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Example 8.3 Let \( F = \mathbb{Q} \) and \( K = \mathbb{Q}\left( \sqrt[3]{2}\right) \) . We will determine the norm and trace of \( \sqrt[3]{2} \) . An \( F \) -basis for \( K \) is \( \{ 1,\sqrt[3]{2},\sqrt[3]{4}\} \) . We can check that \( {L}_{\sqrt[3]{2}}\left( 1\right) = \sqrt[3]{2},{L}_{\sqrt[3]{2}}\left( \sqrt[3... | \[ \left( \begin{array}{lll} 0 & 0 & 2 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right) ,\] so \( {N}_{K/F}\left( \sqrt[3]{2}\right) = 2 \) and \( {T}_{K/F}\left( \sqrt[3]{2}\right) = 0 \) . | Yes |
Lemma 8.5 Let \( K \) be a finite extension of \( F \) with \( n = \left\lbrack {K : F}\right\rbrack \) .\n\n1. If \( a \in K \), then \( {N}_{K/F}\left( a\right) \) and \( {T}_{K/F}\left( a\right) \) lie in \( F \) .\n\n2. The trace map \( {T}_{K/F} \) is an \( F \) -linear transformation.\n\n3. If \( \alpha \in F \),... | Proof. These properties all follow immediately from the definitions and properties of the determinant and trace functions. | No |
Proposition 8.6 Let \( K \) be an extension of \( F \) with \( \left\lbrack {K : F}\right\rbrack = n \) . If \( a \in K \) and \( p\left( x\right) = {x}^{m} + {\alpha }_{m - 1}{x}^{m - 1} + \cdots + {\alpha }_{1}x + {\alpha }_{0} \) is the minimal polynomial of a over \( F \), then \( {N}_{K/F}\left( a\right) = {\left(... | Proof. Let \( \varphi : K \rightarrow {\operatorname{End}}_{F}\left( K\right) \) be the map \( \varphi \left( a\right) = {L}_{a} \) . It is easy to see that \( {L}_{a + b} = {L}_{a} + {L}_{b} \) and \( {L}_{ab} = {L}_{a} \circ {L}_{b} \), so \( \varphi \) is a ring homomorphism. Also, if \( \alpha \in F \) and \( a \in... | Yes |
If \( F \) is any field and if \( K = F\left( \sqrt{d}\right) \) for some \( d \in F - {F}^{2} \), then a short calculation shows that the minimal polynomial of \( a + b\sqrt{d} \) is \( {x}^{2} - {2ax} + \left( {{a}^{2} - {b}^{2}d}\right) \). | Proposition 8.6 yields \( {N}_{K/F}\left( {a + b\sqrt{d}}\right) = {a}^{2} - {b}^{2}d \) and \( {T}_{K/F}\left( {a + b\sqrt{d}}\right) = {2a} \), as we had obtained before. | Yes |
Lemma 8.9 Let \( K \) be a finite extension of \( F \), and let \( S \) be the separable closure of \( F \) in \( K \). Then \( \left\lbrack {S : F}\right\rbrack \) is equal to the number of \( F \)-homomorphisms from \( K \) to an algebraic closure of \( F \). | Proof. Let \( M \) be an algebraic closure of \( F \). We may assume that \( K \subseteq M \). If \( S \) is the separable closure of \( F \) in \( K \), then \( S = F\left( a\right) \) for some \( a \) by the primitive element theorem. If \( r = \left\lbrack {S : F}\right\rbrack \), then there are \( r \) distinct roo... | Yes |
Lemma 8.10 Let \( K \) be a finite dimensional, purely inseparable extension of \( F \). If \( a \in K \), then \( {a}^{\left\lbrack K : F\right\rbrack } \in F \). More generally, if \( N \) is a finite dimensional, Galois extension of \( F \) and if \( a \in {NK} \), then \( {a}^{\left\lbrack K : F\right\rbrack } \in ... | Proof. Let \( K \) be purely inseparable over \( F \), and let \( n = \left\lbrack {K : F}\right\rbrack \). If \( a \in K \), then \( {a}^{\left\lbrack F\left( a\right) : F\right\rbrack } \in F \) by Lemma 4.16. Since \( \left\lbrack {F\left( a\right) : F}\right\rbrack \) divides \( n = \left\lbrack {K : F}\right\rbrac... | Yes |
Lemma 8.11 Suppose that \( F \subseteq L \subseteq K \) are fields with \( \left\lbrack {K : F}\right\rbrack < \infty \) . Then \( {\left\lbrack K : F\right\rbrack }_{i} = {\left\lbrack K : L\right\rbrack }_{i} \cdot {\left\lbrack L : F\right\rbrack }_{i}. | Proof. Let \( {S}_{1} \) be the separable closure of \( F \) in \( L \), let \( {S}_{2} \) be the separable closure of \( L \) in \( K \), and let \( S \) be the separable closure of \( F \) in \( K \) . Since any element of \( K \) that is separable over \( F \) is also separable over \( L \), we see that \( S \subset... | Yes |
Corollary 8.13 If \( K/F \) is Galois with Galois group \( G \), then for all \( a \in K \) , \[ {N}_{K/F}\left( a\right) = \mathop{\prod }\limits_{{\sigma \in G}}\sigma \left( a\right) \;\text{ and }\;{T}_{K/F}\left( a\right) = \mathop{\sum }\limits_{{\sigma \in G}}\sigma \left( a\right) . \] | Example 8.14 Let \( F \) be a field of characteristic not 2, and let \( K = F\left( \sqrt{d}\right) \) for some \( d \in F - {F}^{2} \) . Then \( \operatorname{Gal}\left( {K/F}\right) = \{ \mathrm{{id}},\sigma \} \), where \( \sigma \left( \sqrt{d}\right) = - \sqrt{d} \) . Therefore, \[ {N}_{K/F}\left( {a + b\sqrt{d}}\... | No |
Example 8.14 Let \( F \) be a field of characteristic not 2, and let \( K = F\\left( \\sqrt{d}\\right) \) for some \( d \\in F - {F}^{2} \) . Then \( \\operatorname{Gal}\\left( {K/F}\\right) = \\{ \\mathrm{id},\\sigma \\} \), where \( \\sigma \\left( \\sqrt{d}\\right) = - \\sqrt{d} \) . | Therefore,\n\n\\[ \n{N}_{K/F}\\left( {a + b\\sqrt{d}}\\right) = \\left( {a + b\\sqrt{d}}\\right) \\left( {a - b\\sqrt{d}}\\right) = {a}^{2} - {b}^{2}d, \n\\]\n\n\\[ \n{T}_{K/F}\\left( {a + b\\sqrt{d}}\\right) = \\left( {a + b\\sqrt{d}}\\right) + \\left( {a - b\\sqrt{d}}\\right) = {2a}. \n\\] | No |
Example 8.15 Suppose that \( F \) is a field containing a primitive \( n \) th root of unity \( \omega \), and let \( K \) be an extension of \( F \) of degree \( n \) with \( K = F\left( \alpha \right) \) and \( {\alpha }^{n} = a \in F \) . By the isomorphism extension theorem, there is an automorphism of \( K \) with... | \[ {N}_{K/F}\left( \alpha \right) = {\alpha \sigma }\left( \alpha \right) \cdots {\sigma }^{n - 1}\left( \alpha \right) = \alpha \cdot {\omega \alpha }\cdots {\omega }^{n - 1}\alpha \] \[ = {\omega }^{n\left( {n - 1}\right) /2}{\alpha }^{n} = {\left( -1\right) }^{n}a. \] If \( n \) is odd, then \( n\left( {n - 1}\right... | Yes |
Theorem 8.16 If \( F \subseteq L \subseteq K \) are fields with \( \left\lbrack {K : F}\right\rbrack < \infty \), then\n\n\[ \n{N}_{K/F} = {N}_{L/F} \circ {N}_{K/L}\;\text{ and }\;{T}_{K/F} = {T}_{L/F} \circ {T}_{K/L};\n\]\n\nthat is, \( {N}_{K/F}\left( a\right) = {N}_{L/F}\left( {{N}_{K/L}\left( a\right) }\right) \) a... | Proof. Let \( M \) be an algebraic closure of \( F \), let \( {\sigma }_{1},\ldots ,{\sigma }_{r} \) be the distinct \( F \) -homomorphisms of \( L \) to \( M \), and let \( {\tau }_{1},\ldots ,{\tau }_{s} \) be the distinct \( L \) - homomorphisms of \( K \) to \( M \) . By the isomorphism extension theorem, we can ex... | Yes |
Corollary 8.17 A finite extension \( K/F \) is separable if and only if \( {T}_{K/F} \) is not the zero map; that is, \( K/F \) is separable if and only if there is an \( a \in K \) with \( {T}_{K/F}\left( a\right) \neq 0 \) . | Proof. Suppose that \( K/F \) is not separable. Then \( \operatorname{char}\left( F\right) = p > 0 \) . Let \( S \) be the separable closure of \( F \) in \( K \) . Then \( S \neq K \) and \( K/S \) is a purely inseparable extension. Moreover, \( \left\lbrack {K : S}\right\rbrack = {p}^{t} \) for some \( t \geq 1 \) by... | Yes |
Let \( \omega \) be a primitive fifth root of unity in \( \mathbb{C} \), let \( F = \mathbb{Q}\left( \omega \right) \) , and let \( K = F\left( \sqrt[5]{2}\right) \) . Then \( K \) is the splitting field of \( {x}^{5} - 2 \) over \( F \), so \( K \) is Galois over \( F \) . Also, \( \left\lbrack {F : \mathbb{Q}}\right\... | Let \( \alpha = \sqrt[5]{2} \) . The roots of \( \min \left( {F,\alpha }\right) \) are \( \alpha ,{\omega \alpha },{\omega }^{2}\alpha ,{\omega }^{3}\alpha \), and \( {\omega }^{4}\alpha \) . By the isomorphism extension theorem, there is a \( \sigma \in \operatorname{Gal}\left( {K/F}\right) \) with \( \sigma \left( \a... | Yes |
Lemma 9.4 Let \( F \) be a field containing a primitive nth root of unity \( \omega \) , let \( K/F \) be a cyclic extension of degree \( n \), and let \( \sigma \) be a generator of \( \operatorname{Gal}\left( {K/F}\right) \) . Then there is an \( a \in K \) with \( \omega = \sigma \left( a\right) /a \) . | Proof. The automorphism \( \sigma \) is an \( F \) -linear transformation of \( K \) . We wish to find an \( a \in K \) with \( \sigma \left( a\right) = {\omega a} \) ; that is, we want to show that \( \omega \) is an eigenvalue for \( \sigma \) . To do this, we show that \( \omega \) is a root of the characteristic po... | Yes |
Theorem 9.5 Let \( F \) be a field containing a primitive nth root of unity, and let \( K/F \) be a cyclic Galois extension of degree \( n \) . Then there is an \( a \in K \) with \( K = F\left( a\right) \) and \( {a}^{n} = b \in F \) ; that is, \( K = F\left( \sqrt[n]{b}\right) \) . | Proof. By the lemma, there is an \( a \) with \( \sigma \left( a\right) = {\omega a} \) . Therefore, \( {\sigma }^{i}\left( a\right) = {\omega }^{i}a \) , so \( a \) is fixed by \( {\sigma }^{i} \) only when \( n \) divides \( i \) . Since the order of \( \sigma \) is \( n \), we see that \( a \) is fixed only by id, s... | Yes |
Corollary 9.7 Let \( K/F \) be a cyclic extension of degree \( n \), and suppose that \( F \) contains a primitive nth root of unity. If \( K = F\left( \sqrt[n]{a}\right) \) with \( a \in F \), then any intermediate field of \( K/F \) is of the form \( F\left( \sqrt[m]{a}\right) \) for some divisor \( m \) of \( n \). | Proof. Let \( \sigma \) be a generator for \( \operatorname{Gal}\left( {K/F}\right) \). Then any subgroup of \( \operatorname{Gal}\left( {K/F}\right) \) is of the form \( \left\langle {\sigma }^{t}\right\rangle \) for some divisor \( t \) of \( n \). By the fundamental theorem, the intermediate fields are the fixed fie... | Yes |
Theorem 9.8 Let \( \operatorname{char}\left( F\right) = p \), and let \( K/F \) be a cyclic Galois extension of degree \( p \) . Then \( K = F\left( \alpha \right) \) with \( {\alpha }^{p} - \alpha - a = 0 \) for some \( a \in F \) ; that is, \( K = F\left( {{\wp }^{-1}\left( a\right) }\right) \) . | Proof. Let \( \sigma \) be a generator of \( \operatorname{Gal}\left( {K/F}\right) \), and let \( T \) be the linear transformation \( T = \sigma - \mathrm{{id}} \) . The kernel of \( T \) is\n\n\[ \ker \left( T\right) = \{ b \in K : \sigma \left( b\right) = b\} \]\n\n\[ = F\text{.} \]\n\nAlso, \( {T}^{p} = {\left( \si... | Yes |
Theorem 9.9 Let \( F \) be a field of characteristic \( p \), and let \( a \in F - {\wp }^{-1}\left( F\right) \) . Then \( f\left( x\right) = {x}^{p} - x - a \) is irreducible over \( F \), and the splitting field of \( f \) over \( F \) is a cyclic Galois extension of \( F \) of degree \( p \) . | Proof. Let \( K \) be the splitting field of \( f \) over \( F \) . If \( \alpha \) is a root of \( f \), it is easy to check that \( \alpha + 1 \) is also a root of \( f \) . Hence, the \( p \) roots of \( f \) are \( \alpha ,\alpha + 1,\ldots ,\alpha + p - 1 \) . Therefore, \( K = F\left( \alpha \right) \) . The assu... | Yes |
Example 9.10 Let \( F = {\mathbb{F}}_{p}\left( x\right) \) be the rational function field in one variable over \( {\mathbb{F}}_{p} \). We claim that \( x \notin {\wp }^{-1}\left( F\right) \), so the extension \( F\left( {{\wp }^{-1}\left( x\right) }\right) \) is a cyclic extension of \( F \) of degree \( p \). To prove... | To prove this, suppose instead that \( x \in {\wp }^{-1}\left( F\right) \), so \( x = {a}^{p} - a \) for some \( a \in F \). We can write \( a = f/g \) with \( f, g \in {\mathbb{F}}_{p}\left\lbrack x\right\rbrack \) relatively prime. Then \( x = {f}^{p}/{g}^{p} - f/g \), or \( {g}^{p}x = {f}^{p} - f{g}^{p - 1} \). Solv... | Yes |
Proposition 10.1 Let \( K \) be a Galois extension of \( F \) with Galois group \( G \) , and let \( f : G \rightarrow {K}^{ * } \) be a crossed homomorphism. Then there is an \( a \in K \) with \( f\left( \tau \right) = \tau \left( a\right) /a \) for all \( \sigma \in G \) . | \( \\textbf{Proof. The Dedekind independence lemma shows that }\\mathop{\\sum }\\limits_{{\\sigma \\in G}}f\\left( \\sigma \\right) \\sigma \\left( c\\right) \\neq 0 \\) for some \( c \\in K \\), since each \( f\\left( \\sigma \\right) \\neq 0 \\) . Let \( b = \\mathop{\\sum }\\limits_{{\\sigma \\in G}}f\\left( \\sigma... | Yes |
Theorem 10.2 (Hilbert Theorem 90) Let \( K/F \) be a cyclic Galois extension, and let \( \sigma \) be a generator of \( \operatorname{Gal}\left( {K/F}\right) \). If \( u \in K \), then \( {N}_{K/F}\left( u\right) = 1 \) if and only if \( u = \sigma \left( a\right) /a \) for some \( a \in K \). | Proof. One direction is easy. If \( u = \sigma \left( a\right) /a \), then \( {N}_{K/F}\left( {\sigma \left( a\right) }\right) = {N}_{K/F}\left( a\right) \), so \( N\left( u\right) = 1 \). Conversely, if \( {N}_{K/F}\left( u\right) = 1 \), then define \( f : G \rightarrow {K}^{ * } \) by \( f\left( \mathrm{{id}}\right)... | Yes |
Proposition 10.3 Let \( K/F \) be a Galois extension with Galois group \( G \) , and let \( g : G \rightarrow K \) be a 1-cocycle. Then there is an \( a \in K \) with \( g\left( \tau \right) = \) \( \tau \left( a\right) - a \) for all \( \tau \in G \) . | Proof. Since \( K/F \) is separable, the trace map \( {T}_{K/F} \) is not the zero map. Thus, there is a \( c \in K \) with \( {T}_{K/F}\left( c\right) \neq 0 \) . If \( \alpha = {T}_{K/F}\left( c\right) \), then \( \alpha \in {F}^{ * } \) and \( {T}_{K/F}\left( {{\alpha }^{-1}c}\right) = 1 \) . By replacing \( c \) wi... | Yes |
Theorem 10.5 (Additive Hilbert Theorem 90) Let \( K \) be a cyclic Galois extension of \( F \), and let \( \sigma \) be a generator of \( \operatorname{Gal}\left( {K/F}\right) \) . If \( u \in K \) , then \( {T}_{K/F}\left( u\right) = 0 \) if and only if \( u = \sigma \left( a\right) - a \) for some \( a \in K \) . | Proof. If \( u = \sigma \left( a\right) - a \), then \( {T}_{K/F}\left( u\right) = 0 \) . Conversely, suppose that \( {T}_{K/F}\left( u\right) = 0 \) . Let \( n = \left\lbrack {K : F}\right\rbrack \), and define \( g : G \rightarrow K \) by \( g\left( \mathrm{{id}}\right) = 0, g\left( \sigma \right) = \) \( u \), and f... | Yes |
Theorem 10.5 (Additive Hilbert Theorem 90) Let \( K \) be a cyclic Galois extension of \( F \), and let \( \sigma \) be a generator of \( \operatorname{Gal}\left( {K/F}\right) \) . If \( u \in K \) , then \( {T}_{K/F}\left( u\right) = 0 \) if and only if \( u = \sigma \left( a\right) - a \) for some \( a \in K \) . | Proof. If \( u = \sigma \left( a\right) - a \), then \( {T}_{K/F}\left( u\right) = 0 \) . Conversely, suppose that \( {T}_{K/F}\left( u\right) = 0 \) . Let \( n = \left\lbrack {K : F}\right\rbrack \), and define \( g : G \rightarrow K \) by \( g\left( \mathrm{{id}}\right) = 0, g\left( \sigma \right) = \) \( u \), and f... | Yes |
Let \( {Q}_{8} \) be the quaternion group. Then \( {Q}_{8} = \) \( \{ \pm 1, \pm i, \pm j, \pm k\} \), and the operation on \( {Q}_{8} \) is given by the relations \( {i}^{2} = \) \( {j}^{2} = {k}^{2} = - 1 \) and \( {ij} = k = - {ji} \) . We show that \( {Q}_{8} \) is a group extension of \( M = \langle i\rangle \) by... | First note that \( M \) is an Abelian normal subgroup of \( {Q}_{8} \) and that \( {Q}_{8}/M \cong \mathbb{Z}/2\mathbb{Z} \) . Therefore, \( {Q}_{8} \) is a group extension of \( M \) by \( \mathbb{Z}/2\mathbb{Z} \) . We use 1 and \( j \) as coset representatives of \( M \) in \( {Q}_{8} \) . Our cocycle \( f \) that r... | Yes |
Example 10.11 Let \( \mathbb{H} \) be Hamilton’s quaternions. The ring \( \mathbb{H} \) consists of all symbols \( a + {bi} + {cj} + {dk} \) with \( a, b, c, d \in \mathbb{R} \), and multiplication is given by the relations \( {i}^{2} = {j}^{2} = {k}^{2} = - 1 \) and \( {ij} = k = - {ji} \) . This was the first example... | \[ {x}_{\sigma }\left( {a + {bi}}\right) {x}_{\sigma }^{-1} = j\left( {a + {bi}}\right) {j}^{-1} = a - {bi} = \sigma \left( {a + {bi}}\right) . \] The cocycle \( f \) associated to this algebra is given by \[ f\left( {\mathrm{{id}},\mathrm{{id}}}\right) = {x}_{\mathrm{{id}}}{x}_{\mathrm{{id}}}{x}_{\mathrm{{id}}}^{-1} =... | Yes |
Example 10.12 Let \( K/F \) be a Galois extension of degree \( n \) with Galois group \( G \), and consider the crossed product \( A = \left( {K/F, G,1}\right) \), where 1 represents the trivial cocycle. We will show that \( A \cong {M}_{n}\left( F\right) \), the ring of \( n \times n \) matrices over \( F \) . | First, note that \( A = { \oplus }_{\sigma \in G}K{x}_{\sigma } \), where multiplication on \( A \) is determined by the relations \( {x}_{\sigma }{x}_{\tau } = {x}_{\sigma \tau } \) and \( {x}_{\sigma }a = \sigma \left( a\right) {x}_{\sigma } \) for \( a \in K \) . If \( f = \sum {a}_{\sigma }{x}_{\sigma } \in A \), t... | Yes |
Let \( K = \mathbb{Q}\left( {\sqrt{2},\sqrt{3}}\right) \). The field \( K \) is the splitting field of \( \left( {{x}^{2} - 2}\right) \left( {{x}^{2} - 3}\right) \) over \( \mathbb{Q} \), so \( K \) is a Galois extension of \( \mathbb{Q} \). A short calculation shows that \( \left\lbrack {K : \mathbb{Q}}\right\rbrack =... | \[
\text{id} : \sqrt{2} \rightarrow \sqrt{2},\;\sqrt{3} \rightarrow \sqrt{3}\text{,}
\]
\[
\sigma : \sqrt{2} \rightarrow - \sqrt{2},\;\sqrt{3} \rightarrow \sqrt{3}
\]
\[
\tau : \sqrt{2} \rightarrow \sqrt{2},\;\sqrt{3} \rightarrow - \sqrt{3}
\]
\[
{\sigma \tau } : \sqrt{2} \rightarrow - \sqrt{2},\;\sqrt{3} \rightarro... | Yes |
Lemma 11.8 Let \( B : G \times H \rightarrow C \) be a bilinear pairing. If \( h \in H \), let \( {B}_{h} : G \rightarrow C \) be defined by \( {B}_{h}\left( g\right) = B\left( {g, h}\right) \) . Then the map \( \varphi : h \mapsto {B}_{h} \) is a group homomorphism from \( H \) to \( \hom \left( {G, C}\right) \) . If ... | Proof. The property \( B\left( {g,{h}_{1}{h}_{2}}\right) = B\left( {g,{h}_{1}}\right) B\left( {g,{h}_{2}}\right) \) translates to \( {B}_{{h}_{1}{h}_{2}} = \) \( {B}_{{h}_{1}}{B}_{{h}_{2}} \) . Thus, \( \varphi \left( {{h}_{1}{h}_{2}}\right) = \varphi \left( {h}_{1}\right) \varphi \left( {h}_{2}\right) \), so \( \varph... | Yes |
Proposition 11.9 Let \( K \) be an \( n \) -Kummer extension of \( F \), and let \( B \) : \( \operatorname{Gal}\left( {K/F}\right) \times \operatorname{kum}\left( {K/F}\right) \rightarrow \mu \left( F\right) \) be the associated Kummer pairing. Then \( B \) is nondegenerate. Consequently, \( \operatorname{kum}\left( {... | Proof. First, we show that \( B \) is a bilinear pairing. Let \( \sigma ,\tau \in \operatorname{Gal}\left( {K/F}\right) \) and \( \alpha {F}^{ * } \in \operatorname{kum}\left( {K/F}\right) \) . Then\n\n\[ B\left( {{\sigma \tau },\alpha {F}^{ * }}\right) = \frac{{\sigma \tau }\left( \alpha \right) }{\alpha } = \frac{\si... | Yes |
Proposition 11.10 Let \( K/F \) be an \( n \) -Kummer extension. Then there is an injective group homomorphism \( f : \operatorname{kum}\left( {K/F}\right) \rightarrow {F}^{ * }/{F}^{*n} \), given by \( f\left( {\alpha {F}^{ * }}\right) = {\alpha }^{n}{F}^{*n} \) . The image of \( f \) is then a finite subgroup of \( {... | Proof. It is easy to see that \( f \) is well defined and that \( f \) preserves multiplication. For injectivity, let \( \alpha {F}^{ * } \in \ker \left( f\right) \) . Then \( {\alpha }^{n} \in {F}^{*n} \), so \( {\alpha }^{n} = {a}^{n} \) for some \( a \in F \) . Hence, \( \alpha /a \) is an \( n \) th root of unity, ... | Yes |
Example 11.11 Let \( F = \mathbb{C}\left( {x, y, z}\right) \) be the rational function field in three variables over \( \mathbb{C} \), and let \( K = F\left( {\sqrt[4]{xyz},\sqrt[4]{{y}^{2}z},\sqrt[4]{x{z}^{2}}}\right) \) . Then \( K/F \) is a 4- Kummer extension. The image of \( \operatorname{kum}\left( {K/F}\right) \... | The subgroup \( \langle a, b\rangle \) of \( {F}^{ * }/{F}^{*4} \) generated by \( a \) and \( b \) has order 16, since the 16 elements \( {a}^{i}{b}^{j} \) with \( 1 \leq i, j \leq 4 \) are all distinct. To see this, suppose that \( {a}^{i}{b}^{j} = {a}^{k}{b}^{l} \) . Then there is an \( h \in {F}^{ * } \) with\n\n\[... | Yes |
Lemma 12.3 Let \( F \) be a field with \( \operatorname{char}\left( F\right) \neq 2 \), let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) be an irreducible, separable polynomial, and let \( K \) be the splitting field of \( f\left( x\right) \) over \( F \) . If \( \Delta \) is defined as in Definition 12.2, t... | Proof. Before we prove this, we note that the proof we give is the same as the typical proof that every permutation of \( {S}_{n} \) is either even or odd. In fact, the proof of this result about \( {S}_{n} \) is really about discriminants. It is easy to see that each \( \sigma \in G = \operatorname{Gal}\left( {K/F}\ri... | Yes |
Corollary 12.4 Let \( F, K \), and \( f \) be as in Lemma 12.3, and let \( G = \) \( \operatorname{Gal}\left( {K/F}\right) \) . Then \( G \subseteq {A}_{n} \) if and only if \( \operatorname{disc}\left( f\right) \in {F}^{2} \) . Under the correspondence of the fundamental theorem, the field \( F\left( \Delta \right) \s... | Proof. This follows from the lemma, since \( G \subseteq {A}_{n} \) if and only if each \( \sigma \in G \) is even, and this occurs if and only if \( \sigma \left( \Delta \right) = \Delta \) . Therefore, \( G \subseteq {A}_{n} \) if and only if \( \operatorname{disc}\left( f\right) \in {F}^{2} \) . | Yes |
If \( K \) is a field and \( {\alpha }_{1},\ldots ,{\alpha }_{n} \in K \), then the determinant of the Vandermonde matrix \( V\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) is \( \mathop{\prod }\limits_{{i < j}}\left( {{\alpha }_{j} - {\alpha }_{i}}\right) \). | Let \( A = V\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) . That \( \det \left( A\right) = \mathop{\prod }\limits_{{i < j}}\left( {{\alpha }_{j} - {\alpha }_{i}}\right) \) is a moderately standard fact from linear algebra. For those who have not seen this, we give a proof. Note that if \( {\alpha }_{i} = {\alp... | Yes |
Proposition 12.6 (Newton’s Identities) Let \( f\left( x\right) = {a}_{0} + {a}_{1}x + \cdots + \) \( {a}_{n - 1}{x}^{n - 1} + {x}^{n} \) be a monic polynomial over \( F \) with roots \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) . If \( {t}_{i} = \mathop{\sum }\limits_{j}{\alpha }_{j}^{i}, \) then\n\n\[ \left. \begin{matri... | Proof. An alternative way of stating Newton's identities is to use the elementary symmetric functions \( {s}_{i} \) in the \( {a}_{i} \), instead of the \( {a}_{i} \) . Since \( {s}_{i} = \) \( {\left( -1\right) }^{i}{a}_{n - i} \), Newton’s identities can also be written as\n\n\[ {t}_{m} - {s}_{1}{t}_{m - 1} + {s}_{2}... | Yes |
Example 12.7 Let \( f\left( x\right) = {x}^{2} + {bx} + c \) . Then \( {t}_{0} = 2 \) . Also, Newton’s identities yield \( {t}_{1} + b = 0 \), so \( {t}_{1} = - b \) . For \( {t}_{2} \), we have \( {t}_{2} + b{t}_{1} + {2c} = 0 \), so \( {t}_{2} = - b{t}_{1} - {2c} = {b}^{2} - {2c} \) . | Therefore,\n\n\[\n\operatorname{disc}\left( f\right) = \left| \begin{matrix} 2 & - b \\ - b & {b}^{2} - {2c} \end{matrix}\right| = 2\left( {{b}^{2} - {2c}}\right) - {b}^{2} = {b}^{2} - {4c}\n\]\n\nthe usual discriminant of a monic quadratic. | Yes |
Proposition 12.9 Let \( L = F\left( \alpha \right) \) be a field extension of \( F \) . If \( f\left( x\right) = \) \( \min \left( {F,\alpha }\right) \), then \( \operatorname{disc}\left( f\right) = {\left( -1\right) }^{n\left( {n - 1}\right) /2}{N}_{L/F}\left( {{f}^{\prime }\left( \alpha \right) }\right) \), where \( ... | Proof. Let \( K \) be a splitting field for \( f \) over \( F \), and write \( f\left( x\right) = (x - \) \( \left. {\alpha }_{1}\right) \cdots \left( {x - {\alpha }_{n}}\right) \in K\left\lbrack x\right\rbrack \) . Set \( \alpha = {\alpha }_{1} \) . Then a short calculation shows that \( {f}^{\prime }\left( {\alpha }_... | Yes |
Example 12.10 Let \( p \) be an odd prime, and let \( \omega \) be a primitive \( p \) th root of unity in \( \mathbb{C} \) . We use the previous result to determine \( \operatorname{disc}\left( \omega \right) \) . Let \( K = \) \( \mathbb{Q}\left( \omega \right) \), the \( p \) th cyclotomic extension of \( \mathbb{Q}... | First,\n\n\[ \n{f}^{\prime }\left( x\right) = \frac{p{x}^{p - 1}\left( {x - 1}\right) - \left( {{x}^{p} - 1}\right) }{{\left( x - 1\right) }^{2}} \n\]\n\nso \( {f}^{\prime }\left( \omega \right) = p{\omega }^{p - 1}/\left( {\omega - 1}\right) \) . We claim that \( {N}_{K/\mathbb{Q}}\left( \omega \right) = 1 \) and \( {... | Yes |
Lemma 12.12 Let \( K \) be a separable field extension of \( F \) of degree \( n \), and let \( {\alpha }_{1},\ldots ,{\alpha }_{n} \in K \) . Then \( \operatorname{disc}\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) = \det \left( {{\operatorname{Tr}}_{K/F}\left( {{\alpha }_{i}{\alpha }_{j}}\right) }\right) \) . C... | Proof. Let \( {\sigma }_{1},\ldots ,{\sigma }_{n} \) be the distinct \( F \) -homomorphisms from \( K \) to an algebraic closure of \( F \) . If \( A = \left( {{\sigma }_{i}\left( {\alpha }_{j}\right) }\right) \), then the discriminant of the \( n \) - tuple \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) is the determinant ... | Yes |
Proposition 12.13 Let \( K \) be a separable field extension of \( F \) of degree \( n \) , and let \( {\alpha }_{1},\ldots ,{\alpha }_{n} \in K \) . Then \( \operatorname{disc}\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) = 0 \) if and only if \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) are linearly dependent over... | Proof. Suppose that the \( {\alpha }_{i} \) are linearly dependent over \( F \) . Then one of the \( {\alpha }_{i} \) is an \( F \) -linear combination of the others. If \( {\alpha }_{i} = \mathop{\sum }\limits_{{k \neq i}}{a}_{k}{\alpha }_{k} \) with \( {a}_{j} \in F \), then\n\n\[ \n{\operatorname{Tr}}_{K/F}\left( {{... | Yes |
Proposition 12.14 Let \( \left\{ {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right\} \) and \( \left\{ {{\beta }_{1},\ldots ,{\beta }_{n}}\right\} \) be two \( F \) -bases for \( K \) . Let \( A = \left( {a}_{ij}\right) \) be the \( n \times n \) transition matrix between the two bases; that \( {is},\;{\beta }_{j} = \mathop{... | Proof. Since \( {\beta }_{j} = \mathop{\sum }\limits_{k}{a}_{kj}{\alpha }_{k} \), we have \( {\sigma }_{i}\left( {\beta }_{j}\right) = \mathop{\sum }\limits_{k}{a}_{kj}{\sigma }_{i}\left( {\alpha }_{k}\right) \) . In terms of matrices, this says that\n\n\[ \left( {{\sigma }_{i}\left( {\beta }_{j}\right) }\right) = {\le... | Yes |
In this example, we show that the discriminant of a polynomial is equal to the discriminant of an appropriate field extension. Suppose that \( K = F\left( \alpha \right) \) is an extension of \( F \) of degree \( n \) . Then \( 1,\alpha \) , \( {\alpha }^{2},\ldots ,{\alpha }^{n - 1} \) is a basis for \( K \) . We calc... | We have \( \operatorname{disc}\left( {K/F}\right) = \det {\left( {\sigma }_{i}\left( {\alpha }^{j - 1}\right) \right) }^{2} \) . Consequently, if \( {\alpha }_{i} = {\sigma }_{i}\left( \alpha \right) \) , then\n\n\[ \operatorname{disc}\left( {K/F}\right) = \det {\left( \begin{matrix} 1 & {\sigma }_{1}\left( \alpha \rig... | Yes |
Example 12.16 Let \( K = \mathbb{Q}\left( \sqrt{-1}\right) \). If \( i = \sqrt{-1} \), then using the basis \( 1, i \) of \( K/\mathbb{Q} \), we get | \[ \operatorname{disc}\left( {\mathbb{Q}\left( i\right) /\mathbb{Q}}\right) = \det {\left( \begin{matrix} 1 & i \\ 1 & - i \end{matrix}\right) }^{2} = {\left( -2i\right) }^{2} = - 4. \] | Yes |
We now show that the discriminant of a field extension is the discriminant of the trace form. Let \( K \) be a finite separable extension of \( F \) . Let \( B : K \times K \rightarrow F \) be defined by \( B\left( {a, b}\right) = {T}_{K/F}\left( {ab}\right) \) . Then \( B \) is a bilinear form because the trace is lin... | But, by Lemma 12.12, this is the discriminant of \( K/F \) . Therefore, the previous notions of discriminant are special cases of the notion of discriminant of a bilinear form. | Yes |
Theorem 13.1 Let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) be an irreducible, separable polynomial of degree 3 over \( F \), and let \( K \) be the splitting field of \( f \) over \( F \) . If \( D \) is the discriminant of \( f \), then \( \operatorname{Gal}\left( {K/F}\right) \cong {S}_{3} \) if and onl... | Proof. Let \( G = \operatorname{Gal}\left( {K/F}\right) \) . By Corollary \( {12.4}, G \subseteq {A}_{3} \) if and only if \( D \in {F}^{2} \) . But \( G \cong {S}_{3} \) or \( G \cong {A}_{3} \), so \( G \cong {S}_{3} \) if and only if \( D \) is a square in \( F \) . | Yes |
Consider \( {x}^{3} - {3x} + 1 \) . Then \( \Gamma = - D/{108} = - {81}/{108} = \) \( - 3/4 \) . We have \( p = - 3 \) and \( q = 1 \) . Then \( A = - 1/2 + i\sqrt{3}/2 \) and \( B = - 1/2 - i\sqrt{3}/2 \), so \( A = \exp \left( {{2\pi i}/3}\right) \) and \( B = \exp \left( {-{2\pi i}/3}\right) \) . We can then set \( ... | Suppose that the polynomial \( f\left( x\right) = {x}^{3} + {px} + q \) has real coefficients. If \( \Gamma > 0 \), then \( D < 0 \), so \( D \) is not a square in \( F \) . We can then take the real cube roots of \( A \) and \( B \) for \( u \) and \( v \) . Furthermore, if \( \omega = \left( {-1 + i\sqrt{3}}\right) /... | Yes |
Theorem 13.4 With the notation above, let \( m = \left\lbrack {L : F}\right\rbrack \) . 1. \( G \cong {S}_{4} \) if and only if \( r\left( x\right) \) is irreducible over \( F \) and \( D \notin {F}^{2} \), if and only if \( m = 6 \) . | Proof. We first point out a couple of things. First, \( \left\lbrack {K : L}\right\rbrack \leq 4 \), since \( K = L\left( {\alpha }_{1}\right) \) . This equality follows from the fundamental theorem, since only the identity automorphism fixes \( L\left( {\alpha }_{1}\right) \) . Second, \( r\left( x\right) \) is irredu... | Yes |
Example 13.5 Let \( f\left( x\right) = {x}^{4} + {x}^{3} + {x}^{2} + x + 1 \) . Then \( a = b = c = d = 1 \) , so \( {s}_{1} = {s}_{3} = - 1 \) and \( {s}_{2} = {s}_{4} = 1 \) . Also,\n\n\[ r\left( x\right) = {x}_{3} - {x}_{2} - {3x} + 2 = \left( {x - 2}\right) \left( {{x}^{2} + x - 1}\right) . \] | Set \( {\beta }_{1} = 2 \) . Then \( u = \sqrt{5} \) . Also,\n\n\[ {v}^{2} = \frac{1}{4}{\left( -1 + u\right) }^{2} - 2\left( {2 + {u}^{-1}\left( {-2 + 2}\right) }\right) \]\n\n\[ = \frac{1}{4}\left( {{u}^{2} - {2u} + 1}\right) - 4 = - \frac{5 + u}{2}. \]\n\nThus, \( v = \frac{i}{2}\sqrt{{10} - 2\sqrt{5}} \) . In addit... | Yes |
Theorem 14.1 (Lindemann-Weierstrauss) Let \( {\alpha }_{1},\ldots ,{\alpha }_{m} \) be distinct algebraic numbers. Then the exponentials \( {e}^{{\alpha }_{1}},\ldots ,{e}^{{\alpha }_{m}} \) are linearly independent over \( \mathbb{Q} \) . | Proof of the theorem. Suppose that there are \( {a}_{j} \in \mathbb{Q} \) with\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{m}{a}_{j}{e}^{{\alpha }_{j}} = 0 \]\n\nBy multiplying by a suitable integer, we may assume that each \( {a}_{j} \in \mathbb{Z} \) . Moreover, by eliminating terms if necessary, we may also assume that e... | Yes |
Corollary 14.2 The numbers \( \pi \) and \( e \) are transcendental over \( \mathbb{Q} \) . | Proof of the corollary. Suppose that \( e \) is algebraic over \( \mathbb{Q} \) . Then there are rationals \( {r}_{i} \) with \( \mathop{\sum }\limits_{{i = 0}}^{n}{r}_{i}{e}^{i} = 0 \) . This means that the numbers \( {e}^{0} \) , \( {e}^{1},\ldots ,{e}^{n - 1} \) are linearly dependent over \( \mathbb{Q} \) . By choo... | Yes |
Lemma 15.1 Let \( K \) be a subfield of \( \mathbb{R} \) .\n\n1. The intersection of two lines in \( K \) is either empty or is a point in the plane of \( K \) .\n\n2. The intersection of a line and a circle in \( K \) is either empty or consists of one or two points in the plane of \( K\left( \sqrt{u}\right) \) for so... | Proof. The first statement is an easy calculation. For the remaining two statements, it suffices to prove statement 2, since if \( {x}^{2} + {y}^{2} + {ax} + {by} + c = 0 \) and \( {x}^{2} + {y}^{2} + {a}^{\prime }x + {b}^{\prime }y + {c}^{\prime } = 0 \) are the equations of circles \( C \) and \( {C}^{\prime } \) , r... | Yes |
Theorem 15.2 A real number \( c \) is constructible if and only if there is a tower of fields \( \mathbb{Q} = {K}_{0} \subseteq {K}_{1} \subseteq \cdots \subseteq {K}_{r} \) such that \( c \in {K}_{r} \) and \( \left\lbrack {{K}_{i + 1} : }\right. \) \( \left. {K}_{i}\right\rbrack \leq 2 \) for each \( i \) . Therefore... | Proof. If \( c \) is constructible, then the point \( \left( {c,0}\right) \) can be obtained from a finite sequence of constructions starting from the plane of \( \mathbb{Q} \) . We then obtain a finite sequence of points, each an intersection of constructible lines and circles, ending at \( \left( {c,0}\right) \) . By... | Yes |
Theorem 15.3 It is impossible to trisect a \( {60}^{ \circ } \) angle by ruler and compass construction. | Proof. As noted above, a \( {60}^{ \circ } \) angle can be constructed. If a \( {60}^{ \circ } \) angle can be trisected, then it is possible to construct the number \( \alpha = \cos {20}^{ \circ } \) . However, the triple angle formula \( \cos {3\theta } = 4{\cos }^{3}\theta - 3\cos \theta \) gives \( 4{\alpha }^{3} -... | Yes |
Theorem 15.4 It is impossible to double a cube of length 1 by ruler and compass construction. | Proof. The length of a side of a cube of volume 2 is \( \sqrt[3]{2} \) . The minimal polynomial of \( \sqrt[3]{2} \) over \( \mathbb{Q} \) is \( {x}^{3} - 2 \) . Thus, \( \left\lbrack {\mathbb{Q}\left( \sqrt[3]{2}\right) : \mathbb{Q}}\right\rbrack = 3 \) is not a power of 2, so \( \sqrt[3]{2} \) is not constructible. | Yes |
Theorem 15.5 It is impossible to square a circle of radius 1. | Proof. We are asking whether we can construct a square of area \( \pi \) . To do so requires us to construct a line segment of length \( \sqrt{\pi } \), which is impossible since \( \sqrt{\pi } \) is transcendental over \( \mathbb{Q} \) by the Lindemann-Weierstrauss theorem; hence, \( \sqrt{\pi } \) is not algebraic of... | Yes |
Theorem 15.6 A regular \( n \) -gon is constructible if and only if \( \phi \left( n\right) \) is a power of \( 2 \) . | Proof. We point out that a regular \( n \) -gon is constructible if and only if the central angles \( {2\pi }/n \) are constructible, and this occurs if and only if \( \cos \left( {{2\pi }/n}\right) \) is a constructible number. Let \( \omega = {e}^{{2\pi i}/n} = \cos \left( {{2\pi }/n}\right) + \) \( i\sin \left( {{2\... | Yes |
Lemma 16.6 Let \( K \) be an \( n \)-radical extension of \( F \), and let \( N \) be the normal closure of \( K/F \). Then \( N \) is an \( n \)-radical extension of \( F \). | Proof. Let \( K = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right) \) with \( {\alpha }_{i}^{n} \in F\left( {{\alpha }_{1},\ldots ,{\alpha }_{i - 1}}\right) \). We argue by induction on \( r \). If \( r = 1 \), then \( K = F\left( \alpha \right) \) with \( {\alpha }^{n} = a \in F \). Then \( N = F\left( {{\beta }_{... | Yes |
Example 16.4 If \( K = \mathbb{Q}\left( \sqrt[4]{2}\right) \), then \( K \) is both a 4-radical extension and a 2-radical extension of \( \mathbb{Q} \). | The second statement is true by considering the tower\n\n\[\n\mathbb{Q} \subseteq \mathbb{Q}\left( \sqrt{2}\right) \subseteq \mathbb{Q}\left( \sqrt{2}\right) \left( \sqrt{\sqrt{2}}\right) = \mathbb{Q}\left( \sqrt[4]{2}\right)\n\] | Yes |
Lemma 16.6 Let \( K \) be an \( n \) -radical extension of \( F \), and let \( N \) be the normal closure of \( K/F \) . Then \( N \) is an \( n \) -radical extension of \( F \) . | Proof. Let \( K = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right) \) with \( {\alpha }_{i}^{n} \in F\left( {{\alpha }_{1},\ldots ,{\alpha }_{i - 1}}\right) \) . We argue by induction on \( r \) . If \( r = 1 \), then \( K = F\left( \alpha \right) \) with \( {\alpha }^{n} = a \in F \) . Then \( N = F\left( {{\beta ... | Yes |
Lemma 16.6 Let \( K \) be an \( n \) -radical extension of \( F \), and let \( N \) be the normal closure of \( K/F \) . Then \( N \) is an \( n \) -radical extension of \( F \) . | Proof. Let \( K = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right) \) with \( {\alpha }_{i}^{n} \in F\left( {{\alpha }_{1},\ldots ,{\alpha }_{i - 1}}\right) \) . We argue by induction on \( r \) . If \( r = 1 \), then \( K = F\left( \alpha \right) \) with \( {\alpha }^{n} = a \in F \) . Then \( N = F\left( {{\beta ... | Yes |
Corollary 16.11 Let \( f\left( x\right) \) be the general \( n \) th degree polynomial over a field of characteristic 0 . If \( n \geq 5 \), then \( f \) is not solvable by radicals. | Example 16.12 Let \( f\left( x\right) = {x}^{5} - {4x} + 2 \in \mathbb{Q}\left\lbrack x\right\rbrack \) . By graphing techniques of calculus, we see that this polynomial has exactly two nonreal roots, as indicated in the graph below.\n\n = {x}^{5} - {4x} + 2 \in \mathbb{Q}\left\lbrack x\right\rbrack \) . By graphing techniques of calculus, we see that this polynomial has exactly two nonreal roots, as indicated in the graph below. | Furthermore, \( f \) is irreducible over \( \mathbb{Q} \) by the Eisenstein criterion. Let \( K \) be the splitting field of \( f \) over \( \mathbb{Q} \) . Then \( \left\lbrack {K : \mathbb{Q}}\right\rbrack \) is a multiple of 5, since any root of \( f \) generates a field of dimension 5 over \( \mathbb{Q} \) . Let \(... | Yes |
Let \( f\left( x\right) = {x}^{3} - {3x} + 1 \in \mathbb{Q}\left\lbrack x\right\rbrack \), and let \( K \) be the splitting field of \( f \) over \( \mathbb{Q} \). We show that \( f \) is solvable by radicals but that \( K \) is not a radical extension of \( \mathbb{Q} \). | Since \( f \) has no roots in \( \mathbb{Q} \) and \( \deg \left( f\right) = 3 \), the polynomial \( f \) is irreducible over \( \mathbb{Q} \). The discriminant of \( f \) is \( {81} = {9}^{2} \), so the Galois group of \( K/\mathbb{Q} \) is \( {A}_{3} \) and \( \left\lbrack {K : \mathbb{Q}}\right\rbrack = 3 \), by Cor... | Yes |
Lemma 17.1 If \( {\alpha }_{1},\ldots ,{\alpha }_{n} \in K \), then there is an \( E \in \mathcal{I} \) with \( {\alpha }_{i} \in E \) for all \( i \) . | Proof. Let \( E \subseteq K \) be the splitting field of the minimal polynomials of the \( {\alpha }_{i} \) over \( F \) . Then, as each \( {\alpha }_{i} \) is separable over \( F \), the field \( E \) is normal and separable over \( F \) ; hence, \( E \) is Galois over \( F \) . Since there are finitely many \( {\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 |
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 |
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 |
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 \( K = F\left( {{x}_{1},\ldots ,{x}_{n}}\right) \), 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 |
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 \( L \) ... | 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... | 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 |
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 |
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