Q
stringlengths 27
2.03k
| A
stringlengths 4
2.98k
| Result
stringclasses 2
values |
|---|---|---|
Example 6.3 Let \( F = {\mathbb{F}}_{p} \) . A generator for \( {F}^{ * } \) is often called a primitive root modulo \( p \) . For example,2 is a primitive root modulo 5 . Moreover,2 is not a primitive root modulo 7, while 3 is a primitive root modulo 7 . In general, it is not easy to find a primitive root modulo \( p \), and there is no simple way to find a primitive root in terms of \( p \) . | Null | No |
Corollary 6.4 If \( K/F \) is an extension of finite fields, then \( K \) is a simple extension of \( F \) . | Proof. By the previous corollary, the group \( {K}^{ * } \) is cyclic. Let \( \alpha \) be a generator of the cyclic group \( {K}^{ * } \) . Every nonzero element of \( K \) is a power of \( \alpha \) , so \( K = F\left( \alpha \right) \) . Therefore, \( K \) is a simple extension of \( F \) . | Yes |
Theorem 6.5 Let \( F \) be a finite field with \( \operatorname{char}\left( F\right) = p \), and set \( \left| F\right| = {p}^{n} \) . Then \( F \) is the splitting field of the separable polynomial \( {x}^{{p}^{n}} - x \) over \( {\mathbb{F}}_{p} \) . Thus, \( F/{\mathbb{F}}_{p} \) is Galois. Furthermore, if \( \sigma \) is defined on \( F \) by \( \sigma \left( a\right) = {a}^{p} \), then \( \sigma \) generates the Galois group \( \operatorname{Gal}\left( {F/{\mathbb{F}}_{p}}\right) \), so this Galois group is cyclic. | Proof. Let \( \left| F\right| = {p}^{n} \), so \( \left| {F}^{ * }\right| = {p}^{n} - 1 \) . By Lagrange’s theorem, if \( a \in {F}^{ * } \) , then \( {a}^{{p}^{n} - 1} = 1 \) . Multiplying by \( a \) gives \( {a}^{{p}^{n}} = a \) . This equation also holds for \( a = 0 \) . Therefore, the elements of \( F \) are roots of the polynomial \( {x}^{{p}^{n}} - x \) . However, this polynomial has at most \( {p}^{n} \) roots, so the elements of \( F \) are precisely the roots of \( {x}^{{p}^{n}} - x \) . This proves that \( F \) is the splitting field over \( {\mathbb{F}}_{p} \) of \( {x}^{{p}^{n}} - x \), and so \( F \) is normal over \( {\mathbb{F}}_{p} \) . Moreover, the derivative test shows that \( {x}^{{p}^{n}} - x \) has no repeated roots, so \( {x}^{{p}^{n}} - x \) is separable over \( {\mathbb{F}}_{p} \) . Thus, \( F \) is Galois over \( {\mathbb{F}}_{p} \) .\n\nDefine \( \sigma : F \rightarrow F \) by \( \sigma \left( a\right) = {a}^{p} \) . An easy computation shows that \( \sigma \) is an \( {\mathbb{F}}_{p} \) -homomorphism, and \( \sigma \) is surjective since \( F \) is finite. Hence, \( \sigma \) is an \( {\mathbb{F}}_{p} \) -automorphism of \( F \) . The fixed field of \( \sigma \) is \( \left\{ {a \in F : {a}^{p} = a}\right\} \supseteq {\mathbb{F}}_{p} \) . Each element in \( \mathcal{F}\left( \sigma \right) \) is a root of \( {x}^{p} - x \), so there are at most \( p \) elements in \( \mathcal{F}\left( \sigma \right) \) . This proves that \( {\mathbb{F}}_{p} = \mathcal{F}\left( \sigma \right) \), so \( \operatorname{Gal}\left( {F/{\mathbb{F}}_{p}}\right) \) is the cyclic group generated by \( \sigma \) . | Yes |
Corollary 6.6 Any two finite fields of the same size are isomorphic. | Proof. The proof of Theorem 6.5 shows that any two fields of order \( {p}^{n} \) are splitting fields over \( {\mathbb{F}}_{p} \) of \( {x}^{{p}^{n}} - x \), so the corollary follows from the isomorphic extension theorem. | Yes |
Corollary 6.7 If \( K/F \) is an extension of finite fields, then \( K/F \) is Galois with a cyclic Galois group. Moreover, if \( \operatorname{char}\left( F\right) = p \) and \( \left| F\right| = {p}^{n} \), then \( \operatorname{Gal}\left( {K/F}\right) \) is generated by the automorphism \( \tau \) defined by \( \tau \left( a\right) = {a}^{{p}^{n}} \) . | Proof. Say \( \left\lbrack {K : {\mathbb{F}}_{p}}\right\rbrack = m \) . Then \( \operatorname{Gal}\left( {K/{\mathbb{F}}_{p}}\right) \) is a cyclic group of order \( m \) by Theorem 6.5, so the order of the Frobenius automorphism \( \sigma \) of \( K \) is \( m \) . The group \( \operatorname{Gal}\left( {K/F}\right) \) is a subgroup of \( \operatorname{Gal}\left( {K/{\mathbb{F}}_{p}}\right) \), so it is also cyclic. If \( s = \left| {\operatorname{Gal}\left( {K/F}\right) }\right| \) and \( m = {ns} \), then a generator of \( \operatorname{Gal}\left( {K/F}\right) \) is \( {\sigma }^{n} \) . By induction, we see that the function \( {\sigma }^{n} \) is given by \( {\sigma }^{n}\left( a\right) = {a}^{{p}^{n}} \) . Also, since \( s = \left\lbrack {K : F}\right\rbrack \), we have that \( n = \left\lbrack {F : {\mathbb{F}}_{p}}\right\rbrack \), so \( \left| F\right| = {p}^{n} \) . | Yes |
Theorem 6.8 Let \( N \) be an algebraic closure of \( {\mathbb{F}}_{p} \) . For any positive integer \( n \), there is a unique subfield of \( N \) of order \( {p}^{n} \) . If \( K \) and \( L \) are subfields of \( N \) of orders \( {p}^{m} \) and \( {p}^{n} \), respectively, then \( K \subseteq L \) if and only if \( m \) divides \( n \) . When this occurs, \( L \) is Galois over \( K \) with Galois group generated by \( \tau \) , where \( \tau \left( a\right) = {a}^{{p}^{n}} \) . | Proof. Let \( n \) be a positive integer. The set of roots in \( N \) of the polynomial \( {x}^{{p}^{n}} - x \) has \( {p}^{n} \) elements and is a field. Thus, there is a subfield of \( N \) of order \( {p}^{n} \) . Since any two fields of order \( {p}^{n} \) in \( N \) are splitting fields of \( {x}^{{p}^{n}} - x \) over \( {\mathbb{F}}_{p} \) by Theorem 6.5, any subfield of \( N \) of order \( {p}^{n} \) consists exactly of the roots of \( {x}^{{p}^{n}} - x \) . Therefore, there is a unique subfield of \( N \) of order \( {p}^{n} \) . Let \( K \) and \( L \) be subfields of \( N \), of orders \( {p}^{m} \) and \( {p}^{n} \), respectively. First, suppose that \( K \subseteq L \) . Then \[ n = \left\lbrack {L : {\mathbb{F}}_{p}}\right\rbrack = \left\lbrack {L : K}\right\rbrack \cdot \left\lbrack {K : {\mathbb{F}}_{p}}\right\rbrack = m\left\lbrack {L : K}\right\rbrack \] so \( m \) divides \( n \) . Conversely, suppose that \( m \) divides \( n \) . Each element \( a \) of \( K \) satisfies \( {a}^{{p}^{m}} = a \) . Since \( m \) divides \( n \), each \( a \) also satisfies \( {a}^{{p}^{n}} = a \), so \( a \in L \) . This proves that \( K \subseteq L \) . When this happens \( L \) is Galois over \( K \) by Corollary 6.7. That corollary also shows that \( \operatorname{Gal}\left( {L/K}\right) \) is generated by \( \tau \) , where \( \tau \) is defined by \( \tau \left( a\right) = {a}^{\left| K\right| } \) . | Yes |
Corollary 6.9 Let \( F \) be a finite field, and let \( f\left( x\right) \) be a monic irreducible polynomial over \( F \) of degree \( n \) . 1. If \( a \) is a root of \( f \) in some extension field of \( F \), then \( F\left( a\right) \) is a splitting field for \( f \) over \( F \) . Consequently, if \( K \) is a splitting field for \( f \) over \( F \), then \( \left\lbrack {K : F}\right\rbrack = n \) . 2. If \( \left| F\right| = q \), then the set of roots of \( f \) is \( \left\{ {{a}^{{q}^{r}} : r \geq 1}\right\} \) . | Proof. Let \( K \) be a splitting field of \( f \) over \( F \) . If \( a \in K \) is a root of \( f\left( x\right) \) , then \( F\left( a\right) \) is an \( n \) -dimensional extension of \( F \) inside \( K \) . By Theorem 6.5, \( F\left( a\right) \) is a Galois extension of \( F \) ; hence, \( f\left( x\right) = \min \left( {F, a}\right) \) splits over \( F\left( a\right) \) . Therefore, \( F\left( a\right) \) is a splitting field of \( f \) over \( F \), so \( K = F\left( a\right) \) . This proves the first statement. For the second, we note that \( \operatorname{Gal}\left( {K/F}\right) = \langle \sigma \rangle \), where \( \sigma \left( c\right) = {c}^{q} \) for any \( c \in K \), by Theorem 6.8. Each root of \( f \) is then of the form \( {\sigma }^{r}\left( a\right) = {a}^{{q}^{r}} \) by the isomorphism extension theorem, which shows that the set of roots of \( f \) is \( \left\{ {{a}^{{q}^{r}} : r \geq 1}\right\} \) . | Yes |
Example 6.10 Let \( F = {\mathbb{F}}_{2} \) and \( K = F\left( \alpha \right) \), where \( \alpha \) is a root of \( f\left( x\right) = \) \( {x}^{3} + {x}^{2} + 1 \) . This polynomial has no roots in \( F \), as a quick calculation shows, so it is irreducible over \( F \) and \( \left\lbrack {K : F}\right\rbrack = 3 \) . The field \( K \) is the splitting field of \( f \) over \( F \), and the roots of \( f \) are \( \alpha ,{\alpha }^{2} \), and \( {\alpha }^{4} \), by Corollary 6.9. Since \( f\left( \alpha \right) = 0 \), we see that \( {\alpha }^{3} = {\alpha }^{2} + 1 \), so \( {\alpha }^{4} = {\alpha }^{3} + \alpha = {\alpha }^{2} + \alpha + 1 \) . Therefore, in terms of the basis \( \left\{ {1,\alpha ,{\alpha }^{2}}\right\} \) for \( K/F \), the roots of \( f \) are \( \alpha ,{\alpha }^{2} \), and \( 1 + \alpha + {\alpha }^{2} \) . This shows explicitly that \( F\left( \alpha \right) \) is the splitting field of \( f \) over \( F \) . | Null | No |
Example 6.11 Let \( F = {\mathbb{F}}_{2} \) and \( f\left( x\right) = {x}^{4} + x + 1 \) . By the derivative test, we see that \( f \) has no repeated roots. The polynomial \( f \) is irreducible over \( f \), since \( f \) has no roots in \( F \) and is not divisible by the unique irreducible quadratic \( {x}^{2} + x + 1 \) in \( F\left\lbrack x\right\rbrack \) . If \( \alpha \) is a root of \( f \), then \( {\alpha }^{4} = \alpha + 1 \) ; hence, the roots of \( f \) are \( \alpha ,\alpha + 1,{\alpha }^{2} \), and \( {\alpha }^{2} + 1 \) . | Null | No |
Example 6.12 Let \( f\left( x\right) = {x}^{2} + 1 \) . If \( p \) is an odd prime, then we show that \( f \) is reducible over \( F = {\mathbb{F}}_{p} \) if and only if \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \) . | To prove this, if \( a \in F \) is a root of \( {x}^{2} + 1 \), then \( {a}^{2} = - 1 \), so \( a \) has order 4 in \( {F}^{ * } \) . By Lagrange’s theorem,4 divides \( \left| {F}^{ * }\right| = p - 1 \), so \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \) . Conversely, if \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \), then 4 divides \( p - 1 \), so there is an element \( a \in {F}^{ * } \) of order 4, since \( {F}^{ * } \) is a cyclic group of order \( p - 1 \) . Thus, \( {a}^{4} = 1 \) and \( {a}^{2} \neq 1 \) . This forces \( {a}^{2} = - 1 \), so \( a \) is a root of \( f \) . | Yes |
Proposition 6.14 Let \( n \) be a positive integer. Then \( {x}^{{p}^{n}} - x \) factors over \( {\mathbb{F}}_{p} \) into the product of all monic irreducible polynomials over \( {\mathbb{F}}_{p} \) of degree a divisor of \( n \) . | Proof. Let \( F \) be a field of order \( {p}^{n} \) . Then \( F \) is the splitting field of \( {x}^{{p}^{n}} - x \) over \( {\mathbb{F}}_{p} \) by Theorem 6.5. Recall that \( F \) is exactly the set of roots of \( {x}^{{p}^{n}} - x \) . Let \( a \in F \), and set \( m = \left\lbrack {{\mathbb{F}}_{p}\left( a\right) : {\mathbb{F}}_{p}}\right\rbrack \), a divisor of \( \left\lbrack {F : {\mathbb{F}}_{p}}\right\rbrack \) . The polynomial \( \min \left( {{\mathbb{F}}_{p}, a}\right) \) divides \( {x}^{{p}^{n}} - x \), since \( a \) is a root of \( {x}^{{p}^{n}} - x \) . Conversely, if \( f\left( x\right) \) is a monic irreducible polynomial over \( {\mathbb{F}}_{p} \) of degree \( m \), where \( m \) divides \( n \) , let \( K \) be the splitting field of \( f \) over \( {\mathbb{F}}_{p} \) inside some algebraic closure of \( F \) . If \( a \) is a root of \( f \) in \( K \), then \( K = {\dot{\mathbb{F}}}_{p}\left( a\right) \) by Corollary 6.9. Therefore, \( \left\lbrack {K : {\mathbb{F}}_{p}}\right\rbrack = m \), so \( K \subseteq F \) by Theorem 6.8. Thus, \( a \in F \), so \( a \) is a root of \( {x}^{{p}^{n}} - x \) . Since \( f \) is irreducible over \( {\mathbb{F}}_{p} \), we have \( f = \min \left( {{\mathbb{F}}_{p}, a}\right) \), so \( f \) divides \( {x}^{{p}^{n}} - x \) . Since \( {x}^{{p}^{n}} - x \) has no repeated roots, \( {x}^{{p}^{n}} - x \) factors into distinct irreducible factors over \( {\mathbb{F}}_{p} \) . We have shown that the irreducible factors of \( {x}^{{p}^{n}} - x \) are exactly the irreducible polynomials of degree a divisor of \( n \) ; hence, the proposition is proven. | Yes |
The monic irreducible polynomials of degree 5 over \( {\mathbb{F}}_{2} \) can be determined by factoring \( {x}^{{2}^{5}} - x \), which we see factors as | \[ {x}^{{2}^{5}} - x = x\left( {x + 1}\right) \left( {{x}^{5} + {x}^{3} + 1}\right) \left( {{x}^{5} + {x}^{2} + 1}\right) \times \left( {{x}^{5} + {x}^{4} + {x}^{3} + x + 1}\right) \left( {{x}^{5} + {x}^{4} + {x}^{2} + x + 1}\right) \times \left( {{x}^{5} + {x}^{4} + {x}^{3} + {x}^{2} + 1}\right) \left( {{x}^{5} + {x}^{3} + {x}^{2} + x + 1}\right) . \] This factorization produces the six monic irreducible polynomials of degree 5 over \( {\mathbb{F}}_{2} \) . | Yes |
Proposition 7.2 Suppose that \( \operatorname{char}\left( F\right) \) does not divide \( n \), and let \( K \) be a splitting field of \( {x}^{n} - 1 \) over \( F \) . Then \( K/F \) is Galois, \( K = F\left( \omega \right) \) is generated by any primitive nth root of unity \( \omega \) , and \( \mathrm{{Gal}}\left( {K/F}\right) \) is isomorphic to a subgroup of \( {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ * } \) . Thus, \( \operatorname{Gal}\left( {K/F}\right) \) is Abelian and \( \left\lbrack {K : F}\right\rbrack \) divides \( \phi \left( n\right) \) . | Proof. Since \( \operatorname{char}\left( F\right) \) does not divide \( n \), the derivative test shows that \( {x}^{n} - 1 \) is a separable polynomial over \( F \) . Therefore, \( K \) is both normal and separable over \( F \) ; hence, \( K \) is Galois over \( F \) . Let \( \omega \in K \) be a primitive \( n \) th root of unity. Then all \( n \) th roots of unity are powers of \( \omega \), so \( {x}^{n} - 1 \) splits over \( F\left( \omega \right) \) . This proves that \( K = F\left( \omega \right) \) . Any automorphism of \( K \) that fixes \( F \) is determined by what it does to \( \omega \) . However, any automorphism restricts to a group automorphism of the set of roots of unity, so it maps the set of primitive \( n \) th roots of unity to itself. Any primitive \( n \) th root of unity in \( K \) is of the form \( {\omega }^{t} \) for some \( t \) relatively prime to \( n \) . Therefore, the map \( \theta : \operatorname{Gal}\left( {K/F}\right) \rightarrow {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ * } \) given by \( \sigma \mapsto t + n\mathbb{Z} \), where \( \sigma \left( \omega \right) = {\omega }^{t} \) , is well defined. If \( \sigma ,\tau \in \operatorname{Gal}\left( {K/F}\right) \) with \( \sigma \left( \omega \right) = {\omega }^{t} \) and \( \tau \left( \omega \right) = {\omega }^{s} \), then \( \left( {\sigma \tau }\right) \left( \omega \right) = \sigma \left( {\omega }^{s}\right) = {\omega }^{st} \), so \( \theta \) is a group homomorphism. The kernel of \( \theta \) is the set of all \( \sigma \) with \( \sigma \left( \omega \right) = \omega \) ; that is, \( \ker \left( \theta \right) = \langle \mathrm{{id}}\rangle \) . Thus, \( \theta \) is injective, so \( \operatorname{Gal}\left( {K/F}\right) \) is isomorphic to a subgroup of the Abelian group \( {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ * } \), a group of order \( \phi \left( n\right) \) . This finishes the proof. | Yes |
The structure of \( F \) determines the degree \( \left\lbrack {F\left( \omega \right) : F}\right\rbrack \) or, equivalently, the size of \( \operatorname{Gal}\left( {F\left( \omega \right) /F}\right) \) . For instance, let \( \omega = {e}^{{2\pi i}/8} \) be a primitive eighth root of unity in \( \mathbb{C} \) . Then \( {\omega }^{2} = i \) is a primitive fourth root of unity. The degree of \( \mathbb{Q}\left( \omega \right) \) over \( \mathbb{Q} \) is 4, which we will show below. If \( F = \mathbb{Q}\left( i\right) \), then the degree of \( F\left( \omega \right) \) over \( F \) is 2, since \( \omega \) satisfies the polynomial \( {x}^{2} - i \) over \( F \) and \( \omega \notin F \) . If \( F = \mathbb{R} \), then \( \mathbb{R}\left( \omega \right) = \mathbb{C} \), so \( \left\lbrack {\mathbb{R}\left( \omega \right) : \mathbb{R}}\right\rbrack = 2 \) . In fact, if \( n \geq 3 \) and if \( \tau \) is any primitive \( n \) th root of unity in \( \mathbb{C} \), then \( \mathbb{R}\left( \tau \right) = \mathbb{C} \), so \( \left\lbrack {\mathbb{R}\left( \tau \right) : \mathbb{R}}\right\rbrack = 2 \) . | Null | No |
Let \( F = {\mathbb{F}}_{2} \) . If \( \omega \) is a primitive third root of unity over \( F \) , then \( \omega \) is a root of \( {x}^{3} - 1 = \left( {x - 1}\right) \left( {{x}^{2} + x + 1}\right) \) . Since \( \omega \neq 1 \) and \( {x}^{2} + x + 1 \) is irreducible over \( F \), we have \( \left\lbrack {F\left( \omega \right) : F}\right\rbrack = 2 \) and \( \min \left( {F,\omega }\right) = {x}^{2} + x + 1 \) . | Null | No |
Lemma 7.6 Let \( n \) be any positive integer. Then \( {x}^{n} - 1 = \mathop{\prod }\limits_{{d \mid n}}{\Psi }_{d}\left( x\right) \) . Moreover, \( {\Psi }_{n}\left( x\right) \in \mathbb{Z}\left\lbrack x\right\rbrack \) . | Proof. We know that \( {x}^{n} - 1 = \prod \left( {x - \omega }\right) \), where \( \omega \) ranges over the set of all \( n \) th roots of unity. If \( d \) is the order of \( \omega \) in \( {\mathbb{C}}^{ * } \), then \( d \) divides \( n \), and \( \omega \) is a primitive \( d \) th root of unity. Gathering all the \( d \) th root of unity terms together in this factorization proves the first statement. For the second, we use induction on \( n \) ; the case \( n = 1 \) is clear since \( {\Psi }_{1}\left( x\right) = x - 1 \) . Suppose that \( {\Psi }_{d}\left( x\right) \in \mathbb{Z}\left\lbrack x\right\rbrack \) for all \( d < n \) . Then from the first part, we have\n\n\[ \n{x}^{n} - 1 = \left( {\mathop{\prod }\limits_{{d \mid n, d < n}}{\Psi }_{d}\left( x\right) }\right) \cdot {\Psi }_{n}\left( x\right)\n\]\n\nSince \( {x}^{n} - 1 \) and \( \mathop{\prod }\limits_{{d \mid n}}{\Psi }_{d}\left( x\right) \) are monic polynomials in \( \mathbb{Z}\left\lbrack x\right\rbrack \), the division algorithm, Theorem 3.2 of Appendix A, shows that \( {\Psi }_{n}\left( x\right) \in \mathbb{Z}\left\lbrack x\right\rbrack \) . | Yes |
Corollary 7.8 If \( K \) is a splitting field of \( {x}^{n} - 1 \) over \( \mathbb{Q} \), then \( \left\lbrack {K : \mathbb{Q}}\right\rbrack = \) \( \phi \left( n\right) \) and \( \operatorname{Gal}\left( {K/\mathbb{Q}}\right) \cong {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ * } \) . Moreover, if \( \omega \) is a primitive nth root of unity in \( K \), then \( \operatorname{Gal}\left( {K/\mathbb{Q}}\right) = \left\{ {{\sigma }_{i} : \gcd \left( {i, n}\right) = 1}\right\} \), where \( {\sigma }_{i} \) is determined by \( {\sigma }_{i}\left( \omega \right) = {\omega }^{i} \) . | Proof. The first part of the corollary follows immediately from Proposition 7.2 and Theorem 7.7. The description of \( \operatorname{Gal}\left( {K/\mathbb{Q}}\right) \) is a consequence of the proof of Proposition 7.2. | Yes |
Example 7.9 Let \( K = {\mathbb{Q}}_{7} \), and let \( \omega \) be a primitive seventh root of unity in \( \mathbb{C} \). By Corollary 7.8, \( \operatorname{Gal}\left( {K/\mathbb{Q}}\right) \cong {\left( \mathbb{Z}/7\mathbb{Z}\right) }^{ * } \), which is a cyclic group of order 6. The Galois group of \( K/\mathbb{Q} \) is \( \left\{ {{\sigma }_{1},{\sigma }_{2},{\sigma }_{3},{\sigma }_{4},{\sigma }_{5},{\sigma }_{6}}\right\} \), where \( {\sigma }_{i}\left( \omega \right) = \) \( {\omega }^{i} \). Thus, \( {\sigma }_{1} = \mathrm{{id}} \), and it is easy to check that \( {\sigma }_{3} \) generates this group. Moreover, \( {\sigma }_{i} \circ {\sigma }_{j} = {\sigma }_{ij} \), where the subscripts are multiplied modulo 7. The subgroups of \( \operatorname{Gal}\left( {K/\mathbb{Q}}\right) \) are then | \[ \text{(id),}\left\langle {\sigma }_{3}^{3}\right\rangle ,\left\langle {\sigma }_{3}^{2}\right\rangle ,\left\langle {\sigma }_{3}\right\rangle \] whose orders are \( 1,2,3 \), and 6, respectively. Let us find the corresponding intermediate fields. If \( L = \mathcal{F}\left( {\sigma }_{3}^{3}\right) = \mathcal{F}\left( {\sigma }_{6}\right) \), then \( \left\lbrack {K : L}\right\rbrack = \left| \left\langle {\sigma }_{6}\right\rangle \right| = 2 \) by the fundamental theorem. To find \( L \), we note that \( \omega \) must satisfy a quadratic over \( L \) and that this quadratic is \[ \left( {x - \omega }\right) \left( {x - {\sigma }_{6}\left( \omega \right) }\right) = \left( {x - \omega }\right) \left( {x - {\omega }^{6}}\right) . \] Expanding, this polynomial is \[ {x}^{2} - \left( {\omega + {\omega }^{6}}\right) x + \omega {\omega }^{6} = {x}^{2} - \left( {\omega + {\omega }^{6}}\right) x + 1. \] Therefore, \( \omega + {\omega }^{6} \in L \). If we let \( \omega = \exp \left( {{2\pi i}/7}\right) = \cos \left( {{2\pi }/7}\right) + i\sin \left( {{2\pi }/7}\right) \), then \( \omega + {\omega }^{6} = 2\cos \left( {{2\pi }/7}\right) \). Therefore, \( \omega \) satisfies a quadratic over \( \mathbb{Q}\left( {\cos \left( {{2\pi }/7}\right) }\right) \); hence, \( L \) has degree at most 2 over this field. This forces \( L = \mathbb{Q}\left( {\cos \left( {{2\pi }/7}\right) }\right) \). | Yes |
Example 7.10 Let \( K = {\mathbb{Q}}_{8} \), and let \( \omega = \exp \left( {{2\pi i}/8}\right) = \left( {1 + i}\right) /\sqrt{2} \) . The Galois group of \( K/\mathbb{Q} \) is \( \left\{ {{\sigma }_{1},{\sigma }_{3},{\sigma }_{5},{\sigma }_{7}}\right\} \), and note that each of the three nonidentity automorphisms of \( K \) have order 2 . The subgroups of this Galois group are then \[ \text{(id),}\left\langle {\sigma }_{3}\right\rangle ,\left\langle {\sigma }_{5}\right\rangle ,\left\langle {\sigma }_{7}\right\rangle ,\operatorname{Gal}\left( {K/\mathbb{Q}}\right) \text{.} \] Each of the three proper intermediate fields has degree 2 over \( \mathbb{Q} \) . One is easy to find, since \( {\omega }^{2} = i \) is a primitive fourth root of unity. The group associated to \( \mathbb{Q}\left( i\right) \) is \( \left\langle {\sigma }_{5}\right\rangle \), since \( {\sigma }_{5}\left( {\omega }^{2}\right) = {\omega }^{10} = {\omega }^{2} \) . We could find the two other fields in the same manner as in the previous example: Show that the fixed field of \( {\sigma }_{3} \) is generated over \( \mathbb{Q} \) by \( \omega + {\sigma }_{3}\left( \omega \right) \) . However, we can get this more easily due to the special form of \( \omega \) . Since \( \omega = \left( {1 + i}\right) /\sqrt{2} \) and \( {\omega }^{-1} = \left( {1 - i}\right) /\sqrt{2} \), we see that \( \sqrt{2} = \omega + {\omega }^{-1} \in K \) . The element \( \omega + {\omega }^{-1} = \omega + {\omega }^{7} \) is fixed by \( {\sigma }_{7} \) ; hence, the fixed field of \( {\sigma }_{7} \) is \( \mathbb{Q}\left( \sqrt{2}\right) \) . We know \( i \in K \) and \( \sqrt{2} \in K \), so \( \sqrt{-2} \in K \) . This element must generate the fixed field of \( {\sigma }_{3} \) . The intermediate fields are then \[ K,\mathbb{Q}\left( \sqrt{-2}\right) ,\mathbb{Q}\left( \sqrt{-1}\right) ,\mathbb{Q}\left( \sqrt{2}\right) ,\mathbb{Q}\text{.} \] The description of the intermediate fields also shows that \( K = \mathbb{Q}\left( {\sqrt{2}, i}\right) \) . | Null | No |
Example 8.2 Let \( F \) be any field, and let \( K = F\left( \sqrt{d}\right) \) for some \( d \in F - {F}^{2} \) . A convenient basis for \( K \) is \( \{ 1,\sqrt{d}\} \) . If \( \alpha = a + b\sqrt{d} \) with \( a, b \in F \), we determine the norm and trace of \( \alpha \) . | The linear transformation \( {L}_{\alpha } \) is equal to \( a{L}_{1} + b{L}_{\sqrt{d}} \), so we first need to find the matrix representations for \( {L}_{1} \) and \( {L}_{\sqrt{d}} \) . The identity transformation \( {L}_{1} \) has matrix \( \left( \begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) \) . For \( {L}_{\sqrt{d}} \) , we have\n\n\[ \n{L}_{\sqrt{d}}\left( 1\right) = \sqrt{d} = 0 \cdot 1 + 1 \cdot \sqrt{d} \n\]\n\n\[ \n{L}_{\sqrt{d}}\left( \sqrt{d}\right) = d = d \cdot 1 + 0 \cdot \sqrt{d} \n\]\n\nTherefore, the matrix for \( {L}_{\sqrt{d}} \) is \( \left( \begin{array}{ll} 0 & d \\ 1 & 0 \end{array}\right) \) . The matrix for \( {L}_{\alpha } \) is then\n\n\[ \na\left( \begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) + b\left( \begin{array}{ll} 0 & d \\ 1 & 0 \end{array}\right) = \left( \begin{matrix} a & {bd} \\ b & a \end{matrix}\right) .\n\]\n\nFrom this we obtain \( {N}_{K/F}\left( {a + b\sqrt{d}}\right) = {a}^{2} - {b}^{2}d \) and \( {T}_{K/F}\left( {a + b\sqrt{d}}\right) = {2a} \) . In particular, \( {N}_{K/F}\left( \sqrt{d}\right) = - d \) and \( {T}_{K/F}\left( \sqrt{d}\right) = 0 \) . | Yes |
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 |
Example 8.4 Let \( F \) be a field of characteristic \( p > 0 \), and let \( K/F \) be a purely inseparable extension of degree \( p \) . Say \( K = F\left( \alpha \right) \) with \( {\alpha }^{p} = \) \( a \in F \) . For instance, we could take \( K \) to be the rational function field \( k\left( x\right) \) over a field \( k \) of characteristic \( p \) and \( F = k\left( {x}^{p}\right) \) . A basis for \( K \) is \( \left\{ {1,\alpha ,{\alpha }^{2},\ldots ,{\alpha }^{p - 1}}\right\} \) . With respect to this basis, the matrix for \( {L}_{\alpha } \) is\n\n\[ \left( \begin{matrix} 0 & 0 & \cdots & 0 & a \\ 1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & \ddots & \vdots & \vdots \\ \vdots & \vdots & \ddots & 0 & 0 \\ 0 & 0 & & 1 & 0 \end{matrix}\right) . \]\n\nWe leave it to the reader to check that the matrix for \( {L}_{{\alpha }^{2}} \) is obtained by taking this matrix and shifting the columns to the left, moving the first column to the end. Similar processes yield the matrices \( {L}_{{\alpha }^{i}} \) for each \( i \) . From these matrices, we see that \( {N}_{K/F}\left( \alpha \right) = {\left( -1\right) }^{p}a \) . For traces, each \( {L}_{{\alpha }^{i}} \) has trace 0, including the identity matrix, since \( p \cdot 1 = 0 \) in \( F \) . Therefore, for any \( \beta \in K \) we have \( {\operatorname{Tr}}_{K/F}\left( \beta \right) = 0 \) . The trace map \( {T}_{K/F} \) is thus the zero function. | Null | No |
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 |
Example 8.8 If \( K \) is a purely inseparable extension of \( F \) of characteristic \( p \), then the minimal polynomial of any element of \( K \) is of the form \( {x}^{{p}^{n}} - a \) . From this, it follows that the trace of any element is zero. | Null | No |
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 \( F \) be a field of characteristic not 2, and let \( a \in {F}^{ * } - {F}^{*2} \). If \( K = F\\left( \\sqrt{a}\\right) \), then \( \\operatorname{Gal}\\left( {K/F}\\right) = \\{ \\mathrm{{id}},\\sigma \\} \) where \( \\sigma \\left( \\sqrt{a}\\right) = - \\sqrt{a} \). Thus, \( \\operatorname{Gal}\\left( {K/F}\\right) \\cong \\mathbb{Z}/2\\mathbb{Z} \) is cyclic. | Null | No |
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 |
Example 10.6 Let \( E = {S}_{3} \). If \( M = \langle \left( {123}\right) \rangle \), then \( M \) is isomorphic to \( \mathbb{Z}/3\mathbb{Z} \) and \( M \) is an Abelian normal subgroup of \( E \). The quotient group \( E/M \) is isomorphic to \( \mathbb{Z}/2\mathbb{Z} \). Therefore, \( {S}_{3} \) is a group extension of \( \mathbb{Z}/2\mathbb{Z} \) by \( \mathbb{Z}/3\mathbb{Z} \). | Null | No |
Example 10.7 Let \( E = {D}_{n} \), the dihedral group. One description of \( E \) is by generators and relations. The group \( E \) is generated by elements \( \sigma \) and \( \tau \) satisfying \( {\tau }^{n} = {\sigma }^{2} = e \) and \( {\sigma \tau \sigma } = {\tau }^{-1} \) . Let \( M = \langle \sigma \rangle \), a normal subgroup of \( E \) that is isomorphic to \( \mathbb{Z}/n\mathbb{Z} \) . The quotient \( E/M \) is isomorphic to \( \mathbb{Z}/2\mathbb{Z} \) , so \( E \) is a group extension of \( \mathbb{Z}/2\mathbb{Z} \) by \( \mathbb{Z}/n\mathbb{Z} \) . | Null | No |
Let \( M \) and \( G \) be groups, and let \( \varphi : G \rightarrow \operatorname{End}\left( M\right) \) be a group homomorphism. If \( E \) is the semidirect product \( M{ \times }_{\varphi }G \), then \( {M}^{\prime } = \{ \left( {m, e}\right) : m \in M\} \) is a normal subgroup of \( E \) isomorphic to \( M \), and \( E/{M}^{\prime } \cong G \) . Thus, \( E \) is a group extension of \( M \) by \( G \) . | Null | No |
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 |
If \( F \) is a field that contains a primitive \( n \) th root of unity, and if \( K/F \) is a cyclic extension of degree \( n \), then \( K/F \) is an \( n \) -Kummer extension. If \( F \) also contains a primitive \( m \) th root of unity for some \( m \) that is a multiple of \( n \), then \( K/F \) is also an \( m \) -Kummer extension. Therefore, if an extension is an \( n \) -Kummer extension, the integer \( n \) need not be unique. | Null | No |
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 |
If \( K = \mathbb{Q}\left( {\sqrt{{a}_{1}},\ldots ,\sqrt{{a}_{r}}}\right) \) for some \( {a}_{i} \in \mathbb{Q} \), then \( K/\mathbb{Q} \) is a 2-Kummer extension by Theorem 11.4. The degree of \( K/F \) is no larger than \( {2}^{r} \), but it may be less depending on the choice of the \( {a}_{i} \) . | Null | No |
Example 11.6 Let \( F = \mathbb{Q}\left( i\right) \), where \( i = \sqrt{-1} \), and let \( K = F\left( {\sqrt[4]{12},\sqrt[4]{3}}\right) \) . Since \( i \) is a primitive fourth root of unity, \( K/F \) is a 4-Kummer extension. The degree of \( K/F \) is 8, not 16, since \( K = F\left( {\sqrt{2},\sqrt[4]{3}}\right) \) ; this equality is true because \( \sqrt[4]{12} = \sqrt{2}\sqrt[4]{3} \) . This example shows that if \( K = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) is an \( n \) -Kummer extension of \( F \) with \( {\alpha }_{i}^{n} \in F \), it might be the case that a smaller power of some of the \( {\alpha }_{i} \) is also in \( F \) . | Null | No |
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 |
Example 12.8 Let \( f\left( x\right) = {x}^{3} + {px} + q \) . Then \( {a}_{0} = q,{a}_{1} = p \), and \( {a}_{2} = 0 \) , so by Newton's identities we get\n\n\[ \n{t}_{1} = 0 \]\n\n\[ \n{t}_{2} = - {2p} \]\n\n\[ \n{t}_{3} = - {3q} \]\n\n\[ \n{t}_{4} = 2{p}^{2} \]\n\nTherefore\n\n\[ \n\operatorname{disc}\left( f\right) = \left| \begin{matrix} 3 & 0 & - {2p} \\ 0 & - {2p} & - {3q} \\ - {2p} & - {3q} & 2{p}^{2} \end{matrix}\right| = - 4{p}^{3} - {27}{q}^{2}. \]\n | Null | No |
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 \( F \) . | No |
The polynomial \( {x}^{3} - {3x} + 1 \in \mathbb{Q}\left\lbrack x\right\rbrack \) has discriminant \( {81} = \) \( {9}^{2} \), and it is irreducible over \( \mathbb{Q} \) by an application of the rational root test. Thus, the Galois group of its splitting field over \( \mathbb{Q} \) is \( {A}_{3} \) . | Null | No |
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 |
Example 13.6 Let \( f\left( x\right) = {x}^{4} - 4{x}^{3} + 4{x}^{2} + 6 \) . This polynomial is irreducible by the Eisenstein criterion. Now,\n\n\[ r\left( x\right) = {x}^{3} - 4{x}^{2} - {24x} = x\left( {{x}^{2} - {4x} - {24}}\right) ,\] \n\nso \( L = \mathbb{Q}\left( \sqrt{7}\right) \) . Take \( {\beta }_{1} = 0 \) . Then\n\n\[ h\left( x\right) = \left( {{x}^{2} + 6}\right) \left( {{x}^{2} - {4x} + 4}\right) = \left( {{x}^{2} + 6}\right) {\left( x - 2\right) }^{2}. \] \n\nSince \( h \) does not split over \( L \), we see that the Galois group of \( f \) is isomorphic to \( {D}_{4} \) . | Null | No |
Example 13.7 Let \( p \) be a prime, and let \( f\left( x\right) = {x}^{4} + {px} + p \) . Then \( r\left( x\right) = {x}^{3} - {4px} - {p}^{2} \) . To test for roots of \( r\left( x\right) \) in \( \mathbb{Q} \), we only need to check \( \pm 1, \pm p, \pm {p}^{2} \) . We see that \( \pm 1 \) and \( \pm {p}^{2} \) are never roots, but \( r\left( p\right) = {p}^{2}\left( {p - 5}\right) \) and \( r\left( {-p}\right) = {p}^{2}\left( {3 - p}\right) \) . Therefore, for \( p \neq 3,5 \), the resolvent \( r \) has no roots in \( \mathbb{Q} \) ; hence, \( r \) is irreducible over \( \mathbb{Q} \) . The discriminant \( D = {p}^{3}\left( {{256} - {27p}}\right) \) is not a square in \( \mathbb{Q} \), since if \( p \) is odd, then \( p \) does not divide \( {256} - {27p} \), and \( D = {1616} \notin {\mathbb{Q}}^{2} \) for \( p = 2 \) . Let \( G \) be the Galois group of \( f \) . Then \( G \cong {S}_{4} \) for \( p \neq 3,5 \) . If \( p = 3 \), let \( {\beta }_{1} = - 3 \) . Then \( r\left( x\right) = \left( {x + 3}\right) \left( {{x}^{2} - {3x} - 3}\right) \), so \( L = \mathbb{Q}\left( \sqrt{21}\right) \) . Then \( h\left( x\right) = \left( {{x}^{2} + {3x} + 3}\right) \left( {{x}^{2} + 3}\right) \) does not split over \( L \), so \( G \cong {D}_{4} \) . If \( p = 5 \), then \( r\left( x\right) = \left( {x - 5}\right) \left( {{x}^{2} + {5x} + 5}\right) \), so \( L = \mathbb{Q}\left( \sqrt{5}\right) \) . As \( h\left( x\right) = \left( {{x}^{2} - {5x} + 5}\right) \left( {{x}^{2} - 5}\right), h \) splits over \( L \), so \( G \cong {C}_{4} \) . | Null | No |
Example 13.8 Let \( l \in \mathbb{Q} \), and let \( f\left( x\right) = {x}^{4} - l \) . Then the resolvent of \( f \) is \( r\left( x\right) = {x}^{3} + {4lx} = x\left( {{x}^{2} + {4l}}\right) \) . If \( - l \) is not a square in \( \mathbb{Q} \), then \( r\left( x\right) \) has exactly one root in \( \mathbb{Q} \) . Moreover, \( h\left( x\right) = {x}^{2}\left( {{x}^{2} + l}\right) \) does not factor completely over \( \mathbb{Q} \), so the Galois group \( G \) of \( f \) is \( {D}_{4} \) by Theorem 13.4. On the other hand, if \( - l \) is a square in \( \mathbb{Q} \), then \( r \) factors completely over \( \mathbb{Q} \), so \( G \cong V \) . For example, the Galois group of \( {x}^{4} + 4 \) is \( V \) . The splitting field of \( {x}^{4} + 4 \) over \( \mathbb{Q} \) is then \( \mathbb{Q}\left( \sqrt[4]{-4}\right) \) . | Null | No |
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 |
Proposition 16.8 Let \( G \) be a group and \( N \) be a normal subgroup of \( G \) . Then \( G \) is solvable if and only if \( N \) and \( G/N \) are solvable. | Null | No |
Proposition 16.9 If \( n \geq 5 \), then \( {S}_{n} \) is not solvable. | Null | No |
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. | No |
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 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.