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Lemma 17.1 If \( {\alpha }_{1},\ldots ,{\alpha }_{n} \in K \), then there is an \( E \in \mathcal{I} \) with \( {\alpha }_{i} \in E \) for all \( i \) . | Proof. Let \( E \subseteq K \) be the splitting field of the minimal polynomials of the \( {\alpha }_{i} \) over \( F \) . Then, as each \( {\alpha }_{i} \) is separable over \( F \), the field \( E \) is normal and separable over \( F \) ; hence, \( E \) is Galois over \( F \) . Since there are finitely many \( {\alpha }_{i} \), we have \( \left\lbrack {E : F}\right\rbrack < \infty \), so \( E \in \mathcal{I} \) . | Yes |
Lemma 17.2 Let \( N \in \mathcal{N} \), and set \( N = \operatorname{Gal}\left( {K/E}\right) \) with \( E \in \mathcal{I} \) . Then \( E = \mathcal{F}\left( N\right) \) and \( N \) is normal in \( G \) . Moreover, \( G/N \cong \operatorname{Gal}\left( {E/F}\right) \) . Thus, \( \left| {G/N}\right| = \left| {\operatorname{Gal}\left( {E/F}\right) }\right| = \left\lbrack {E : F}\right\rbrack < \infty \). | Proof. Since \( K \) is normal and separable over \( F \), the field \( K \) is also normal and separable over \( E \), so \( K \) is Galois over \( E \) . Therefore, \( E = \mathcal{F}\left( N\right) \) . As in the proof of the fundamental theorem, the map \( \theta : G \rightarrow \operatorname{Gal}\left( {E/F}\right) \) given by \( \sigma \mapsto {\left. \sigma \right| }_{E} \) is a group homomorphism with kernel \( \operatorname{Gal}\left( {K/E}\right) = N \) . Proposition 3.28 shows that \( \theta \) is surjective. The remaining statements then follow. | Yes |
Lemma 17.3 We have \( \mathop{\bigcap }\limits_{{N \in \mathcal{N}}}N = \{ \mathrm{{id}}\} \) . Furthermore, \( \mathop{\bigcap }\limits_{{N \in \mathcal{N}}}{\sigma N} = \{ \sigma \} \) for all \( \sigma \in G \) . | Proof. Let \( \tau \in \mathop{\bigcap }\limits_{{N \in \mathcal{N}}}N \) and let \( a \in K \) . By Lemma 17.1, there is an \( E \in \mathcal{I} \) with \( a \in E \) . Set \( N = \operatorname{Gal}\left( {K/E}\right) \in \mathcal{N} \) . The automorphism \( \tau \) fixes \( E \) since \( \tau \in N \), so \( \tau \left( a\right) = a \) . Thus, \( \tau = \mathrm{{id}} \), so \( \mathop{\bigcap }\limits_{{N \in \mathcal{N}}}N = \{ \mathrm{{id}}\} \) . For the second statement, if \( \tau \in {\sigma N} \) for all \( N \), then \( {\sigma }^{-1}\tau \in N \) for all \( N \) ; thus, \( {\sigma }^{-1}\tau = \mathrm{{id}} \) by the first part. This yields \( \tau = \sigma \), so \( \mathop{\bigcap }\limits_{{N \in \mathcal{N}}}{\sigma N} = \{ \sigma \} \) . | Yes |
Lemma 17.4 Let \( {N}_{1},{N}_{2} \in \mathcal{N} \) . Then \( {N}_{1} \cap {N}_{2} \in \mathcal{N} \) . | Proof. Let \( {N}_{i} = \operatorname{Gal}\left( {K/{E}_{i}}\right) \) with \( {E}_{i} \in \mathcal{I} \) . Each \( {E}_{i} \) is finite Galois over \( F \) ; hence, \( {E}_{1}{E}_{2} \) is also finite Galois over \( F \), so \( {E}_{1}{E}_{2} \in \mathcal{I} \) . However, \( \operatorname{Gal}\left( {K/{E}_{1}{E}_{2}}\right) = {N}_{1} \cap {N}_{2} \) ; to see this, we note that \( \sigma \in {N}_{1} \cap {N}_{2} \) if and only if \( {\left. \sigma \right| }_{{E}_{1}} = \) id and \( {\left. \sigma \right| }_{{E}_{2}} = \) id, if and only if \( {E}_{1} \subseteq \mathcal{F}\left( \sigma \right) \) and \( {E}_{2} \subseteq \mathcal{F}\left( \sigma \right) \) , and if and only if \( {E}_{1}{E}_{2} \subseteq \mathcal{F}\left( \sigma \right) \) . This last condition is true if and only if \( \sigma \in \operatorname{Gal}\left( {K/{E}_{1}{E}_{2}}\right) \) . Thus, \( {N}_{1} \cap {N}_{2} = \operatorname{Gal}\left( {K/{E}_{1}{E}_{2}}\right) \in \mathcal{N} \) . | Yes |
Theorem 17.7 Let \( H \) be a subgroup of \( G \), and let \( {H}^{\prime } = \operatorname{Gal}\left( {K/\mathcal{F}\left( H\right) }\right) \) . Then \( {H}^{\prime } = \bar{H} \), the closure of \( H \) in the topology of \( G \) . | Proof. It is clear that \( H \subseteq {H}^{\prime } \), so it suffices to show that \( {H}^{\prime } \) is closed and that \( {H}^{\prime } \subseteq \bar{H} \) . To show that \( {H}^{\prime } \) is closed, take any \( \sigma \in G - {H}^{\prime } \) . Then there is an \( \alpha \in \mathcal{F}\left( H\right) \) with \( \sigma \left( \alpha \right) \neq \alpha \) . Take \( E \in \mathcal{I} \) with \( \alpha \in E \), and let \( N = \operatorname{Gal}\left( {K/E}\right) \in \mathcal{N} \) . Then, for any \( \tau \in N \), we have \( \tau \left( \alpha \right) = \alpha \), so \( {\sigma \tau }\left( \alpha \right) = \sigma \left( \alpha \right) \neq \alpha \) . Hence, \( {\sigma N} \) is an open neighborhood of \( \sigma \) disjoint from \( {H}^{\prime } \) . Therefore, \( G - {H}^{\prime } \) is open, so \( {H}^{\prime } \) is closed. To prove the inclusion \( {H}^{\prime } \subseteq \) \( \bar{H} \), we first set \( L = \mathcal{F}\left( H\right) \) . Let \( \sigma \in {H}^{\prime } \) and \( N \in \mathcal{N} \) . Set \( E = \mathcal{F}\left( N\right) \in \mathcal{I} \) , and let \( {H}_{0} = \left\{ {{\left. \rho \right| }_{E} : \rho \in H}\right\} \), a subgroup of the finite group \( \operatorname{Gal}\left( {E/F}\right) \) . Since \( \mathcal{F}\left( {H}_{0}\right) = \mathcal{F}\left( H\right) \cap E = L \cap E \), the fundamental theorem for finite Galois extensions shows that \( {H}_{0} = \operatorname{Gal}\left( {E/\left( {E \cap L}\right) }\right) \) . Since \( \sigma \in {H}^{\prime } \), we have \( {\left. \sigma \right| }_{L} = \) id, so \( {\left. \sigma \right| }_{E} \in {H}_{0} \) . Therefore, there is a \( \rho \in H \) with \( {\left. \rho \right| }_{E} = {\left. \sigma \right| }_{E} \) . Thus, \( {\sigma }^{-1}\rho \in \operatorname{Gal}\left( {K/E}\right) = N \), so \( \rho \in {\sigma N} \cap H \) . This shows that every basic open neighborhood \( {\sigma N} \) of \( \sigma \in {H}^{\prime } \) meets \( H \), so \( \sigma \in \bar{H} \) . This proves the inclusion \( {H}^{\prime } \subseteq \bar{H} \) and finishes the proof. | Yes |
Theorem 17.8 (Fundamental Theorem of Infinite Galois Theory) Let \( K \) be a Galois extension of \( F \), and let \( G = \operatorname{Gal}\left( {K/F}\right) \) . With the Krull topology on \( G \), the maps \( L \mapsto \operatorname{Gal}\left( {K/L}\right) \) and \( H \mapsto \mathcal{F}\left( H\right) \) give an inclusion reversing correspondence between the fields \( L \) with \( F \subseteq L \subseteq K \) and the closed subgroups \( H \) of \( G \) . Furthermore, if \( L \leftrightarrow H \), then \( \left| {G : H}\right| < \infty \) if and only if \( \left\lbrack {L : F}\right\rbrack < \infty \), if and only if \( H \) is open. When this occurs, \( \left| {G : H}\right| = \left\lbrack {L : F}\right\rbrack \) . Also, \( H \) is normal in \( G \) if and only if \( L \) is Galois over \( F \), and when this occurs, there is a group isomorphism \( \operatorname{Gal}\left( {L/F}\right) \cong G/N \) . If \( G/N \) is given the quotient topology, this isomorphism is also a homeomorphism. | Proof. If \( L \) is a subfield of \( K \) containing \( F \), then \( K \) is normal and separable over \( L \), so \( K \) is Galois over \( L \) . Thus, \( L = \mathcal{F}\left( {\operatorname{Gal}\left( {K/L}\right) }\right) \) . If \( H \) is a subgroup of \( G \), then Theorem 17.7 shows that \( H = \operatorname{Gal}\left( {K/\mathcal{F}\left( H\right) }\right) \) if and only if \( H \) is closed. The two maps \( L \mapsto \operatorname{Gal}\left( {K/L}\right) \) and \( H \mapsto \mathcal{F}\left( H\right) \) then give an inclusion reversing correspondence between the set of intermediate fields of \( K/F \) and the set of closed subgroups of \( G \) .\n\nLet \( L \) be an intermediate field of \( K/F \), and let \( H = \operatorname{Gal}\left( {K/L}\right) \) . Suppose that \( \left| {G : H}\right| < \infty \) . Then \( G - H \) is a finite union of cosets of \( H \), each of which is closed, since \( H \) is closed. Thus, \( G - H \) is closed, so \( H \) is open. Conversely, if \( H \) is open, then \( H \) contains some basic neighborhood of id, so \( N \subseteq H \) for some \( N \in \mathcal{N} \) . If \( E = \mathcal{F}\left( N\right) \), then \( L \subseteq E \), so \( \left\lbrack {L : F}\right\rbrack < \infty \) . Finally, if \( \left\lbrack {L : F}\right\rbrack < \infty \), then choose an \( E \in \mathcal{I} \) with \( L \subseteq E \), possible by Lemma 17.1. Let \( N = \operatorname{Gal}\left( {K/E}\right) \) . Then \( N \subseteq H \), since \( L \subseteq E \), so \( \left| {G : H}\right| \leq \left| {G : N}\right| < \infty \) . By Lemma 17.2, we have \( G/N \cong \operatorname{Gal}\left( {E/F}\right) \) via the map \( {\sigma N} \mapsto {\left. \sigma \right| }_{E} \) . Thus, \( H/N \) maps to \( \left\{ {{\left. \rho \right| }_{E} : \rho \in H}\right\} \), a subgroup of \( \operatorname{Gal}\left( {E/F}\right) \) with fixed field \( L \cap E = L \) . By the fundamental theorem for finite extensions, the order of this group is \( \left\lbrack {E : L}\right\rbrack \) . Therefore,\n\n\[ \left| {G : H}\right| = \left| {G/N : H/N}\right| = \frac{\left| G/N\right| }{\left| H/N\right| } = \frac{\left\lbrack E : F\right\rbrack }{\left\lbrack E : L\right\rbrack } = \left\lbrack {L : F}\right\rbrack \]\n\nFor the statement about normality, we continue to assume that \( H = \) \( \operatorname{Gal}\left( {K/L}\right) \) . Suppose that \( H \) is a normal subgroup of \( G \) . Let \( a \in L \), and let \( f\left( x\right) = \min \left( {F, a}\right) \) . If \( b \in K \) is any root of \( f \), by the isomorphism extension theorem there is a \( \sigma \in G \) with \( \sigma \left( a\right) = b \) . To see that \( b \in L \), take \( \tau \in H \) . Then\n\n\[ \tau \left( b\right) = {\sigma }^{-1}\left( {{\sigma \tau }{\sigma }^{-1}\left( a\right) }\right) = {\sigma }^{-1}\left( a\right) = b \] since \( {\sigma \tau }{\sigma }^{-1} \in H \), as \( H \) is normal in \( G \) . Thus, \( b \in \mathcal{F}\left( H\right) | Yes |
Example 17.9 Let \( K/F \) be a Galois extension with \( \left\lbrack {K : F}\right\rbrack < \infty \) . Then the Krull topology on \( \operatorname{Gal}\left( {K/F}\right) \) is the discrete topology; hence, every subgroup of \( \operatorname{Gal}\left( {K/F}\right) \) is closed. Thus, we recover the original fundamental theorem of Galois theory from Theorem 17.8. | Null | No |
Let \( K = \mathbb{Q}\left( \left\{ {{e}^{{2\pi ik}/n} : k, n \in \mathbb{N}}\right\} \right) \) be the field generated over \( \mathbb{Q} \) by all roots of unity in \( \mathbb{C} \) . Then \( K \) is the splitting field over \( \mathbb{Q} \) of the set \( \left\{ {{x}^{n} - 1 : n \in \mathbb{N}}\right\} \), so \( K/\mathbb{Q} \) is Galois. If \( L \) is a finite Galois extension of \( \mathbb{Q} \) contained in \( K \), then \( L \) is contained in a cyclotomic extension of \( \mathbb{Q} \) . The Galois group of a cyclotomic extension is Abelian. Consequently, \( \operatorname{Gal}\left( {L/F}\right) \) is Abelian. To see that \( \operatorname{Gal}\left( {K/F}\right) \) is Abelian, by the proof of Theorem 17.8 the Galois group \( \operatorname{Gal}\left( {K/F}\right) \) is isomorphic to a subgroup of the direct product of the \( \operatorname{Gal}\left( {L/F}\right) \) as \( L \) ranges over finite Galois subextensions of \( \mathbb{Q} \) , so \( \operatorname{Gal}\left( {K/F}\right) \) is Abelian. As a consequence of this fact, any subextension of \( K/\mathbb{Q} \) is a Galois extension of \( \mathbb{Q} \). | We give an alternate proof that \( \operatorname{Gal}\left( {K/F}\right) \) is Abelian that does not use the proof of Theorem 17.8. Take \( \sigma ,\tau \in \operatorname{Gal}\left( {K/\mathbb{Q}}\right) \) . If \( a \in K \), then there is an intermediate field \( L \) of \( K/\mathbb{Q} \) that is Galois over \( \mathbb{Q} \) and that \( a \in L \) . The restrictions \( {\left. \sigma \right| }_{L},{\left. \tau \right| }_{L} \) are elements of \( \operatorname{Gal}\left( {L/\mathbb{Q}}\right) \), and this group is Abelian by the previous paragraph. Thus,\n\n\[ \sigma \left( {\tau \left( a\right) }\right) = {\left. \sigma \right| }_{L}\left( {{\left. \tau \right| }_{L}\left( a\right) }\right) = {\left. \tau \right| }_{L}\left( {{\left. \sigma \right| }_{L}\left( a\right) }\right) = \tau \left( {\sigma \left( a\right) }\right) . \]\n\nConsequently, \( {\sigma \tau } = {\tau \sigma } \), so \( \operatorname{Gal}\left( {K/\mathbb{Q}}\right) \) is Abelian. | Yes |
Example 17.11 Let \( K \) be an algebraic closure of \( {\mathbb{F}}_{p} \) . Since \( {\mathbb{F}}_{p} \) is perfect, \( K \) is separable, and hence \( K \) is Galois over \( {\mathbb{F}}_{p} \) . Let \( \sigma : K \rightarrow K \) be defined by \( \sigma \left( a\right) = {a}^{p} \) . Then \( \sigma \in G = \operatorname{Gal}\left( {K/{\mathbb{F}}_{p}}\right) \), and the fixed field of the cyclic subgroup \( H \) of \( G \) generated by \( \sigma \) is \( {\mathbb{F}}_{p} \) . However, we prove that \( H \neq G \) by constructing an automorphism of \( K \) that is not in \( H \) . | To see this, pick an integer \( {n}_{r} \) for each \( r \in \mathbb{N} \) such that if \( r \) divides \( s \), then \( {n}_{s} \equiv {n}_{r}\left( {\;\operatorname{mod}\;r}\right) \) . If \( {F}_{r} \) is the subfield of \( K \) containing \( {p}^{r} \) elements, then define \( \tau \) by \( \tau \left( a\right) = {\sigma }^{{n}_{r}}\left( a\right) \) if \( a \in {F}_{r} \) . The conditions on the \( {n}_{r} \) show that \( \tau \) is well defined, and an easy argument shows that \( \tau \) is an automorphism of \( K \) that fixes \( {\mathbb{F}}_{p} \) . For a specific example of a choice of the \( {n}_{r} \), for \( r \in \mathbb{N} \), write \( r = {p}^{m}q \) with \( q \) not a multiple of \( p \) . Let \( {n}_{r} \) satisfy\n\n\[ \n{n}_{r} \equiv 1 + p + \cdots + {p}^{m - 1}\left( {\;\operatorname{mod}\;{p}^{m}}\right) , \n\]\n\n\[ \n{n}_{r} \equiv 0\left( {\;\operatorname{mod}\;q}\right) \n\]\n\nSuch integers exist by the Chinese remainder theorem of number theory, since \( {p}^{m} \) and \( q \) are relatively prime. If \( \tau = {\sigma }^{t} \) for some \( t \), then for all \( r,{\left. \tau \right| }_{{F}_{r}} = {\left. {\sigma }^{t}\right| }_{{F}_{r}} \), so \( {n}_{r} \equiv t\left( {\;\operatorname{mod}\;r}\right) \), as \( \operatorname{Gal}\left( {{F}_{r}/{\mathbb{F}}_{p}}\right) \) is the cyclic group generated by \( {\left. \sigma \right| }_{{F}_{r}} \), which has order \( r \) . This cannot happen as \( {n}_{{p}^{m}} \rightarrow \infty \) as \( m \rightarrow \infty \) . Therefore, \( \tau \notin H \), so \( H \) is not a closed subgroup of \( G \) . The group \( G \) is obtained topologically from \( H \), since \( G = \bar{H} \) by Theorem 17.7. The argument that \( G = \operatorname{im}\left( f\right) \) in the proof of Theorem 17.6 shows that any element of \( G \) is obtained by the construction above, for an appropriate choice of the \( {n}_{r} \) . This gives a description of the Galois group \( G \) as\n\n\[ \n\operatorname{Gal}\left( {K/{\mathbb{F}}_{p}}\right) \cong \left\{ {\left\{ {n}_{r}\right\} \in \mathop{\prod }\limits_{r}{\mathbb{F}}_{{p}^{r}} : \text{ if }r\text{ divides }s,\text{ then }{n}_{s} \equiv {n}_{r}\left( {\;\operatorname{mod}\;r}\right) }\right\} .\n\] | Yes |
Proposition 18.1 Let \( {F}_{s} \) be the separable closure of the field \( F \) . Then \( {F}_{s} \) is Galois over \( F \) with \( \operatorname{Gal}\left( {{F}_{s}/F}\right) \cong \operatorname{Gal}\left( {{F}_{ac}/F}\right) \) . Moreover, \( {F}_{s} \) is a maximal separable extension of \( F \), meaning that \( {F}_{s} \) is not properly contained in any separable extension of \( F \) . Thus, \( {F}_{s} \) is separably closed. | Proof. The field \( {F}_{s} \) is Galois over \( F \), and \( \operatorname{Gal}\left( {{F}_{s}/F}\right) = \operatorname{Gal}\left( {{F}_{ac}/F}\right) \) by Theorem 4.23. Suppose that \( {F}_{s} \subseteq L \) with \( L/F \) separable. Then we can embed \( L \subseteq {F}_{ac} \), and then \( L = {F}_{s} \), since \( {F}_{s} \) is the set of all separable elements over \( F \) in \( {F}_{ac} \) . Finally, if \( L \) is a separable extension of \( {F}_{s} \), then by transitivity of separability, \( L \) is a separable extension of \( F \), so \( L = {F}_{s} \) . Therefore, \( {F}_{s} \) is separably closed. | Yes |
Proposition 18.2 Let \( F \) be a field with \( \operatorname{char}\left( F\right) \neq 2 \) . Then the quadratic closure \( {F}_{q} \) of \( F \) is the composite inside a fixed algebraic closure of \( F \) of all Galois extensions of \( F \) of degree a power of 2. | Proof. Let \( K \) be the composite inside a fixed algebraic closure of \( F \) of all Galois extensions of \( F \) of degree a power of 2 . Then \( K \) is Galois over \( F \) . To show that \( G = \operatorname{Gal}\left( {K/F}\right) \) is a pro-2-group, let \( N \) be an open normal subgroup of \( G \) . If \( L = \mathcal{F}\left( H\right) \), then \( \left\lbrack {L : F}\right\rbrack = \left\lbrack {G : N}\right\rbrack \) by the fundamental theorem. The intermediate field \( L \) is a finite extension of \( F \) ; hence, \( L \) lies in a composite of finitely many Galois extensions of \( F \) of degree a power of 2. Any such composite has degree over \( F \) a power of 2 by the theorem of natural irrationalities, so \( \left\lbrack {L : F}\right\rbrack \) is a power of 2 . Thus, \( \left\lbrack {G : N}\right\rbrack \) is a power of 2, so \( G \) is a pro-2-group.\n\nTo see that \( K \) is quadratically closed, suppose that \( L/K \) is a quadratic extension, and say \( L = K\left( \sqrt{a}\right) \) for some \( a \in K \) . Then \( a \in E \) for some finite Galois subextension \( E \) . By the argument above, we have \( \left\lbrack {E : F}\right\rbrack = {2}^{r} \) for some \( r \) . The extension \( E\left( \sqrt{a}\right) /E \) has degree at most 2 . If \( \sqrt{a} \in E \), then \( L = K \) and we are done. If not, consider the polynomial\n\n\[\n\mathop{\prod }\limits_{{\sigma \in \operatorname{Gal}\left( {E/F}\right) }}\left( {{x}^{2} - \sigma \left( a\right) }\right) \in F\left\lbrack x\right\rbrack\n\]\n\nThe splitting field \( N \) over \( F \) of this polynomial is \( N = F(\{ \sqrt{\sigma \left( a\right) } : \sigma \in \) Gal \( \left( {E/F}\right) \} \) ). Hence, \( N \) is a 2-Kummer extension of \( F \), so \( \left\lbrack {N : F}\right\rbrack \) is a power of 2 . The field \( N \) is a Galois extension of \( F \) of degree a power of 2, so \( N \subseteq K \) . Moreover, \( \sqrt{a} \in N \) . This shows that \( \sqrt{a} \in K \), so \( L = K \) . Thus, \( K \) is quadratically closed. | Yes |
Proposition 18.3 Let \( F \) be a field of characteristic with \( \operatorname{char}\left( F\right) \neq 2 \) . We define fields \( \left\{ {F}_{n}\right\} \) by recursion by setting \( {F}_{0} = F \) and \( {F}_{n + 1} = {F}_{n}(\{ \sqrt{a} \) : \( \left. \left. {a \in {F}_{n}}\right\} \right) \) . Then the quadratic closure of \( F \) is the union \( \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{F}_{n} \) . | Proof. Let \( K = \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{F}_{n} \) . Then \( K \) is a field, since \( \left\{ {F}_{n}\right\} \) is a totally ordered collection of fields. We show that \( K \) is quadratically closed. If \( a \in K \), then \( a \in {F}_{n} \) for some \( n \), so \( \sqrt{a} \in {F}_{n + 1} \subseteq K \) . Thus, \( K\left( \sqrt{a}\right) = K \), so \( K \) is indeed quadratically closed. Let \( {F}_{q} \) be the quadratic closure of \( F \) . Then \( \sqrt{a} \in {F}_{q} \) for each \( a \in {F}_{q} \), so we see that \( {F}_{1} \subseteq {F}_{q} \) . Suppose that \( {F}_{n} \subseteq {F}_{q} \) . The reasoning we used to show that \( K \) is quadratically closed shows also that \( {F}_{n + 1} \subseteq {F}_{q} \), so \( K \subseteq {F}_{q} \) . To see that this inclusion is an equality, let \( E \) be a Galois extension of \( F \) of degree a power of 2 . Then \( {EK}/K \) has degree a power of 2 by natural irrationalities. If \( \left\lbrack {{EK} : K}\right\rbrack > 1 \), then the group \( \operatorname{Gal}\left( {{EK}/K}\right) \) has a subgroup of index 2 by the theory of \( p \) -groups. If \( L \) is the fixed field of this subgroup, then \( \left\lbrack {L : K}\right\rbrack = 2 \) . However, this is impossible, since \( K \) is quadratically closed. This forces \( {EK} = K \), so \( E \subseteq K \) . Since \( {F}_{q} \) is the composite of all such \( E \), we see that \( {F}_{q} \subseteq K \), so \( K = {F}_{q} \) . | Yes |
Lemma 18.4 Let \( p \) be a prime, and let \( F \) be a field with \( \operatorname{char}\left( F\right) \neq p \) . If \( L \) is an intermediate field of \( {F}_{ac}/F \) with \( \left\lbrack {L : F}\right\rbrack < \infty \), then \( L \subseteq {F}_{p} \) if and only if \( L \) lies in a Galois extension of \( F \) of degree a power of \( p \) . In particular, any finite intermediate field of \( {F}_{p}/F \) has degree over \( F \) a power of \( p \) . | Proof. If \( L \) is a field lying inside some Galois extension \( E \) of \( F \) with \( \left\lbrack {E : F}\right\rbrack \) a power of \( p \), then \( E \subseteq {F}_{p} \), so \( L \subseteq {F}_{p} \) . Conversely, suppose that \( L \subseteq {F}_{p} \) and \( \left\lbrack {L : F}\right\rbrack < \infty \) . Then \( L = F\left( {{a}_{1},\ldots ,{a}_{n}}\right) \) for some \( {a}_{i} \in L \) . From the definition of \( {F}_{p} \), for each \( i \) there is a Galois extension \( {E}_{i}/F \) such that \( {a}_{i} \in {E}_{i} \) and \( \left\lbrack {{E}_{i} : F}\right\rbrack \) is a power of \( p \) . The composition of the \( {E}_{i} \) is a Galois extension of \( F \), whose degree over \( F \) is also a power of \( p \) by natural irrationalities. | Yes |
Proposition 18.6 Suppose that \( F \) contains a primitive pth root of unity. Define a sequence of fields \( \left\{ {F}_{n}\right\} \) by recursion by setting \( {F}_{0} = F \) and \( {F}_{n + 1} = \) \( {F}_{n}\left( \left\{ {\sqrt[p]{a} : a \in {F}_{n}}\right\} \right) \) . Then the p-closure of \( F \) is \( \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{F}_{n} \) . | Proof. The proof is essentially the same as that for the quadratic closure, so we only outline the proof. If \( {F}_{n} \subseteq {F}_{p} \) and \( a \in {F}_{n} \), then either \( {F}_{n}\left( \sqrt[p]{a}\right) = {F}_{n} \) , or \( {F}_{n}\left( \sqrt[p]{a}\right) /{F}_{n} \) is a Galois extension of degree \( p \), by Proposition 9.6. In either case, \( {F}_{n}\left( \sqrt[p]{a}\right) \subseteq {F}_{p} \) by the previous proposition. This shows that \( \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{F}_{n} \subseteq {F}_{p} \) . To get the reverse inclusion, let \( E/F \) be a Galois extension of degree a power of \( p \) . By the theory of \( p \) -groups and the fundamental theorem of Galois theory, there is a chain of intermediate fields\n\n\[ F = {E}_{0} \subseteq {E}_{1} \subseteq \cdots \subseteq {E}_{n} = E \]\n\nwith \( {E}_{i + 1}/{E}_{i} \) Galois of degree \( p \) . Since \( F \) contains a primitive \( p \) th root of unity, \( {E}_{i + 1} = {E}_{i}\left( \sqrt[p]{{a}_{i}}\right) \) for some \( {a}_{i} \in {E}_{i} \) by Theorem 9.5. By induction, we can see that \( {E}_{i} \subseteq {F}_{i} \), so \( E \subseteq \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{F}_{n} \) . Since \( {F}_{p} \) is the composite of all such \( E \), this gives the reverse inclusion we want. | Yes |
Proposition 18.7 Let \( F \) be a field, let \( p \) be a prime, and let \( K \) be a maximal prime to p extension of F . Then any finite extension of \( K \) has degree a power of \( p \), and if \( L \) is an intermediate field of \( K/F \) with \( \left\lbrack {L : F}\right\rbrack < \infty \), then \( \left\lbrack {L : F}\right\rbrack \) is relatively prime to p. Moreover, any separable field extension \( L \) of \( F \) with \( \left\lbrack {L : F}\right\rbrack \) relatively prime to \( p \) is contained in some maximal prime to \( p \) extension of \( F \) . | Proof. Recall that if \( U \) is an open subgroup of a \( p \) -Sylow subgroup \( P \) of \( G = \operatorname{Gal}\left( {{F}_{s}/F}\right) \), then \( \left\lbrack {P : U}\right\rbrack \) is a power of \( p \), and if \( V \) is open in \( G \) with \( P \subseteq V \subseteq G \), then \( \left\lbrack {G : V}\right\rbrack \) is relatively prime to \( p \) . Suppose that \( M \) is a finite extension of \( K \) . If \( H = \operatorname{Gal}\left( {{F}_{s}/M}\right) \), then by the fundamental theorem, we have \( \left\lbrack {P : H}\right\rbrack = \left\lbrack {M : K}\right\rbrack < \infty \), so \( H \) is an open subgroup of \( P \) . Thus, \( \left\lbrack {P : H}\right\rbrack \) is a power of \( p \), so \( \left\lbrack {M : K}\right\rbrack \) is a power of \( p \) .\n\nFor the second statement, let \( L \) be an intermediate field of \( K/F \) with \( \left\lbrack {L : F}\right\rbrack < \infty \) . If \( A = \operatorname{Gal}\left( {{F}_{s}/L}\right) \), then \( P \subseteq A \) and \( \left\lbrack {G : A}\right\rbrack = \left\lbrack {L : F}\right\rbrack \) is finite, by the fundamental theorem. Since \( \left\lbrack {G : A}\right\rbrack \) is relatively prime to \( p \), we see that \( \left\lbrack {L : F}\right\rbrack \) is relatively prime to \( p \) .\n\nLet \( L/F \) be an extension with \( \left\lbrack {L : F}\right\rbrack \) relatively prime to \( p \) . Let \( {F}_{s} \) be the separable closure of \( F \), and let \( G = \operatorname{Gal}\left( {{F}_{s}/F}\right) \) . Set \( H = \operatorname{Gal}\left( {{F}_{s}/L}\right) \), a closed subgroup of \( G \), and let \( {P}^{\prime } \) be a \( p \) -Sylow subgroup of \( H \) . There is a \( p \) -Sylow subgroup \( P \) of \( G \) that contains \( {P}^{\prime } \) . Note that \( \left\lbrack {G : H}\right\rbrack = \left\lbrack {L : F}\right\rbrack \) is relatively prime to \( p \) . Moreover, we have\n\n\[ \left\lbrack {G : {P}^{\prime }}\right\rbrack = \left\lbrack {G : H}\right\rbrack \cdot \left\lbrack {H : {P}^{\prime }}\right\rbrack \]\n\n\[ = \left\lbrack {G : P}\right\rbrack \cdot \left\lbrack {P : {P}^{\prime }}\right\rbrack \]\n\nBoth \( \left\lbrack {G : H}\right\rbrack \) and \( \left\lbrack {H : {P}^{\prime }}\right\rbrack \) are supernatural numbers not divisible by \( p \), so \( \left\lbrack {P : {P}^{\prime }}\right\rbrack \) is not divisible by \( p \) . But, since \( P \) is a pro- \( p \) -group, \( \left\lbrack {P : {P}^{\prime }}\right\rbrack \) is a power of \( p \) . This forces \( \left\lbrack {P : {P}^{\prime }}\right\rbrack = 1 \), so \( {P}^{\prime } = P \) . Therefore, \( P \subseteq H \), and so \( L = \mathcal{F}\left( H\right) \) is contained in \( \mathcal{F}\left( P\right) \), a maximal prime to \( p \) extension of \( F \) . | Yes |
Example 18.8 The maximal prime to \( p \) extension of a field \( F \) need not be the composite of all finite extensions of degree relatively prime to \( p \) . For example, if \( F = \mathbb{Q} \) and \( p = 3 \), then \( \mathbb{Q}\left( \sqrt[3]{5}\right) \) and \( \mathbb{Q}\left( {\omega \sqrt[3]{5}}\right) \) are both of degree 3 over \( \mathbb{Q} \), where \( \omega \) is a primitive third root of unity, but their composite is \( \mathbb{Q}\left( {\omega ,\sqrt[3]{5}}\right) \), which has degree 6 over \( \mathbb{Q} \) . Therefore, these fields are not both contained in a common maximal prime to \( p \) extension of \( \mathbb{Q} \) . | Null | No |
Proposition 18.9 Let \( {F}_{a} \) be the maximal Abelian extension of a field \( F \) . Then \( {F}_{a}/F \) is a Galois extension and \( \operatorname{Gal}\left( {{F}_{a}/F}\right) \) is an Abelian group. The field \( {F}_{a} \) has no extensions that are Abelian Galois extensions of \( F \) . Moreover, \( {F}_{a} \) is the composite in \( {F}_{s} \) of all finite Abelian Galois extensions of \( F \) . | Proof. The commutator subgroup \( {G}^{\prime } \) of \( G \) is a normal subgroup, so the closure \( \overline{{G}^{\prime }} \) of \( {G}^{\prime } \) is a closed normal subgroup of \( G \) (see Problem 17.8). Thus, by the fundamental theorem, \( {F}_{a} = \mathcal{F}\left( \overset{⏜}{{G}^{\prime }}\right) \) is a Galois extension of \( F \) and \( \operatorname{Gal}\left( {{F}_{a}/F}\right) \cong G/\overline{{G}^{\prime }} \) . The group \( G/\overline{{G}^{\prime }} \) is a homomorphic image of the Abelian group \( G/{G}^{\prime } \), so \( G/\overline{{G}^{\prime }} \) is also Abelian.\n\nIf \( L \supseteq {F}_{a} \) is an Abelian Galois extension of \( F \), then \( L \subseteq {F}_{s} \) . Let \( H = \) \( \operatorname{Gal}\left( {{F}_{s}/L}\right) \), a subgroup of \( {G}^{\prime } \) . However, \( G/H \cong \operatorname{Gal}\left( {L/F}\right) \), so \( G/H \) is Abelian. Thus, \( {G}^{\prime } \subseteq H \), so \( H = {G}^{\prime } \) . Therefore, \( {F}_{a} \) is not properly contained in any Abelian extension of \( F \).\n\nFor the final statement, if \( K/F \) is finite Abelian Galois, then \( K{F}_{a}/{F}_{a} \) is Abelian Galois by natural irrationalities. Thus, \( K{F}_{a} = {F}_{a} \), so \( K \subseteq {F}_{a} \) . Since every element of \( {F}_{a} \) lies in a finite Galois extension of \( F \), to show that \( {F}_{a} \) is the composite of all finite Abelian Galois extensions of \( F \) it suffices to show that every finite Galois extension of \( F \) inside \( {F}_{a} \) is an Abelian extension. Let \( E \) be such an extension. If \( H = \operatorname{Gal}\left( {{F}_{s}/E}\right) \), then \( H \) is a normal subgroup of \( G \) containing \( {G}^{\prime } \) ; hence, \( G/H \) is Abelian. But, by the fundamental theorem, we have \( \operatorname{Gal}\left( {E/F}\right) \cong G/H \), so \( E/F \) is an Abelian Galois extension. | Yes |
If \( F \) is a field containing a primitive \( n \) th root of unity for all \( n \), then the maximal Abelian extension of \( F \) is \( F\left( {\{ \sqrt[n]{a} : a \in F, n \in \mathbb{N}\} }\right) \) . | This follows from Kummer theory (see Problem 11.6 for part of this claim). | No |
If \( K = F\left( {{x}_{1},\ldots ,{x}_{n}}\right) \) is the field of rational functions over \( F \) in \( n \) variables, then \( \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \) is algebraically independent over \( F \) . Moreover, if \( {r}_{1},\ldots ,{r}_{n} \) are any positive integers, then \( \left\{ {{x}_{1}^{{r}_{1}},\ldots ,{x}_{n}^{{r}_{n}}}\right\} \) is also algebraically independent over \( F \) . | Null | No |
Example 19.3 By convention, the empty set \( \varnothing \) is algebraically independent over any field. The singleton sets \( \{ e\} ,\{ \pi \} \), and \( \left\{ {4{e}^{-1}}\right\} \) are all algebraically independent over \( \mathbb{Q} \) . The set \( \left\{ {e,{e}^{2}}\right\} \) is not algebraically independent over \( \mathbb{Q} \), since \( f\left( {e,{e}^{2}}\right) = 0 \) if \( f\left( {{x}_{1},{x}_{2}}\right) = {x}_{1}^{2} - {x}_{2} \) . It is unknown whether \( \{ e,\pi \} \) is algebraically independent over \( \mathbb{Q} \) . | Null | No |
Example 19.4 Let \( F \subseteq K \subseteq L \) be fields, and let \( S \) be a subset of \( L \) . If \( S \) is algebraically independent over \( F \), then \( S \) is also algebraically independent over \( K \) . This is clear from the definition of algebraic independence. Moreover, if \( T \) is any subset of \( S \) and if \( S \) is algebraically independent over \( F \) , then \( T \) is also algebraically independent over \( F \) . The converse of the first statement is false in general. Suppose that \( K = F\left( x\right) = L \) . Then \( \{ x\} \) is algebraically independent over \( F \), but \( \{ x\} \) is algebraically dependent over \( K \) . | Null | No |
Lemma 19.5 Let \( K \) be a field extension of \( F \) . If \( {t}_{1},\ldots ,{t}_{n} \in K \) are algebraically independent over \( F \), then \( F\left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \) and \( F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) are \( F \) - isomorphic rings, and so \( F\left( {{t}_{1},\ldots ,{t}_{n}}\right) \) and \( F\left( {{x}_{1},\ldots ,{x}_{n}}\right) \) are \( F \) -isomorphic fields. | Proof. Define \( \varphi : F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \rightarrow K \) by \( \varphi \left( {f\left( {{x}_{1},\ldots ,{x}_{n}}\right) }\right) = f\left( {{t}_{1},\ldots ,{t}_{n}}\right) \) . Then \( \varphi \) is an \( F \) -homomorphism of rings. The algebraic independence of the \( {t}_{i} \) shows that \( \varphi \) is injective, and the image of \( \varphi \) is \( F\left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \) . Therefore, \( F\left\lbrack {{t}_{1}\ldots ,{t}_{n}}\right\rbrack \) and \( F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) are isomorphic. This map induces an \( F \) -isomorphism of quotient fields, which finishes the proof. | Yes |
Lemma 19.7 Let \( K \) be a field extension of \( F \), and let \( {t}_{1},\ldots ,{t}_{n} \in K \) . Then the following statements are equivalent:\n\n1. The set \( \left\{ {{t}_{1},\ldots ,{t}_{n}}\right\} \) is algebraically independent over \( F \) .\n\n2. For each \( i,{t}_{i} \) is transcendental over \( F\left( {{t}_{1},\ldots ,{t}_{i - 1},{t}_{i + 1},\ldots ,{t}_{n}}\right) \) .\n\n3. For each \( i,{t}_{i} \) is transcendental over \( F\left( {{t}_{1},\ldots ,{t}_{i - 1}}\right) \) . | Proof. (1) \( \Rightarrow \) (2): Suppose that there are \( {a}_{j} \in F\left( {{t}_{1},\ldots ,{t}_{i - 1},{t}_{i + 1},\ldots ,{t}_{n}}\right) \) such that \( {a}_{0} + {a}_{1}{t}_{i} + \cdots + {t}_{i}^{m} = 0 \) . We may write \( {a}_{j} = {b}_{j}/c \) with \( {b}_{1,}\ldots ,{b}_{n}, c \in \) \( F\left\lbrack {{t}_{1},\ldots ,{t}_{i - 1},{t}_{i + 1},\ldots ,{t}_{n}}\right\rbrack \), and so \( {b}_{0} + {b}_{1}{t}_{i} + \cdots + {b}_{m}{t}_{i}^{m} = 0 \) . If \( {b}_{j} = \) \( {g}_{j}\left( {{t}_{1},\ldots ,{t}_{i - 1},{t}_{i + 1},\ldots ,{t}_{n}}\right) \), then \( f = \mathop{\sum }\limits_{j}{g}_{j}\left( {{x}_{1},\ldots ,{x}_{i - 1},{x}_{i + 1},\ldots ,{x}_{n}}\right) {x}_{i}^{j} \) is a polynomial and \( f\left( {{t}_{1},\ldots ,{t}_{n}}\right) = 0 \) . Since \( \left\{ {{t}_{1},\ldots ,{t}_{n}}\right\} \) is algebraically independent over \( F \), the polynomial \( f \) must be 0 . Consequently, each \( {a}_{j} = \) 0, so \( {t}_{i} \) is transcendental over \( F\left( {{t}_{1},\ldots ,{t}_{i - 1},{t}_{i + 1},\ldots ,{t}_{n}}\right) \) .\n\n\( \left( 2\right) \Rightarrow \left( 3\right) \) : If \( {t}_{i} \) is transcendental over \( F\left( {{t}_{1},\ldots ,{t}_{i - 1},{t}_{i + 1},\ldots ,{t}_{n}}\right) \), then \( {t}_{i} \) clearly is transcendental over the smaller field \( F\left( {{t}_{1},\ldots ,{t}_{i - 1}}\right) \) .\n\n\( \left( 3\right) \Rightarrow \left( 1\right) \) : Suppose that the \( {t}_{i} \) are not algebraically independent over \( F \) . Choose \( m \) minimal such that there is a nonzero \( f\left( {{x}_{1},\ldots ,{x}_{m}}\right) \in \) \( F\left\lbrack {{x}_{1},\ldots ,{x}_{m}}\right\rbrack \) with \( f\left( {{t}_{1},\ldots ,{t}_{m}}\right) = 0 \) . Say \( f = \mathop{\sum }\limits_{j}{g}_{j}{x}_{m}^{j} \) with \( {g}_{j} \in \) \( F\left\lbrack {{x}_{1},\ldots ,{x}_{m - 1}}\right\rbrack \), and let \( {a}_{j} = g\left( {{t}_{1},\ldots ,{t}_{m - 1}}\right) \) . Then \( {a}_{0} + {a}_{1}{t}_{m} + \cdots + {a}_{r}{t}_{m}^{r} = \) 0 . If the \( {a}_{j} \) are not all zero, then \( {t}_{m} \) is algebraic over \( F\left( {{t}_{1},\ldots ,{t}_{m}}\right) \), a contradiction. Thus, \( {a}_{j} = 0 \) for each \( j \) . By the minimality of \( m \), the \( {t}_{1},\ldots ,{t}_{m - 1} \) are algebraically independent over \( F \), which implies that all \( {g}_{j} = 0 \), so \( f = 0 \) . This proves that \( \left\{ {{t}_{1},\ldots ,{t}_{n}}\right\} \) is algebraically independent over \( F \) . | Yes |
If \( K/F \) is a field extension, then \( \varnothing \) is a transcendence basis for \( K/F \) if and only if \( K/F \) is algebraic. | Null | No |
If \( K = F\left( {{x}_{1},\ldots ,{x}_{n}}\right) \), then \( \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \) is a transcendence basis for \( K/F \) . Moreover, if \( {r}_{1},\ldots ,{r}_{n} \) are positive integers, then we show that \( \left\{ {{x}_{1}^{{r}_{1}},\ldots ,{x}_{n}^{{r}_{n}}}\right\} \) is also a transcendence basis for \( K/F \) . | We saw in Example 19.2 that \( \left\{ {{x}_{1}^{{r}_{1}},\ldots ,{x}_{n}^{{r}_{n}}}\right\} \) is algebraically independent over \( F \) . We need to show that \( K \) is algebraic over \( L = F\left( {{x}_{1}^{{r}_{1}},\ldots ,{x}_{n}^{{r}_{n}}}\right) \) . This is true because for each \( i \) the element \( {x}_{i} \) satisfies the polynomial \( {t}^{{r}_{i}} - {x}_{i}^{{r}_{i}} \in L\left\lbrack t\right\rbrack \) . | Yes |
We show that \( \{ u\} \) is a transcendence basis for \( K/k \) . Since \( {v}^{2} = {u}^{3} - u \), the field \( K \) is algebraic over \( k\left( u\right) \) . We then need to show that \( u \) is transcendental over \( k \) . | If this is false, then \( u \) is algebraic over \( k \), so \( K \) is algebraic over \( k \) . We claim that this forces \( A = k\left\lbrack {u, v}\right\rbrack \) to be a field. To prove this, take \( t \in A \) . Then \( {t}^{-1} \in K \) is algebraic over \( k \), so \( {t}^{-n} + {\alpha }_{n - 1}{t}^{n - 1} + \cdots + {\alpha }_{0} = 0 \) for some \( {\alpha }_{i} \in k \) with \( {\alpha }_{0} \neq 0 \) . Multiplying by \( {t}^{n - 1} \) gives\n\n\[ \n{t}^{-1} = - \left( {{\alpha }_{n - 1} + {\alpha }_{n - 2}t + \cdots + {\alpha }_{0}{t}^{n - 1}}\right) \in A, \n\]\n\nproving that \( A \) is a field. However, \( A = k\left\lbrack {x, y}\right\rbrack /\left( f\right) \) is a field if and only if \( \left( f\right) \) is a maximal ideal of \( k\left\lbrack {x, y}\right\rbrack \) . The ring \( A \) cannot be a field, since \( \left( f\right) \) is properly contained in the ideal \( \left( {x, y}\right) \) of \( k\left\lbrack {x, y}\right\rbrack \) . Thus, \( u \) is not algebraic over \( k \), so \( \{ u\} \) is a transcendence basis for \( K/k \) . | Yes |
Example 19.12 We give a generalization of the previous example. Let \( k \) be a field and let \( f \in k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) be an irreducible polynomial. Then \( A = k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack /\left( f\right) \) is an integral domain. Let \( K \) be the quotient field of A. We may write\n\n\[ f = {g}_{m}{x}_{n}^{m} + {g}_{m - 1}{x}_{n}^{m - 1} + \cdots + {g}_{0} \]\n\nwith each \( {g}_{i} \in k\left\lbrack {{x}_{1},\ldots ,{x}_{n - 1}}\right\rbrack \) . Let us assume that \( m > 0 \), so that \( f \) does involve the variable \( {x}_{n} \) . If \( {t}_{i} = {x}_{i} + \left( f\right) \) is the image of \( {x}_{i} \) in \( A \), we claim that \( \left\{ {{t}_{1},\ldots ,{t}_{n - 1}}\right\} \) is a transcendence basis for \( K/k \) . | To see this, the equation for \( f \) above shows that \( {t}_{n} \) is algebraic over \( k\left( {{t}_{1},\ldots ,{t}_{n - 1}}\right) \), so we only need to show that \( \left\{ {{t}_{1},\ldots ,{t}_{n - 1}}\right\} \) is algebraically independent over \( k \) . Suppose that there is a polynomial \( h \in k\left\lbrack {{x}_{1},\ldots ,{x}_{n - 1}}\right\rbrack \) with \( h\left( {{t}_{1},\ldots ,{t}_{n - 1}}\right) = 0 \) . Then \( h\left( {{x}_{1},\ldots ,{x}_{n - 1}}\right) \in \left( f\right) \), so \( f \) divides \( h \) . Thus, \( h = {fg} \) for some \( g \in k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . However, the polynomial \( h \) does not involve the variable \( {x}_{n} \) while \( f \) does, so comparing degrees in \( {x}_{n} \) of \( h \) and \( {fg} \) shows that \( h = 0 \) . Therefore, \( \left\{ {{t}_{1},\ldots ,{t}_{n - 1}}\right\} \) is algebraically independent over \( k \), so \( \left\{ {{t}_{1},\ldots ,{t}_{n - 1}}\right\} \) is a transcendence basis for \( K/k \) . | Yes |
Lemma 19.13 Let \( K \) be a field extension of \( F \), and let \( S \subseteq K \) be algebraically independent over \( F \) . If \( t \in K \) is transcendental over \( F\left( S\right) \), then \( S \cup \{ t\} \) is algebraically independent over \( F \) . | Proof. Suppose that the lemma is false. Then there is a nonzero polynomial \( f \in F\left\lbrack {{x}_{1},\ldots ,{x}_{n}, y}\right\rbrack \) with \( f\left( {{s}_{1},\ldots ,{s}_{n}, t}\right) = 0 \) for some \( {s}_{i} \in S \) . This polynomial must involve \( y \), since \( S \) is algebraically independent over \( F \) . Write \( f = \mathop{\sum }\limits_{{j = 0}}^{m}{g}_{j}{y}^{j} \) with \( {g}_{j} \in F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . Since \( {g}_{m} \neq 0 \), the element \( t \) is algebraic over \( F\left( S\right) \), a contradiction. Thus, \( S \cup \{ t\} \) is algebraically independent over \( F \) . | Yes |
Theorem 19.14 Let \( K \) be a field extension of \( F \). 1. There exists a transcendence basis for \( K/F \). 2. If \( T \subseteq K \) such that \( K/F\left( T\right) \) is algebraic, then \( T \) contains a transcendence basis for \( K/F \). 3. If \( S \subseteq K \) is algebraically independent over \( F \), then \( S \) is contained in a transcendence basis of \( K/F \). 4. If \( S \subseteq T \subseteq K \) such that \( S \) is algebraically independent over \( F \) and \( K/F\left( T\right) \) is algebraic, then there is a transcendence basis \( X \) for \( K/F \) with \( S \subseteq X \subseteq T \). | Proof. We first mention why statement 4 implies the first three statements. If statement 4 is true, then statements 2 and 3 are true by setting \( S = \varnothing \) and \( T = K \), respectively. Statement 1 follows from statement 4 by setting \( S = \varnothing \) and \( T = K \). To prove statement 4, let \( \mathcal{S} \) be the set of all algebraically independent subsets of \( T \) that contain \( S \). Then \( \mathcal{S} \) is nonempty, since \( S \in \mathcal{S} \). Ordering \( \mathcal{S} \) by inclusion, a Zorn’s lemma argument shows that \( \mathcal{S} \) contains a maximal element \( M \). If \( K \) is not algebraic over \( F\left( M\right) \), then \( F\left( T\right) \) is not algebraic over \( F\left( M\right) \), since \( K \) is algebraic over \( F\left( T\right) \). Thus, there is a \( t \in T \) with \( t \) transcendental over \( F\left( M\right) \). But by Lemma 19.13, \( M \cup \{ t\} \) is algebraically independent over \( F \) and is a subset of \( T \), contradicting maximality of \( M \). Thus, \( K \) is algebraic over \( F\left( M\right) \), so \( M \) is a transcendence basis of \( K/F \) contained in \( X \). | Yes |
Theorem 19.15 Let \( K \) be a field extension of \( F \). If \( S \) and \( T \) are transcendence bases for \( K/F \), then \( \left| S\right| = \left| T\right| \). | Proof. We first prove this in the case where \( S = \left\{ {{s}_{1},\ldots ,{s}_{n}}\right\} \) is finite. Since \( S \) is a transcendence basis for \( K/F \), the field \( K \) is not algebraic over \( F\left( {S - \left\{ {s}_{1}\right\} }\right) \). As \( K \) is algebraic over \( F\left( T\right) \), some \( t \in T \) must be transcendental over \( F\left( {S - \left\{ {s}_{1}\right\} }\right) \). Hence, by Lemma 19.13, \( \left\{ {{s}_{2},\ldots ,{s}_{n}, t}\right\} \) is algebraically independent over \( F \). Furthermore, \( {s}_{1} \) is algebraic over \( F\left( {{s}_{2},\ldots ,{s}_{n}, t}\right) \), or else \( \left\{ {{s}_{1},\ldots ,{s}_{n}, t}\right\} \) is algebraically independent, which is false. Thus, \( \left\{ {{s}_{2},\ldots ,{s}_{n}, t}\right\} \) is a transcendence basis for \( K/F \). Set \( {t}_{1} = t \). Assuming we have found \( {t}_{i} \in T \) for all \( i \) with \( 1 \leq i < m \leq n \) such that \( \left\{ {{t}_{1},\ldots ,{t}_{m - 1},{s}_{m},\ldots ,{s}_{n}}\right\} \) is a transcendence basis for \( K/F \), by replacing \( S \) by this set, the argument above shows that there is a \( {t}^{\prime } \in T \) such that \( \left\{ {{t}_{1},\ldots ,{t}_{m - 1},{t}^{\prime },\ldots ,{s}_{n}}\right\} \) is a transcendence basis for \( K/F \). Setting \( {t}_{m} = {t}^{\prime } \) and continuing in this way, we get a transcendence basis \( \left\{ {{t}_{1},\ldots ,{t}_{n}}\right\} \subseteq T \) of \( K/F \). Since \( T \) is a transcendence basis for \( K/F \), we see that \( \left\{ {{t}_{1},\ldots ,{t}_{n}}\right\} = T \), so \( \left| T\right| = n \).\n\nFor the general case, by the previous argument we may suppose that \( S \) and \( T \) are both infinite. Each \( t \in T \) is algebraic over \( F\left( S\right) \); hence, there is a finite subset \( {S}_{t} \subseteq S \) with \( t \) algebraic over \( F\left( {S}_{t}\right) \). If \( {S}^{\prime } = \mathop{\bigcup }\limits_{{t \in T}}{S}_{t} \), then each \( t \in T \) is algebraic over \( F\left( {S}^{\prime }\right) \). Since \( K \) is algebraic over \( F\left( T\right) \), we see\n\nthat \( K \) is algebraic over \( F\left( {S}^{\prime }\right) \). Thus, \( S = {S}^{\prime } \) since \( {S}^{\prime } \subseteq S \) and \( S \) is a transcendence basis for \( K/F \). We then have\n\n\[ \left| S\right| = \left| {S}^{\prime }\right| = \left| {\mathop{\bigcup }\limits_{{t \in T}}{S}_{t}}\right| \leq \left| {T \times \mathbb{N}}\right| = \left| T\right| \]\n\nwhere the last equality is true since \( T \) is infinite. Reversing the argument, we see that \( \left| T\right| \leq \left| S\right| \), so \( \left| S\right| = \left| T\right| \). | Yes |
Corollary 19.17 Let \( {t}_{1},\ldots ,{t}_{n} \in K \) . Then the fields \( F\left( {{t}_{1},\ldots ,{t}_{n}}\right) \) and \( F\left( {{x}_{1},\ldots ,{x}_{n}}\right) \) are \( F \) -isomorphic if and only if \( \left\{ {{t}_{1},\ldots ,{t}_{n}}\right\} \) is an algebraically independent set over \( F \) . | Proof. If \( \left\{ {{t}_{1},\ldots ,{t}_{n}}\right\} \) is algebraically independent over \( F \), then \( F\left( {{t}_{1},\ldots ,{t}_{n}}\right) \) and \( F\left( {{x}_{1},\ldots ,{x}_{n}}\right) \) are \( F \) -isomorphic fields by Lemma 19.5. Conversely, if \( F\left( {{t}_{1},\ldots ,{t}_{n}}\right) \cong F\left( {{x}_{1},\ldots ,{x}_{n}}\right) \), suppose that \( \left\{ {{t}_{1},\ldots ,{t}_{n}}\right\} \) is algebraically dependent over \( F \) . By the previous theorem, there is a subset \( S \) of \( \left\{ {{t}_{1},\ldots ,{t}_{n}}\right\} \) such that \( S \) is a transcendence basis for \( F\left( {{t}_{1},\ldots ,{t}_{n}}\right) /F \) . However, the transcendence degree of this extension is \( n \), which forces \( \left| S\right| = n \), so \( S = \left\{ {{t}_{1},\ldots ,{t}_{n}}\right\} \) . Thus, \( \left\{ {{t}_{1},\ldots ,{t}_{n}}\right\} \) is algebraically independent over \( F \) . | Yes |
Proposition 19.18 Let \( F \subseteq L \subseteq K \) be fields. Then\n\n\[ \operatorname{trdeg}\left( {K/F}\right) = \operatorname{trdeg}\left( {K/L}\right) + \operatorname{trdeg}\left( {L/F}\right) . \] | Proof. Let \( S \) be a transcendence basis for \( L/F \), and let \( T \) be a transcendence basis for \( K/L \) . We show that \( S \cup T \) is a transcendence basis for \( K/F \), which will prove the result because \( S \cap T = \varnothing \) . Since \( T \) is algebraically independent over \( L \), the set \( T \) is also algebraically independent over \( F\left( S\right) \subseteq L \) , so \( S \cup T \) is algebraically independent over \( F \) . To show that \( K \) is algebraic over \( F\left( {S \cup T}\right) \), we know that \( K/L\left( T\right) \) and \( L/F\left( S\right) \) are algebraic. Therefore, \( L\left( T\right) \) is algebraic over \( F\left( {S \cup T}\right) = F\left( S\right) \left( T\right) \), since each \( t \in T \) is algebraic over \( F\left( {S \cup T}\right) \) . Thus, by transitivity, \( K \) is algebraic over \( F\left( {S \cup T}\right) \), so \( S \cup T \) is a transcendence basis for \( K/F \) . This proves the proposition. | Yes |
Example 19.19 Let \( K = k\left( {{x}_{1},\ldots ,{x}_{n}}\right) \) be the field of rational functions in \( n \) variables over a field \( k \), and let \( F = k\left( {{s}_{1},\ldots ,{s}_{n}}\right) \) be the subfield of \( K \) generated over \( k \) by the elementary symmetric functions \( {s}_{1},\ldots ,{s}_{n} \) . In Example 3.9, we saw that \( K \) is an algebraic extension of \( F \) with \( \left\lbrack {K : F}\right\rbrack = \) \( n \) !. Therefore, \( \left\{ {{s}_{1},\ldots ,{s}_{n}}\right\} \) contains a transcendence basis of \( K/k \) . However, \( \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \) is a transcendence basis for \( K/k \), so \( \operatorname{trdeg}\left( {K/k}\right) = n \) . This forces the \( {s}_{i} \) to be algebraically independent over \( k \) ; hence, they form a transcendence basis for \( K/k \) . In particular, this shows that \( k\left( {{s}_{1},\ldots ,{s}_{n}}\right) \cong \) \( k\left( {{x}_{1},\ldots ,{x}_{n}}\right) . | Null | No |
Consider the field extension \( \mathbb{C}/\mathbb{Q} \). Since \( \mathbb{Q} \) is countable and \( \mathbb{C} \) is uncountable, the transcendence degree of \( \mathbb{C}/\mathbb{Q} \) must be infinite (in fact, uncountable), for if \( {t}_{1},\ldots ,{t}_{n} \) form a transcendence basis for \( \mathbb{C}/\mathbb{Q} \), then \( \mathbb{C} \) is algebraic over \( \mathbb{Q}\left( {{t}_{1},\ldots ,{t}_{n}}\right) \), so \( \mathbb{C} \) and \( \mathbb{Q} \) have the same cardinality, since they are infinite fields. However, one can show that \( \mathbb{Q}\left( {{t}_{1},\ldots ,{t}_{n}}\right) \) is countable. This would give a contradiction to the uncountability of \( \mathbb{C} \). Thus, any transcendence basis \( T \) of \( \mathbb{C}/\mathbb{Q} \) is infinite. | Let \( T \) be any transcendence basis of \( \mathbb{C}/\mathbb{Q} \). Since \( \mathbb{C} \) is algebraic over \( \mathbb{Q}\left( T\right) \) and is algebraically closed, \( \mathbb{C} \) is an algebraic closure of \( \mathbb{Q}\left( T\right) \). Let \( \sigma \) be a permutation of \( T \). Then \( \sigma \) induces an automorphism of \( \mathbb{Q}\left( T\right) \) that is trivial on \( \mathbb{Q} \); hence, \( \sigma \) extends to an automorphism of \( \mathbb{C} \) by the isomorphism extension theorem. Since there are infinitely many such \( \sigma \), we see that \( \left| {\operatorname{Aut}\left( \mathbb{C}\right) }\right| = \infty \). Because any automorphism of \( \mathbb{R} \) is the identity, the only automorphisms of \( \mathbb{C} \) that map \( \mathbb{R} \) to \( \mathbb{R} \) are the identity map and complex conjugation. Thus, there are infinitely many \( \sigma \in \operatorname{Aut}\left( \mathbb{C}\right) \) with \( \sigma \left( \mathbb{R}\right) \neq \mathbb{R} \). We can easily show that \( \left\lbrack {\mathbb{C} : \sigma \left( \mathbb{R}\right) }\right\rbrack = 2 \). This means that there are infinitely many subfields \( F \) of \( \mathbb{C} \) with \( \left\lbrack {\mathbb{C} : F}\right\rbrack = 2 \). It is a whole different question to try to construct such fields. Note that in order to get these automorphisms of \( \mathbb{C} \), we invoked Zorn's lemma twice, once for the existence of a transcendence basis of \( \mathbb{C}/\mathbb{Q} \) and the second time indirectly by using the isomorphism extension theorem. | "No" |
Proposition 20.2 Let \( K \) and \( L \) be field extensions of a field \( F \) . Then \( K \) and \( L \) are linearly disjoint over \( F \) if and only if the map \( \varphi : K{ \otimes }_{F}L \rightarrow K\left\lbrack L\right\rbrack \) given on generators by \( a \otimes b \mapsto {ab} \) is an isomorphism of \( F \) -vector spaces. | Proof. The natural map \( \varphi : K{ \otimes }_{F}L \rightarrow K\left\lbrack L\right\rbrack \) is surjective by the description of \( K\left\lbrack L\right\rbrack \) given above. So, we need to show that \( K \) and \( L \) are linearly disjoint over \( F \) if and only if \( \varphi \) is injective. Suppose first that \( K \) and \( L \) are linearly disjoint over \( F \) . Let \( {\left\{ {k}_{i}\right\} }_{i \in I} \) be a basis for \( K \) as an \( F \) -vector space. Each element of \( K{ \otimes }_{F}L \) has a unique representation in the form \( \sum {k}_{i} \otimes {l}_{i} \), with the \( {l}_{i} \in L \) . Suppose that \( \sum {k}_{i} \otimes {l}_{i} \in \ker \left( \varphi \right) \), so \( \sum {k}_{i}{l}_{i} = 0 \) . Then each \( {l}_{i} = 0 \), since \( K \) and \( L \) are linearly disjoint over \( F \) and \( \left\{ {k}_{i}\right\} \) is \( F \) -linearly independent. Thus, \( \varphi \) is injective, and so \( \varphi \) is an isomorphism.\n\nConversely, suppose that the map \( \varphi \) is an isomorphism. Let \( {\left\{ {a}_{j}\right\} }_{j \in J} \) be an \( F \) -linearly independent subset of \( K \) . By enlarging \( J \), we may assume that the set \( \left\{ {a}_{j}\right\} \) is a basis for \( K \) . If \( \left\{ {a}_{j}\right\} \) is not \( L \) -linearly independent, then there are \( {l}_{j} \in L \) with \( \sum {a}_{j}{l}_{j} = 0 \), a finite sum. Then \( \sum {a}_{j} \otimes {l}_{j} \in \ker \left( \varphi \right) \) , so \( \sum {a}_{j} \otimes {l}_{j} = 0 \) by the injectivity of \( \varphi \) . However, elements of \( K{ \otimes }_{F}L \) can be represented uniquely in the form \( \sum {a}_{j} \otimes {m}_{j} \) with \( {m}_{j} \in L \) . Therefore, each \( {l}_{j} = 0 \), which forces the set \( \left\{ {a}_{j}\right\} \) to be \( L \) -linearly independent. Thus, \( K \) and \( L \) are linearly disjoint over \( F \) . | Yes |
Corollary 20.3 The definition of linear disjointness is symmetric; that is, \( K \) and \( L \) are linearly disjoint over \( F \) if and only if \( L \) and \( K \) are linearly disjoint over \( F \) . | Proof. This follows from Proposition 20.2. The map \( \varphi : K{ \otimes }_{F}L \rightarrow K\left\lbrack L\right\rbrack \) is an isomorphism if and only if \( \tau : L{ \otimes }_{F}K \rightarrow L\left\lbrack K\right\rbrack = K\left\lbrack L\right\rbrack \) is an isomorphism, since \( \tau = i \circ \varphi \), where \( i \) is the canonical isomorphism \( K{ \otimes }_{F}L \rightarrow L{ \otimes }_{F}K \) that sends \( a \otimes b \) to \( b \otimes a \) . | Yes |
Lemma 20.4 Suppose that \( K \) and \( L \) are finite extensions of \( F \) . Then \( K \) and \( L \) are linearly disjoint over \( F \) if and only if \( \left\lbrack {{KL} : F}\right\rbrack = \left\lbrack {K : F}\right\rbrack \cdot \left\lbrack {L : F}\right\rbrack \) . | Proof. The natural map \( \varphi : K{ \otimes }_{F}L \rightarrow K\left\lbrack L\right\rbrack \) that sends \( k \otimes l \) to \( {kl} \) is surjective and\n\n\[ \dim \left( {K{ \otimes }_{F}L}\right) = \left\lbrack {K : F}\right\rbrack \cdot \left\lbrack {L : F}\right\rbrack . \]\n\nThus, \( \varphi \) is an isomorphism if and only if \( \left\lbrack {{KL} : F}\right\rbrack = \left\lbrack {K : F}\right\rbrack \cdot \left\lbrack {L : F}\right\rbrack \) . The lemma then follows from Proposition 20.2. | Yes |
Example 20.5 Suppose that \( K \) and \( L \) are extensions of \( F \) with \( \left\lbrack {K : F}\right\rbrack \) and \( \left\lbrack {L : F}\right\rbrack \) relatively prime. Then \( K \) and \( L \) are linearly disjoint over \( F \) . | To see this, note that both \( \left\lbrack {K : F}\right\rbrack \) and \( \left\lbrack {L : F}\right\rbrack \) divide \( \left\lbrack {{KL} : F}\right\rbrack \), so their product divides \( \left\lbrack {{KL} : F}\right\rbrack \) since these degrees are relatively prime. The linear disjointness of \( K \) and \( L \) over \( F \) follows from the lemma. | No |
Example 20.6 Let \( K \) be a finite Galois extension of \( F \) . If \( L \) is any extension of \( F \), then \( K \) and \( L \) are linearly disjoint over \( F \) if and only if \( K \cap L = F \) . | This follows from the previous example and the theorem of natural irrationalities, since\n\n\[ \left\lbrack {{KL} : F}\right\rbrack = \left\lbrack {L : F}\right\rbrack \left\lbrack {K : K \cap L}\right\rbrack \]\n\nso \( \left\lbrack {{KL} : F}\right\rbrack = \left\lbrack {K : F}\right\rbrack \left\lbrack {L : F}\right\rbrack \) if and only if \( K \cap L = F \) . | Yes |
Lemma 20.8 Suppose that \( F \) is a field, and \( F \subseteq A \subseteq {A}^{\prime } \) and \( F \subseteq B \subseteq {B}^{\prime } \) are all subrings of a field \( C \) . If \( {A}^{\prime } \) and \( {B}^{\prime } \) are linearly disjoint over \( F \), then \( A \) and \( B \) are linearly disjoint over \( F \) . | Proof. This follows immediately from properties of tensor products. There is a natural injective homomorphism \( i : A{ \otimes }_{F}B \rightarrow {A}^{\prime }{ \otimes }_{F}{B}^{\prime } \) sending \( a \otimes b \) to \( a \otimes b \) for \( a \in A \) and \( B \in B \) . If the natural map \( {\varphi }^{\prime } : {A}^{\prime }{ \otimes }_{F}{B}^{\prime } \rightarrow {A}^{\prime }\left\lbrack {B}^{\prime }\right\rbrack \) is injective, then restricting \( \varphi \) to the image of \( i \) shows that the map \( \varphi \) : \( A{ \otimes }_{F}B \rightarrow A\left\lbrack B\right\rbrack \) is also injective. | Yes |
Example 20.9 Let \( K \) and \( L \) be extensions of a field \( F \) . If \( K \cap L \) is larger than \( F \), then \( K \) and \( L \) are not linearly disjoint over \( F \) by the preceding lemma since \( K \cap L \) is not linearly disjoint to itself over \( F \) . However, \( K \) and \( L \) may not be linearly disjoint over \( F \) even if \( K \cap L = F \) . As an example, let \( F = \mathbb{Q}, K = F\left( \sqrt[3]{2}\right) \), and \( L = F\left( {\omega \sqrt[3]{2}}\right) \), where \( \omega \) is a primitive third root of unity. Then \( K \cap L = F \), but \( {KL} = F\left( {\sqrt[3]{2},\omega }\right) \) has dimension 6 over \( F \), whereas \( K{ \otimes }_{F}L \) has dimension 9, so the map \( K{ \otimes }_{F}L \rightarrow {KL} \) is not injective. | Null | No |
Lemma 20.10 Suppose that \( A \) and \( B \) are subrings of a field \( C \), each containing a field \( F \), with quotient fields \( K \) and \( L \), respectively. Then \( A \) and \( B \) are linearly disjoint over \( F \) if and only if \( K \) and \( L \) are linearly disjoint over \( F \) . | Proof. If \( K \) and \( L \) are linearly disjoint over \( F \), then \( A \) and \( B \) are also linearly disjoint over \( F \) by the previous lemma. Conversely, suppose that \( A \) and \( B \) are linearly disjoint over \( F \) . Let \( \left\{ {{k}_{1},\ldots ,{k}_{n}}\right\} \subseteq K \) be an \( F \) -linearly independent set, and suppose that there are \( {l}_{i} \in L \) with \( \sum {k}_{i}{l}_{i} = 0 \) . There are nonzero \( s \in A \) and \( t \in B \) with \( s{k}_{i} \in A \) and \( t{l}_{i} \in B \) for each \( i \) . The set \( \left\{ {{a}_{1},\ldots ,{a}_{n}}\right\} \) is also \( F \) -linearly independent; consequently, \( \sum {a}_{i} \otimes {b}_{i} \neq 0 \) , since it maps to the nonzero element \( \sum {a}_{i} \otimes {b}_{i} \in K{ \otimes }_{F}L \) under the natural map \( A{ \otimes }_{F}B \rightarrow K{ \otimes }_{F}B \) . However, \( \sum {a}_{i} \otimes {b}_{i} \) is in the kernel of the map \( A{ \otimes }_{F}B \rightarrow A\left\lbrack B\right\rbrack \) ; hence, it is zero by the assumption that \( A \) and \( B \) are linearly disjoint over \( F \) . This shows that \( \left\{ {k}_{i}\right\} \) is \( L \) -linearly independent, so \( K \) and \( L \) are linearly disjoint over \( F \) . | Yes |
Suppose that \( K/F \) is an algebraic extension and that \( L/F \) is a purely transcendental extension. Then \( K \) and \( L \) are linearly disjoint over \( F \). | To see this, let \( X \) be an algebraically independent set over \( F \) with \( L = F\left( X\right) \). From the previous lemma, it suffices to show that \( K \) and \( F\left\lbrack X\right\rbrack \) are linearly disjoint over \( F \). We can view \( F\left\lbrack X\right\rbrack \) as a polynomial ring in the variables \( x \in X \). The ring generated by \( K \) and \( F\left\lbrack X\right\rbrack \) is the polynomial ring \( K\left\lbrack X\right\rbrack \). The standard homomorphism \( K{ \otimes }_{F}F\left\lbrack X\right\rbrack \rightarrow K\left\lbrack X\right\rbrack \) is an isomorphism because there is a ring homomorphism \( \tau : K\left\lbrack X\right\rbrack \rightarrow K{ \otimes }_{F}F\left\lbrack X\right\rbrack \) induced by \( x \mapsto 1 \otimes x \) for each \( x \in X \), and this is the inverse of \( \varphi \). Thus, \( K \) and \( F\left\lbrack X\right\rbrack \) are linearly disjoint over \( F \), so \( K \) and \( L \) are linearly disjoint over \( F \). | Yes |
Theorem 20.12 Let \( K \) and \( L \) be extension fields of \( F \), and let \( E \) be a field with \( F \subseteq E \subseteq K \) . Then \( K \) and \( L \) are linearly disjoint over \( F \) if and only if \( E \) and \( L \) are linearly disjoint over \( F \) and \( K \) and \( {EL} \) are linearly disjoint over \( E \) . | Proof. We have the following tower of fields.\n\n\n\nConsider the sequence of homomorphisms\n\n\[ K{ \otimes }_{F}L\overset{f}{ \rightarrow }K{ \otimes }_{E}\left( {E{ \otimes }_{F}L}\right) \overset{{\varphi }_{1}}{ \rightarrow }K{ \otimes }_{E}{EL}\overset{{\varphi }_{2}}{ \rightarrow }K\left\lbrack L\right\rbrack ,\]\n\nwhere the maps \( f,{\varphi }_{1} \), and \( {\varphi }_{2} \) are given on generators by\n\n\[ f\left( {k \otimes l}\right) = k \otimes \left( {1 \otimes l}\right) \]\n\n\[ {\varphi }_{1}\left( {k \otimes \left( {e \otimes l}\right) }\right) = k \otimes {el} \]\n\n\[ {\varphi }_{2}\left( {k\otimes \sum {e}_{i}{l}_{i}}\right) = \sum k{e}_{i}{l}_{i} \]\n\nrespectively. Each can be seen to be well defined by the universal mapping property of tensor products. The map \( f \) is an isomorphism by counting dimensions. Moreover, \( {\varphi }_{1} \) and \( {\varphi }_{2} \) are surjective. The composition of these three maps is the standard map \( \varphi : K{ \otimes }_{F}L \rightarrow K\left\lbrack L\right\rbrack \) . First, suppose that \( K \) and \( L \) are linearly disjoint over \( F \) . Then \( \varphi \) is an isomorphism by Proposition 20.2. This forces both \( {\varphi }_{1} \) and \( {\varphi }_{2} \) to be isomorphisms, since all maps in question are surjective. The injectivity of \( {\varphi }_{2} \) implies that \( K \) and \( {EL} \) are linearly disjoint over \( E \) . If \( \sigma : E{ \otimes }_{F}L \rightarrow E\left\lbrack L\right\rbrack \) is the standard map, then \( {\varphi }_{1} \) is given on generators by \( {\varphi }_{1}\left( {k \otimes \left( {e \otimes l}\right) }\right) = k \otimes \sigma \left( {e \otimes l}\right) \) ; hence, \( \sigma \) is also injective. This shows that \( E \) and \( L \) are linearly disjoint over \( F \) .\n\nConversely, suppose that \( E \) and \( L \) are linearly disjoint over \( F \) and that \( K \) and \( {EL} \) are linearly disjoint over \( E \) . Then \( {\varphi }_{2} \) and \( \sigma \) are isomorphisms by Proposition 20.2. The map \( {\varphi }_{1} \) is also an isomorphism; this follows from the relation between \( {\varphi }_{1} \) and \( \sigma \) above. Then \( \varphi \) is a composition of three isomorphisms; hence, \( \varphi \) is an isomorphism. Using Proposition 20.2 again, we see that \( K \) and \( L \) are linearly disjoint over \( F \) . | Yes |
Let \( K/F \) be a separable extension, and let \( L/F \) be a purely inseparable extension. Then \( K \) and \( L \) are linearly disjoint over \( F \). | To prove this, note that if \( \operatorname{char}\left( F\right) = 0 \), then \( L = F \), and the result is trivial. So, suppose that \( \operatorname{char}\left( F\right) = p > 0 \) . We first consider the case where \( K/F \) is a finite extension. By the primitive element theorem, we may write \( K = F\left( a\right) \) for some \( a \in K \) . Let \( f\left( x\right) = \min \left( {F, a}\right) \) and \( g\left( x\right) = \min \left( {L, a}\right) \) . Then \( g \) divides \( f \) in \( L\left\lbrack x\right\rbrack \) . If \( g\left( x\right) = {\alpha }_{0} + \cdots + {\alpha }_{n - 1}{x}^{n - 1} + {x}^{n} \), then for each \( i \) there is a positive integer \( {r}_{i} \) with \( {\alpha }_{i}^{{p}^{{r}_{i}}} \in F \) . If \( r \) is the maximum of the \( {r}_{i} \), then \( {\alpha }_{i}^{{p}^{r}} \in F \) for each \( i \), so \( g{\left( x\right) }^{{p}^{r}} \in F\left\lbrack x\right\rbrack \) . Consequently, \( g{\left( x\right) }^{{p}^{r}} \) is a polynomial over \( F \) for which \( a \) is a root. Thus, \( f \) divides \( {g}^{{p}^{r}} \) in \( F\left\lbrack x\right\rbrack \) . Viewing these two divisibilities in \( L\left\lbrack x\right\rbrack \), we see that the only irreducible factor of \( f \) in \( L\left\lbrack x\right\rbrack \) is \( g \), so \( f \) is a power of \( g \) . The field extension \( K/F \) is separable; hence, \( f \) has no irreducible factors in any extension field of \( F \) . This forces \( f = g \), so\n\n\[ \left\lbrack {{KL} : L}\right\rbrack = \left\lbrack {L\left( a\right) : L}\right\rbrack = \deg \left( g\right) \]\n\n\[ = \deg \left( f\right) = \left\lbrack {K : F}\right\rbrack \]\n\nFrom this, we obtain \( \left\lbrack {{KL} : F}\right\rbrack = \left\lbrack {K : F}\right\rbrack \cdot \left\lbrack {L : F}\right\rbrack \), so \( K \) and \( L \) are linearly disjoint over \( F \) by Lemma 20.4.\n\nIf \( K/F \) is not necessarily finite, suppose that \( \varphi : K{ \otimes }_{F}L \rightarrow {KL} \) is not injective. Then there are \( {k}_{1},\ldots ,{k}_{n} \in K \) and \( {l}_{1},\ldots ,{l}_{n} \in L \) with \( \varphi \left( {\sum {k}_{i} \otimes }\right. \) \( \left. {l}_{i}\right) = 0 \) . If \( {K}_{0} \) is the field generated over \( F \) by the \( {k}_{i} \), then the restriction of \( \varphi \) to \( {K}_{0}{ \otimes }_{F}L \) is not injective, which is false by the finite dimensional case. Thus, \( \varphi \) is injective, so \( K \) and \( L \) are linearly disjoint over \( F \). | Yes |
Let \( K = F\left( x\right) \) be the rational function field in one variable over a field \( F \) of characteristic \( p \) . Then \( \{ x\} \) is a separating transcendence basis for \( K/F \) . However, \( \left\{ {x}^{p}\right\} \) is also a transcendence basis, but \( K/F\left( {x}^{p}\right) \) is not separable. This example shows that even if \( K/F \) is separably generated, not all transcendence bases of \( K/F \) are separating transcendence bases. | Null | No |
Example 20.17 If \( K/F \) is algebraic, then \( K \) is separable over \( F \) if and only if \( K/F \) is separably generated, so the definition of separably generated agrees with the definition of separable for algebraic extensions. | Null | No |
Corollary 20.20 If \( K/F \) is separably generated, then \( K/F \) is separable. Conversely, if \( K/F \) is separable and finitely generated, then \( K/F \) is separably generated. | Null | No |
Corollary 20.21 Suppose that \( K = F\left( {{a}_{1},\ldots ,{a}_{n}}\right) \) is finitely generated and separable over \( F \) . Then there is a subset \( Y \) of \( \left\{ {{a}_{1},\ldots ,{a}_{n}}\right\} \) that is a separating transcendence basis of \( K/F \) . | Proof. This corollary is more accurately a consequence of the proof of (3) \( \Rightarrow \) (1) in Theorem 20.18, since the argument of that step is to show that if \( K \) is finitely generated over \( F \), then any finite generating set contains a separating transcendence basis. | Yes |
Corollary 20.22 Let \( F \) be a perfect field. Then any finitely generated extension of \( F \) is separably generated. | Proof. This follows immediately from part 3 of Theorem 20.18, since \( {F}^{1/{p}^{\infty }} = F \) if \( F \) is perfect. | Yes |
Corollary 20.23 Let \( F \subseteq E \subseteq K \) be fields.\n\n1. If \( K/F \) is separable, then \( E/F \) is separable.\n\n2. If \( E/F \) and \( K/E \) are separable, then \( K/F \) is separable.\n\n3. If \( K/F \) is separable and \( E/F \) is algebraic, then \( K/E \) is separable. | Proof. Part 1 is an immediate consequence of condition 2 of Theorem 20.18. For part 2 we use Theorems 20.18 and 20.12. If \( E/F \) and \( K/E \) are separable, then \( E \) and \( {F}^{1/p} \) are linearly disjoint over \( F \), and \( K \) and \( {E}^{1/p} \) are linearly disjoint over \( E \) . However, it follows from the definition that \( {F}^{1/p} \subseteq {E}^{1/p} \), so \( E{F}^{1/p} \subseteq {E}^{1/p} \) . Thus, \( K \) and \( E{F}^{1/p} \) are linearly disjoint over \( E \) . Theorem 20.12 then shows that \( K \) and \( {F}^{1/p} \) are linearly disjoint over \( F \), so \( K \) is separable over \( F \) .\n\nTo prove part 3, suppose that \( K/F \) is separable and \( E/F \) is algebraic. We know that \( E/F \) is separable by part 1 . Let \( L = E\left( {{a}_{1},\ldots ,{a}_{n}}\right) \) be a finitely generated subextension of \( K/E \) . If \( {L}^{\prime } = F\left( {{a}_{1},\ldots ,{a}_{n}}\right) \), then by the separability of \( K/F \) there is a separating transcendence basis \( \left\{ {{t}_{1},\ldots ,{t}_{m}}\right\} \) for \( {L}^{\prime }/F \) . Because \( E/F \) is separable algebraic, \( E{L}^{\prime } = L \) is separable over \( {L}^{\prime } \), so by transitivity, \( L \) is separable over \( F\left( {{t}_{1},\ldots ,{t}_{m}}\right) \) . Thus, \( L \) is separable over \( E\left( {{t}_{1},\ldots ,{t}_{m}}\right) \), so \( \left\{ {{t}_{1},\ldots ,{t}_{m}}\right\} \) is a separating transcendence basis for \( L/E \) . We have shown that \( L/E \) is separably generated for every finitely generated subextension of \( K/E \), which proves that \( K/E \) is separable. | Yes |
Example 20.24 Let \( F \) be a field of characteristic \( p \), let \( K = F\\left( x\\right) \), the rational function field in one variable over \( F \), and let \( E = F\\left( {x}^{p}\\right) \). Then \( K/F \) is separable, but \( K/E \) is not separable. This example shows the necessity for the assumption that \( E/F \) be algebraic in the previous corollary. | Null | No |
Example 20.25 Here is an example of a separable extension that is not separably generated. Let \( F \) be a field of characteristic \( p \), let \( x \) be transcendental over \( F \), and let \( K = F\left( x\right) \left( \left\{ {{x}^{1/{p}^{n}} : n \geq 1}\right\} \right) \) . Then \( K \) is the union of the fields \( F\left( {x}^{1/{p}^{n}}\right) \), each of which is purely transcendental over \( F \), and hence is separably generated. Any finitely generated subextension \( E \) is a subfield of \( F\left( {x}^{1/{p}^{n}}\right) \) for some \( n \) and hence is separably generated over \( F \) by the previous corollary. Therefore, \( K/F \) is separable. But \( K \) is not separably generated over \( F \), since given any \( f \in K \), there is an \( n \) with \( f \in F\left( {x}^{1/{p}^{n}}\right) \), so \( K/F\left( f\right) \) is not separable, since \( K/F\left( {x}^{1/{p}^{n}}\right) \) is a nontrivial purely inseparable extension. | Null | No |
Example 21.3 Let \( f\left( {x, y}\right) = y - {x}^{2} \) . Then \( Z\left( f\right) = \left\{ {\left( {a,{a}^{2}}\right) : a \in C}\right\} \), a \( k \) -variety for any \( k \subseteq C \) . | Null | No |
Example 21.4 Let \( f\left( {x, y}\right) = {y}^{2} - \left( {{x}^{3} - x}\right) \) . Then \( Z\left( f\right) \) is a \( k \) -variety for any \( k \subseteq C \) . This variety is an example of an elliptic curve, a class of curves of great importance in number theory. | Null | No |
Example 21.6 Let \( V = \left\{ {\left( {{t}^{2},{t}^{3}}\right) : t \in C}\right\} \) . Then \( V \) is the \( k \) -variety \( Z\left( {y}^{2}\right. - \) \( {x}^{3} \) ). The description of \( V \) as the set of points of the form \( \left( {{t}^{2},{t}^{3}}\right) \) is called a parameterization of \( V \) . | Null | No |
Example 21.7 Let \( V = \left\{ {\left( {{t}^{3},{t}^{4},{t}^{5}}\right) : t \in C}\right\} \) . Then \( V \) is a \( k \) -variety, since \( V \) is the zero set of \( \left\{ {{y}^{2} - {xz},{z}^{2} - {x}^{2}y}\right\} \) . | To verify this, note that each point of \( V \) does satisfy these two polynomials. Conversely, suppose that \( \left( {a, b, c}\right) \in {C}^{3} \) is a zero of these three polynomials. If \( a = 0 \), then a quick check of the polynomials shows that \( b = c = 0 \), so \( \left( {a, b, c}\right) \in V \) . If \( a \neq 0 \) , then define \( t = b/a \) . From \( {b}^{2} = {ac} \), we see that \( c = {t}^{2}a \) . Finally, the equation \( {c}^{2} = {a}^{2}b \) yields \( {t}^{4}{a}^{2} = {a}^{3}t \), so \( a = {t}^{3} \) . Thus, \( \left( {a, b, c}\right) = \left( {{t}^{3},{t}^{4},{t}^{5}}\right) \in V \) . | Yes |
Example 21.8 Let \( {S}^{n} = \left\{ {\left( {{a}_{1},\ldots ,{a}_{n}}\right) \in {C}^{n} : \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}^{2} = 1}\right\} \) . Then \( V = \) \( Z\left( {-1 + \mathop{\sum }\limits_{{i = 1}}^{n}{x}_{i}^{2}}\right) \), so \( V \) is a \( k \) -variety. | Null | No |
Example 21.9 Let \( V \) be a \( C \) -vector subspace of \( {C}^{n} \) . We can find a matrix \( A \) such that \( V \) is the nullspace of \( A \) . If \( A = \left( {\alpha }_{ij}\right) \), then a point \( \left( {{a}_{1},\ldots ,{a}_{n}}\right) \) is in \( V \) if and only if \( \mathop{\sum }\limits_{j}{\alpha }_{ij}{a}_{j} = 0 \) for each \( i \) . Thus, \( V \) is the zero set of the set of linear polynomials \( \mathop{\sum }\limits_{j}{\alpha }_{ij}{x}_{j} \), so \( V \) is a \( C \) -variety. If each \( {\alpha }_{ij} \) lies in a subfield \( k \), then \( V \) is a \( k \) -variety. | Null | No |
Example 21.10 Let \( S{L}_{n}\left( C\right) \) be the set of all \( n \times n \) matrices over \( C \) of determinant 1 . We view the set of all \( n \times n \) matrices over \( C \) as the set \( {C}^{{n}^{2}} \) of \( {n}^{2} \) -tuples over \( C \) . The determinant \( \det = \det \left( {x}_{ij}\right) \) is a polynomial in the \( {n}^{2} \) variables \( {x}_{ij} \), and the coefficients of the determinant polynomial are \( \pm 1 \) . We then see that \( S{L}_{n}\left( C\right) = Z\left( {\det - 1}\right) \) is a \( k \) -variety for any subfield \( k \) of \( C \) . For instance, if \( n = 2 \), then \[ S{L}_{2}\left( C\right) = \left\{ {\left( {a, b, c, d}\right) \in {C}^{4} : {ad} - {bc} - 1 = 0}\right\} . | Null | No |
Lemma 21.11 The sets \( \left\{ {Z\left( S\right) : S \subseteq k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack }\right\} \) are the closed sets of a topology on \( {C}^{n} \) ; that is,\n\n1. \( {C}^{n} = Z\left( {\{ 0\} }\right) \) and \( \varnothing = Z\left( {\{ 1\} }\right) \) .\n\n2. If \( S \) and \( T \) are subsets of \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \), then \( Z\left( S\right) \cup Z\left( T\right) = Z\left( {ST}\right) \), where \( {ST} = \{ {fg} : f \in S, t \in T\} \) .\n\n3. If \( \left\{ {S}_{\alpha }\right\} \) is an arbitrary collection of subsets of \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \), then \( \mathop{\bigcap }\limits_{\alpha }Z\left( {S}_{\alpha }\right) = Z\left( {\mathop{\bigcup }\limits_{\alpha }{S}_{\alpha }}\right) \n | Proof. The first two parts are clear from the definitions. For the third, let \( P \in Z\left( S\right) \) . Then \( f\left( P\right) = 0 \) for all \( f \in S \), so \( \left( {fg}\right) \left( P\right) = 0 \) for all \( {fg} \in {ST} \) . Thus, \( Z\left( S\right) \subseteq Z\left( {ST}\right) \) . Similarly, \( Z\left( T\right) \subseteq Z\left( {ST}\right) \), so \( Z\left( S\right) \cup Z\left( T\right) \subseteq Z\left( {ST}\right) \) . For the reverse inclusion, let \( P \in Z\left( {ST}\right) \) . If \( P \notin Z\left( S\right) \), then there is an \( f \in S \) with \( f\left( P\right) \neq 0 \) . If \( g \in T \), then \( 0 = \left( {fg}\right) \left( P\right) = f\left( P\right) g\left( P\right) \), so \( g\left( P\right) = 0 \) , which forces \( P \in Z\left( T\right) \) . Thus, \( Z\left( {ST}\right) \subseteq Z\left( S\right) \cup Z\left( T\right) \) . This proves that \( Z\left( S\right) \cup Z\left( T\right) = Z\left( {ST}\right) \n\nFor the fourth part, the inclusion \( Z\left( {\mathop{\bigcup }\limits_{\alpha }{S}_{\alpha }}\right) \subseteq \mathop{\bigcap }\limits_{\alpha }Z\left( {S}_{\alpha }\right) \) follows from part 1. For the reverse inclusion, take \( P \in \mathop{\bigcap }\limits_{\alpha }Z\left( {S}_{\alpha }\right) \) . Then \( P \in Z\left( {S}_{\alpha }\right) \) for each \( \alpha \), so \( f\left( P\right) = 0 \) for each \( f \in {S}_{\alpha } \) . Thus, \( P \in Z\left( {\mathop{\bigcup }\limits_{\alpha }{S}_{\alpha }}\right) \) . | Yes |
Example 21.12 Let \( G{L}_{n}\left( C\right) \) be the set of all invertible \( n \times n \) matrices over \( C \) . Then \( G{L}_{n}\left( C\right) \) is the complement of the zero set \( Z \) (det), so \( G{L}_{n}\left( C\right) \) is an open subset of \( {C}^{{n}^{2}} \) with respect to the \( k \) -Zariski topology. We can view \( G{L}_{n}\left( C\right) \) differently in order to view it as an algebraic variety. Let \( t \) be a new variable, and consider the zero set \( Z\left( {t\det - 1}\right) \) in \( {C}^{{n}^{2} + 1} \) . Then the map \( G{L}_{n}\left( C\right) \rightarrow Z\left( {t\det - 1}\right) \) given by \( P \mapsto \left( {P,1/\det \left( P\right) }\right) \) is a bijection between \( G{L}_{n}\left( C\right) \) and \( Z\left( {t\det - 1}\right) \) . If we introduce the definition of a morphism of varieties, this map would turn out to be an isomorphism. | Null | No |
Lemma 21.14 If \( V \) is any subset of \( {C}^{n} \), then \( I\left( V\right) \) is a radical ideal of \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . | Proof. Let \( f \in k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) with \( {f}^{r} \in I\left( V\right) \) for some \( r \) . Then \( {f}^{r}\left( P\right) = 0 \) for all \( P \in V \) . But \( {f}^{r}\left( P\right) = {\left( f\left( P\right) \right) }^{r} \), so \( f\left( P\right) = 0 \) . Therefore, \( f \in I\left( V\right) \) ; hence, \( I\left( V\right) \) is equal to its radical, so \( I\left( V\right) \) is a radical ideal. | Yes |
Lemma 21.15 The following statements are some properties of ideals of subsets of \( {C}^{n} \) . | Proof. The first two parts of the lemma are clear from the definition of \( I\left( V\right) \) . For the third, let \( V \) be a subset of \( {C}^{n} \) . If \( f \in I\left( V\right) \), then \( f\left( P\right) = 0 \) for all \( P \in V \), so \( P \in Z\left( {I\left( V\right) }\right) \), which shows that \( V \subseteq Z\left( {I\left( V\right) }\right) \) . Suppose that \( V = Z\left( S\right) \) for some subset \( S \in k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . Then \( S \subseteq I\left( V\right) \), so \( Z\left( {I\left( V\right) }\right) \subseteq \) \( Z\left( S\right) = V \) by the previous lemma. Thus, \( V = Z\left( {I\left( V\right) }\right) \) . Conversely, if \( V = Z\left( {I\left( V\right) }\right) \), then \( V \) is a \( k \) -variety by definition. | Yes |
Theorem 21.16 (Nullstellensatz) Let \( J \) be an ideal of \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) , and let \( V = Z\left( J\right) \) . Then \( I\left( V\right) = \sqrt{J} \) . | Proof. For a proof of the Nullstellensatz, see Atiyah and Macdonald [2, p. 85] or Kunz [19, p. 16]. | No |
Corollary 21.17 There is a 1-1 inclusion reversing correspondence between the \( k \) -varieties in \( {C}^{n} \) and the radical ideals of \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) given by \( V \mapsto I\left( V\right) \) . The inverse correspondence is given by \( J \mapsto Z\left( J\right) \) . | Proof. If \( V \) is a \( k \) -variety, then the previous lemma shows that \( V = \) \( Z\left( {I\left( V\right) }\right) \) . Also, the Nullstellensatz shows that if \( I \) is a radical ideal, then \( J = I\left( {Z\left( J\right) }\right) \) . These two formulas tell us that the association \( V \mapsto I\left( V\right) \) is a bijection and that its inverse is given by \( J \mapsto Z\left( J\right) \) . | Yes |
Let \( f \in k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) be a polynomial, and let \( V = Z\left( f\right) \) . If \( f = {p}_{1}^{{r}_{1}}\cdots {p}_{t}^{{r}_{t}} \) is the irreducible factorization of \( f \), then \( I\left( V\right) = \sqrt{\left( f\right) } \) by the Nullstellensatz. However, we show that \( \sqrt{\left( f\right) } = \left( {{p}_{1}\cdots {p}_{t}}\right) \) | for, if \( g \in \sqrt{\left( f\right) } \), then \( {g}^{m} = {fh} \) for some \( h \in k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) and some \( m > 0 \) . Each \( {p}_{i} \) then divides \( {g}^{m} \) ; hence, each \( {p}_{i} \) divides \( g \) . Thus, \( g \in \left( {{p}_{1}\cdots {p}_{t}}\right) \) . For the reverse inclusion, \( {p}_{1}\cdots {p}_{t} \in \sqrt{\left( f\right) } \), since if \( r \) is the maximum of the \( {r}_{i} \), then \( {\left( {p}_{1}\cdots {p}_{t}\right) }^{r} \in \left( f\right) \) | Yes |
Let \( V \) be an irreducible \( k \) -variety. By taking complements, we see that the definition of irreducibility is equivalent to the condition that any two nonempty open sets have a nonempty intersection. Therefore, if \( U \) and \( {U}^{\prime } \) are nonempty open subsets of \( V \), then \( U \cap {U}^{\prime } \neq \varnothing \) . One consequence of this fact is that any nonempty open subset of \( V \) is dense in \( V \), as we now prove. | If \( U \) is a nonempty open subset of \( V \), and if \( C \) is the closure of \( U \) in \( V \), then \( U \cap \left( {V - C}\right) = \varnothing \) . The set \( V - C \) is open, so one of \( U \) or \( V - C \) is empty. Since \( U \) is nonempty, this forces \( V - C = \varnothing \) , so \( C = V \) . But then the closure of \( U \) in \( V \) is all of \( V \), so \( U \) is dense in \( V \). | Yes |
Proposition 21.21 Let \( V \) be a \( k \) -variety. Then \( V \) is irreducible if and only if \( I\left( V\right) \) is a prime ideal, if and only if the coordinate ring \( k\left\lbrack V\right\rbrack \) is an integral domain. | Proof. First suppose that \( V \) is irreducible. Let \( f, g \in k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) with \( {fg} \in I\left( V\right) \) . Then \( I = I\left( V\right) + \left( f\right) \) and \( J = I\left( V\right) + \left( g\right) \) are ideals of \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) containing \( I\left( V\right) \) ; hence, their zero sets \( Y = Z\left( I\right) \) and \( Z = Z\left( J\right) \) are contained in \( Z\left( {I\left( V\right) }\right) = V \) . Moreover, \( {IJ} \subseteq I\left( V\right) \), since \( {fg} \in I\left( V\right) \), so \( Y \cup Z = Z\left( {IJ}\right) \) contains \( V \) . This forces \( V = Y \cup Z \), so either \( Y = V \) or \( Z = V \), since \( V \) is irreducible. If \( Y = V \), then \( I \subseteq I\left( Y\right) = I\left( V\right) \) , and if \( Z = V \), then \( J \subseteq I\left( Z\right) = I\left( V\right) \) . Thus, either \( f \in I\left( V\right) \) or \( g \in I\left( V\right) \) , so \( I\left( V\right) \) is a prime ideal of \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) .\n\nConversely, suppose that \( I\left( V\right) \) is prime. If \( V = Y \cup Z \) for some \( k \) -varieties \( Y \) and \( Z \), let \( I = I\left( Y\right) \) and \( J = I\left( Z\right) \) . Then \( {IJ} \subseteq I\left( {Y \cup Z}\right) = I\left( V\right) \), so either \( I \subseteq I\left( V\right) \) or \( J \subseteq I\left( V\right) \) . This means that \( V \subseteq Z\left( I\right) = Y \) or \( V \subseteq Z\left( J\right) = Z \) . Therefore, \( Y = V \) or \( Z = V \), so \( V \) is irreducible. | Yes |
Proposition 21.23 Let \( V \) be a \( k \) -variety. Then \( \dim \left( V\right) \) is the maximum nonnegative integer \( n \) such that there is a chain\n\n\[ \n{P}_{0} \subset {P}_{1} \subset \cdots \subset {P}_{n} \n\]\n\nof prime ideals of \( k\left\lbrack V\right\rbrack \) . | Proof. Suppose that \( {Y}_{0} \subset {Y}_{1} \subset \cdots \subset {Y}_{n} \subseteq V \) is a chain of closed irreducible subsets of \( V \) . Then\n\n\[ \nI\left( V\right) \subseteq I\left( {Y}_{n}\right) \subset \cdots \subset I\left( {Y}_{0}\right) \n\]\n\nis a chain of prime ideals of \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) by the previous proposition. Moreover, the inclusions are proper by the Nullstellensatz. By taking images in the quotient ring \( k\left\lbrack V\right\rbrack = k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack /I\left( V\right) \), we get a chain of prime ideals of length \( n \) . However, if we have a chain of prime ideals of \( k\left\lbrack V\right\rbrack \) of length \( n \), then we get a chain \( I\left( V\right) \subseteq {Q}_{0} \subset {Q}_{1} \subset \cdots \subset {Q}_{n} \) of prime ideals of \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . Taking zero sets gives a chain\n\n\[ \nZ\left( {Q}_{n}\right) \subset \cdots \subset Z\left( {Q}_{0}\right) \subseteq Z\left( {I\left( V\right) }\right) = V \n\]\n\nof irreducible \( k \) -subvarieties in \( V \) . The maximum length of a chain of irreducible \( k \) -subvarieties of \( V \) is then the maximum length of a chain of prime ideals of \( k\left\lbrack V\right\rbrack \) . | Yes |
Let \( V = Z\left( {y - {x}^{2}}\right) \). Then the coordinate ring of \( V \) is \( k\left\lbrack {x, y}\right\rbrack /\left( {y - {x}^{2}}\right) \), which is isomorphic to the polynomial ring \( k\left\lbrack t\right\rbrack \) by sending \( t \) to the coset of \( x \) in \( k\left\lbrack V\right\rbrack \). | Therefore, the function field of \( V \) is the rational function field \( k\left( t\right) \). | No |
Let \( V = Z\left( {{y}^{2} - {x}^{3}}\right) \). Then \( k\left( V\right) \) is the field \( k\left( {s, t}\right) \), where \( s \) and \( t \) are the images of \( x \) and \( y \) in \( k\left\lbrack V\right\rbrack = k\left\lbrack {x, y}\right\rbrack /\left( {{y}^{2} - {x}^{3}}\right) \), respectively. Note that \( {t}^{2} = {s}^{3} \). | Let \( z = t/s \). Substituting this equation into \( {t}^{2} = {s}^{3} \) and simplifying shows that \( s = {z}^{2} \), and so \( t = {z}^{3} \). Thus, \( k\left( V\right) = k\left( z\right) \). The element \( z \) is transcendental over \( k \), since if \( k\left( V\right) /k \) is algebraic, then \( k\left\lbrack V\right\rbrack \) is a field by the argument in Example 19.11, so \( \left( {{y}^{2} - {x}^{3}}\right) \) is a maximal ideal of \( k\left\lbrack {x, y}\right\rbrack \). However, this is not true, since \( \left( {{y}^{2} - {x}^{3}}\right) \) is properly contained in the ideal \( \left( {x, y}\right) \). Thus, \( k\left( V\right) \) is a rational function field in one variable over \( k \). Note that \( k\left\lbrack V\right\rbrack \) is isomorphic to \( k\left\lbrack {{x}^{2},{x}^{3}}\right\rbrack \), a ring that is not isomorphic to a polynomial ring in one variable over \( k \). | Yes |
If \( V \) is an irreducible \( k \) -variety, then \( V \) gives rise to a field extension \( k\left( V\right) \) of \( k \) . We can reverse this construction. Let \( K \) be a finitely generated field extension of \( k \) . Say \( K = k\left( {{a}_{1},\ldots ,{a}_{n}}\right) \) for some \( {a}_{i} \in K \) . Let\n\n\[ P = \left\{ {f \in k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack : f\left( {{a}_{1},\ldots ,{a}_{n}}\right) = 0}\right\} . | Then \( P \) is the kernel of the ring homomorphism \( \varphi : k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \rightarrow K \) that sends \( {x}_{i} \) to \( {a}_{i} \), so \( P \) is a prime ideal. If \( V = Z\left( P\right) \), then \( V \) is an irreducible \( k \) -variety with coordinate ring \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack /P \cong k\left\lbrack {{a}_{1},\ldots ,{a}_{n}}\right\rbrack \) | Yes |
5.3 Lemma.\n\n(a) For every fixed nonzero \( u \in {\mathbb{R}}^{n} \), the hyperplane\n\n(*)\n\n\[ {H}_{K}\left( u\right) \mathrel{\text{:=}} \left\{ {x\mid \langle x, u\rangle = {h}_{K}\left( u\right) }\right\} \]\n\nis a supporting hyperplane of \( K \) (Figure 9b).\n\n(b) Every supporting hyperplane of \( K \) has a representation of the form \( \left( *\right) \) . | Proof.\n\n(a) Since \( K \) is compact and \( \langle \cdot, u\rangle \) is continuous, for some \( {x}_{0} \in K \),\n\n\[ \left\langle {{x}_{0}, u}\right\rangle = {h}_{K}\left( u\right) = \mathop{\sup }\limits_{{x \in K}}\langle x, u\rangle .\n\]\nFor an arbitrary \( y \in K \), it follows that \( \langle y, u\rangle \leq \left\langle {{x}_{0}, u}\right\rangle \) ; hence, \( K \subset {H}_{K}^{ - }\left( u\right) \). This proves (a).\n\n(b) Let \( H = \left\{ {x\mid \langle x, u\rangle = \left\langle {{x}_{0}, u}\right\rangle }\right\} \) be a supporting hyperplane of \( K \) at \( {x}_{0} \). We choose \( u \neq 0 \) such that \( K \subset {H}^{ - } \). Then, \( \left\langle {{x}_{0}, u}\right\rangle = \mathop{\sup }\limits_{{x \in K}}\langle x, u\rangle = {h}_{K}\left( u\right) \) which implies (b). | Yes |
6.4 Theorem. Let \( K \) be a convex body with \( 0 \in \operatorname{int}K \) . Then,\n\n(a)\n\[ \n{K}^{* * } = K \n\]\n\n(b) The distance function of \( K \) equals the support function of \( {K}^{ * } \), and, conversely,\n\n\[ \n{d}_{K} = {h}_{{K}^{ * }},\;{d}_{{K}^{ * }} = {h}_{K}. \n\] | Proof.\n\n(a) By definition of \( {H}_{u} \), for every \( u \neq 0 \) of \( K \) ,\n\n\[ \n{H}_{u}^{ - } = \{ x \mid \langle u, x\rangle \leq 1\} \n\]\n\nTherefore,(using the obvious notation \( \langle K, x\rangle \leq 1 \) )\n\n\[ \n{K}^{ * } = \{ x \mid \langle K, x\rangle \leq 1\} \;\text{ and }\;{K}^{* * } = \left\{ {y \mid \left\langle {{K}^{ * }, y}\right\rangle \leq 1}\right\} .\n\]\n\nIf \( y \in K \), then, the definition of \( {K}^{ * } \) yields \( \left\langle {y,{K}^{ * }}\right\rangle \leq 1 \) and, thus, \( K \subset {K}^{* * } \) . Suppose \( K \neq {K}^{* * } \) . Then, let \( x \in {K}^{* * } \smallsetminus K \) . For\n\n\[ \n{x}^{\prime } \mathrel{\text{:=}} {p}_{K}\left( x\right) \;\text{ and }\;u \mathrel{\text{:=}} \frac{x - {x}^{\prime }}{\left\langle {x}^{\prime }, x - {x}^{\prime }\right\rangle },\n\]\n\nLemma 3.5 yields\n\n\[ \nx \in {H}_{u}^{ + } \smallsetminus {H}_{u},\;\text{ but also }K \subset {H}_{u}^{ - },\n\]\n\nwhence \( u \in {K}^{ * } \) . Since \( x \in {K}^{* * } \), it follows that \( \langle u, x\rangle \leq 1 \), i.e., \( x \in {H}_{u}^{ - } \), a contradiction. | No |
6.8 Theorem. Every positive homogeneous and convex function \( h : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) is the support function \( h = {h}_{K} \) of a unique convex body \( K \) (whose dimension is possibly less than \( n \) ). | Proof. Let us write \( {\mathbb{R}}^{n} = U \oplus {U}^{ \bot } \), where \( U \) is the maximal linear subspace of \( {\mathbb{R}}^{n} \) on which \( h \) is linear. Then, there exists \( a \in U \) such that, for \( \left( {x,{x}^{\prime }}\right) \in U \oplus {U}^{ \bot } \) ,\n\n(*)\n\n\[ h\left( {x,{x}^{\prime }}\right) = \langle x, a\rangle + {\left. h\right| }_{{U}^{ \bot }}\left( {x}^{\prime }\right) . \]\n\nMoreover, \( {\Gamma }^{ + }\left( {\left. h\right| }_{{U}^{ \bot }}\right) \) is a cone with apex 0 in \( {U}^{ \bot } \oplus \mathbb{R} \) (see Lemma 5.7). Thus, there exists some \( b \in {U}^{ \bot } \) such that the hyperplane \( H \mathrel{\text{:=}} \left\{ {\left( {y,\langle y, b\rangle }\right) \mid y \in {U}^{ \bot }}\right\} \) in \( {U}^{ \bot } \oplus \mathbb{R} \) intersects \( {\Gamma }^{ + }\left( {\left. h\right| }_{{U}^{ \bot }}\right) \) only in the apex. Now the set\n\n\[ {K}_{0} + \left( {0,1}\right) \mathrel{\text{:=}} \left( {{U}^{ \bot }\times \{ 1\} }\right) \cap {\Gamma }^{ + }\left( {{\left. h\right| }_{{U}^{ \bot }}-\langle \cdot, b\rangle }\right) \]\n\nis a convex body and, by Lemma 5.2, \( {\left. h\right| }_{{U}^{ \bot }} - \langle \cdot, b\rangle \) the support function of \( {K}_{0} - b \) . Finally, \( \left( *\right) \) and Lemma 5.2 yield that \( h \) is the support function of \( K \mathrel{\text{:=}} {K}_{0} - b + \) a. | Yes |
Suppose \( \left\lbrack {p, q}\right\rbrack \) is such an edge of a 4-dimensional polytope \( P \) that \( \left\lbrack {p, q}\right\rbrack \) is the intersection of three facets of \( P \) which are combinatorially isomorphic to triangular prisms. Also suppose that \( p \) and \( q \) are contained each in only one more facet which is a simplex (Figure 3a). Then, \( {P}^{ * } \) contains two facets \( {p}^{ * },{q}^{ * } \) , and \( {\left\lbrack p, q\right\rbrack }^{ * } \mathrel{\text{:=}} {p}^{ * } \cap {q}^{ * } = {P}^{ * } \cap \pi \left( {\operatorname{aff}\left\lbrack {p, q}\right\rbrack }\right) \) is a triangle (Figure 3b). | Null | No |
2.8 Theorem. Let \( F \) be a proper face of the polytope \( P \), and let \( {P}^{ * } \) be the polar polytope of \( P \) with respect to a polarity \( \pi \) . For an affine subspace \( U \) of \( {\mathbb{R}}^{n} \), let \( {\pi }_{U} \) denote the restriction of \( \pi \) to \( U \), and set \( {\pi }_{U}\left( {P/F}\right) \mathrel{\text{:=}} {\pi }_{U}\left( {\operatorname{aff}\left( {P/F}\right) }\right) \) .\n\n(a) For any face figure \( P/F \) of \( F \) ,\n\n\[{\pi }_{U}\left( {P/F}\right) \approx {F}^{ * }\]\n\n(b) Any two face figures of \( P \) with respect to \( F \) are combinatorially isomorphic. Therefore, we can consider \( P/F \) to be the equivalence class of these face figures. | Proof. Let \( k \mathrel{\text{:=}} \dim F \), so that \( \dim U = n - k - 1 \) for \( U \) as in 2.6. For any face \( G \) which contains \( F \) properly, we set\n\n\[G\underset{\varphi }{ \mapsto }\;{\pi }_{U}\left( {G \cap U}\right)\]\n\nIf \( g \mathrel{\text{:=}} \dim \left( {G \cap U}\right) \), so that \( g = \dim G - k - 1 \), then, by the construction in Lemma 2.6,\n\n\( \dim \varphi \left( G\right) = \left( {n - k - 1}\right) - g - 1 = n - \left( {g + 1 + k}\right) - 1 = n - \dim G - 1. \)\n\nTherefore \( \dim \varphi \left( G\right) = \dim {G}^{ * } \) . The mapping \( \varphi \) is inclusion-reversing as is the polarity \( \pi \) . So, by\n\n\[ \varphi \left( G\right) \; \mapsto \;{G}^{ * } \in \mathcal{B}\left( {F}^{ * }\right) \]\n\nwe obtain an isomorphism\n\n\[{\mathcal{B}}_{0}\left( {{\pi }_{U}\left( {P/F}\right) }\right) \; \rightarrow {\mathcal{B}}_{0}\left( {F}^{ * }\right)\]\n\nwhence (a) follows. Note that, by Lemma 2.6(c), only faces \( G \) including \( F \) properly contribute to \( \mathcal{B}\left( {{\pi }_{U}\left( {P/F}\right) }\right) \) .\n\nPart (b) of the theorem is a consequence of (a). | Yes |
Example 5. If \( F \) is a facet of any \( n \) -polytope \( P \), then, \( P/F \) is a point. | Null | No |
Let \( U, W \) be subspaces of \( V = {\mathbb{R}}^{4} = U \oplus W \) where \( U = \operatorname{lin}\left\{ {{e}_{1},{e}_{2}}\right\} \) , \( W = \operatorname{lin}\left\{ {{e}_{3},{e}_{4}}\right\} ,{e}_{1},\ldots ,{e}_{4} \) the canonical basis of \( {\mathbb{R}}^{4} \) . We write the coordinates of the elements of \( U, V, W \) as column vectors and those of the elements of \( {V}^{ * } \) , \( {U}^{ * },{W}^{ * } \) as row vectors, respectively. Applying a dual vector of \( {V}^{ * },{U}^{ * } \), or \( {W}^{ * } \) to a vector of \( V, U, W \), respectively, can be carried out as (matrix) multiplication of a row vector and a column vector. | Let\n\n\[ X = \left( {{x}_{1}{x}_{2}{x}_{3}{x}_{4}}\right) = \left( \begin{array}{llll} 1 & 0 & 1 & 2 \\ 0 & 1 & 2 & 1 \end{array}\right) \]\n\n\[ \left( {{b}_{1}{b}_{2}{b}_{3}{b}_{4}}\right) = \left( \begin{array}{llll} 1 & 0 & 1 & 2 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right) \]\n\n\( {L}_{1},{L}_{2} \) are given as matrices\n\n\[ {L}_{1} = \left( \begin{array}{llll} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array}\right) = \left( {{E}_{2} \mid O}\right) \]\n\n\[ {L}_{2} = \left( \begin{array}{ll} 0 & 0 \\ 0 & 0 \\ 0 & 0 \\ 0 & 1 \end{array}\right) = \left( \frac{O}{{E}_{2}}\right) \]\n\nThen,\n\n\[ \left( \begin{array}{l} {b}_{1}^{ * } \\ {b}_{2}^{ * } \\ {b}_{3}^{ * } \\ {b}_{4}^{ * } \end{array}\right) = \left( \begin{matrix} 1 & 0 & - 1 & - 2 \\ 0 & 1 & - 2 & - 1 \\ & & & \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}\right) \]\n\n\( {L}_{1}\left( {b}_{i}\right) = {x}_{i}, i = 1,\ldots ,4 \) and im \( {L}_{2} = \ker {L}_{1} \) are readily checked. We obtain (Figure 7)\n\n\[ \left( \begin{array}{l} {\bar{x}}_{1} \\ {\bar{x}}_{2} \\ {\bar{x}}_{3} \\ {\bar{x}}_{4} \end{array}\right) = \left( \begin{array}{l} {L}_{2}^{ * }\left( {b}_{1}^{ * }\right) \\ {L}_{2}^{ * }\left( {b}_{2}^{ * }\right) \\ {L}_{2}^{ * }\left( {b}_{3}^{ * }\right) \\ {L}_{2}^{ * }\left( {b}_{4}^{ * }\right) \end{array}\right) = \left( \begin{array}{l} {b}_{1}^{ * } \\ {b}_{2}^{ * } \\ {b}_{3}^{ * } \\ {b}_{4}^{ * } \end{array}\right) {L}_{2} = \left( \begin{array}{rr} - 1 & - 2 \\ - 2 & - 1 \\ 1 & 0 \\ 0 & 1 \end{array}\right) . \] | Yes |
In Example 2, replace the elements \( {x}_{3},{x}_{4} \) by \( \frac{1}{3}{x}_{3},\frac{1}{3}{x}_{4} \), respectively. Then the linear transform obtained by the same calculation as in Example 1 is | \[ \left( \begin{array}{l} {\bar{x}}_{1} \\ {\bar{x}}_{2} \\ {\bar{x}}_{3} \\ {\bar{x}}_{4} \end{array}\right) = \left( \begin{array}{l} {L}_{2}^{ * }\left( {b}_{1}^{ * }\right) \\ {L}_{2}^{ * }\left( {b}_{2}^{ * }\right) \\ {L}_{2}^{ * }\left( {b}_{3}^{ * }\right) \\ {L}_{4}^{ * }\left( {b}_{4}^{ * }\right) \end{array}\right) = \left( \begin{array}{rr} - \frac{1}{3} & - \frac{2}{3} \\ - \frac{2}{3} & - \frac{1}{3} \\ 1 & 0 \\ 0 & 1 \end{array}\right) \] | Yes |
Example 1. We consider a triangular prism \( P \) in \( {\mathbb{R}}^{3} \) and wish to find a Gale transform of \( X \mathrel{\text{:=}} \) vert \( P \) (see Figure 9). Since \( v = 6 \) and \( n = 3 \), the elements of a Gale transform lie in a two-dimensional space. By Theorem 5.3, it is sufficient to find two independent affine dependencies of \( {x}_{1},\ldots ,{x}_{6} \) . Since \( {x}_{4} - {x}_{1},{x}_{5} - {x}_{2} \) , and \( {x}_{6} - {x}_{3} \) coincide, | \[ {\left( {a}_{1}{a}_{2}\right) }^{t} = \left( \begin{array}{rrrrrr} - 1 & 1 & 0 & 1 & - 1 & 0 \\ - 1 & 0 & 1 & 1 & 0 & - 1 \end{array}\right) . \] By Theorem 5.3, the rows of \( \left( {{a}_{1}{a}_{2}}\right) \) provide a Gale transform. | Yes |
5.4 Theorem. A finite sequence \( X \) in \( U = \operatorname{aff}X \) consists of all points of the vertex set of a polytope \( P \) in \( U \) if and only if one (and, thus, every) Gale transform \( {\bar{X}}_{\widehat{U}} \) of \( X \) satisfies the following condition:\n\n(4) For every hyperplane \( H \) in \( \operatorname{lin}{\bar{X}}_{\widehat{U}} \) which contains 0, each of the associated half-spaces \( {H}^{ + } \) and \( {H}^{ - } \) includes at least two points of \( {\bar{X}}_{\widehat{U}} \) in its interior. | Proof. In appropriate coordinates for \( \widehat{U} \), we may identify \( U \) with \( U \times \{ 1\} \subset \widehat{U} \) . First, suppose that \( X \) comes from a polytope. Then, each element \( {\widehat{x}}_{i} \) of \( {X}_{i\prime } \) is a face. By Theorem 4.14, \( {\bar{X}}_{\widehat{U}} \smallsetminus \left\{ {\overline{\widehat{x}}}_{i}\right\} \) satisfies\n\n(5)\n\n\[ 0 \in \text{ relint pos }\left( {{\bar{X}}_{\widehat{U}} \smallsetminus \left\{ {\overline{\widehat{x}}}_{i}\right\} }\right) \]\n\nLet \( H \) be a hyperplane in \( \operatorname{lin}{\bar{X}}_{\widehat{U}} \) with \( 0 \in H \) . Let \( {k}^{ + },{k}^{ - } \) be the number of elements \( {\overline{\widehat{x}}}_{i} \) in int \( {H}^{ + } \), int \( {H}^{ - } \), respectively. If \( {k}^{ + } = {k}^{ - } = 0, H = \operatorname{lin}{\bar{X}}_{\widehat{U}} \) so that \( H \) is not a hyperplane. \( {k}^{ + } \leq 1 \) and \( {k}^{ - } > 0 \) or \( {k}^{ - } \leq 1 \) and \( {k}^{ + } > 0 \) imply a contradiction to (5). So, (4) follows.\n\nConversely, if (4) holds, then, (5) readily follows (since, otherwise, 0 would lie on the boundary of \( \sigma \mathrel{\text{:=}} \operatorname{pos}\left( {{\bar{X}}_{\widehat{U}} \smallsetminus \left\{ {\overline{\widehat{x}}}_{i}\right\} }\right) \), so that, for a supporting hyperplane \( H \) of \( \sigma \) with \( 0 \in H \) and \( \sigma \subset {H}^{ + } \), the half-space \( {H}^{ - } \) would not contain two of the \( {\widehat{x}}_{i} \) ’s in its interior). Therefore, by Theorem 4.14, \( {x}_{i} \in \operatorname{vert}P \) for \( i = 1,\ldots, v \) . | Yes |
The vertex set \( X = \left( {{x}_{1},\ldots ,{x}_{n + 1}}\right) \) of an \( n \) -simplex in \( {\mathbb{R}}^{n} \) is characterized by \( {\overline{\widehat{x}}}_{1} = \cdots = {\overline{\widehat{x}}}_{n} = 0 \) for a Gale transform \( {\bar{X}}_{\widehat{U}} \) of \( X \). | This also follows directly from \( \dim \operatorname{lin}{\bar{X}}_{\widehat{U}} = n + 1 - n - 1 = 0 \) by Lemma 4.5. | Yes |
6.5 Theorem. An \( n \) -dimensional polytope \( P = \operatorname{conv}\left\{ {{x}_{1},\ldots ,{x}_{v}}\right\} \) is simplicial if and only if, for a Gale diagram \( \bar{X} \) of \( X = \left( {\operatorname{vert}P}\right) \), the following condition holds: (1) For any hyperplane \( H \) in \( \operatorname{lin}\bar{X} \), with \( 0 \in H,0 \notin \) relint conv \( \left( {\bar{X} \cap H}\right) \) . | Proof. Suppose \( 0 \in \) relint conv \( \left( {\bar{X} \cap H}\right) \) for some hyperplane \( H \) . Since \( \dim P = n \) and \( P \) lies in a hyperplane of \( U \) which does not contain 0, we have \( \dim U = \operatorname{rank}X = n + 1 \), so, \( \operatorname{rank}\bar{X} = v - \operatorname{rank}X = v - n - 1 \) (Lemma 4.5).\n\n\n\n\n\nThen, by Carathéodory’s theorem (I, Theorem 2.3), for \( k = \operatorname{rank}\left( {H \cap \bar{X}}\right) \leq \) \( v - n - 2 \), there exist \( k + 1 \leq v - n - 1 \) affinely independent elements \( {\bar{x}}_{1},\ldots ,{\bar{x}}_{k + 1} \in \bar{X} \cap H \) such that \( 0 \in \) relint conv \( \left\{ {{\bar{x}}_{1},\ldots ,{\bar{x}}_{k + 1}}\right\} \) . Therefore, \( \left\{ {{\bar{x}}_{1},\ldots ,{\bar{x}}_{k + 1}}\right\} \) is a coface of \( X \), and the corresponding (proper) face has at least \( v - \left( {v - n - 1}\right) = n + 1 \) vertices and cannot be a simplex.\n\nThe converse is also true. | Yes |
Example 2. \( n = 3 \) : Then, \( \left\lbrack {\frac{1}{4}{n}^{2}}\right\rbrack = 2 \) . Two nonisomorphic 3-polytopes with five vertices are represented below. Figure 10a shows a bipyramid over a triangle, Figure 10b a quadrangular pyramid. The two possible triples \( \left( {r + 1, t, s + 1}\right) \) are \( \left( {2,0,3}\right) \) and \( \left( {2,1,2}\right) \) ; by Theorem 5.5, the second one belongs to Figure 10b. | Null | No |
6.5 Theorem. An \( n \) -dimensional polytope \( P = \operatorname{conv}\left\{ {{x}_{1},\ldots ,{x}_{v}}\right\} \) is simplicial if and only if, for a Gale diagram \( \bar{X} \) of \( X = \left( {\operatorname{vert}P}\right) \), the following condition holds: (1) For any hyperplane \( H \) in \( \operatorname{lin}\bar{X} \), with \( 0 \in H,0 \notin \) relint conv \( \left( {\bar{X} \cap H}\right) \) . | Proof. Suppose \( 0 \in \) relint conv \( \left( {\bar{X} \cap H}\right) \) for some hyperplane \( H \) . Since \( \dim P = n \) and \( P \) lies in a hyperplane of \( U \) which does not contain 0, we have \( \dim U = \operatorname{rank}X = n + 1 \), so, \( \operatorname{rank}\bar{X} = v - \operatorname{rank}X = v - n - 1 \) (Lemma 4.5).\n\n\n\n\n\nFigure 11a, b, c, d. (a) Double simplex. (b) Cyclic polytope \( C\left( {6,4}\right) \) . (c) Pyramid over a double simplex. (d) 2-Fold pyramid.\n\nThen, by Carathéodory’s theorem (I, Theorem 2.3), for \( k = \operatorname{rank}\left( {H \cap \bar{X}}\right) \leq \) \( v - n - 2 \), there exist \( k + 1 \leq v - n - 1 \) affinely independent elements \( {\bar{x}}_{1},\ldots ,{\bar{x}}_{k + 1} \in \bar{X} \cap H \) such that \( 0 \in \) relint conv \( \left\{ {{\bar{x}}_{1},\ldots ,{\bar{x}}_{k + 1}}\right\} \) . Therefore, \( \left\{ {{\bar{x}}_{1},\ldots ,{\bar{x}}_{k + 1}}\right\} \) is a coface of \( X \), and the corresponding (proper) face has at least \( v - \left( {v - n - 1}\right) = n + 1 \) vertices and cannot be a simplex.\n\nThe converse is also true. | Yes |
1.16 Lemma. Let \( F,{F}^{\prime } \) be cones in \( {\mathbb{R}}^{n} \) such that \( F \cdot {F}^{\prime } \) is defined. | Proof. (a) For \( x \in F \cdot {F}^{\prime } \), we may set\n\n\[ x = {\lambda }_{1}{y}_{1} + \cdots + {\lambda }_{k}{y}_{k} + {\lambda }_{k + 1}{y}_{k + 1} + \cdots + {\lambda }_{m}{y}_{m} \]\n\nwhere \( {y}_{1},\ldots ,{y}_{k} \in F,{y}_{k + 1},\ldots ,{y}_{m} \in {F}^{\prime } \), and \( {\lambda }_{1},\ldots ,{\lambda }_{m} \geq 0 \) . Then, for \( y \mathrel{\text{:=}} {\lambda }_{1}{y}_{1} + \cdots + {\lambda }_{k}{y}_{k},{y}^{\prime } \mathrel{\text{:=}} {\lambda }_{k + 1}{y}_{k + 1} + \cdots + {\lambda }_{m}{y}_{m} \), we have \( x = y + {y}^{\prime } \) . Hence, \( F \cdot {F}^{\prime } \subset F + {F}^{\prime } \) . The inclusion \( F + {F}^{\prime } \subset F \cdot {F}^{\prime } \) is trivially true, so (a) follows.\n\n(b) By definition, \( F \cdot {F}^{\prime } \) is a cone. Suppose \( \{ 0\} \) is not an apex of \( F \cdot {F}^{\prime } \) . Then, there exists \( x \neq 0 \) such that \( x \in F \cdot {F}^{\prime } \) and \( - x \in F \cdot {F}^{\prime } \) . By (a), we have \( x = y + {y}^{\prime } \) and \( - x = z + {z}^{\prime } \) where \( y, z \in F \) and \( {y}^{\prime },{z}^{\prime } \in {F}^{\prime } \) . Now \( 0 = x - x = y + z + {y}^{\prime } + {z}^{\prime } \) , so that \( y + z = - \left( {{y}^{\prime } + {z}^{\prime }}\right) \) . By the assumption \( \left( {\operatorname{lin}F}\right) \cap \left( {\operatorname{lin}{F}^{\prime }}\right) = \{ 0\} \), we obtain \( y + z = 0 = {y}^{\prime } + {z}^{\prime } \), and, hence, \( z = - y,{z}^{\prime } = - {y}^{\prime } \) . Since 0 is an apex of \( F \) and of \( {F}^{\prime } \), we find \( z = y = {z}^{\prime } = {y}^{\prime } = 0 \) and, therefore, \( x = 0 \), a contradiction.\n\n(c) Since, by (a) and (b), \( F \cdot {F}^{\prime } \) is a cone with apex 0, we can find a hyperplane \( H \) which does not pass through 0 and intersects \( F \cdot {F}^{\prime } \) in a polytope \( \left( {F \cdot {F}^{\prime }}\right) \cap H \neq \varnothing \) . By central projection from 0, we see that \( \left( {F \cdot {F}^{\prime }}\right) \cap H \) and \( \left( {F \cdot {F}^{\prime }}\right) \cap S \) are homeomorphic. It is readily seen that \( \left( {F \cdot {F}^{\prime }}\right) \cap H = \left( {F \cap H}\right) \cdot \left( {{F}^{\prime } \cap H}\right) \), so that (c) follows. | Yes |
1.16 Lemma. Let \( F,{F}^{\prime } \) be cones in \( {\mathbb{R}}^{n} \) such that \( F \cdot {F}^{\prime } \) is defined.\n\n(a) \( F \cdot {F}^{\prime } = F + {F}^{\prime } \) (vector sum).\n\n(b) \( F \cdot {F}^{\prime } \) is a cone with apex 0 .\n\n(c) If \( S \) is the unit sphere of \( {\mathbb{R}}^{n} \), then,\n\n\[ \left( {F \cap S}\right) \cdot \left( {{F}^{\prime } \cap S}\right) = \left( {F \cdot {F}^{\prime }}\right) \cap S \] | Proof. (a) For \( x \in F \cdot {F}^{\prime } \), we may set\n\n\[ x = {\lambda }_{1}{y}_{1} + \cdots + {\lambda }_{k}{y}_{k} + {\lambda }_{k + 1}{y}_{k + 1} + \cdots + {\lambda }_{m}{y}_{m} \]\n\nwhere \( {y}_{1},\ldots ,{y}_{k} \in F,{y}_{k + 1},\ldots ,{y}_{m} \in {F}^{\prime } \), and \( {\lambda }_{1},\ldots ,{\lambda }_{m} \geq 0 \) . Then, for \( y \mathrel{\text{:=}} {\lambda }_{1}{y}_{1} + \cdots + {\lambda }_{k}{y}_{k},{y}^{\prime } \mathrel{\text{:=}} {\lambda }_{k + 1}{y}_{k + 1} + \cdots + {\lambda }_{m}{y}_{m} \), we have \( x = y + {y}^{\prime } \) . Hence, \( F \cdot {F}^{\prime } \subset F + {F}^{\prime } \) . The inclusion \( F + {F}^{\prime } \subset F \cdot {F}^{\prime } \) is trivially true, so (a) follows.\n\n(b) By definition, \( F \cdot {F}^{\prime } \) is a cone. Suppose \( \{ 0\} \) is not an apex of \( F \cdot {F}^{\prime } \) . Then, there exists \( x \neq 0 \) such that \( x \in F \cdot {F}^{\prime } \) and \( - x \in F \cdot {F}^{\prime } \) . By (a), we have \( x = y + {y}^{\prime } \) and \( - x = z + {z}^{\prime } \) where \( y, z \in F \) and \( {y}^{\prime },{z}^{\prime } \in {F}^{\prime } \) . Now \( 0 = x - x = y + z + {y}^{\prime } + {z}^{\prime } \) , so that \( y + z = - \left( {{y}^{\prime } + {z}^{\prime }}\right) \) . By the assumption \( \left( {\operatorname{lin}F}\right) \cap \left( {\operatorname{lin}{F}^{\prime }}\right) = \{ 0\} \), we obtain \( y + z = 0 = {y}^{\prime } + {z}^{\prime } \), and, hence, \( z = - y,{z}^{\prime } = - {y}^{\prime } \) . Since 0 is an apex of \( F \) and of \( {F}^{\prime } \), we find \( z = y = {z}^{\prime } = {y}^{\prime } = 0 \) and, therefore, \( x = 0 \), a contradiction.\n\n(c) Since, by (a) and (b), \( F \cdot {F}^{\prime } \) is a cone with apex 0, we can find a hyperplane \( H \) which does not pass through 0 and intersects \( F \cdot {F}^{\prime } \) in a polytope \( \left( {F \cdot {F}^{\prime }}\right) \cap H \neq \varnothing \) . By central projection from 0, we see that \( \left( {F \cdot {F}^{\prime }}\right) \cap H \) and \( \left( {F \cdot {F}^{\prime }}\right) \cap S \) are homeomorphic. It is readily seen that \( \left( {F \cdot {F}^{\prime }}\right) \cap H = \left( {F \cap H}\right) \cdot \left( {{F}^{\prime } \cap H}\right) \), so that (c) follows. | Yes |
1.16 Lemma. Let \( F,{F}^{\prime } \) be cones in \( {\mathbb{R}}^{n} \) such that \( F \cdot {F}^{\prime } \) is defined.\n\n(a) \( F \cdot {F}^{\prime } = F + {F}^{\prime } \) (vector sum).\n\n(b) \( F \cdot {F}^{\prime } \) is a cone with apex 0 .\n\n(c) If \( S \) is the unit sphere of \( {\mathbb{R}}^{n} \), then,\n\n\[ \left( {F \cap S}\right) \cdot \left( {{F}^{\prime } \cap S}\right) = \left( {F \cdot {F}^{\prime }}\right) \cap S \] | Proof. (a) For \( x \in F \cdot {F}^{\prime } \), we may set\n\n\[ x = {\lambda }_{1}{y}_{1} + \cdots + {\lambda }_{k}{y}_{k} + {\lambda }_{k + 1}{y}_{k + 1} + \cdots + {\lambda }_{m}{y}_{m} \]\n\nwhere \( {y}_{1},\ldots ,{y}_{k} \in F,{y}_{k + 1},\ldots ,{y}_{m} \in {F}^{\prime } \), and \( {\lambda }_{1},\ldots ,{\lambda }_{m} \geq 0 \) . Then, for \( y \mathrel{\text{:=}} {\lambda }_{1}{y}_{1} + \cdots + {\lambda }_{k}{y}_{k},{y}^{\prime } \mathrel{\text{:=}} {\lambda }_{k + 1}{y}_{k + 1} + \cdots + {\lambda }_{m}{y}_{m} \), we have \( x = y + {y}^{\prime } \) . Hence, \( F \cdot {F}^{\prime } \subset F + {F}^{\prime } \) . The inclusion \( F + {F}^{\prime } \subset F \cdot {F}^{\prime } \) is trivially true, so (a) follows.\n\n(b) By definition, \( F \cdot {F}^{\prime } \) is a cone. Suppose \( \{ 0\} \) is not an apex of \( F \cdot {F}^{\prime } \) . Then, there exists \( x \neq 0 \) such that \( x \in F \cdot {F}^{\prime } \) and \( - x \in F \cdot {F}^{\prime } \) . By (a), we have \( x = y + {y}^{\prime } \) and \( - x = z + {z}^{\prime } \) where \( y, z \in F \) and \( {y}^{\prime },{z}^{\prime } \in {F}^{\prime } \) . Now \( 0 = x - x = y + z + {y}^{\prime } + {z}^{\prime } \) , so that \( y + z = - \left( {{y}^{\prime } + {z}^{\prime }}\right) \) . By the assumption \( \left( {\operatorname{lin}F}\right) \cap \left( {\operatorname{lin}{F}^{\prime }}\right) = \{ 0\} \), we obtain \( y + z = 0 = {y}^{\prime } + {z}^{\prime } \), and, hence, \( z = - y,{z}^{\prime } = - {y}^{\prime } \) . Since 0 is an apex of \( F \) and of \( {F}^{\prime } \), we find \( z = y = {z}^{\prime } = {y}^{\prime } = 0 \) and, therefore, \( x = 0 \), a contradiction.\n\n(c) Since, by (a) and (b), \( F \cdot {F}^{\prime } \) is a cone with apex 0, we can find a hyperplane \( H \) which does not pass through 0 and intersects \( F \cdot {F}^{\prime } \) in a polytope \( \left( {F \cdot {F}^{\prime }}\right) \cap H \neq \varnothing \) . By central projection from 0, we see that \( \left( {F \cdot {F}^{\prime }}\right) \cap H \) and \( \left( {F \cdot {F}^{\prime }}\right) \cap S \) are homeomorphic. It is readily seen that \( \left( {F \cdot {F}^{\prime }}\right) \cap H = \left( {F \cap H}\right) \cdot \left( {{F}^{\prime } \cap H}\right) \), so that (c) follows. | Yes |
If \( F \) is a point \( p \), then, \( s\left( {p;\{ p\} }\right) \mathcal{C} \) and \( \mathcal{C} \) have the same 0 -cells, though, in general, \( s\left( {p;\{ p\} }\right) \mathcal{C} \neq \mathcal{C} \). | Null | No |
If \( F \) is a cell of maximal dimension in \( \mathcal{C} \), then, \( \operatorname{st}\left( {F,\mathcal{C}}\right) = \{ F\} \) and \( s\left( {p;F}\right) \mathcal{C} \) splits only \( F \) and leaves all other cells of \( \mathcal{C} \) unchanged (Figure 2). | Null | No |
Let \( P \) be a polytope and \( s\left( {p;F}\right) \mathcal{B}\left( P\right) \) a stellar subdivision of its boundary complex. Then, there exists a polytope \( {P}^{\prime } \) such that \[ s\left( {p;F}\right) \mathcal{B}\left( P\right) \approx \mathcal{B}\left( {P}^{\prime }\right) \] | We may assume \( \dim P = n \) . For \( \dim F = n - 1 \), we obtain a \( {P}^{\prime } \) by placing a sufficiently \ | No |
2.2 Theorem. Let \( P \) be a polytope and \( s\left( {p;F}\right) \mathcal{B}\left( P\right) \) a stellar subdivision of its boundary complex. Then, there exists a polytope \( {P}^{\prime } \) such that \n\n\[ \n s\left( {p;F}\right) \mathcal{B}\left( P\right) \approx \mathcal{B}\left( {P}^{\prime }\right) \n\] | Proof. We may assume \( \dim P = n \) . For \( \dim F = n - 1 \), we obtain a \( {P}^{\prime } \) by placing a sufficiently \ | No |
2.6 Theorem. Let \( \mathcal{C} \) be a cell complex and let \( {Z}_{1},\ldots ,{Z}_{r} \) be the 1-cells (in case \( \mathcal{C} \) is a complex of cones) or 0-cells otherwise (in some order). For \( {v}_{i} \in \) relint \( {Z}_{i}\left( \left\{ {v}_{i}\} = {Z}_{i}\text{in case of 0 -cells}), i = 1,\ldots, r\text{, we set}{\mathcal{C}}_{0} \mathrel{\text{:=}} \mathcal{C}\text{and}\right\} \right. \) \( {\mathcal{C}}_{i} = s\left( {{v}_{i};{Z}_{i}}\right) {\mathcal{C}}_{i - 1}, i = 1,\ldots, r \) . Then, \( {\mathcal{C}}_{r} \) is a simplicial complex which has the same 1-cells (in case of a complex of cones) or 0-cells (in case of a compact complex) as \( \mathcal{C} \) has. Furthermore, \( \left| {\mathcal{C}}_{r}\right| = \left| \mathcal{C}\right| \) . | Proof. By definition, \( s\left( {{v}_{i};{Z}_{i}}\right) \) does not add a 1-cell (in case of a cone complex) or a 0-cell (otherwise) to \( {\mathcal{C}}_{i - 1}, i = 1,\ldots, r \) . Suppose \( F \in {\mathcal{C}}_{r} \) is not a simplex cone or a simplex. Then, there exist \( {Z}_{j} \subset F,{Z}_{k} \subset F \) such that \( \operatorname{conv}\left( {{Z}_{j} \cup {Z}_{k}}\right) \) is not a cell of \( {\mathcal{C}}_{r} \) . But \( s\left( {{v}_{j};{Z}_{j}}\right) {\mathcal{C}}_{j - 1} \) contains \( \operatorname{conv}\left( {{Z}_{j} \cup {Z}_{k}}\right) \) (by definition), a contradiction. | Yes |
2.6 Theorem. Let \( \mathcal{C} \) be a cell complex and let \( {Z}_{1},\ldots ,{Z}_{r} \) be the 1-cells (in case \( \mathcal{C} \) is a complex of cones) or 0-cells otherwise (in some order). For \( {v}_{i} \in \) relint \( {Z}_{i}\left( \left\{ {v}_{i}\} = {Z}_{i}\text{in case of 0 -cells}), i = 1,\ldots, r\text{, we set}{\mathcal{C}}_{0} \mathrel{\text{:=}} \mathcal{C}\text{and}\right\} \right. \) \( {\mathcal{C}}_{i} = s\left( {{v}_{i};{Z}_{i}}\right) {\mathcal{C}}_{i - 1}, i = 1,\ldots, r \) . Then, \( {\mathcal{C}}_{r} \) is a simplicial complex which has the same 1-cells (in case of a complex of cones) or 0-cells (in case of a compact complex) as \( \mathcal{C} \) has. Furthermore, \( \left| {\mathcal{C}}_{r}\right| = \left| \mathcal{C}\right| \) . | Proof. By definition, \( s\left( {{v}_{i};{Z}_{i}}\right) \) does not add a 1-cell (in case of a cone complex) or a 0-cell (otherwise) to \( {\mathcal{C}}_{i - 1}, i = 1,\ldots, r \) . Suppose \( F \in {\mathcal{C}}_{r} \) is not a simplex cone or a simplex. Then, there exist \( {Z}_{j} \subset F,{Z}_{k} \subset F \) such that \( \operatorname{conv}\left( {{Z}_{j} \cup {Z}_{k}}\right) \) is not a cell of \( {\mathcal{C}}_{r} \) . But \( s\left( {{v}_{j};{Z}_{j}}\right) {\mathcal{C}}_{j - 1} \) contains \( \operatorname{conv}\left( {{Z}_{j} \cup {Z}_{k}}\right) \) (by definition), a contradiction. | No |
In \( {\mathbb{R}}^{2} \), let \( K \mathrel{\text{:=}} \operatorname{conv}\left\{ {2{e}_{1}, - 2{e}_{1},2{e}_{2}}\right\}, L \mathrel{\text{:=}} \operatorname{conv}\left\{ {{e}_{1}, - {e}_{1},{e}_{1} + }\right. \) \( \left. {2{e}_{2}, - {e}_{1} + 2{e}_{2}}\right\} \) . As is seen from Figure \( 4, d\left( {K, L}\right) = 1 \) . | Proof of Theorem 2.3.\n\n(1) and (3) are true by definition of \( d \) .\n\n(2) If \( d\left( {K, L}\right) = 0, K + 0 \cdot B \supset L \), and \( L + 0 \cdot B \supset K \) ; hence, \( K = L \) . Clearly, \( d\left( {K, K}\right) = 0 \) .\n\n(4) We set \( r \mathrel{\text{:=}} d\left( {K, M}\right), s \mathrel{\text{:=}} d\left( {M, L}\right) \), and \( t \mathrel{\text{:=}} d\left( {K, L}\right) \) . Then, \( K + {rB} \supset M \) , \( M + {rB} \supset K \) and \( M + {sB} \supset L, L + {sB} \supset M \) imply\n\n\[ K + \left( {r + s}\right) B \supset M + {sB} \supset L, \]\n\n\[ L + \left( {r + s}\right) B \supset M + {rB} \supset K, \]\n\nhence, \( r + s \geq t \) . | Yes |
Example 1. For planar convex bodies \( {K}_{1},{K}_{2} \) we have \( V\left( {{K}_{1} + {K}_{2}}\right) = V\left( {K}_{1}\right) + \) \( V\left( {K}_{2}\right) + {2V}\left( {{K}_{1},{K}_{2}}\right) \) . | Null | No |
Let \( S \mathrel{\text{:=}} \operatorname{conv}\left\{ {0,{e}_{1},{e}_{2},{e}_{3}}\right\} \) be a simplex in \( {\mathbb{R}}^{3} \), and \( {I}_{1} \mathrel{\text{:=}} \left\lbrack {0,{e}_{1}}\right\rbrack \) , \( {I}_{2} \mathrel{\text{:=}} \left\lbrack {0,{e}_{2}}\right\rbrack \) line segments. We wish to calculate \( V\left( {S,{I}_{1},{I}_{2}}\right) \) . | By Theorem 3.7,\n\n\[ \n{6V}\left( {S,{I}_{1},{I}_{2}}\right) = V\left( {S + {I}_{1} + {I}_{2}}\right) - V\left( {S + {I}_{1}}\right) - V\left( {S + {I}_{2}}\right) \n\]\n\n\[ \n- V\left( {{I}_{1} + {I}_{2}}\right) + V\left( S\right) + V\left( {I}_{1}\right) + V\left( {I}_{2}\right) . \n\]\n\nClearly \( V\left( {I}_{1}\right) = V\left( {I}_{2}\right) = V\left( {{I}_{1} + {I}_{2}}\right) = 0 \) since \( V\left( \cdot \right) \) denotes three-dimensional volume. For reasons of symmetry, \( V\left( {S + {I}_{1}}\right) = V\left( {S + {I}_{2}}\right) \) . We know \( V\left( S\right) = \frac{1}{6} \), so we must only calculate \( V\left( {S + {I}_{1}}\right) \) and \( V\left( {S + {I}_{1} + {I}_{2}}\right) \) . We obtain \( S + {I}_{1} \) from \( S \) by joining a triangular prism to it (see Figure 10) whose volume is evidently \( \frac{1}{2} \) . Therefore, \( V\left( {S + {I}_{1}}\right) = V\left( {S + {I}_{2}}\right) = \frac{2}{3} \). From Figure 10, it is readily seen that \( V\left( {S + {I}_{1} + {I}_{2}}\right) = 2\frac{1}{6} \). Therefore \( {6V}\left( {S,{I}_{1},{I}_{2}}\right) = 2\frac{1}{6} - \frac{4}{3} + \frac{1}{6} = 1 \), hence, \( V\left( {S,{I}_{1},{I}_{2}}\right) = \frac{1}{6} \). | Yes |
Example 1. Let \( P \) be a triangular prism with vertices \( 0,{e}_{1},{e}_{2},{e}_{3},{e}_{1} + {e}_{3},{e}_{2} + {e}_{3} \) . It is readily seen that \( {3V}\left( {B, P, P}\right) = 3 + \sqrt{2} \) and \( {3V}\left( {B, B, P}\right) = \left( {2 + \frac{1}{2}\sqrt{2}}\right) \pi \) . | Null | No |
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