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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 \( {\alph... | 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| {\operatorna... | 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... | 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 \le... | 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}}... | 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 ... | 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 i... | 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... | 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 fundamen... | 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/\... | 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 \( \mat... | 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 = \operato... | 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) = {... | 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... | 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 \(... | 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 = ... | 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... | 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} \su... | 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 \( F \) ... | 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 \( \left\lbrack {L : F}\right\rbrack < \infty \) . The... | 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 \( \matho... | 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 \( p \),... | 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 {... | 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\... | 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 \( \mathbb{Q}\left( {\omega \sqrt[3]{5}}\right) \) ar... | 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, \( {F}_{a} \)... | 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 G... | 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}... | 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 ... | 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... | 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... | 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} \... | 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( {... | 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}... | 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 transcend... | 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}... | 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} + \... | 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 quo... | 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\lbrac... | 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 \... | 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 ... | 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 \( \mathca... | 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 \i... | 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\lef... | 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... | 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}... | 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}... | 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 \( \sigma \) induce... | Yes |
Proposition 20.2 Let \( K \) and \( L \) be field extensions of a field \( F \) . Then \( K \) and \( L \) are linearly disjoint over \( F \) if and only if the map \( \varphi : K{ \otimes }_{F}L \rightarrow K\left\lbrack L\right\rbrack \) given on generators by \( a \otimes b \mapsto {ab} \) is an isomorphism of \( F ... | 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 t... | 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 \( \tau = i \circ \varphi \), whe... | 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 isomor... | 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 \( L \) ... | Yes |
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... | 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 } ... | 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... | 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 \( F \) -linea... | 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 varia... | 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 ... | 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}}{... | 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\rig... | 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. ... | 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 f... | 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 \(... | 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 field... | 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 ... | 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 }_... | 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 p... | 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 \( T \) a... | 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\l... | 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 topolog... | 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 ... | 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 \su... | 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\... | 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 th... | 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, \( g \in \left( {{p}_{1}\cdots {p}_{t}}\right) \) . Fo... | 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 ... | 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 th... | 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}}\rig... | 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},... | 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 \( k\left( t\right) \). | Yes |
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 \(... | 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... | 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... | 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\lbr... | 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 ... | 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... | 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}^{* * } = \lef... | Yes |
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 }}\... | 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 mo... | 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... | 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 = \... | 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 ... | 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... | 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... | 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 indep... | \[ {\left( {a}_{1}{a}_{2}\right) }^{t} = \left( \begin{array}{rrrrrr} - 1 & 1 & 0 & 1 & - 1 & 0 \\ - 1 & 0 & 1 & 1 & 0 & - 1 \end{array}\right) . \] 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 \( \op... | 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 \l... | 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}\ba... | 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... | 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( ... | 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}\ba... | 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... | 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 ,{\la... | 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,... | 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 ,{\la... | 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,... | 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 ,{\la... | 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{, w... | 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 \( {Z}_{j} \subset F,{Z}_{k} \subset F \) s... | 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{, w... | 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 \( {Z}_{j} \subset F,{Z}_{k} \subset F \) s... | Yes |
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... | 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{... | 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... | 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}... | 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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