}_{p} \) . Since \( {\mathbb{F}}_{p} \) is perfect, \( K \) is separable, and hence \( K \) is Galois over \( {\mathbb{F}}_{p} \) . Let \( \sigma : K \rightarrow K \) be defined by \( \sigma \left( a\right) = {a}^{p} \) . Then \( \sigma \in G = \operatorname{Gal}\left( {K/{\mathbb{F}}_{p}}\right) \), and the fixed field of the cyclic subgroup \( H \) of \( G \) generated by \( \sigma \) is \( {\mathbb{F}}_{p} \) . However, we prove that \( H \neq G \) by constructing an automorphism of \( K \) that is not in \( H \) . | To see this, pick an integer \( {n}_{r} \) for each \( r \in \mathbb{N} \) such that if \( r \) divides \( s \), then \( {n}_{s} \equiv {n}_{r}\left( {\;\operatorname{mod}\;r}\right) \) . If \( {F}_{r} \) is the subfield of \( K \) containing \( {p}^{r} \) elements, then define \( \tau \) by \( \tau \left( a\right) = {\sigma }^{{n}_{r}}\left( a\right) \) if \( a \in {F}_{r} \) . The conditions on the \( {n}_{r} \) show that \( \tau \) is well defined, and an easy argument shows that \( \tau \) is an automorphism of \( K \) that fixes \( {\mathbb{F}}_{p} \) . For a specific example of a choice of the \( {n}_{r} \), for \( r \in \mathbb{N} \), write \( r = {p}^{m}q \) with \( q \) not a multiple of \( p \) . Let \( {n}_{r} \) satisfy\n\n\[ \n{n}_{r} \equiv 1 + p + \cdots + {p}^{m - 1}\left( {\;\operatorname{mod}\;{p}^{m}}\right) , \n\]\n\n\[ \n{n}_{r} \equiv 0\left( {\;\operatorname{mod}\;q}\right) \n\]\n\nSuch integers exist by the Chinese remainder theorem of number theory, since \( {p}^{m} \) and \( q \) are relatively prime. If \( \tau = {\sigma }^{t} \) for some \( t \), then for all \( r,{\left. \tau \right| }_{{F}_{r}} = {\left. {\sigma }^{t}\right| }_{{F}_{r}} \), so \( {n}_{r} \equiv t\left( {\;\operatorname{mod}\;r}\right) \), as \( \operatorname{Gal}\left( {{F}_{r}/{\mathbb{F}}_{p}}\right) \) is the cyclic group generated by \( {\left. \sigma \right| }_{{F}_{r}} \), which has order \( r \) . This cannot happen as \( {n}_{{p}^{m}} \rightarrow \infty \) as \( m \rightarrow \infty \) . Therefore, \( \tau \notin H \), so \( H \) is not a closed subgroup of \( G \) . The group \( G \) is obtained topologically from \( H \), since \( G = \bar{H} \) by Theorem 17.7. The argument that \( G = \operatorname{im}\left( f\right) \) in the proof of Theorem 17.6 shows that any element of \( G \) is obtained by the construction above, for an appropriate choice of the \( {n}_{r} \) . This gives a description of the Galois group \( G \) as\n\n\[ \n\operatorname{Gal}\left( {K/{\mathbb{F}}_{p}}\right) \cong \left\{ {\left\{ {n}_{r}\right\} \in \mathop{\prod }\limits_{r}{\mathbb{F}}_{{p}^{r}} : \text{ if }r\text{ divides }s,\text{ then }{n}_{s} \equiv {n}_{r}\left( {\;\operatorname{mod}\;r}\right) }\right\} .\n\] | Yes |
Proposition 18.1 Let \( {F}_{s} \) be the separable closure of the field \( F \) . Then \( {F}_{s} \) is Galois over \( F \) with \( \operatorname{Gal}\left( {{F}_{s}/F}\right) \cong \operatorname{Gal}\left( {{F}_{ac}/F}\right) \) . Moreover, \( {F}_{s} \) is a maximal separable extension of \( F \), meaning that \( {F}_{s} \) is not properly contained in any separable extension of \( F \) . Thus, \( {F}_{s} \) is separably closed. | Proof. The field \( {F}_{s} \) is Galois over \( F \), and \( \operatorname{Gal}\left( {{F}_{s}/F}\right) = \operatorname{Gal}\left( {{F}_{ac}/F}\right) \) by Theorem 4.23. Suppose that \( {F}_{s} \subseteq L \) with \( L/F \) separable. Then we can embed \( L \subseteq {F}_{ac} \), and then \( L = {F}_{s} \), since \( {F}_{s} \) is the set of all separable elements over \( F \) in \( {F}_{ac} \) . Finally, if \( L \) is a separable extension of \( {F}_{s} \), then by transitivity of separability, \( L \) is a separable extension of \( F \), so \( L = {F}_{s} \) . Therefore, \( {F}_{s} \) is separably closed. | Yes |
Proposition 18.2 Let \( F \) be a field with \( \operatorname{char}\left( F\right) \neq 2 \) . Then the quadratic closure \( {F}_{q} \) of \( F \) is the composite inside a fixed algebraic closure of \( F \) of all Galois extensions of \( F \) of degree a power of 2. | Proof. Let \( K \) be the composite inside a fixed algebraic closure of \( F \) of all Galois extensions of \( F \) of degree a power of 2 . Then \( K \) is Galois over \( F \) . To show that \( G = \operatorname{Gal}\left( {K/F}\right) \) is a pro-2-group, let \( N \) be an open normal subgroup of \( G \) . If \( L = \mathcal{F}\left( H\right) \), then \( \left\lbrack {L : F}\right\rbrack = \left\lbrack {G : N}\right\rbrack \) by the fundamental theorem. The intermediate field \( L \) is a finite extension of \( F \) ; hence, \( L \) lies in a composite of finitely many Galois extensions of \( F \) of degree a power of 2. Any such composite has degree over \( F \) a power of 2 by the theorem of natural irrationalities, so \( \left\lbrack {L : F}\right\rbrack \) is a power of 2 . Thus, \( \left\lbrack {G : N}\right\rbrack \) is a power of 2, so \( G \) is a pro-2-group.\n\nTo see that \( K \) is quadratically closed, suppose that \( L/K \) is a quadratic extension, and say \( L = K\left( \sqrt{a}\right) \) for some \( a \in K \) . Then \( a \in E \) for some finite Galois subextension \( E \) . By the argument above, we have \( \left\lbrack {E : F}\right\rbrack = {2}^{r} \) for some \( r \) . The extension \( E\left( \sqrt{a}\right) /E \) has degree at most 2 . If \( \sqrt{a} \in E \), then \( L = K \) and we are done. If not, consider the polynomial\n\n\[\n\mathop{\prod }\limits_{{\sigma \in \operatorname{Gal}\left( {E/F}\right) }}\left( {{x}^{2} - \sigma \left( a\right) }\right) \in F\left\lbrack x\right\rbrack\n\]\n\nThe splitting field \( N \) over \( F \) of this polynomial is \( N = F(\{ \sqrt{\sigma \left( a\right) } : \sigma \in \) Gal \( \left( {E/F}\right) \} \) ). Hence, \( N \) is a 2-Kummer extension of \( F \), so \( \left\lbrack {N : F}\right\rbrack \) is a power of 2 . The field \( N \) is a Galois extension of \( F \) of degree a power of 2, so \( N \subseteq K \) . Moreover, \( \sqrt{a} \in N \) . This shows that \( \sqrt{a} \in K \), so \( L = K \) . Thus, \( K \) is quadratically closed. | Yes |
Proposition 18.3 Let \( F \) be a field of characteristic with \( \operatorname{char}\left( F\right) \neq 2 \) . We define fields \( \left\{ {F}_{n}\right\} \) by recursion by setting \( {F}_{0} = F \) and \( {F}_{n + 1} = {F}_{n}(\{ \sqrt{a} \) : \( \left. \left. {a \in {F}_{n}}\right\} \right) \) . Then the quadratic closure of \( F \) is the union \( \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{F}_{n} \) . | Proof. Let \( K = \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{F}_{n} \) . Then \( K \) is a field, since \( \left\{ {F}_{n}\right\} \) is a totally ordered collection of fields. We show that \( K \) is quadratically closed. If \( a \in K \), then \( a \in {F}_{n} \) for some \( n \), so \( \sqrt{a} \in {F}_{n + 1} \subseteq K \) . Thus, \( K\left( \sqrt{a}\right) = K \), so \( K \) is indeed quadratically closed. Let \( {F}_{q} \) be the quadratic closure of \( F \) . Then \( \sqrt{a} \in {F}_{q} \) for each \( a \in {F}_{q} \), so we see that \( {F}_{1} \subseteq {F}_{q} \) . Suppose that \( {F}_{n} \subseteq {F}_{q} \) . The reasoning we used to show that \( K \) is quadratically closed shows also that \( {F}_{n + 1} \subseteq {F}_{q} \), so \( K \subseteq {F}_{q} \) . To see that this inclusion is an equality, let \( E \) be a Galois extension of \( F \) of degree a power of 2 . Then \( {EK}/K \) has degree a power of 2 by natural irrationalities. If \( \left\lbrack {{EK} : K}\right\rbrack > 1 \), then the group \( \operatorname{Gal}\left( {{EK}/K}\right) \) has a subgroup of index 2 by the theory of \( p \) -groups. If \( L \) is the fixed field of this subgroup, then \( \left\lbrack {L : K}\right\rbrack = 2 \) . However, this is impossible, since \( K \) is quadratically closed. This forces \( {EK} = K \), so \( E \subseteq K \) . Since \( {F}_{q} \) is the composite of all such \( E \), we see that \( {F}_{q} \subseteq K \), so \( K = {F}_{q} \) . | Yes |
Lemma 18.4 Let \( p \) be a prime, and let \( F \) be a field with \( \operatorname{char}\left( F\right) \neq p \) . If \( L \) is an intermediate field of \( {F}_{ac}/F \) with \( \left\lbrack {L : F}\right\rbrack < \infty \), then \( L \subseteq {F}_{p} \) if and only if \( L \) lies in a Galois extension of \( F \) of degree a power of \( p \) . In particular, any finite intermediate field of \( {F}_{p}/F \) has degree over \( F \) a power of \( p \) . | Proof. If \( L \) is a field lying inside some Galois extension \( E \) of \( F \) with \( \left\lbrack {E : F}\right\rbrack \) a power of \( p \), then \( E \subseteq {F}_{p} \), so \( L \subseteq {F}_{p} \) . Conversely, suppose that \( L \subseteq {F}_{p} \) and \( \left\lbrack {L : F}\right\rbrack < \infty \) . Then \( L = F\left( {{a}_{1},\ldots ,{a}_{n}}\right) \) for some \( {a}_{i} \in L \) . From the definition of \( {F}_{p} \), for each \( i \) there is a Galois extension \( {E}_{i}/F \) such that \( {a}_{i} \in {E}_{i} \) and \( \left\lbrack {{E}_{i} : F}\right\rbrack \) is a power of \( p \) . The composition of the \( {E}_{i} \) is a Galois extension of \( F \), whose degree over \( F \) is also a power of \( p \) by natural irrationalities. | Yes |
Proposition 18.6 Suppose that \( F \) contains a primitive pth root of unity. Define a sequence of fields \( \left\{ {F}_{n}\right\} \) by recursion by setting \( {F}_{0} = F \) and \( {F}_{n + 1} = \) \( {F}_{n}\left( \left\{ {\sqrt[p]{a} : a \in {F}_{n}}\right\} \right) \) . Then the p-closure of \( F \) is \( \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{F}_{n} \) . | Proof. The proof is essentially the same as that for the quadratic closure, so we only outline the proof. If \( {F}_{n} \subseteq {F}_{p} \) and \( a \in {F}_{n} \), then either \( {F}_{n}\left( \sqrt[p]{a}\right) = {F}_{n} \) , or \( {F}_{n}\left( \sqrt[p]{a}\right) /{F}_{n} \) is a Galois extension of degree \( p \), by Proposition 9.6. In either case, \( {F}_{n}\left( \sqrt[p]{a}\right) \subseteq {F}_{p} \) by the previous proposition. This shows that \( \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{F}_{n} \subseteq {F}_{p} \) . To get the reverse inclusion, let \( E/F \) be a Galois extension of degree a power of \( p \) . By the theory of \( p \) -groups and the fundamental theorem of Galois theory, there is a chain of intermediate fields\n\n\[ F = {E}_{0} \subseteq {E}_{1} \subseteq \cdots \subseteq {E}_{n} = E \]\n\nwith \( {E}_{i + 1}/{E}_{i} \) Galois of degree \( p \) . Since \( F \) contains a primitive \( p \) th root of unity, \( {E}_{i + 1} = {E}_{i}\left( \sqrt[p]{{a}_{i}}\right) \) for some \( {a}_{i} \in {E}_{i} \) by Theorem 9.5. By induction, we can see that \( {E}_{i} \subseteq {F}_{i} \), so \( E \subseteq \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{F}_{n} \) . Since \( {F}_{p} \) is the composite of all such \( E \), this gives the reverse inclusion we want. | Yes |
Proposition 18.7 Let \( F \) be a field, let \( p \) be a prime, and let \( K \) be a maximal prime to p extension of F . Then any finite extension of \( K \) has degree a power of \( p \), and if \( L \) is an intermediate field of \( K/F \) with \( \left\lbrack {L : F}\right\rbrack < \infty \), then \( \left\lbrack {L : F}\right\rbrack \) is relatively prime to p. Moreover, any separable field extension \( L \) of \( F \) with \( \left\lbrack {L : F}\right\rbrack \) relatively prime to \( p \) is contained in some maximal prime to \( p \) extension of \( F \) . | Proof. Recall that if \( U \) is an open subgroup of a \( p \) -Sylow subgroup \( P \) of \( G = \operatorname{Gal}\left( {{F}_{s}/F}\right) \), then \( \left\lbrack {P : U}\right\rbrack \) is a power of \( p \), and if \( V \) is open in \( G \) with \( P \subseteq V \subseteq G \), then \( \left\lbrack {G : V}\right\rbrack \) is relatively prime to \( p \) . Suppose that \( M \) is a finite extension of \( K \) . If \( H = \operatorname{Gal}\left( {{F}_{s}/M}\right) \), then by the fundamental theorem, we have \( \left\lbrack {P : H}\right\rbrack = \left\lbrack {M : K}\right\rbrack < \infty \), so \( H \) is an open subgroup of \( P \) . Thus, \( \left\lbrack {P : H}\right\rbrack \) is a power of \( p \), so \( \left\lbrack {M : K}\right\rbrack \) is a power of \( p \) .\n\nFor the second statement, let \( L \) be an intermediate field of \( K/F \) with \( \left\lbrack {L : F}\right\rbrack < \infty \) . If \( A = \operatorname{Gal}\left( {{F}_{s}/L}\right) \), then \( P \subseteq A \) and \( \left\lbrack {G : A}\right\rbrack = \left\lbrack {L : F}\right\rbrack \) is finite, by the fundamental theorem. Since \( \left\lbrack {G : A}\right\rbrack \) is relatively prime to \( p \), we see that \( \left\lbrack {L : F}\right\rbrack \) is relatively prime to \( p \) .\n\nLet \( L/F \) be an extension with \( \left\lbrack {L : F}\right\rbrack \) relatively prime to \( p \) . Let \( {F}_{s} \) be the separable closure of \( F \), and let \( G = \operatorname{Gal}\left( {{F}_{s}/F}\right) \) . Set \( H = \operatorname{Gal}\left( {{F}_{s}/L}\right) \), a closed subgroup of \( G \), and let \( {P}^{\prime } \) be a \( p \) -Sylow subgroup of \( H \) . There is a \( p \) -Sylow subgroup \( P \) of \( G \) that contains \( {P}^{\prime } \) . Note that \( \left\lbrack {G : H}\right\rbrack = \left\lbrack {L : F}\right\rbrack \) is relatively prime to \( p \) . Moreover, we have\n\n\[ \left\lbrack {G : {P}^{\prime }}\right\rbrack = \left\lbrack {G : H}\right\rbrack \cdot \left\lbrack {H : {P}^{\prime }}\right\rbrack \]\n\n\[ = \left\lbrack {G : P}\right\rbrack \cdot \left\lbrack {P : {P}^{\prime }}\right\rbrack \]\n\nBoth \( \left\lbrack {G : H}\right\rbrack \) and \( \left\lbrack {H : {P}^{\prime }}\right\rbrack \) are supernatural numbers not divisible by \( p \), so \( \left\lbrack {P : {P}^{\prime }}\right\rbrack \) is not divisible by \( p \) . But, since \( P \) is a pro- \( p \) -group, \( \left\lbrack {P : {P}^{\prime }}\right\rbrack \) is a power of \( p \) . This forces \( \left\lbrack {P : {P}^{\prime }}\right\rbrack = 1 \), so \( {P}^{\prime } = P \) . Therefore, \( P \subseteq H \), and so \( L = \mathcal{F}\left( H\right) \) is contained in \( \mathcal{F}\left( P\right) \), a maximal prime to \( p \) extension of \( F \) . | Yes |
Proposition 18.9 Let \( {F}_{a} \) be the maximal Abelian extension of a field \( F \) . Then \( {F}_{a}/F \) is a Galois extension and \( \operatorname{Gal}\left( {{F}_{a}/F}\right) \) is an Abelian group. The field \( {F}_{a} \) has no extensions that are Abelian Galois extensions of \( F \) . Moreover, \( {F}_{a} \) is the composite in \( {F}_{s} \) of all finite Abelian Galois extensions of \( F \) . | Proof. The commutator subgroup \( {G}^{\prime } \) of \( G \) is a normal subgroup, so the closure \( \overline{{G}^{\prime }} \) of \( {G}^{\prime } \) is a closed normal subgroup of \( G \) (see Problem 17.8). Thus, by the fundamental theorem, \( {F}_{a} = \mathcal{F}\left( \overset{⏜}{{G}^{\prime }}\right) \) is a Galois extension of \( F \) and \( \operatorname{Gal}\left( {{F}_{a}/F}\right) \cong G/\overline{{G}^{\prime }} \) . The group \( G/\overline{{G}^{\prime }} \) is a homomorphic image of the Abelian group \( G/{G}^{\prime } \), so \( G/\overline{{G}^{\prime }} \) is also Abelian.\n\nIf \( L \supseteq {F}_{a} \) is an Abelian Galois extension of \( F \), then \( L \subseteq {F}_{s} \) . Let \( H = \) \( \operatorname{Gal}\left( {{F}_{s}/L}\right) \), a subgroup of \( {G}^{\prime } \) . However, \( G/H \cong \operatorname{Gal}\left( {L/F}\right) \), so \( G/H \) is Abelian. Thus, \( {G}^{\prime } \subseteq H \), so \( H = {G}^{\prime } \) . Therefore, \( {F}_{a} \) is not properly contained in any Abelian extension of \( F \).\n\nFor the final statement, if \( K/F \) is finite Abelian Galois, then \( K{F}_{a}/{F}_{a} \) is Abelian Galois by natural irrationalities. Thus, \( K{F}_{a} = {F}_{a} \), so \( K \subseteq {F}_{a} \) . Since every element of \( {F}_{a} \) lies in a finite Galois extension of \( F \), to show that \( {F}_{a} \) is the composite of all finite Abelian Galois extensions of \( F \) it suffices to show that every finite Galois extension of \( F \) inside \( {F}_{a} \) is an Abelian extension. Let \( E \) be such an extension. If \( H = \operatorname{Gal}\left( {{F}_{s}/E}\right) \), then \( H \) is a normal subgroup of \( G \) containing \( {G}^{\prime } \) ; hence, \( G/H \) is Abelian. But, by the fundamental theorem, we have \( \operatorname{Gal}\left( {E/F}\right) \cong G/H \), so \( E/F \) is an Abelian Galois extension. | Yes |
If \( F \) is a field containing a primitive \( n \) th root of unity for all \( n \), then the maximal Abelian extension of \( F \) is \( F\left( {\{ \sqrt[n]{a} : a \in F, n \in \mathbb{N}\} }\right) \) . | This follows from Kummer theory (see Problem 11.6 for part of this claim). | No |
Lemma 19.5 Let \( K \) be a field extension of \( F \) . If \( {t}_{1},\ldots ,{t}_{n} \in K \) are algebraically independent over \( F \), then \( F\left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \) and \( F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) are \( F \) - isomorphic rings, and so \( F\left( {{t}_{1},\ldots ,{t}_{n}}\right) \) and \( F\left( {{x}_{1},\ldots ,{x}_{n}}\right) \) are \( F \) -isomorphic fields. | Proof. Define \( \varphi : F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \rightarrow K \) by \( \varphi \left( {f\left( {{x}_{1},\ldots ,{x}_{n}}\right) }\right) = f\left( {{t}_{1},\ldots ,{t}_{n}}\right) \) . Then \( \varphi \) is an \( F \) -homomorphism of rings. The algebraic independence of the \( {t}_{i} \) shows that \( \varphi \) is injective, and the image of \( \varphi \) is \( F\left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \) . Therefore, \( F\left\lbrack {{t}_{1}\ldots ,{t}_{n}}\right\rbrack \) and \( F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) are isomorphic. This map induces an \( F \) -isomorphism of quotient fields, which finishes the proof. | Yes |
Lemma 19.7 Let \( K \) be a field extension of \( F \), and let \( {t}_{1},\ldots ,{t}_{n} \in K \) . Then the following statements are equivalent:\n\n1. The set \( \left\{ {{t}_{1},\ldots ,{t}_{n}}\right\} \) is algebraically independent over \( F \) .\n\n2. For each \( i,{t}_{i} \) is transcendental over \( F\left( {{t}_{1},\ldots ,{t}_{i - 1},{t}_{i + 1},\ldots ,{t}_{n}}\right) \) .\n\n3. For each \( i,{t}_{i} \) is transcendental over \( F\left( {{t}_{1},\ldots ,{t}_{i - 1}}\right) \) . | Proof. (1) \( \Rightarrow \) (2): Suppose that there are \( {a}_{j} \in F\left( {{t}_{1},\ldots ,{t}_{i - 1},{t}_{i + 1},\ldots ,{t}_{n}}\right) \) such that \( {a}_{0} + {a}_{1}{t}_{i} + \cdots + {t}_{i}^{m} = 0 \) . We may write \( {a}_{j} = {b}_{j}/c \) with \( {b}_{1,}\ldots ,{b}_{n}, c \in \) \( F\left\lbrack {{t}_{1},\ldots ,{t}_{i - 1},{t}_{i + 1},\ldots ,{t}_{n}}\right\rbrack \), and so \( {b}_{0} + {b}_{1}{t}_{i} + \cdots + {b}_{m}{t}_{i}^{m} = 0 \) . If \( {b}_{j} = \) \( {g}_{j}\left( {{t}_{1},\ldots ,{t}_{i - 1},{t}_{i + 1},\ldots ,{t}_{n}}\right) \), then \( f = \mathop{\sum }\limits_{j}{g}_{j}\left( {{x}_{1},\ldots ,{x}_{i - 1},{x}_{i + 1},\ldots ,{x}_{n}}\right) {x}_{i}^{j} \) is a polynomial and \( f\left( {{t}_{1},\ldots ,{t}_{n}}\right) = 0 \) . Since \( \left\{ {{t}_{1},\ldots ,{t}_{n}}\right\} \) is algebraically independent over \( F \), the polynomial \( f \) must be 0 . Consequently, each \( {a}_{j} = \) 0, so \( {t}_{i} \) is transcendental over \( F\left( {{t}_{1},\ldots ,{t}_{i - 1},{t}_{i + 1},\ldots ,{t}_{n}}\right) \) .\n\n\( \left( 2\right) \Rightarrow \left( 3\right) \) : If \( {t}_{i} \) is transcendental over \( F\left( {{t}_{1},\ldots ,{t}_{i - 1},{t}_{i + 1},\ldots ,{t}_{n}}\right) \), then \( {t}_{i} \) clearly is transcendental over the smaller field \( F\left( {{t}_{1},\ldots ,{t}_{i - 1}}\right) \) .\n\n\( \left( 3\right) \Rightarrow \left( 1\right) \) : Suppose that the \( {t}_{i} \) are not algebraically independent over \( F \) . Choose \( m \) minimal such that there is a nonzero \( f\left( {{x}_{1},\ldots ,{x}_{m}}\right) \in \) \( F\left\lbrack {{x}_{1},\ldots ,{x}_{m}}\right\rbrack \) with \( f\left( {{t}_{1},\ldots ,{t}_{m}}\right) = 0 \) . Say \( f = \mathop{\sum }\limits_{j}{g}_{j}{x}_{m}^{j} \) with \( {g}_{j} \in \) \( F\left\lbrack {{x}_{1},\ldots ,{x}_{m - 1}}\right\rbrack \), and let \( {a}_{j} = g\left( {{t}_{1},\ldots ,{t}_{m - 1}}\right) \) . Then \( {a}_{0} + {a}_{1}{t}_{m} + \cdots + {a}_{r}{t}_{m}^{r} = \) 0 . If the \( {a}_{j} \) are not all zero, then \( {t}_{m} \) is algebraic over \( F\left( {{t}_{1},\ldots ,{t}_{m}}\right) \), a contradiction. Thus, \( {a}_{j} = 0 \) for each \( j \) . By the minimality of \( m \), the \( {t}_{1},\ldots ,{t}_{m - 1} \) are algebraically independent over \( F \), which implies that all \( {g}_{j} = 0 \), so \( f = 0 \) . This proves that \( \left\{ {{t}_{1},\ldots ,{t}_{n}}\right\} \) is algebraically independent over \( F \) . | Yes |
If \( K = F\left( {{x}_{1},\ldots ,{x}_{n}}\right) \), then \( \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \) is a transcendence basis for \( K/F \) . Moreover, if \( {r}_{1},\ldots ,{r}_{n} \) are positive integers, then we show that \( \left\{ {{x}_{1}^{{r}_{1}},\ldots ,{x}_{n}^{{r}_{n}}}\right\} \) is also a transcendence basis for \( K/F \) . | We saw in Example 19.2 that \( \left\{ {{x}_{1}^{{r}_{1}},\ldots ,{x}_{n}^{{r}_{n}}}\right\} \) is algebraically independent over \( F \) . We need to show that \( K \) is algebraic over \( L = F\left( {{x}_{1}^{{r}_{1}},\ldots ,{x}_{n}^{{r}_{n}}}\right) \) . This is true because for each \( i \) the element \( {x}_{i} \) satisfies the polynomial \( {t}^{{r}_{i}} - {x}_{i}^{{r}_{i}} \in L\left\lbrack t\right\rbrack \) . | Yes |
We show that \( \{ u\} \) is a transcendence basis for \( K/k \) . Since \( {v}^{2} = {u}^{3} - u \), the field \( K \) is algebraic over \( k\left( u\right) \) . We then need to show that \( u \) is transcendental over \( k \) . | If this is false, then \( u \) is algebraic over \( k \), so \( K \) is algebraic over \( k \) . We claim that this forces \( A = k\left\lbrack {u, v}\right\rbrack \) to be a field. To prove this, take \( t \in A \) . Then \( {t}^{-1} \in K \) is algebraic over \( k \), so \( {t}^{-n} + {\alpha }_{n - 1}{t}^{n - 1} + \cdots + {\alpha }_{0} = 0 \) for some \( {\alpha }_{i} \in k \) with \( {\alpha }_{0} \neq 0 \) . Multiplying by \( {t}^{n - 1} \) gives\n\n\[ \n{t}^{-1} = - \left( {{\alpha }_{n - 1} + {\alpha }_{n - 2}t + \cdots + {\alpha }_{0}{t}^{n - 1}}\right) \in A, \n\]\n\nproving that \( A \) is a field. However, \( A = k\left\lbrack {x, y}\right\rbrack /\left( f\right) \) is a field if and only if \( \left( f\right) \) is a maximal ideal of \( k\left\lbrack {x, y}\right\rbrack \) . The ring \( A \) cannot be a field, since \( \left( f\right) \) is properly contained in the ideal \( \left( {x, y}\right) \) of \( k\left\lbrack {x, y}\right\rbrack \) . Thus, \( u \) is not algebraic over \( k \), so \( \{ u\} \) is a transcendence basis for \( K/k \) . | Yes |
Example 19.12 We give a generalization of the previous example. Let \( k \) be a field and let \( f \in k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) be an irreducible polynomial. Then \( A = k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack /\left( f\right) \) is an integral domain. Let \( K \) be the quotient field of A. We may write\n\n\[ f = {g}_{m}{x}_{n}^{m} + {g}_{m - 1}{x}_{n}^{m - 1} + \cdots + {g}_{0} \]\n\nwith each \( {g}_{i} \in k\left\lbrack {{x}_{1},\ldots ,{x}_{n - 1}}\right\rbrack \) . Let us assume that \( m > 0 \), so that \( f \) does involve the variable \( {x}_{n} \) . If \( {t}_{i} = {x}_{i} + \left( f\right) \) is the image of \( {x}_{i} \) in \( A \), we claim that \( \left\{ {{t}_{1},\ldots ,{t}_{n - 1}}\right\} \) is a transcendence basis for \( K/k \) . | To see this, the equation for \( f \) above shows that \( {t}_{n} \) is algebraic over \( k\left( {{t}_{1},\ldots ,{t}_{n - 1}}\right) \), so we only need to show that \( \left\{ {{t}_{1},\ldots ,{t}_{n - 1}}\right\} \) is algebraically independent over \( k \) . Suppose that there is a polynomial \( h \in k\left\lbrack {{x}_{1},\ldots ,{x}_{n - 1}}\right\rbrack \) with \( h\left( {{t}_{1},\ldots ,{t}_{n - 1}}\right) = 0 \) . Then \( h\left( {{x}_{1},\ldots ,{x}_{n - 1}}\right) \in \left( f\right) \), so \( f \) divides \( h \) . Thus, \( h = {fg} \) for some \( g \in k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . However, the polynomial \( h \) does not involve the variable \( {x}_{n} \) while \( f \) does, so comparing degrees in \( {x}_{n} \) of \( h \) and \( {fg} \) shows that \( h = 0 \) . Therefore, \( \left\{ {{t}_{1},\ldots ,{t}_{n - 1}}\right\} \) is algebraically independent over \( k \), so \( \left\{ {{t}_{1},\ldots ,{t}_{n - 1}}\right\} \) is a transcendence basis for \( K/k \) . | Yes |
Lemma 19.13 Let \( K \) be a field extension of \( F \), and let \( S \subseteq K \) be algebraically independent over \( F \) . If \( t \in K \) is transcendental over \( F\left( S\right) \), then \( S \cup \{ t\} \) is algebraically independent over \( F \) . | Proof. Suppose that the lemma is false. Then there is a nonzero polynomial \( f \in F\left\lbrack {{x}_{1},\ldots ,{x}_{n}, y}\right\rbrack \) with \( f\left( {{s}_{1},\ldots ,{s}_{n}, t}\right) = 0 \) for some \( {s}_{i} \in S \) . This polynomial must involve \( y \), since \( S \) is algebraically independent over \( F \) . Write \( f = \mathop{\sum }\limits_{{j = 0}}^{m}{g}_{j}{y}^{j} \) with \( {g}_{j} \in F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . Since \( {g}_{m} \neq 0 \), the element \( t \) is algebraic over \( F\left( S\right) \), a contradiction. Thus, \( S \cup \{ t\} \) is algebraically independent over \( F \) . | Yes |
Theorem 19.14 Let \( K \) be a field extension of \( F \). 1. There exists a transcendence basis for \( K/F \). 2. If \( T \subseteq K \) such that \( K/F\left( T\right) \) is algebraic, then \( T \) contains a transcendence basis for \( K/F \). 3. If \( S \subseteq K \) is algebraically independent over \( F \), then \( S \) is contained in a transcendence basis of \( K/F \). 4. If \( S \subseteq T \subseteq K \) such that \( S \) is algebraically independent over \( F \) and \( K/F\left( T\right) \) is algebraic, then there is a transcendence basis \( X \) for \( K/F \) with \( S \subseteq X \subseteq T \). | Proof. We first mention why statement 4 implies the first three statements. If statement 4 is true, then statements 2 and 3 are true by setting \( S = \varnothing \) and \( T = K \), respectively. Statement 1 follows from statement 4 by setting \( S = \varnothing \) and \( T = K \). To prove statement 4, let \( \mathcal{S} \) be the set of all algebraically independent subsets of \( T \) that contain \( S \). Then \( \mathcal{S} \) is nonempty, since \( S \in \mathcal{S} \). Ordering \( \mathcal{S} \) by inclusion, a Zorn’s lemma argument shows that \( \mathcal{S} \) contains a maximal element \( M \). If \( K \) is not algebraic over \( F\left( M\right) \), then \( F\left( T\right) \) is not algebraic over \( F\left( M\right) \), since \( K \) is algebraic over \( F\left( T\right) \). Thus, there is a \( t \in T \) with \( t \) transcendental over \( F\left( M\right) \). But by Lemma 19.13, \( M \cup \{ t\} \) is algebraically independent over \( F \) and is a subset of \( T \), contradicting maximality of \( M \). Thus, \( K \) is algebraic over \( F\left( M\right) \), so \( M \) is a transcendence basis of \( K/F \) contained in \( X \). | Yes |
Theorem 19.15 Let \( K \) be a field extension of \( F \). If \( S \) and \( T \) are transcendence bases for \( K/F \), then \( \left| S\right| = \left| T\right| \). | Proof. We first prove this in the case where \( S = \left\{ {{s}_{1},\ldots ,{s}_{n}}\right\} \) is finite. Since \( S \) is a transcendence basis for \( K/F \), the field \( K \) is not algebraic over \( F\left( {S - \left\{ {s}_{1}\right\} }\right) \). As \( K \) is algebraic over \( F\left( T\right) \), some \( t \in T \) must be transcendental over \( F\left( {S - \left\{ {s}_{1}\right\} }\right) \). Hence, by Lemma 19.13, \( \left\{ {{s}_{2},\ldots ,{s}_{n}, t}\right\} \) is algebraically independent over \( F \). Furthermore, \( {s}_{1} \) is algebraic over \( F\left( {{s}_{2},\ldots ,{s}_{n}, t}\right) \), or else \( \left\{ {{s}_{1},\ldots ,{s}_{n}, t}\right\} \) is algebraically independent, which is false. Thus, \( \left\{ {{s}_{2},\ldots ,{s}_{n}, t}\right\} \) is a transcendence basis for \( K/F \). Set \( {t}_{1} = t \). Assuming we have found \( {t}_{i} \in T \) for all \( i \) with \( 1 \leq i < m \leq n \) such that \( \left\{ {{t}_{1},\ldots ,{t}_{m - 1},{s}_{m},\ldots ,{s}_{n}}\right\} \) is a transcendence basis for \( K/F \), by replacing \( S \) by this set, the argument above shows that there is a \( {t}^{\prime } \in T \) such that \( \left\{ {{t}_{1},\ldots ,{t}_{m - 1},{t}^{\prime },\ldots ,{s}_{n}}\right\} \) is a transcendence basis for \( K/F \). Setting \( {t}_{m} = {t}^{\prime } \) and continuing in this way, we get a transcendence basis \( \left\{ {{t}_{1},\ldots ,{t}_{n}}\right\} \subseteq T \) of \( K/F \). Since \( T \) is a transcendence basis for \( K/F \), we see that \( \left\{ {{t}_{1},\ldots ,{t}_{n}}\right\} = T \), so \( \left| T\right| = n \).\n\nFor the general case, by the previous argument we may suppose that \( S \) and \( T \) are both infinite. Each \( t \in T \) is algebraic over \( F\left( S\right) \); hence, there is a finite subset \( {S}_{t} \subseteq S \) with \( t \) algebraic over \( F\left( {S}_{t}\right) \). If \( {S}^{\prime } = \mathop{\bigcup }\limits_{{t \in T}}{S}_{t} \), then each \( t \in T \) is algebraic over \( F\left( {S}^{\prime }\right) \). Since \( K \) is algebraic over \( F\left( T\right) \), we see\n\nthat \( K \) is algebraic over \( F\left( {S}^{\prime }\right) \). Thus, \( S = {S}^{\prime } \) since \( {S}^{\prime } \subseteq S \) and \( S \) is a transcendence basis for \( K/F \). We then have\n\n\[ \left| S\right| = \left| {S}^{\prime }\right| = \left| {\mathop{\bigcup }\limits_{{t \in T}}{S}_{t}}\right| \leq \left| {T \times \mathbb{N}}\right| = \left| T\right| \]\n\nwhere the last equality is true since \( T \) is infinite. Reversing the argument, we see that \( \left| T\right| \leq \left| S\right| \), so \( \left| S\right| = \left| T\right| \). | Yes |
Corollary 19.17 Let \( {t}_{1},\ldots ,{t}_{n} \in K \) . Then the fields \( F\left( {{t}_{1},\ldots ,{t}_{n}}\right) \) and \( F\left( {{x}_{1},\ldots ,{x}_{n}}\right) \) are \( F \) -isomorphic if and only if \( \left\{ {{t}_{1},\ldots ,{t}_{n}}\right\} \) is an algebraically independent set over \( F \) . | Proof. If \( \left\{ {{t}_{1},\ldots ,{t}_{n}}\right\} \) is algebraically independent over \( F \), then \( F\left( {{t}_{1},\ldots ,{t}_{n}}\right) \) and \( F\left( {{x}_{1},\ldots ,{x}_{n}}\right) \) are \( F \) -isomorphic fields by Lemma 19.5. Conversely, if \( F\left( {{t}_{1},\ldots ,{t}_{n}}\right) \cong F\left( {{x}_{1},\ldots ,{x}_{n}}\right) \), suppose that \( \left\{ {{t}_{1},\ldots ,{t}_{n}}\right\} \) is algebraically dependent over \( F \) . By the previous theorem, there is a subset \( S \) of \( \left\{ {{t}_{1},\ldots ,{t}_{n}}\right\} \) such that \( S \) is a transcendence basis for \( F\left( {{t}_{1},\ldots ,{t}_{n}}\right) /F \) . However, the transcendence degree of this extension is \( n \), which forces \( \left| S\right| = n \), so \( S = \left\{ {{t}_{1},\ldots ,{t}_{n}}\right\} \) . Thus, \( \left\{ {{t}_{1},\ldots ,{t}_{n}}\right\} \) is algebraically independent over \( F \) . | Yes |
Proposition 19.18 Let \( F \subseteq L \subseteq K \) be fields. Then\n\n\[ \operatorname{trdeg}\left( {K/F}\right) = \operatorname{trdeg}\left( {K/L}\right) + \operatorname{trdeg}\left( {L/F}\right) . \] | Proof. Let \( S \) be a transcendence basis for \( L/F \), and let \( T \) be a transcendence basis for \( K/L \) . We show that \( S \cup T \) is a transcendence basis for \( K/F \), which will prove the result because \( S \cap T = \varnothing \) . Since \( T \) is algebraically independent over \( L \), the set \( T \) is also algebraically independent over \( F\left( S\right) \subseteq L \) , so \( S \cup T \) is algebraically independent over \( F \) . To show that \( K \) is algebraic over \( F\left( {S \cup T}\right) \), we know that \( K/L\left( T\right) \) and \( L/F\left( S\right) \) are algebraic. Therefore, \( L\left( T\right) \) is algebraic over \( F\left( {S \cup T}\right) = F\left( S\right) \left( T\right) \), since each \( t \in T \) is algebraic over \( F\left( {S \cup T}\right) \) . Thus, by transitivity, \( K \) is algebraic over \( F\left( {S \cup T}\right) \), so \( S \cup T \) is a transcendence basis for \( K/F \) . This proves the proposition. | Yes |
Consider the field extension \( \mathbb{C}/\mathbb{Q} \). Since \( \mathbb{Q} \) is countable and \( \mathbb{C} \) is uncountable, the transcendence degree of \( \mathbb{C}/\mathbb{Q} \) must be infinite (in fact, uncountable), for if \( {t}_{1},\ldots ,{t}_{n} \) form a transcendence basis for \( \mathbb{C}/\mathbb{Q} \), then \( \mathbb{C} \) is algebraic over \( \mathbb{Q}\left( {{t}_{1},\ldots ,{t}_{n}}\right) \), so \( \mathbb{C} \) and \( \mathbb{Q} \) have the same cardinality, since they are infinite fields. However, one can show that \( \mathbb{Q}\left( {{t}_{1},\ldots ,{t}_{n}}\right) \) is countable. This would give a contradiction to the uncountability of \( \mathbb{C} \). Thus, any transcendence basis \( T \) of \( \mathbb{C}/\mathbb{Q} \) is infinite. | Let \( T \) be any transcendence basis of \( \mathbb{C}/\mathbb{Q} \). Since \( \mathbb{C} \) is algebraic over \( \mathbb{Q}\left( T\right) \) and is algebraically closed, \( \mathbb{C} \) is an algebraic closure of \( \mathbb{Q}\left( T\right) \). Let \( \sigma \) be a permutation of \( T \). Then \( \sigma \) induces an automorphism of \( \mathbb{Q}\left( T\right) \) that is trivial on \( \mathbb{Q} \); hence, \( \sigma \) extends to an automorphism of \( \mathbb{C} \) by the isomorphism extension theorem. Since there are infinitely many such \( \sigma \), we see that \( \left| {\operatorname{Aut}\left( \mathbb{C}\right) }\right| = \infty \). Because any automorphism of \( \mathbb{R} \) is the identity, the only automorphisms of \( \mathbb{C} \) that map \( \mathbb{R} \) to \( \mathbb{R} \) are the identity map and complex conjugation. Thus, there are infinitely many \( \sigma \in \operatorname{Aut}\left( \mathbb{C}\right) \) with \( \sigma \left( \mathbb{R}\right) \neq \mathbb{R} \). We can easily show that \( \left\lbrack {\mathbb{C} : \sigma \left( \mathbb{R}\right) }\right\rbrack = 2 \). This means that there are infinitely many subfields \( F \) of \( \mathbb{C} \) with \( \left\lbrack {\mathbb{C} : F}\right\rbrack = 2 \). It is a whole different question to try to construct such fields. Note that in order to get these automorphisms of \( \mathbb{C} \), we invoked Zorn's lemma twice, once for the existence of a transcendence basis of \( \mathbb{C}/\mathbb{Q} \) and the second time indirectly by using the isomorphism extension theorem. | Yes |
Proposition 20.2 Let \( K \) and \( L \) be field extensions of a field \( F \) . Then \( K \) and \( L \) are linearly disjoint over \( F \) if and only if the map \( \varphi : K{ \otimes }_{F}L \rightarrow K\left\lbrack L\right\rbrack \) given on generators by \( a \otimes b \mapsto {ab} \) is an isomorphism of \( F \) -vector spaces. | Proof. The natural map \( \varphi : K{ \otimes }_{F}L \rightarrow K\left\lbrack L\right\rbrack \) is surjective by the description of \( K\left\lbrack L\right\rbrack \) given above. So, we need to show that \( K \) and \( L \) are linearly disjoint over \( F \) if and only if \( \varphi \) is injective. Suppose first that \( K \) and \( L \) are linearly disjoint over \( F \) . Let \( {\left\{ {k}_{i}\right\} }_{i \in I} \) be a basis for \( K \) as an \( F \) -vector space. Each element of \( K{ \otimes }_{F}L \) has a unique representation in the form \( \sum {k}_{i} \otimes {l}_{i} \), with the \( {l}_{i} \in L \) . Suppose that \( \sum {k}_{i} \otimes {l}_{i} \in \ker \left( \varphi \right) \), so \( \sum {k}_{i}{l}_{i} = 0 \) . Then each \( {l}_{i} = 0 \), since \( K \) and \( L \) are linearly disjoint over \( F \) and \( \left\{ {k}_{i}\right\} \) is \( F \) -linearly independent. Thus, \( \varphi \) is injective, and so \( \varphi \) is an isomorphism.\n\nConversely, suppose that the map \( \varphi \) is an isomorphism. Let \( {\left\{ {a}_{j}\right\} }_{j \in J} \) be an \( F \) -linearly independent subset of \( K \) . By enlarging \( J \), we may assume that the set \( \left\{ {a}_{j}\right\} \) is a basis for \( K \) . If \( \left\{ {a}_{j}\right\} \) is not \( L \) -linearly independent, then there are \( {l}_{j} \in L \) with \( \sum {a}_{j}{l}_{j} = 0 \), a finite sum. Then \( \sum {a}_{j} \otimes {l}_{j} \in \ker \left( \varphi \right) \) , so \( \sum {a}_{j} \otimes {l}_{j} = 0 \) by the injectivity of \( \varphi \) . However, elements of \( K{ \otimes }_{F}L \) can be represented uniquely in the form \( \sum {a}_{j} \otimes {m}_{j} \) with \( {m}_{j} \in L \) . Therefore, each \( {l}_{j} = 0 \), which forces the set \( \left\{ {a}_{j}\right\} \) to be \( L \) -linearly independent. Thus, \( K \) and \( L \) are linearly disjoint over \( F \) . | Yes |
Corollary 20.3 The definition of linear disjointness is symmetric; that is, \( K \) and \( L \) are linearly disjoint over \( F \) if and only if \( L \) and \( K \) are linearly disjoint over \( F \) . | Proof. This follows from Proposition 20.2. The map \( \varphi : K{ \otimes }_{F}L \rightarrow K\left\lbrack L\right\rbrack \) is an isomorphism if and only if \( \tau : L{ \otimes }_{F}K \rightarrow L\left\lbrack K\right\rbrack = K\left\lbrack L\right\rbrack \) is an isomorphism, since \( \tau = i \circ \varphi \), where \( i \) is the canonical isomorphism \( K{ \otimes }_{F}L \rightarrow L{ \otimes }_{F}K \) that sends \( a \otimes b \) to \( b \otimes a \) . | Yes |
Lemma 20.4 Suppose that \( K \) and \( L \) are finite extensions of \( F \) . Then \( K \) and \( L \) are linearly disjoint over \( F \) if and only if \( \left\lbrack {{KL} : F}\right\rbrack = \left\lbrack {K : F}\right\rbrack \cdot \left\lbrack {L : F}\right\rbrack \) . | Proof. The natural map \( \varphi : K{ \otimes }_{F}L \rightarrow K\left\lbrack L\right\rbrack \) that sends \( k \otimes l \) to \( {kl} \) is surjective and\n\n\[ \dim \left( {K{ \otimes }_{F}L}\right) = \left\lbrack {K : F}\right\rbrack \cdot \left\lbrack {L : F}\right\rbrack . \]\n\nThus, \( \varphi \) is an isomorphism if and only if \( \left\lbrack {{KL} : F}\right\rbrack = \left\lbrack {K : F}\right\rbrack \cdot \left\lbrack {L : F}\right\rbrack \) . The lemma then follows from Proposition 20.2. | Yes |
Example 20.5 Suppose that \( K \) and \( L \) are extensions of \( F \) with \( \left\lbrack {K : F}\right\rbrack \) and \( \left\lbrack {L : F}\right\rbrack \) relatively prime. Then \( K \) and \( L \) are linearly disjoint over \( F \) . | To see this, note that both \( \left\lbrack {K : F}\right\rbrack \) and \( \left\lbrack {L : F}\right\rbrack \) divide \( \left\lbrack {{KL} : F}\right\rbrack \), so their product divides \( \left\lbrack {{KL} : F}\right\rbrack \) since these degrees are relatively prime. The linear disjointness of \( K \) and \( L \) over \( F \) follows from the lemma. | No |
Example 20.6 Let \( K \) be a finite Galois extension of \( F \) . If \( L \) is any extension of \( F \), then \( K \) and \( L \) are linearly disjoint over \( F \) if and only if \( K \cap L = F \) . | This follows from the previous example and the theorem of natural irrationalities, since\n\n\[ \left\lbrack {{KL} : F}\right\rbrack = \left\lbrack {L : F}\right\rbrack \left\lbrack {K : K \cap L}\right\rbrack \]\n\nso \( \left\lbrack {{KL} : F}\right\rbrack = \left\lbrack {K : F}\right\rbrack \left\lbrack {L : F}\right\rbrack \) if and only if \( K \cap L = F \) . | Yes |
Lemma 20.8 Suppose that \( F \) is a field, and \( F \subseteq A \subseteq {A}^{\prime } \) and \( F \subseteq B \subseteq {B}^{\prime } \) are all subrings of a field \( C \) . If \( {A}^{\prime } \) and \( {B}^{\prime } \) are linearly disjoint over \( F \), then \( A \) and \( B \) are linearly disjoint over \( F \) . | Proof. This follows immediately from properties of tensor products. There is a natural injective homomorphism \( i : A{ \otimes }_{F}B \rightarrow {A}^{\prime }{ \otimes }_{F}{B}^{\prime } \) sending \( a \otimes b \) to \( a \otimes b \) for \( a \in A \) and \( B \in B \) . If the natural map \( {\varphi }^{\prime } : {A}^{\prime }{ \otimes }_{F}{B}^{\prime } \rightarrow {A}^{\prime }\left\lbrack {B}^{\prime }\right\rbrack \) is injective, then restricting \( \varphi \) to the image of \( i \) shows that the map \( \varphi \) : \( A{ \otimes }_{F}B \rightarrow A\left\lbrack B\right\rbrack \) is also injective. | Yes |
Lemma 20.10 Suppose that \( A \) and \( B \) are subrings of a field \( C \), each containing a field \( F \), with quotient fields \( K \) and \( L \), respectively. Then \( A \) and \( B \) are linearly disjoint over \( F \) if and only if \( K \) and \( L \) are linearly disjoint over \( F \) . | Proof. If \( K \) and \( L \) are linearly disjoint over \( F \), then \( A \) and \( B \) are also linearly disjoint over \( F \) by the previous lemma. Conversely, suppose that \( A \) and \( B \) are linearly disjoint over \( F \) . Let \( \left\{ {{k}_{1},\ldots ,{k}_{n}}\right\} \subseteq K \) be an \( F \) -linearly independent set, and suppose that there are \( {l}_{i} \in L \) with \( \sum {k}_{i}{l}_{i} = 0 \) . There are nonzero \( s \in A \) and \( t \in B \) with \( s{k}_{i} \in A \) and \( t{l}_{i} \in B \) for each \( i \) . The set \( \left\{ {{a}_{1},\ldots ,{a}_{n}}\right\} \) is also \( F \) -linearly independent; consequently, \( \sum {a}_{i} \otimes {b}_{i} \neq 0 \) , since it maps to the nonzero element \( \sum {a}_{i} \otimes {b}_{i} \in K{ \otimes }_{F}L \) under the natural map \( A{ \otimes }_{F}B \rightarrow K{ \otimes }_{F}B \) . However, \( \sum {a}_{i} \otimes {b}_{i} \) is in the kernel of the map \( A{ \otimes }_{F}B \rightarrow A\left\lbrack B\right\rbrack \) ; hence, it is zero by the assumption that \( A \) and \( B \) are linearly disjoint over \( F \) . This shows that \( \left\{ {k}_{i}\right\} \) is \( L \) -linearly independent, so \( K \) and \( L \) are linearly disjoint over \( F \) . | Yes |
Suppose that \( K/F \) is an algebraic extension and that \( L/F \) is a purely transcendental extension. Then \( K \) and \( L \) are linearly disjoint over \( F \). | To see this, let \( X \) be an algebraically independent set over \( F \) with \( L = F\left( X\right) \). From the previous lemma, it suffices to show that \( K \) and \( F\left\lbrack X\right\rbrack \) are linearly disjoint over \( F \). We can view \( F\left\lbrack X\right\rbrack \) as a polynomial ring in the variables \( x \in X \). The ring generated by \( K \) and \( F\left\lbrack X\right\rbrack \) is the polynomial ring \( K\left\lbrack X\right\rbrack \). The standard homomorphism \( K{ \otimes }_{F}F\left\lbrack X\right\rbrack \rightarrow K\left\lbrack X\right\rbrack \) is an isomorphism because there is a ring homomorphism \( \tau : K\left\lbrack X\right\rbrack \rightarrow K{ \otimes }_{F}F\left\lbrack X\right\rbrack \) induced by \( x \mapsto 1 \otimes x \) for each \( x \in X \), and this is the inverse of \( \varphi \). Thus, \( K \) and \( F\left\lbrack X\right\rbrack \) are linearly disjoint over \( F \), so \( K \) and \( L \) are linearly disjoint over \( F \). | Yes |
Theorem 20.12 Let \( K \) and \( L \) be extension fields of \( F \), and let \( E \) be a field with \( F \subseteq E \subseteq K \) . Then \( K \) and \( L \) are linearly disjoint over \( F \) if and only if \( E \) and \( L \) are linearly disjoint over \( F \) and \( K \) and \( {EL} \) are linearly disjoint over \( E \) . | Proof. We have the following tower of fields.\n\n\n\nConsider the sequence of homomorphisms\n\n\[ K{ \otimes }_{F}L\overset{f}{ \rightarrow }K{ \otimes }_{E}\left( {E{ \otimes }_{F}L}\right) \overset{{\varphi }_{1}}{ \rightarrow }K{ \otimes }_{E}{EL}\overset{{\varphi }_{2}}{ \rightarrow }K\left\lbrack L\right\rbrack ,\]\n\nwhere the maps \( f,{\varphi }_{1} \), and \( {\varphi }_{2} \) are given on generators by\n\n\[ f\left( {k \otimes l}\right) = k \otimes \left( {1 \otimes l}\right) \]\n\n\[ {\varphi }_{1}\left( {k \otimes \left( {e \otimes l}\right) }\right) = k \otimes {el} \]\n\n\[ {\varphi }_{2}\left( {k\otimes \sum {e}_{i}{l}_{i}}\right) = \sum k{e}_{i}{l}_{i} \]\n\nrespectively. Each can be seen to be well defined by the universal mapping property of tensor products. The map \( f \) is an isomorphism by counting dimensions. Moreover, \( {\varphi }_{1} \) and \( {\varphi }_{2} \) are surjective. The composition of these three maps is the standard map \( \varphi : K{ \otimes }_{F}L \rightarrow K\left\lbrack L\right\rbrack \) . First, suppose that \( K \) and \( L \) are linearly disjoint over \( F \) . Then \( \varphi \) is an isomorphism by Proposition 20.2. This forces both \( {\varphi }_{1} \) and \( {\varphi }_{2} \) to be isomorphisms, since all maps in question are surjective. The injectivity of \( {\varphi }_{2} \) implies that \( K \) and \( {EL} \) are linearly disjoint over \( E \) . If \( \sigma : E{ \otimes }_{F}L \rightarrow E\left\lbrack L\right\rbrack \) is the standard map, then \( {\varphi }_{1} \) is given on generators by \( {\varphi }_{1}\left( {k \otimes \left( {e \otimes l}\right) }\right) = k \otimes \sigma \left( {e \otimes l}\right) \) ; hence, \( \sigma \) is also injective. This shows that \( E \) and \( L \) are linearly disjoint over \( F \) .\n\nConversely, suppose that \( E \) and \( L \) are linearly disjoint over \( F \) and that \( K \) and \( {EL} \) are linearly disjoint over \( E \) . Then \( {\varphi }_{2} \) and \( \sigma \) are isomorphisms by Proposition 20.2. The map \( {\varphi }_{1} \) is also an isomorphism; this follows from the relation between \( {\varphi }_{1} \) and \( \sigma \) above. Then \( \varphi \) is a composition of three isomorphisms; hence, \( \varphi \) is an isomorphism. Using Proposition 20.2 again, we see that \( K \) and \( L \) are linearly disjoint over \( F \) . | Yes |
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