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Example 6.3 Let \( F = {\mathbb{F}}_{p} \) . A generator for \( {F}^{ * } \) is often called a primitive root modulo \( p \) . For example,2 is a primitive root modulo 5 . Moreover,2 is not a primitive root modulo 7, while 3 is a primitive root modulo 7 . In general, it is not easy to find a primitive root modulo \( p ... | Null | No |
Corollary 6.4 If \( K/F \) is an extension of finite fields, then \( K \) is a simple extension of \( F \) . | Proof. By the previous corollary, the group \( {K}^{ * } \) is cyclic. Let \( \alpha \) be a generator of the cyclic group \( {K}^{ * } \) . Every nonzero element of \( K \) is a power of \( \alpha \) , so \( K = F\left( \alpha \right) \) . Therefore, \( K \) is a simple extension of \( F \) . | Yes |
Theorem 6.5 Let \( F \) be a finite field with \( \operatorname{char}\left( F\right) = p \), and set \( \left| F\right| = {p}^{n} \) . Then \( F \) is the splitting field of the separable polynomial \( {x}^{{p}^{n}} - x \) over \( {\mathbb{F}}_{p} \) . Thus, \( F/{\mathbb{F}}_{p} \) is Galois. Furthermore, if \( \sigma... | Proof. Let \( \left| F\right| = {p}^{n} \), so \( \left| {F}^{ * }\right| = {p}^{n} - 1 \) . By Lagrange’s theorem, if \( a \in {F}^{ * } \) , then \( {a}^{{p}^{n} - 1} = 1 \) . Multiplying by \( a \) gives \( {a}^{{p}^{n}} = a \) . This equation also holds for \( a = 0 \) . Therefore, the elements of \( F \) are roots... | Yes |
Corollary 6.6 Any two finite fields of the same size are isomorphic. | Proof. The proof of Theorem 6.5 shows that any two fields of order \( {p}^{n} \) are splitting fields over \( {\mathbb{F}}_{p} \) of \( {x}^{{p}^{n}} - x \), so the corollary follows from the isomorphic extension theorem. | Yes |
Corollary 6.7 If \( K/F \) is an extension of finite fields, then \( K/F \) is Galois with a cyclic Galois group. Moreover, if \( \operatorname{char}\left( F\right) = p \) and \( \left| F\right| = {p}^{n} \), then \( \operatorname{Gal}\left( {K/F}\right) \) is generated by the automorphism \( \tau \) defined by \( \tau... | Proof. Say \( \left\lbrack {K : {\mathbb{F}}_{p}}\right\rbrack = m \) . Then \( \operatorname{Gal}\left( {K/{\mathbb{F}}_{p}}\right) \) is a cyclic group of order \( m \) by Theorem 6.5, so the order of the Frobenius automorphism \( \sigma \) of \( K \) is \( m \) . The group \( \operatorname{Gal}\left( {K/F}\right) \)... | Yes |
Theorem 6.8 Let \( N \) be an algebraic closure of \( {\mathbb{F}}_{p} \) . For any positive integer \( n \), there is a unique subfield of \( N \) of order \( {p}^{n} \) . If \( K \) and \( L \) are subfields of \( N \) of orders \( {p}^{m} \) and \( {p}^{n} \), respectively, then \( K \subseteq L \) if and only if \(... | Proof. Let \( n \) be a positive integer. The set of roots in \( N \) of the polynomial \( {x}^{{p}^{n}} - x \) has \( {p}^{n} \) elements and is a field. Thus, there is a subfield of \( N \) of order \( {p}^{n} \) . Since any two fields of order \( {p}^{n} \) in \( N \) are splitting fields of \( {x}^{{p}^{n}} - x \) ... | Yes |
Corollary 6.9 Let \( F \) be a finite field, and let \( f\left( x\right) \) be a monic irreducible polynomial over \( F \) of degree \( n \) . 1. If \( a \) is a root of \( f \) in some extension field of \( F \), then \( F\left( a\right) \) is a splitting field for \( f \) over \( F \) . Consequently, if \( K \) is a ... | Proof. Let \( K \) be a splitting field of \( f \) over \( F \) . If \( a \in K \) is a root of \( f\left( x\right) \) , then \( F\left( a\right) \) is an \( n \) -dimensional extension of \( F \) inside \( K \) . By Theorem 6.5, \( F\left( a\right) \) is a Galois extension of \( F \) ; hence, \( f\left( x\right) = \mi... | Yes |
Example 6.10 Let \( F = {\mathbb{F}}_{2} \) and \( K = F\left( \alpha \right) \), where \( \alpha \) is a root of \( f\left( x\right) = \) \( {x}^{3} + {x}^{2} + 1 \) . This polynomial has no roots in \( F \), as a quick calculation shows, so it is irreducible over \( F \) and \( \left\lbrack {K : F}\right\rbrack = 3 \... | Null | No |
Example 6.11 Let \( F = {\mathbb{F}}_{2} \) and \( f\left( x\right) = {x}^{4} + x + 1 \) . By the derivative test, we see that \( f \) has no repeated roots. The polynomial \( f \) is irreducible over \( f \), since \( f \) has no roots in \( F \) and is not divisible by the unique irreducible quadratic \( {x}^{2} + x ... | Null | No |
Example 6.12 Let \( f\left( x\right) = {x}^{2} + 1 \) . If \( p \) is an odd prime, then we show that \( f \) is reducible over \( F = {\mathbb{F}}_{p} \) if and only if \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \) . | To prove this, if \( a \in F \) is a root of \( {x}^{2} + 1 \), then \( {a}^{2} = - 1 \), so \( a \) has order 4 in \( {F}^{ * } \) . By Lagrange’s theorem,4 divides \( \left| {F}^{ * }\right| = p - 1 \), so \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \) . Conversely, if \( p \equiv 1\left( {\;\operatorname{mo... | Yes |
Proposition 6.14 Let \( n \) be a positive integer. Then \( {x}^{{p}^{n}} - x \) factors over \( {\mathbb{F}}_{p} \) into the product of all monic irreducible polynomials over \( {\mathbb{F}}_{p} \) of degree a divisor of \( n \) . | Proof. Let \( F \) be a field of order \( {p}^{n} \) . Then \( F \) is the splitting field of \( {x}^{{p}^{n}} - x \) over \( {\mathbb{F}}_{p} \) by Theorem 6.5. Recall that \( F \) is exactly the set of roots of \( {x}^{{p}^{n}} - x \) . Let \( a \in F \), and set \( m = \left\lbrack {{\mathbb{F}}_{p}\left( a\right) :... | Yes |
The monic irreducible polynomials of degree 5 over \( {\mathbb{F}}_{2} \) can be determined by factoring \( {x}^{{2}^{5}} - x \), which we see factors as | \[ {x}^{{2}^{5}} - x = x\left( {x + 1}\right) \left( {{x}^{5} + {x}^{3} + 1}\right) \left( {{x}^{5} + {x}^{2} + 1}\right) \times \left( {{x}^{5} + {x}^{4} + {x}^{3} + x + 1}\right) \left( {{x}^{5} + {x}^{4} + {x}^{2} + x + 1}\right) \times \left( {{x}^{5} + {x}^{4} + {x}^{3} + {x}^{2} + 1}\right) \left( {{x}^{5} + {x}^... | Yes |
Proposition 7.2 Suppose that \( \operatorname{char}\left( F\right) \) does not divide \( n \), and let \( K \) be a splitting field of \( {x}^{n} - 1 \) over \( F \) . Then \( K/F \) is Galois, \( K = F\left( \omega \right) \) is generated by any primitive nth root of unity \( \omega \) , and \( \mathrm{{Gal}}\left( {K... | Proof. Since \( \operatorname{char}\left( F\right) \) does not divide \( n \), the derivative test shows that \( {x}^{n} - 1 \) is a separable polynomial over \( F \) . Therefore, \( K \) is both normal and separable over \( F \) ; hence, \( K \) is Galois over \( F \) . Let \( \omega \in K \) be a primitive \( n \) th... | Yes |
The structure of \( F \) determines the degree \( \left\lbrack {F\left( \omega \right) : F}\right\rbrack \) or, equivalently, the size of \( \operatorname{Gal}\left( {F\left( \omega \right) /F}\right) \) . For instance, let \( \omega = {e}^{{2\pi i}/8} \) be a primitive eighth root of unity in \( \mathbb{C} \) . Then \... | Null | No |
Let \( F = {\mathbb{F}}_{2} \) . If \( \omega \) is a primitive third root of unity over \( F \) , then \( \omega \) is a root of \( {x}^{3} - 1 = \left( {x - 1}\right) \left( {{x}^{2} + x + 1}\right) \) . Since \( \omega \neq 1 \) and \( {x}^{2} + x + 1 \) is irreducible over \( F \), we have \( \left\lbrack {F\left( ... | Null | No |
Lemma 7.6 Let \( n \) be any positive integer. Then \( {x}^{n} - 1 = \mathop{\prod }\limits_{{d \mid n}}{\Psi }_{d}\left( x\right) \) . Moreover, \( {\Psi }_{n}\left( x\right) \in \mathbb{Z}\left\lbrack x\right\rbrack \) . | Proof. We know that \( {x}^{n} - 1 = \prod \left( {x - \omega }\right) \), where \( \omega \) ranges over the set of all \( n \) th roots of unity. If \( d \) is the order of \( \omega \) in \( {\mathbb{C}}^{ * } \), then \( d \) divides \( n \), and \( \omega \) is a primitive \( d \) th root of unity. Gathering all t... | Yes |
Corollary 7.8 If \( K \) is a splitting field of \( {x}^{n} - 1 \) over \( \mathbb{Q} \), then \( \left\lbrack {K : \mathbb{Q}}\right\rbrack = \) \( \phi \left( n\right) \) and \( \operatorname{Gal}\left( {K/\mathbb{Q}}\right) \cong {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ * } \) . Moreover, if \( \omega \) is a primi... | Proof. The first part of the corollary follows immediately from Proposition 7.2 and Theorem 7.7. The description of \( \operatorname{Gal}\left( {K/\mathbb{Q}}\right) \) is a consequence of the proof of Proposition 7.2. | No |
Example 7.9 Let \( K = {\mathbb{Q}}_{7} \), and let \( \omega \) be a primitive seventh root of unity in \( \mathbb{C} \). By Corollary 7.8, \( \operatorname{Gal}\left( {K/\mathbb{Q}}\right) \cong {\left( \mathbb{Z}/7\mathbb{Z}\right) }^{ * } \), which is a cyclic group of order 6. The Galois group of \( K/\mathbb{Q} \... | \[ \text{(id),}\left\langle {\sigma }_{3}^{3}\right\rangle ,\left\langle {\sigma }_{3}^{2}\right\rangle ,\left\langle {\sigma }_{3}\right\rangle \] whose orders are \( 1,2,3 \), and 6, respectively. Let us find the corresponding intermediate fields. If \( L = \mathcal{F}\left( {\sigma }_{3}^{3}\right) = \mathcal{F}\lef... | Yes |
Example 7.10 Let \( K = {\mathbb{Q}}_{8} \), and let \( \omega = \exp \left( {{2\pi i}/8}\right) = \left( {1 + i}\right) /\sqrt{2} \) . The Galois group of \( K/\mathbb{Q} \) is \( \left\{ {{\sigma }_{1},{\sigma }_{3},{\sigma }_{5},{\sigma }_{7}}\right\} \), and note that each of the three nonidentity automorphisms of ... | Null | No |
Example 8.2 Let \( F \) be any field, and let \( K = F\left( \sqrt{d}\right) \) for some \( d \in F - {F}^{2} \) . A convenient basis for \( K \) is \( \{ 1,\sqrt{d}\} \) . If \( \alpha = a + b\sqrt{d} \) with \( a, b \in F \), we determine the norm and trace of \( \alpha \) . | The linear transformation \( {L}_{\alpha } \) is equal to \( a{L}_{1} + b{L}_{\sqrt{d}} \), so we first need to find the matrix representations for \( {L}_{1} \) and \( {L}_{\sqrt{d}} \) . The identity transformation \( {L}_{1} \) has matrix \( \left( \begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) \) . For \( {L}_... | Yes |
Example 8.3 Let \( F = \mathbb{Q} \) and \( K = \mathbb{Q}\left( \sqrt[3]{2}\right) \) . We will determine the norm and trace of \( \sqrt[3]{2} \) . An \( F \) -basis for \( K \) is \( \{ 1,\sqrt[3]{2},\sqrt[3]{4}\} \) . We can check that \( {L}_{\sqrt[3]{2}}\left( 1\right) = \sqrt[3]{2},{L}_{\sqrt[3]{2}}\left( \sqrt[3... | \[ \left( \begin{array}{lll} 0 & 0 & 2 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right) ,\] so \( {N}_{K/F}\left( \sqrt[3]{2}\right) = 2 \) and \( {T}_{K/F}\left( \sqrt[3]{2}\right) = 0 \) . | Yes |
Example 8.4 Let \( F \) be a field of characteristic \( p > 0 \), and let \( K/F \) be a purely inseparable extension of degree \( p \) . Say \( K = F\left( \alpha \right) \) with \( {\alpha }^{p} = \) \( a \in F \) . For instance, we could take \( K \) to be the rational function field \( k\left( x\right) \) over a fi... | Null | No |
Lemma 8.5 Let \( K \) be a finite extension of \( F \) with \( n = \left\lbrack {K : F}\right\rbrack \) .\n\n1. If \( a \in K \), then \( {N}_{K/F}\left( a\right) \) and \( {T}_{K/F}\left( a\right) \) lie in \( F \) .\n\n2. The trace map \( {T}_{K/F} \) is an \( F \) -linear transformation.\n\n3. If \( \alpha \in F \),... | 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. | No |
Example 8.8 If \( K \) is a purely inseparable extension of \( F \) of characteristic \( p \), then the minimal polynomial of any element of \( K \) is of the form \( {x}^{{p}^{n}} - a \) . From this, it follows that the trace of any element is zero. | Null | No |
Lemma 8.9 Let \( K \) be a finite extension of \( F \), and let \( S \) be the separable closure of \( F \) in \( K \). Then \( \left\lbrack {S : F}\right\rbrack \) is equal to the number of \( F \)-homomorphisms from \( K \) to an algebraic closure of \( F \). | Proof. Let \( M \) be an algebraic closure of \( F \). We may assume that \( K \subseteq M \). If \( S \) is the separable closure of \( F \) in \( K \), then \( S = F\left( a\right) \) for some \( a \) by the primitive element theorem. If \( r = \left\lbrack {S : F}\right\rbrack \), then there are \( r \) distinct 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 \( F \) be a field of characteristic not 2, and let \( a \in {F}^{ * } - {F}^{*2} \). If \( K = F\\left( \\sqrt{a}\\right) \), then \( \\operatorname{Gal}\\left( {K/F}\\right) = \\{ \\mathrm{{id}},\\sigma \\} \) where \( \\sigma \\left( \\sqrt{a}\\right) = - \\sqrt{a} \). Thus, \( \\operatorname{Gal}\\left( {K/F}\\... | Null | No |
Let \( \omega \) be a primitive fifth root of unity in \( \mathbb{C} \), let \( F = \mathbb{Q}\left( \omega \right) \) , and let \( K = F\left( \sqrt[5]{2}\right) \) . Then \( K \) is the splitting field of \( {x}^{5} - 2 \) over \( F \), so \( K \) is Galois over \( F \) . Also, \( \left\lbrack {F : \mathbb{Q}}\right\... | 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 |
Example 10.6 Let \( E = {S}_{3} \). If \( M = \langle \left( {123}\right) \rangle \), then \( M \) is isomorphic to \( \mathbb{Z}/3\mathbb{Z} \) and \( M \) is an Abelian normal subgroup of \( E \). The quotient group \( E/M \) is isomorphic to \( \mathbb{Z}/2\mathbb{Z} \). Therefore, \( {S}_{3} \) is a group extension... | Null | No |
Example 10.7 Let \( E = {D}_{n} \), the dihedral group. One description of \( E \) is by generators and relations. The group \( E \) is generated by elements \( \sigma \) and \( \tau \) satisfying \( {\tau }^{n} = {\sigma }^{2} = e \) and \( {\sigma \tau \sigma } = {\tau }^{-1} \) . Let \( M = \langle \sigma \rangle \)... | Null | No |
Let \( M \) and \( G \) be groups, and let \( \varphi : G \rightarrow \operatorname{End}\left( M\right) \) be a group homomorphism. If \( E \) is the semidirect product \( M{ \times }_{\varphi }G \), then \( {M}^{\prime } = \{ \left( {m, e}\right) : m \in M\} \) is a normal subgroup of \( E \) isomorphic to \( M \), an... | Null | No |
Let \( {Q}_{8} \) be the quaternion group. Then \( {Q}_{8} = \) \( \{ \pm 1, \pm i, \pm j, \pm k\} \), and the operation on \( {Q}_{8} \) is given by the relations \( {i}^{2} = \) \( {j}^{2} = {k}^{2} = - 1 \) and \( {ij} = k = - {ji} \) . We show that \( {Q}_{8} \) is a group extension of \( M = \langle i\rangle \) by... | 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 |
If \( F \) is a field that contains a primitive \( n \) th root of unity, and if \( K/F \) is a cyclic extension of degree \( n \), then \( K/F \) is an \( n \) -Kummer extension. If \( F \) also contains a primitive \( m \) th root of unity for some \( m \) that is a multiple of \( n \), then \( K/F \) is also an \( m... | Null | No |
Let \( K = \mathbb{Q}\left( {\sqrt{2},\sqrt{3}}\right) \). The field \( K \) is the splitting field of \( \left( {{x}^{2} - 2}\right) \left( {{x}^{2} - 3}\right) \) over \( \mathbb{Q} \), so \( K \) is a Galois extension of \( \mathbb{Q} \). A short calculation shows that \( \left\lbrack {K : \mathbb{Q}}\right\rbrack =... | \[
\text{id} : \sqrt{2} \rightarrow \sqrt{2},\;\sqrt{3} \rightarrow \sqrt{3}\text{,}
\]
\[
\sigma : \sqrt{2} \rightarrow - \sqrt{2},\;\sqrt{3} \rightarrow \sqrt{3}
\]
\[
\tau : \sqrt{2} \rightarrow \sqrt{2},\;\sqrt{3} \rightarrow - \sqrt{3}
\]
\[
{\sigma \tau } : \sqrt{2} \rightarrow - \sqrt{2},\;\sqrt{3} \rightarro... | Yes |
If \( K = \mathbb{Q}\left( {\sqrt{{a}_{1}},\ldots ,\sqrt{{a}_{r}}}\right) \) for some \( {a}_{i} \in \mathbb{Q} \), then \( K/\mathbb{Q} \) is a 2-Kummer extension by Theorem 11.4. The degree of \( K/F \) is no larger than \( {2}^{r} \), but it may be less depending on the choice of the \( {a}_{i} \) . | Null | No |
Example 11.6 Let \( F = \mathbb{Q}\left( i\right) \), where \( i = \sqrt{-1} \), and let \( K = F\left( {\sqrt[4]{12},\sqrt[4]{3}}\right) \) . Since \( i \) is a primitive fourth root of unity, \( K/F \) is a 4-Kummer extension. The degree of \( K/F \) is 8, not 16, since \( K = F\left( {\sqrt{2},\sqrt[4]{3}}\right) \)... | Null | No |
Lemma 11.8 Let \( B : G \times H \rightarrow C \) be a bilinear pairing. If \( h \in H \), let \( {B}_{h} : G \rightarrow C \) be defined by \( {B}_{h}\left( g\right) = B\left( {g, h}\right) \) . Then the map \( \varphi : h \mapsto {B}_{h} \) is a group homomorphism from \( H \) to \( \hom \left( {G, C}\right) \) . If ... | 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 |
Example 12.8 Let \( f\left( x\right) = {x}^{3} + {px} + q \) . Then \( {a}_{0} = q,{a}_{1} = p \), and \( {a}_{2} = 0 \) , so by Newton's identities we get\n\n\[ \n{t}_{1} = 0 \]\n\n\[ \n{t}_{2} = - {2p} \]\n\n\[ \n{t}_{3} = - {3q} \]\n\n\[ \n{t}_{4} = 2{p}^{2} \]\n\nTherefore\n\n\[ \n\operatorname{disc}\left( f\right)... | Null | No |
Proposition 12.9 Let \( L = F\left( \alpha \right) \) be a field extension of \( F \) . If \( f\left( x\right) = \) \( \min \left( {F,\alpha }\right) \), then \( \operatorname{disc}\left( f\right) = {\left( -1\right) }^{n\left( {n - 1}\right) /2}{N}_{L/F}\left( {{f}^{\prime }\left( \alpha \right) }\right) \), where \( ... | 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. | No |
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 |
The polynomial \( {x}^{3} - {3x} + 1 \in \mathbb{Q}\left\lbrack x\right\rbrack \) has discriminant \( {81} = \) \( {9}^{2} \), and it is irreducible over \( \mathbb{Q} \) by an application of the rational root test. Thus, the Galois group of its splitting field over \( \mathbb{Q} \) is \( {A}_{3} \) . | Null | No |
Consider \( {x}^{3} - {3x} + 1 \) . Then \( \Gamma = - D/{108} = - {81}/{108} = \) \( - 3/4 \) . We have \( p = - 3 \) and \( q = 1 \) . Then \( A = - 1/2 + i\sqrt{3}/2 \) and \( B = - 1/2 - i\sqrt{3}/2 \), so \( A = \exp \left( {{2\pi i}/3}\right) \) and \( B = \exp \left( {-{2\pi i}/3}\right) \) . We can then set \( ... | 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 |
Example 13.6 Let \( f\left( x\right) = {x}^{4} - 4{x}^{3} + 4{x}^{2} + 6 \) . This polynomial is irreducible by the Eisenstein criterion. Now,\n\n\[ r\left( x\right) = {x}^{3} - 4{x}^{2} - {24x} = x\left( {{x}^{2} - {4x} - {24}}\right) ,\] \n\nso \( L = \mathbb{Q}\left( \sqrt{7}\right) \) . Take \( {\beta }_{1} = 0 \) ... | Null | No |
Example 13.7 Let \( p \) be a prime, and let \( f\left( x\right) = {x}^{4} + {px} + p \) . Then \( r\left( x\right) = {x}^{3} - {4px} - {p}^{2} \) . To test for roots of \( r\left( x\right) \) in \( \mathbb{Q} \), we only need to check \( \pm 1, \pm p, \pm {p}^{2} \) . We see that \( \pm 1 \) and \( \pm {p}^{2} \) are ... | Null | No |
Example 13.8 Let \( l \in \mathbb{Q} \), and let \( f\left( x\right) = {x}^{4} - l \) . Then the resolvent of \( f \) is \( r\left( x\right) = {x}^{3} + {4lx} = x\left( {{x}^{2} + {4l}}\right) \) . If \( - l \) is not a square in \( \mathbb{Q} \), then \( r\left( x\right) \) has exactly one root in \( \mathbb{Q} \) . M... | Null | No |
Theorem 14.1 (Lindemann-Weierstrauss) Let \( {\alpha }_{1},\ldots ,{\alpha }_{m} \) be distinct algebraic numbers. Then the exponentials \( {e}^{{\alpha }_{1}},\ldots ,{e}^{{\alpha }_{m}} \) are linearly independent over \( \mathbb{Q} \) . | Proof of the theorem. Suppose that there are \( {a}_{j} \in \mathbb{Q} \) with\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{m}{a}_{j}{e}^{{\alpha }_{j}} = 0 \]\n\nBy multiplying by a suitable integer, we may assume that each \( {a}_{j} \in \mathbb{Z} \) . Moreover, by eliminating terms if necessary, we may also assume that 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\[\n\mathbb{Q} \subseteq \mathbb{Q}\left( \sqrt{2}\right) \subseteq \mathbb{Q}\left( \sqrt{2}\right) \left( \sqrt{\sqrt{2}}\right) = \mathbb{Q}\left( \sqrt[4]{2}\right)\n\] | No |
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 |
Proposition 16.8 Let \( G \) be a group and \( N \) be a normal subgroup of \( G \) . Then \( G \) is solvable if and only if \( N \) and \( G/N \) are solvable. | Null | No |
Proposition 16.9 If \( n \geq 5 \), then \( {S}_{n} \) is not solvable. | Null | No |
Corollary 16.11 Let \( f\left( x\right) \) be the general \( n \) th degree polynomial over a field of characteristic 0 . If \( n \geq 5 \), then \( f \) is not solvable by radicals. | Example 16.12 Let \( f\left( x\right) = {x}^{5} - {4x} + 2 \in \mathbb{Q}\left\lbrack x\right\rbrack \) . By graphing techniques of calculus, we see that this polynomial has exactly two nonreal roots, as indicated in the graph below.\n\n = {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 |
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