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Theorem 17 (Kronecker). Let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \), where \( F \) is a field. There exists a field \( E \) containing \( F \) over which \( f\left( x\right) \) splits. | Proof. The proof is by induction on \( \partial f \) . If \( \partial f = 1 \), then \( f\left( x\right) \) is linear and we can choose \( E = F \) . If \( \partial f > 1 \), write \( f\left( x\right) = p\left( x\right) g\left( x\right) \), where \( p\left( x\right) \) is irreducible. If \( p\left( x\right) \) is linea... | Yes |
Theorem 18. If \( F \) is a field, then its prime field is isomorphic to either \( \mathbb{Q} \) or \( {\mathbb{Z}}_{p} \) for some prime \( p \) . | Proof. Define \( \chi : \mathbb{Z} \rightarrow F \) by \( n \mapsto {n1} \) (where 1 is the \ | No |
Theorem 19 (Galois). For every prime \( p \) and every positive integer \( n \) , there exists a field having exactly \( {p}^{n} \) elements. | Proof. If there were a field \( K \) with \( \left| K\right| = {p}^{n} = q \), then \( {K}^{\# } = K - \{ 0\} \) would be a multiplicative group of order \( q - 1 \) ; by Lagrange’s theorem (Theorem A3), \( {a}^{q - 1} = 1 \) for all \( a \in {K}^{\# } \) . It follows that every element of \( K \) would be a root of th... | No |
Lemma 20 (Gauss). The product of two primitive polynomials \( f\left( x\right), g\left( x\right) \) is itself primitive. | Proof. Assume that \( f\left( x\right) g\left( x\right) = \left( {\sum {a}_{i}{x}^{i}}\right) \left( {\sum {b}_{j}{x}^{j}}\right) = \sum {c}_{k}{x}^{k} \) is not primitive, so that there is some prime \( p \) dividing each \( {c}_{k} \) . Let \( {a}_{i} \) and \( {b}_{j} \) be the first coefficients of \( f\left( x\rig... | Yes |
Lemma 21. Every nonzero \( f\left( x\right) \in \mathbb{Q}\left\lbrack x\right\rbrack \) has a unique factorization\n\n\[ f\left( x\right) = c\left( f\right) {f}^{ * }\left( x\right) \]\n\nwhere \( c\left( f\right) \in \mathbb{Q} \) is positive and \( {f}^{ * }\left( x\right) \in \mathbb{Z}\left\lbrack x\right\rbrack \... | Proof. Let \( f\left( x\right) = \left( {{a}_{0}/{b}_{0}}\right) + \left( {{a}_{1}/{b}_{1}}\right) x + \cdots + \left( {{a}_{n}/{b}_{n}}\right) {x}^{n} \in \mathbb{Q}\left\lbrack x\right\rbrack \) . Define \( B = {b}_{0}\cdots {b}_{n} \), so that \( f\left( x\right) = \left( {1/B}\right) g\left( x\right) \) for \( g\le... | Yes |
Lemma 22. If \( f\left( x\right) \in \mathbb{Q}\left\lbrack x\right\rbrack \) factors as \( f\left( x\right) = g\left( x\right) h\left( x\right) \), then\n\n\[ c\left( f\right) = c\left( g\right) c\left( h\right) \;\text{ and }\;{f}^{ * }\left( x\right) = {g}^{ * }\left( x\right) {h}^{ * }\left( x\right) . \] | Proof. We have \( f\left( x\right) = g\left( x\right) h\left( x\right) = \left\lbrack {c\left( g\right) {g}^{ * }\left( x\right) }\right\rbrack \left\lbrack {c\left( h\right) {h}^{ * }\left( x\right) }\right\rbrack = \) \( c\left( g\right) c\left( h\right) {g}^{ * }\left( x\right) {h}^{ * }\left( x\right) \) . Since \(... | Yes |
Theorem 23. If \( f\left( x\right) \in \mathbb{Z}\left\lbrack x\right\rbrack \) factors in \( \mathbb{Q}\left\lbrack x\right\rbrack \), then it also factors in \( \mathbb{Z}\left\lbrack x\right\rbrack \) (into polynomials of the same degree as over \( \mathbb{Q} \) ). Equivalently, if \( f\left( x\right) \in \) \( \mat... | Proof. Assume that \( f\left( x\right) = g\left( x\right) h\left( x\right) \) in \( \mathbb{Q}\left\lbrack x\right\rbrack \) . Then \( f\left( x\right) = \) \( c\left( g\right) c\left( h\right) {g}^{ * }\left( x\right) {h}^{ * }\left( x\right) \) in \( \mathbb{Q}\left\lbrack x\right\rbrack \), where \( {g}^{ * },{h}^{ ... | Yes |
Theorem 24 (Eisenstein Criterion). Let \( f\left( x\right) = {a}_{0} + {a}_{1}x + \cdots + {a}_{n}{x}^{n} \in \) \( \mathbb{Z}\left\lbrack x\right\rbrack \) . If there is a prime \( p \) dividing \( {a}_{i} \) for all \( i < n \), but with \( p \) not dividing \( {a}_{n} \) and \( {p}^{2} \) not dividing \( {a}_{0} \),... | Proof. Let \( f\left( x\right) = \left( {{b}_{0} + {b}_{1}x + \cdots + {b}_{m}{x}^{m}}\right) \left( {{c}_{0} + {c}_{1}x + \cdots + {c}_{k}{x}^{k}}\right) \) ; by Theorem 23, we may assume that both factors lie in \( \mathbb{Z}\left\lbrack x\right\rbrack \) . Now \( p \mid {a}_{0} = {b}_{0}{c}_{0} \) so that, by Euclid... | Yes |
Corollary 25. The pth cyclotomic polynomial is irreducible over \( \mathbb{Q} \) for every prime \( p \) . | Proof. Recall Exercise 69: a polynomial \( f\left( x\right) \) is irreducible if and only if \( f\left( {x + c}\right) \) is irreducible, where \( c \) is a constant. In particular, \( {\Phi }_{p}\left( x\right) = \) \( \left( {{x}^{p} - 1}\right) /\left( {x - 1}\right) \) is irreducible if and only if \( {\Phi }_{p}\l... | No |
Corollary 26. If \( a \neq \pm 1 \) is a square-free integer, then \( {x}^{n} - a \) is irreducible over \( \mathbb{Q} \) for every \( n \geq 2 \) . | Proof. Since \( a \neq \pm 1 \), there is some prime \( p \) dividing \( a \), and Eisenstein’s criterion applies with this prime. | Yes |
Theorem 27. Let \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) be an irreducible polynomial of degree \( d \) . Then \( E = F\left\lbrack x\right\rbrack /\left( {p\left( x\right) }\right) \) is an extension field of \( F \) of degree \( d \) . | Proof. Denote \( \left( {p\left( x\right) }\right) \) by \( I \), and denote \( x + I \) in \( E \) by \( \alpha \) ; it suffices to prove that \( \left\{ {1,\alpha ,{\alpha }^{2},\ldots ,{\alpha }^{d - 1}}\right\} \) is a basis of \( E \) over \( F \) . If, for \( 0 \leq i \leq d - 1 \), there are \( {a}_{i} \in F \) ... | Yes |
Theorem 28. Let \( E/F \) be an extension field, and let \( \alpha \in E \) be algebraic over \( F \). (i) There is a monic irreducible polynomial \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) having \( \alpha \) as a root; (ii) \( p\left( x\right) \) is the monic polynomial of least degree in \( F\left\lbrac... | Proof. Choose \( p\left( x\right) \) as the monic polynomial of least degree in \( F\left\lbrack x\right\rbrack \) having \( \alpha \) as a root \( \left( {p\left( x\right) \text{exists because}\alpha \text{is algebraic}}\right) \) . The evaluation map \( F\left\lbrack x\right\rbrack \rightarrow \) \( F\left( \alpha \r... | Yes |
Corollary 29. Let \( \sigma : F \rightarrow {F}^{\prime } \) be an isomorphism of fields, let \( {\sigma }^{ * } : F\left\lbrack x\right\rbrack \rightarrow \) \( {F}^{\prime }\left\lbrack x\right\rbrack \) (defined by \( \sum {r}_{i}{x}^{i} \mapsto \sum \sigma \left( {r}_{i}\right) {x}^{i} \) ) be the corresponding iso... | Proof. The isomorphism \( {\sigma }^{ * } : F\left\lbrack x\right\rbrack \rightarrow {F}^{\prime }\left\lbrack x\right\rbrack \) carries the ideal \( \left( {p\left( x\right) }\right) \) onto the ideal \( \left( {{p}^{ * }\left( x\right) }\right) \), and so the existence of \( \breve{\sigma } \) follows from the theore... | No |
Theorem 30. Every polynomial \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) has a splitting field. | Proof. By Kronecker's theorem (Theorem 17), there is an extension field \( K/F \) over which \( f\left( x\right) \) splits. Define \( E = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \), where \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) are the roots of \( f\left( x\right) \) in \( K \) . It is plain that \( f\le... | Yes |
Lemma 31. If \( F \subset B \subset E \) are fields with \( \left\lbrack {E : B}\right\rbrack \) and \( \left\lbrack {B : F}\right\rbrack \) finite, then \( E/F \) is finite and \[ \left\lbrack {E : F}\right\rbrack = \left\lbrack {E : B}\right\rbrack \left\lbrack {B : F}\right\rbrack . \] | Proof. Let \( \left\{ {{\alpha }_{1},\ldots ,{\alpha }_{m}}\right\} \) be a basis of \( E/B \), and let \( \left\{ {{\beta }_{1},\ldots ,{\beta }_{n}}\right\} \) be a basis of \( B/F \) . It suffices to prove that \( \left\{ {{\beta }_{j}{\alpha }_{i} : 1 \leq i \leq m,1 \leq j \leq n}\right\} \) is a basis of \( E/F \... | Yes |
Theorem 32. Let \( \sigma : F \rightarrow {F}^{\prime } \) be an isomorphism of fields, let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) , and let \( {f}^{ * }\left( x\right) = {\sigma }^{ * }{\left( f\left( x\right) \right) }^{3} \) be the corresponding polynomial in \( {F}^{\prime }\left\lbrack x\right\rbr... | Proof. (i) The proof is by induction on \( \left\lbrack {E : F}\right\rbrack \) . If \( \left\lbrack {E : F}\right\rbrack = 1 \), then \( E = F \) and \( f\left( x\right) \) is a product of linear factors in \( F\left\lbrack x\right\rbrack \) ; it follows that \( {f}^{ * }\left( x\right) \) is also a product of linear ... | Yes |
Corollary 33. If \( f\left( x\right) \in F\left\lbrack x\right\rbrack \), then any two splitting fields of \( f\left( x\right) \) over \( F \) are isomorphic by an isomorphism fixing \( F \) pointwise. | Proof. Choose \( F = {F}^{\prime } \) and \( \sigma \) the identity on \( F \) . | No |
Corollary 34 (E.H. Moore). Any two finite fields of order \( {p}^{n} \) are isomorphic. | Proof. Any field \( F \) of order \( {p}^{n} \) is the splitting field of \( {x}^{q} - x \) over \( {\mathbb{Z}}_{p} \), where \( q = {p}^{n} \) | No |
Lemma 35. Let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) and let \( E/F \) be the splitting field of \( f\left( x\right) \) over \( F \) . If \( \sigma : E \rightarrow E \) is an automorphism (an isomorphism of \( E \) with itself) fixing \( F \) pointwise and if \( \alpha \) is a root of \( f\left( x\righ... | Proof. Let \( f\left( x\right) = {a}_{0} + {a}_{1}x + \cdots + {a}_{n}{x}^{n} \), so that \( {a}_{0} + {a}_{1}\alpha + \cdots + {a}_{n}{\alpha }^{n} = 0 \) . Applying \( \sigma \) gives \( \sigma \left( {a}_{0}\right) + \sigma \left( {a}_{1}\right) \sigma \left( \alpha \right) + \cdots + \sigma \left( {a}_{n}\right) \s... | Yes |
Theorem 36. If \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) has \( n \) distinct roots in its splitting field \( E \) , then \( \operatorname{Gal}\left( {E/F}\right) \) is isomorphic to a subgroup of the symmetric group \( {S}_{n} \) . | Proof. Let \( X = \left\{ {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right\} \) be the set of all the roots of \( f\left( x\right) \) in \( E \) . By Lemma 35, if \( \sigma \in \operatorname{Gal}\left( {E/F}\right) \), then \( \sigma \left( X\right) = X \) . The map \( \operatorname{Gal}\left( {E/F}\right) \rightarrow {S}_{... | No |
Theorem 37. If \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) is a separable polynomial, and if \( E/F \) is its splitting field, then \( \left| {\operatorname{Gal}\left( {E/F}\right) }\right| = \left\lbrack {E : F}\right\rbrack \) . | Proof. By Theorem 32(ii) with \( F = {F}^{\prime }, E = {E}^{\prime } \), and \( \sigma : F \rightarrow F \) the identity, there are exactly \( \left\lbrack {E : F}\right\rbrack \) automorphisms of \( E \) that fix \( F \) . | Yes |
Lemma 38. Let \( F \subset B \subset E \) be a tower of fields with \( B/F \) the splitting field of some polynomial \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) . If \( \sigma \in \operatorname{Gal}\left( {E/F}\right) \), then \( \sigma \mid B \in \operatorname{Gal}\left( {B/F}\right) \) . | Proof. It suffices to prove that \( \sigma \left( B\right) = B \) . If \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) are the roots of \( f\left( x\right) \), then \( B = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) . Now \( \sigma \left( F\right) = F \), and \( \sigma \left( {\alpha }_{i}\right) \in B \) for all... | No |
Theorem 39. Let \( F \subset B \subset E \) be a tower of fields with \( B/F \) the splitting field of some polynomial \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) and \( E/F \) the splitting field of some \( g\left( x\right) \in F\left\lbrack x\right\rbrack \) . Then \( \operatorname{Gal}\left( {E/B}\right)... | Proof. Define \( \psi : \operatorname{Gal}\left( {E/F}\right) \rightarrow \operatorname{Gal}\left( {B/F}\right) \) by \( \sigma \mapsto \sigma \mid B \) ; Lemma 38 says that \( \psi \) does take its values in \( \operatorname{Gal}\left( {B/F}\right) \) . It is easily seen that \( \psi \) is a homomorphism with kernel \... | Yes |
Theorem 40. Let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) have degree \( n \), and assume that \( F \) contains all the pth roots of unity for all primes \( p \) dividing \( n \) !. If \( f\left( x\right) \) is solvable by radicals, then its Galois group is a solvable group. | Proof. Since \( f\left( x\right) \) is solvable by radicals, there is a radical extension \( F = \) \( {B}_{0} \subset {B}_{1} \subset \ldots \subset {B}_{t} \) with \( E \subset {B}_{t} \), where \( E \) is the splitting field of \( f\left( x\right) \) over \( F \) . By Exercise 78, we may assume each \( \left\lbrack ... | No |
Lemma 41. If \( C = \langle a\rangle \) is a cyclic group of order \( n \) and generator \( a \), then \( C \) has a unique subgroup of order \( d \) for each divisor \( d \) of \( n \) . | Proof. If \( n = {dc} \), we show that \( {a}^{c} \) has order \( d \) (and so \( \left\langle {a}^{c}\right\rangle \) is a subgroup of order \( d \) ). Clearly \( {\left( {a}^{c}\right) }^{d} = 1 \) ; we claim that \( d \) is the smallest such power. If \( {\left( {a}^{c}\right) }^{r} = 1 \), then \( n \mid {cr} \) ; ... | Yes |
Theorem 42. If \( n \) is a positive integer, then\n\n\[ n = \mathop{\sum }\limits_{\substack{{d \mid n} \\ {1 \leq d \leq n} }}\varphi \left( d\right) \] | Proof. If \( G \) is a group, then it is easy to see that it is the disjoint union\n\n\[ G = \cup g\left( C\right) \]\n\nwhere \( C \) ranges over all the cyclic subgroups of \( G \) . If \( G \) is a cyclic group of order \( n \), then counting gives\n\n\[ n = \sum \left| {g\left( C\right) }\right| = \sum \varphi \lef... | Yes |
Theorem 43. A group \( G \) of order \( n \) is cyclic if and only if, for each divisor \( d \) of \( n \), there is at most one cyclic subgroup of order \( d \) . | Proof. If \( G \) is cyclic, then the result follows from Lemma 41. Conversely, write \( G \) as a disjoint union (as in the preceding proof): \( G = \mathcal{G}g\left( C\right) \) . Hence \( n = \left| G\right| = \sum \left| {g\left( C\right) }\right| \), where the summation is over all cyclic subgroups \( C \) of \( ... | Yes |
Theorem 44. If \( F \) is a field with multiplicative group \( {F}^{\# } = F - \{ 0\} \), then every finite subgroup \( G \) of \( {F}^{\# } \) is cyclic. | Proof. Suppose \( \left| G\right| = n \) and \( d \mid n \) . If \( C \) is a cyclic subgroup of \( G \) of order \( d \), then Lagrange’s theorem gives \( {x}^{d} = 1 \) for each of the \( d \) elements \( x \in C \) . Were there a second cyclic subgroup of order \( d \), then \( G \) would contain at least \( d + 1 \... | Yes |
Lemma 47. If \( \alpha \) is a primitive element of \( {GF}\left( {p}^{n}\right) \), then \( \alpha \) is a root of an irreducible polynomial of degree \( n \) . | Proof. If the irreducible polynomial of \( \alpha \) over \( {\mathbb{Z}}_{p} \) has degree \( d \), then \( {\mathbb{Z}}_{p}\left( \alpha \right) \) has order \( {p}^{d} \) . But this subfield is all of \( {GF}\left( {p}^{n}\right) \) because \( \alpha \) is a primitive element; hence \( d = n \) . | Yes |
Theorem 48. \( \operatorname{Gal}\left( {{GF}\left( {p}^{n}\right) /{GF}\left( p\right) }\right) \cong {\mathbb{Z}}_{n} \) with generator \( u \mapsto {u}^{p} \) . | Proof. Denote \( {GF}\left( {p}^{n}\right) \) by \( K \) and denote the Galois group by \( G \) . If \( \alpha \) is a primitive element, then its irreducible polynomial \( p\left( x\right) \) has degree \( n \) (Lemma 47), and so \( K \) contains at most \( n \) of its roots. If \( \sigma \in G \), then \( \sigma \) i... | Yes |
Lemma 49. Let \( n \) be a positive integer and let \( F \) be a field. If the characteristic of \( F \) is either 0 or is a prime not dividing \( n \), then \( {x}^{n} - 1 \) has \( n \) distinct roots in a splitting field. | Proof. If \( f\left( x\right) = {x}^{n} - 1 \), then its derivative \( {f}^{\prime }\left( x\right) = n{x}^{n - 1} \) . By hypothesis, this is not zero, and so the \( \gcd \left( {f,{f}^{\prime }}\right) = 1 \) ; therefore, \( f\left( x\right) \) has no repeated roots. | Yes |
Theorem 50. If \( F \) is a field and \( E = F\left( \alpha \right) \), where \( \alpha \) is a primitive nth root of unity, then \( \operatorname{Gal}\left( {E/F}\right) \) is abelian. | Proof. Note that \( E \) is the splitting field of \( {x}^{n} - 1 \) because \( \alpha \) is a primitive \( n \) th root of unity. Now \( \sigma \left( \alpha \right) = {\alpha }^{i} \) for every \( \sigma \in \operatorname{Gal}\left( {E/F}\right) \) ; moreover, Theorem A2(ii) says \( i \) must be relatively prime to \... | "No" |
Theorem 51. Let \( F \) contain a primitive nth root of unity, and let \( f\left( x\right) = \) \( {x}^{n} - a \) . If \( E/F \) is a splitting field of \( f\left( x\right) \), then restriction gives an injection\n\n\[ G = \operatorname{Gal}\left( {E/F}\right) \rightarrow {\mathbb{Z}}_{n} \]\n\nMoreover, \( f\left( x\r... | Proof. If \( \omega \) is a primitive \( n \) th root of unity and if \( \alpha \) is a root of \( f\left( x\right) \) , then the list of all the roots of \( f\left( x\right) \) is: \( \alpha ,{\alpha \omega },\ldots ,\alpha {\omega }^{n - 1} \) . If \( \sigma \in G \), then \( \sigma \left( \alpha \right) = \alpha {\o... | Yes |
Corollary 52. Let \( p \) be a prime and let \( F \) be a field containing a primitive pth root of unity. If \( a \in F \), then \( {x}^{p} - a \) either splits or is irreducible. | Proof. Consider the map \( \operatorname{Gal}\left( {E/F}\right) \rightarrow {\mathbb{Z}}_{p} \) of the theorem. If \( f\left( x\right) \) splits, then its image is trivial; if \( f\left( x\right) \) does not split, then its image is a nontrivial subgroup of \( {\mathbb{Z}}_{p} \) . But \( {\mathbb{Z}}_{p} \) has no pr... | Yes |
Theorem 53. Let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) be solvable by radicals over a field \( F \), and let \( E/F \) be its splitting field. Then \( \operatorname{Gal}\left( {E/F}\right) \) is a solvable group. | Proof. By hypothesis, there is a radical extension\n\n\[ F = {B}_{0} \subset {B}_{1} \subset \ldots \subset {B}_{t} \]\n\nwith \( E \subset {B}_{t} \) . Only finitely many roots of unity have been adjoined to \( F \) , say, the \( {k}_{1} \) th, \( {k}_{2} \) th, \( \ldots ,{k}_{s} \) th roots of unity. If \( k \) is t... | Yes |
Theorem 54 (Abel-Ruffini). There exists a quintic polynomial \( f\left( x\right) \in \) \( \mathbb{Q}\left\lbrack x\right\rbrack \) that is not solvable by radicals. | Proof. Let \( f\left( x\right) = {x}^{5} - {4x} + 2;f\left( x\right) \) is irreducible over \( \mathbb{Q} \), by Eisenstein’s criterion. Let \( E/\mathbb{Q} \) be the splitting field of \( f\left( x\right) \) contained in \( \mathbb{C},{}^{5} \) and let \( G = \operatorname{Gal}\left( {E/\mathbb{Q}}\right) \) . If \( \... | Yes |
Lemma 55 (Dedekind). Every set \( \left\{ {{\sigma }_{1},\ldots ,{\sigma }_{n}}\right\} \) of distinct characters of a group \( G \) in a field \( E \) is independent. | Proof. The proof is by induction on \( n \) . If \( n = 1 \), then \( {a}_{1}{\sigma }_{1}\left( x\right) = 0 \) implies that \( {a}_{1} = 0 \) because \( {\sigma }_{1}\left( x\right) \neq 0 \) . Let \( n > 1 \) and assume there is an equation\n\n(1)\n\n\[ \n{a}_{1}{\sigma }_{1}\left( x\right) + \cdots + {a}_{n}{\sigma... | Yes |
Corollary 56. Every set \( \left\{ {{\sigma }_{1},\ldots ,{\sigma }_{n}}\right\} \) of distinct automorphisms of a field \( E \) is independent. | Proof. An automorphism \( \sigma \) of \( E \) restricts to a (group) homomorphism \( \sigma : {E}^{\# } \rightarrow {E}^{\# } \), hence is a character. | No |
Lemma 57. If \( G = \left\{ {{\sigma }_{1},\ldots {\sigma }_{n}}\right\} \) is a set of automorphisms of \( E \), then\n\n\[ \left\lbrack {E : {E}^{G}}\right\rbrack \geq n. \] | Proof. Otherwise \( \left\lbrack {E : {E}^{G}}\right\rbrack = r < n \) ; let \( \left\{ {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right\} \) be a basis of \( E/{E}^{G} \) .\n\nConsider the linear system over \( E \) of \( r \) equations in \( n \) unknowns:\n\n\[ {\sigma }_{1}\left( {\alpha }_{1}\right) {x}_{1} + \cdots + ... | Yes |
Corollary 59. If \( G, H \) are subgroups of \( \operatorname{Aut}\left( E\right) \) with \( {E}^{G} = {E}^{H} \), then \( G = H \) . | Proof. If \( \sigma \in G \), then clearly \( \sigma \) fixes \( {E}^{G} \) . To prove the converse, suppose \( \sigma \) fixes \( {E}^{G} \) and \( \sigma \notin G \) . Then \( {E}^{G} \) is fixed by the \( n + 1 \) elements in \( G \cup \{ \sigma \} \) , so Lemma 57 and Theorem 58 give the contradiction:\n\n\[ n = \l... | Yes |
Theorem 60. The following conditions are equivalent for a finite extension \( E/F \) with Galois group \( G = \operatorname{Gal}\left( {E/F}\right) \) .\n\n(i) \( F = {E}^{G} \) ;\n\n(ii) every irreducible \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) with one root in \( E \) is separable and has all its root... | Proof. (i) \( \Rightarrow \) (ii) Let \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) be an irreducible polynomial having a root \( \alpha \) in \( E \), and let the distinct elements of the set \( \{ \sigma \left( \alpha \right) : \sigma \in G\} \) be \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) . Define \( g\lef... | Yes |
Lemma 61. Let \( E/F \) be a Galois extension, and let \( B \) be an intermediate field. The following conditions are equivalent.\n\n(i) \( B \) has no conjugates (other than \( B \) itself);\n\n(ii) If \( \sigma \in \operatorname{Gal}\left( {E/F}\right) \), then \( \sigma \mid B \in \operatorname{Gal}\left( {B/F}\righ... | Proof. (i) \( \Rightarrow \) (ii) Obvious.\n\n(ii) \( \Rightarrow \) (iii) Let \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) be an irreducible polynomial having a root \( \beta \) in \( B \) . Since \( B \subset E \) and \( E/F \) is Galois, all the roots of \( p\left( x\right) \) lie in \( E \) . Suppose the... | Yes |
Lemma 62. If \( L \) and \( {L}^{\prime } \) are lattices and \( \gamma : L \rightarrow {L}^{\prime } \) is an order reversing bijection \( \left( {a \leq b\text{implies}\gamma \left( b\right) \leq \gamma \left( a\right) }\right) \), then\n\n\[ \gamma \left( {a \vee b}\right) = \gamma \left( a\right) \land \gamma \left... | Proof. Now \( a, b \leq a \vee b \) implies \( \gamma \left( a\right) ,\gamma \left( b\right) \geq \gamma \left( {a \vee b}\right) \) ; that is, \( \gamma \left( {a \vee b}\right) \) is a lower bound of \( \gamma \left( a\right) ,\gamma \left( b\right) \) . It follows that \( \gamma \left( a\right) \land \gamma \left( ... | Yes |
Theorem 63 (Fundamental Theorem of Galois Theory). Let \( E/F \) be a Galois extension with Galois group \( G = \operatorname{Gal}\left( {E/F}\right) \). (i) The function \( \gamma : \operatorname{Sub}\left( G\right) \rightarrow \operatorname{Lat}\left( {E/F}\right) \), defined by \( H \mapsto {E}^{H} \), is an order r... | Proof. (i) It is easy to see that \( \gamma \) is order reversing: \( K \leq H \) implies \( {E}^{H} \leq {E}^{K} \). That \( \gamma \) is injective is precisely the statement of Corollary 59. To see that \( \gamma \) is surjective, consider the composite \[ \operatorname{Lat}\left( {E/F}\right) \overset{\delta }{ \rig... | Yes |
Corollary 64. A Galois extension \( E/F \) has only finitely many intermediate fields. | Proof. Its Galois group is finite, hence has only finitely many subgroups. 口 | Yes |
Theorem 65 (Steinitz). A finite extension \( E/F \) is simple if and only if it has only finitely many intermediate fields. | Proof. Assume that \( E = F\left( \alpha \right) \) and let \( p\left( x\right) \) be the irreducible polynomial of \( \alpha \) over \( F \) . Given an intermediate field \( B \), let \( g\left( x\right) \) be the irreducible polynomial of \( \alpha \) over \( B \) . If \( {B}^{\prime } \) is the subfield of \( B \) g... | Yes |
Corollary 67 (Theorem of the Primitive Element). Every Galois extension \( E/F \) is simple. | Proof. Immediate from Corollary 64 and Theorem 65. | No |
Corollary 68. The Galois field \( {GF}\left( {p}^{n}\right) \) has exactly one subfield of order \( {p}^{d} \) for every divisor \( d \) of \( n \) . | Proof. We have seen in Theorem 48 that \( \operatorname{Gal}\left( {{GF}\left( {p}^{n}\right) /{GF}\left( p\right) }\right) \cong {\mathbb{Z}}_{n} \) ; moreover, Lemma 41 shows that a cyclic group of order \( n \) has exactly one subgroup of order \( d \) for every divisor \( d \) of \( n \) . Now a subgroup of order \... | Yes |
Corollary 69. If \( E/F \) is a Galois extension whose Galois group \( \operatorname{Gal}\left( {E/F}\right) \) is abelian, then every intermediate field \( B/F \) is a Galois extension. | Proof. Every subgroup of an abelian group is a normal subgroup. | No |
Corollary 70. Let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) be a separable polynomial, and let \( E/F \) be a splitting field. Let \( f\left( x\right) = g\left( x\right) h\left( x\right) \) in \( F\left\lbrack x\right\rbrack \), and let \( B/F \) and \( C/F \) be splitting fields of \( g\left( x\right), h... | Proof. Recall that if \( H \) and \( K \) are subgroups of a group \( G \), then \( G \) is their direct product, denoted by \( G = H \times K \), if both \( H \) and \( K \) are normal, \( H \cap K = \) \( \{ 1\} \), and \( H \vee K = {HK} = G \) . Let \( G = \operatorname{Gal}\left( {E/F}\right) \) . Since \( B/F \) ... | Yes |
Theorem 71 (Fundamental Theorem of Algebra). Every nonconstant \( f\left( x\right) \in \mathbb{C}\left\lbrack x\right\rbrack \) has a complex root. | Proof. If \( f\left( x\right) \in \mathbb{C}\left\lbrack x\right\rbrack \), then \( f\left( x\right) \bar{f}\left( x\right) \in \mathbb{R}\left\lbrack x\right\rbrack \), where \( \bar{f}\left( x\right) \) is obtained from \( f\left( x\right) \) by taking the complex conjugate of every coefficient. Since \( f\left( x\ri... | Yes |
Lemma 72 (Accessory Irrationalities). Let \( E/F \) be a splitting field of \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) with Galois group \( G = \operatorname{Gal}\left( {E/F}\right) \) . If \( {F}^{ * }/F \) is an extension and \( {E}^{ * }/{F}^{ * } \) is a splitting field of \( f\left( x\right) \) contai... | Proof. The hypothesis gives \( E = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) and \( {E}^{ * } = {F}^{ * }\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) , where \( \left\{ {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right\} \) are the roots of \( f\left( x\right) \) . If \( \sigma \in \operatorname{Gal}\... | No |
Lemma 73 (Hilbert’s Theorem 90). Let \( E/F \) be a Galois extension whose Galois group \( G = \operatorname{Gal}\left( {E/F}\right) \) is cyclic of order \( n \) ; let \( \sigma \) be a generator of \( G \) . Then \( N\left( \alpha \right) = 1 \) if and only if there exists \( \beta \in E \) with\n\n\[ \alpha = {\beta... | Proof. If \( \alpha = {\beta \sigma }{\left( \beta \right) }^{-1} \), then\n\n\[ N\left( \alpha \right) = N\left( {{\beta \sigma }{\left( \beta \right) }^{-1}}\right) = N\left( \beta \right) N\left( {\sigma {\left( \beta \right) }^{-1}}\right) \]\n\n\[ = N\left( \beta \right) N{\left( \sigma \left( \beta \right) \right... | Yes |
Corollary 74. Let \( E/F \) be a Galois extension of prime degree \( p \) . If \( F \) has a primitive pth root of unity, then \( E = F\left( \beta \right) \), where \( {\beta }^{p} \in F \), and so \( E/F \) is a pure extension. | Proof. If \( \omega \) is a primitive \( p \) th root of unity, then \( N\left( \omega \right) = {\omega }^{p} = 1 \), because \( \omega \in F \) . Now \( G = \operatorname{Gal}\left( {E/F}\right) \) has order \( p \), hence is cyclic; let \( \sigma \) be a generator. By Hilbert’s Theorem 90, we have \( \omega = {\beta... | Yes |
Theorem 75 (Galois). Let \( F \) be a field of characteristic 0, let \( E/F \) be a Galois extension, and let \( G = \operatorname{Gal}\left( {E/F}\right) \) be a solvable group. Then \( E \) can be imbedded in a radical extension of \( F \) . | Proof. The proof is by induction on \( \left\lbrack {E : F}\right\rbrack \) . The base step is trivially true. Since \( G \) is solvable, Corollary A17 provides a normal subgroup \( H \) of prime index, say, \( p \) . Let \( \omega \) be a primitive \( p \) th root of unity (which exists because \( F \) has characteris... | Yes |
Theorem 76. (i) A polynomial \( f\left( x\right) \) and its corresponding reduced polynomial \( \widetilde{f}\left( x\right) \) have the same discriminant. | Proof. (i) If the roots of \( f\left( x\right) \) are \( {\alpha }_{1},\ldots ,{\alpha }_{n} \), then the roots of \( \widetilde{f}\left( x\right) \) are \( {\beta }_{1},\ldots ,{\beta }_{n} \), where \( {\beta }_{i} = {\alpha }_{i} + {a}_{n - 1}/n \) . Therefore\n\n\[ \mathop{\prod }\limits_{{i < j}}\left( {{\alpha }_... | Yes |
Lemma 77. Let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) have discriminant \( D = {\Delta }^{2} \) and Galois group \( G = \operatorname{Gal}\left( {E/F}\right) \) . If \( H = G \cap {A}_{n} \), then \( {E}^{H} = F\left( \Delta \right) \) ; moreover, \( \sqrt{D} \in F \) if and only if \( G \) is a subgrou... | Proof. Clearly \( F\left( \Delta \right) \subset {E}^{H} \) and \( \left\lbrack {{E}^{H} : F}\right\rbrack = \left\lbrack {G : H}\right\rbrack \leq 2 \) ; it suffices to prove that \( \left\lbrack {F\left( \Delta \right) : F}\right\rbrack = \left\lbrack {G : H}\right\rbrack \) . If \( \left\lbrack {G : H}\right\rbrack ... | Yes |
Theorem 78. Let \( f\left( x\right) \in \mathbb{Q}\left\lbrack x\right\rbrack \) be an irreducible cubic with Galois group \( G \) and discriminant \( D \). (i) \( f\left( x\right) \) has exactly one real root if and only if \( D < 0 \), in which case \( G \cong {S}_{3} \). (ii) \( f\left( x\right) \) has three real ro... | Proof. Note that \( D \neq 0 \) because \( \mathbb{Q} \), having characteristic 0, is perfect, hence irreducible polynomials have no repeated roots. If \( f\left( x\right) \) has three real roots, then \( \Delta \) is real and \( D = {\Delta }^{2} > 0 \). Conversely assume \( f\left( x\right) \) has one real root \( \a... | Yes |
Theorem 79. If \( g\left( x\right) \) is the resolvent cubic of \( f\left( x\right) = {x}^{4} + q{x}^{2} + {rx} + s \) , then\n\n\[ g\left( x\right) = {x}^{3} - {2q}{x}^{2} + \left( {{q}^{2} - {4s}}\right) x + {r}^{2}. \] | Proof. In our discussion of the classical quartic formula, we saw that \( f\left( x\right) = \left( {{x}^{2} + {kx} + \ell }\right) \left( {{x}^{2} - {kx} + m}\right) \) and \( {k}^{2} \) is a root of\n\n\[ h\left( x\right) = {x}^{3} + {2q}{x}^{2} + \left( {{q}^{2} - {4s}}\right) x - {r}^{2}, \]\n\na polynomial differi... | Yes |
Theorem 80. Let \( f\left( x\right) \in \mathbb{Q}\left\lbrack x\right\rbrack \) be an irreducible quartic with Galois group \( G \), and let \( m \) be the order of the Galois group of its resolvent cubic.\n\n(i) If \( m = 6 \), then \( G \cong {S}_{4} \).\n\n(ii) If \( m = 3 \), then \( G \cong {A}_{4} \).\n\n(iii) I... | Proof. We have seen that \( \mathbb{Q}\left( {u, v, w}\right) \) is the fixed field of \( V \cap G \) . By the Fundamental Theorem,\n\n\[ \left| {G/V \cap G}\right| = \left\lbrack {G : V \cap G}\right\rbrack = \left\lbrack {\mathbb{Q}\left( {u, v, w}\right) : \mathbb{Q}}\right\rbrack = \left| {\operatorname{Gal}\left( ... | No |
Corollary 4 (Theorem of Primitive Element). If \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) are the roots of \( f\left( x\right) \in F\left\lbrack x\right\rbrack \), then there exists \( \eta \) with \( F\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) = F\left( \eta \right) \) . | Proof. Let \( h\left( {{x}_{1},\ldots ,{x}_{n}}\right) \in F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) be an \( n \) !-valued function; for each \( i \), define \( {g}_{i}\left( {{x}_{1},\ldots ,{x}_{n}}\right) \in F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) by \( {g}_{i}\left( {{x}_{1},\ldots ,... | Yes |
Theorem 1. Let \( \mathrm{E}/\Phi ,\mathrm{P}/\Phi \) be fields over \( \Phi \) and let \( {\mathfrak{L}}_{\Phi }\left( {\mathrm{E},\mathrm{P}}\right) \) be the right vector space over \( \mathrm{P} \) of linear mappings of \( \mathrm{E}/\Phi \) into \( \mathrm{P}/\Phi \) . Then \( \left\lbrack {\mathrm{E} : \Phi }\rig... | Proof. Let \( {\eta }_{1},{\eta }_{2},\cdots ,{\eta }_{n} \) be elements of \( \mathrm{E} \) which are linearly independent over \( \Phi \) . Then we may imbed this set in a basis \( \left\{ {\eta }_{\alpha }\right\} \) for \( \widehat{\mathrm{E}} \) over \( \Phi \) (Vol. II, p. 239). If we choose a correspondent \( {\... | Yes |
Theorem 2 (Jacobson-Bourbaki). Let \( \mathrm{P} \) be a field and \( \mathfrak{A} \) a set of endomorphisms of \( \left( {\mathrm{P}, + }\right) \) such that:\n\n(i) \( \mathfrak{A} \) is a subring of \( \mathfrak{L}\left( {\mathrm{P},\mathrm{P}}\right) \) the ring of endomorphisms of \( \left( {\mathrm{P}, + }\right)... | Proof (Hochschild). The verification that \( \Phi \) is a subfield is immediate and will be omitted. Next we apply the lemma of \( §1 \) to obtain elements \( {\rho }_{1},{\rho }_{2},\cdots ,{\rho }_{n} \) in \( \mathrm{P} \) and a right basis \( \left( {{E}_{1},{E}_{2}}\right. \) , \( \left. {\cdots ,{E}_{n}}\right) \... | No |
Theorem 3 (Dedekind). Let \( \mathrm{E} \) and \( \mathrm{P} \) be fields and let \( {s}_{1},{s}_{2},\cdots \) , \( {s}_{n} \) be distinct isomorphisms of \( \mathbf{E} \) into \( \mathrm{P} \) . Then the \( {s}_{i} \) are right linearly independent over \( \mathrm{P} : \sum {s}_{i}{\rho }_{i} = 0,{\rho }_{i}\varepsilo... | Proof. If the assertion is false, then we have a shortest relation, which by suitable ordering reads:\n\n(3)\n\n\[ \n{s}_{1}{\rho }_{1} + {s}_{2}{\rho }_{2} + \cdots + {s}_{r}{\rho }_{r} = 0, \n\] \n\nwhere every \( {\rho }_{i} \neq 0 \) . Suppose \( r > 1 \) . Since \( {s}_{1} \neq {s}_{2} \) there exists \( {\eta \va... | Yes |
Theorem 4. Let \( \mathrm{E} \) and \( \mathrm{P} \) be fields, \( {s}_{1},{s}_{2},\cdots ,{s}_{n} \) isomorphisms of \( \mathrm{E} \) into \( \mathrm{P} \), and let \( \mathfrak{A} \) be the right \( \mathrm{P} \) -subspace of \( \mathfrak{L}\left( {\mathrm{E},\mathrm{P}}\right) \) of endomorphisms \( \sum {s}_{i}{\rh... | Proof. It is clear that \( \left\{ {\mathop{\sum }\limits_{{j = 1}}^{r}{s}_{ij}{\rho }_{ij} \mid {\rho }_{ij}\text{e P}}\right\} \subseteq \mathfrak{B} \) . To prove the opposite inclusion it suffices to show that, if \( \mathop{\sum }\limits_{1}^{n}{s}_{i}{\rho }_{i}\varepsilon \mathfrak{B} \), then the \( {s}_{i} \) ... | Yes |
Theorem 5. Let \( \mathrm{P} \) be a field and let \( \mathcal{A} \) be the collection of finite groups of automorphisms in \( \mathrm{P},\mathcal{I} \) the collection of subfields of \( \mathrm{P} \) which are Galois and of finite co-dimension in \( \mathrm{P} \) . If \( {\Phi \varepsilon }\mathcal{I} \), let \( A\lef... | Proof. (i)-(ii). If \( {G\varepsilon }\mathcal{A} \) and \( \mathfrak{A} = \left\{ {\sum {s}_{i}{\rho }_{i} \mid {s}_{i}{\varepsilon G},{\rho }_{i}\varepsilon \mathrm{P}}\right\} \), then \( \left\lbrack {\mathrm{P} : I\left( G\right) }\right\rbrack = {\left\lbrack \mathfrak{A} : \mathrm{P}\right\rbrack }_{R} = \left( ... | Yes |
Lemma 1. (1) If \( \mathrm{P}/\Phi \) is a splitting field of \( f\left( x\right) {\varepsilon \Phi }\left\lbrack x\right\rbrack \) and \( \sum /\Phi \) is a subfield of \( \mathrm{P}/\Phi \), then \( \mathrm{P}/\sum \) is a splitting field of \( f\left( x\right) \) . (2) If \( \mathrm{P}/\sum \) is a splitting field f... | Proof. (1) This is an immediate consequence of the definition. (2) By assumption we have \( \mathrm{P} = \mathbf{\sum }\left( {{\rho }_{1},\cdots ,{\rho }_{n}}\right) \) where (5) holds in \( \mathrm{P}\left\lbrack x\right\rbrack \) . Also \( \sum = \Phi \left( {{\sigma }_{1},\cdots ,{\sigma }_{r}}\right) \) and \( f\l... | Yes |
Theorem 6. Any polynomial \( f\left( x\right) {\varepsilon \Phi }\left\lbrack x\right\rbrack \) of positive degree has a splitting field \( \mathrm{P}/\Phi \) . | Proof. Let \( f\left( x\right) = {f}_{1}\left( x\right) {f}_{2}\left( x\right) \cdots {f}_{k}\left( x\right) \) be the factorization of \( f\left( x\right) \) into irreducible factors (with leading coefficients 1). Evidently \( k \leq n = \deg f\left( x\right) \) . We use induction on \( n - k \) . If \( n - k \) \( = ... | Yes |
Lemma 2. Let \( \mathrm{P} = \Phi \left( {{\rho }_{1},{\rho }_{2},\cdots ,{\rho }_{m}}\right) \) and assume that \( {\rho }_{i} \) is algebraic over \( \Phi \left( {{\rho }_{1},{\rho }_{2},\cdots ,{\rho }_{i - 1}}\right), i = 1,2,\cdots, m \) . Then \( \left\lbrack {\mathrm{P} : \Phi }\right\rbrack < \infty \) and \( \... | Proof. We have seen that this holds for \( m = 1 \) . Suppose \( m \) \( > 1 \) and assume the result holds for \( r < m \) . Then \( \Phi \left( {{\rho }_{1},\cdots ,{\rho }_{r}}\right) \) \( = \Phi \left\lbrack {{\rho }_{1},\cdots ,{\rho }_{r}}\right\rbrack \) and this is finite dimensional over \( \Phi \) . Since \(... | Yes |
Theorem 7. Let \( \alpha \rightarrow \bar{\alpha } \) be an isomorphism of a field \( \Phi \) onto the field \( \Phi \) and let \( f\left( x\right) \) be a polynomial of positive degree with leading coefficient \( 1, f\left( x\right) \) in \( \Phi \left\lbrack x\right\rbrack \), and let \( \bar{f}\left( x\right) \) be ... | Proof. Both assertions will be proved by induction on \( \left\lbrack {\mathrm{P} : \Phi }\right\rbrack \) . If \( \left\lbrack {\mathrm{P} : \Phi }\right\rbrack = 1,\mathrm{P} = \Phi \) and \( f\left( x\right) = \Pi \left( {x - {\rho }_{i}}\right) \) in \( \Phi \left\lbrack x\right\rbrack \) . Applying the isomorphism... | Yes |
Theorem 8. If \( f\left( x\right) {\varepsilon \Phi }\left\lbrack x\right\rbrack \) and \( \deg f > 0 \), then all the roots of \( f \) (in its splitting field) are simple if and only if \( \left( {f,{f}^{\prime }}\right) = 1 \) (that is, 1 is the highest common factor of \( f \) and \( {f}^{\prime } \) ). | Proof. Let \( d\left( x\right) \) be the highest common factor \( \left( {f,{f}^{\prime }}\right) \) of \( f \) and \( {f}^{\prime } \) in \( \Phi \left\lbrack x\right\rbrack \) (cf. Vol. I, p. 100, p. 122). Suppose \( f\left( x\right) \) has a multiple root in \( \mathrm{P}\left\lbrack x\right\rbrack \), so \( f\left(... | Yes |
Lemma 1. Let \( \mathrm{P} \supseteq \mathrm{E} \supseteq \Phi \) where \( \mathrm{E} \) and \( \Phi \) are subfields of \( \mathrm{P} \) and \( \mathrm{E}/\Phi \) is finite dimensional Galois. Then any element \( \theta \mathrm{e}\mathrm{P} \) which is separable algebraic over \( \mathrm{E} \) is separable algebraic o... | Proof. Let \( g\left( x\right) \) be the minimum polynomial of \( \theta \) over \( \mathbf{E} \) . If \( {s\varepsilon G} \) the Galois group of \( \mathbf{E}/\Phi \), then \( s \) has a unique extension to \( \mathrm{E}\left\lbrack x\right\rbrack \) satisfying \( {x}^{s} = x \) . Let \( {g}^{{s}_{1}}\left( x\right) ,... | Yes |
Theorem 11. If \( \mathrm{A}/\Phi \) is algebraic, then the set \( \sum \) of elements of \( \mathrm{A} \) which are separable over \( \Phi \) is a subfield containing \( \Phi \) . Moreover, \( \sum \) contains every element of \( \mathrm{A} \) which is separable algebraic over \( \sum \) . | Proof. Let \( \rho ,\sigma \) e \( \sum \) and let \( g\left( x\right) \) and \( h\left( x\right) \) be the minimum polynomials over \( \Phi \) of \( \rho \) and \( \sigma \) respectively. Then \( f\left( x\right) = g\left( x\right) h\left( x\right) \) is separable. If \( \Delta \) is a splitting field over \( \Phi \le... | Yes |
Theorem 14. Let \( \Phi \) be an infinite field and let \( \mathrm{P} = \Phi \left( {\xi ,\eta }\right) \) be a field generated over \( \Phi \) by a separable algebraic element \( \xi \) and an algebraic element \( \eta \) . Then \( \mathrm{P}/\Phi \) has a primitive element. | Proof. Let \( f\left( x\right) \) and \( g\left( x\right) \) be the minimum polynomial over \( \Phi \) of \( \xi \) and \( \eta \) respectively and let \( \Delta /\mathrm{P} \) be a splitting field of \( f\left( x\right) g\left( x\right) \) . Then \( \Delta /\Phi \) is a splitting field of \( f\left( x\right) g\left( x... | Yes |
Theorem 15 (Artin). Let \( \Phi \) be an infinite field and \( \mathrm{P} \) a finite dimensional extension field of \( \Phi \) . Then \( \mathrm{P}/\Phi \) is a simple extension if and only if there are only a finite number of intermediate fields between \( \mathrm{P} \) and \( \Phi \) . | Proof. Suppose first that \( \mathrm{P} = \Phi \left( \theta \right) \) and let \( \mathrm{E} \) be an intermediate field. Let \( g\left( x\right) \) be the minimum polynomial of \( \theta \) over \( \mathrm{E} \) and let \( {\mathrm{E}}^{\prime }/\Phi \) be the field generated by the coefficients of \( g\left( x\right... | Yes |
Theorem 16. Let \( \mathrm{P} \) be finite dimensional Galois over an infinite field \( \Phi ,\mathbf{E} \) a subfield of \( \mathrm{P}/\Phi \), and \( \Omega \) an arbitrary extension field of P. Let \( {s}_{1},{s}_{2},\cdots ,{s}_{m} \) be the different isomorphisms of \( \mathbf{E} \) over \( \Phi \) into P over \( ... | Proof. We recall that the number \( m \) of isomorphisms is \( \left\lbrack {\mathrm{E} : \Phi }\right\rbrack \) (§ 7). We note next that, if \( \left( {{\epsilon }_{1},{\epsilon }_{2},\cdots ,{\epsilon }_{m}}\right) \) is a basis for \( \mathrm{E}/\Phi \) , then the determinant det \( \left( {{\epsilon }_{i}{}^{{s}_{j... | Yes |
Theorem 17. Let \( \mathrm{P} \) be finite dimensional Galois over an infinite \( \Phi \) . Then \( \mathrm{P}/\Phi \) has a normal basis. | Proof. Let \( G = \left\{ {{s}_{1},\cdots ,{s}_{n}}\right\} \) be the Galois group of \( \mathrm{P}/\Phi \) . We have just seen that, if \( \left( {{\rho }_{1},\cdots ,{\rho }_{n}}\right) \) is a basis of \( \mathrm{P} \) over \( \Phi \), then det \( \left( {{\rho }_{i}{}^{{s}_{i}}}\right) \neq 0 \) . Conversely, this ... | Yes |
Lemma 1. Any finite subgroup \( A \) of the multiplicative group of a field is cyclic. | Proof. Let \( m \) be the order of \( A \) and let \( {m}^{\prime } \) be the highest order for the elements of \( A \) . It is known that, if \( a \) and \( b \) are two elements of a finite commutative group, then there exists a \( c \) in the group whose order is the least common multiple of the orders of \( a \) an... | Yes |
Lemma 2. Any cyclic extension \( \mathrm{P}/\Phi \) has a normal basis over \( \Phi \) . | Proof. Let \( s \) be a generator of the Galois \( G \) group of \( \mathrm{P}/\Phi \) . We consider \( s \) as a linear transformation in \( \mathrm{P} \) over \( \Phi \) and let \( \mu \left( x\right) \varepsilon \) \( \Phi \left\lbrack x\right\rbrack \) be its minimum polynomial. Now Dedekind’s independence theorem ... | Yes |
Theorem 19. Let \( s \rightarrow {\mu }_{s} \) be a mapping of \( G \) into \( {\mathrm{P}}^{ * } \) such that \( {\mu }_{st} = {\mu }_{s}{}^{t}{\mu }_{t}, s, t \) e \( G \) . Then there exists a non-zero element \( \gamma \) in \( \mathrm{P} \) such that \( {\mu }_{s} = \gamma {\left( {\gamma }^{s}\right) }^{-1} \) . | Proof. Since the \( {\mu }_{s} \) are \( \neq 0 \) and the automorphisms are right linearly independent over \( \mathrm{P} \), we see that the operator \( {\sum s}{\mu }_{s}( \equiv \)\n\n\( \left. {{\sum s}{\mu }_{sR}}\right) \) is \( \neq 0 \) . Thus we can find a \( {\beta \varepsilon }\mathrm{P} \) such that \( \ga... | Yes |
Theorem 20. Let \( {\delta }_{\varepsilon },{s\varepsilon G} \), be elements of \( \mathrm{P} \) satisfying (63). Then there exists a \( \gamma \) e \( \mathrm{P} \) such that \( {\delta }_{s} = \gamma - {\gamma }^{s} \) . | Proof. We choose an element \( {\rho \varepsilon }\mathrm{P} \) such that \( {T}_{\mathrm{P} \mid \Phi }\left( \rho \right) = \sum {\rho }^{s} \neq \) 0. This can be done since \( \mathop{\sum }\limits_{{s \neq G}}s \neq 0 \) by the Dedekind independence theorem. Set \( \gamma = \mathop{\sum }\limits_{{s \in G}}T{\left... | Yes |
Theorem 21. Let \( \mathrm{E}/\Phi \) and \( \mathrm{P}/\Phi \) be fields such that \( \left\lbrack {\mathrm{P} : \Phi }\right\rbrack < \infty \) and let \( \mathfrak{J} \) be a maximal ideal in \( \mathrm{E}{ \otimes }_{\Phi }\mathrm{P} \) . Let \( s \) be the mapping \( \epsilon \rightarrow \) \( \epsilon \otimes 1 +... | Proof. If \( \Im \) is a maximal ideal in \( \mathrm{E} \otimes \mathrm{P} \), then \( \epsilon \rightarrow \epsilon \otimes 1 \) is a homomorphism into \( \mathrm{E} \otimes \mathrm{P} \) so \( s : \epsilon \rightarrow \epsilon \otimes 1 + \Im \) is a homomorphism into \( \Gamma = \left( {\mathrm{E} \otimes \mathrm{P}... | Yes |
Theorem 1. Let \( \Phi \) be a field of characteristic \( \neq 2 \) and \( f\left( x\right) \) a nonzero polynomial \( \mathrm{e}\Phi \left\lbrack x\right\rbrack \) without multiple roots. Let \( \mathrm{P}/\Phi \) be a splitting field of \( f\left( x\right) ,{\rho }_{1},{\rho }_{2},\cdots ,{\rho }_{m} \) its roots, \(... | Proof. We recall a standard characterization of the alternating group. For this one considers the ring \( \Phi \left\lbrack {{x}_{1},{x}_{2},\cdots ,{x}_{m}}\right\rbrack ,{x}_{i} \) in-determinates. If \( i \rightarrow {i}^{\sigma } \) is a permutation of \( 1,2,\cdots, m \), then we have the automorphism \( A\left( \... | Yes |
Theorem 2. Let \( f\left( x\right) \) e \( \Phi \left\lbrack x\right\rbrack \) have no multiple roots in its splitting field \( \mathrm{P} \) . Then \( f\left( x\right) \) is irreducible in \( \Phi \left\lbrack x\right\rbrack \) if and only if the Galois group \( {G}_{f} \) of \( f\left( x\right) = 0 \) over \( \Phi \)... | Proof. We recall that a transformation group of a set \( M \) is called transitive if given any pair \( \left( {x, y}\right), x, y \) e \( M \) there exists a \( \sigma \) in the group such that \( {x}^{\sigma } = y \) . Suppose first that \( f\left( x\right) \) is irreducible in \( \Phi \left\lbrack x\right\rbrack \) ... | Yes |
If the characteristic of \( \Phi \) is not a divisor of \( n \) ( \( 0 \) included), then the Galois group \( G \) of the cyclotomic field \( \mathrm{P}/\Phi \) of order \( n \) is isomorphic to a subgroup of the multiplicative group \( U\left( n\right) \) of units in \( I/\left( n\right), I \) the ring of integers. | Proof. As in \( §1 \) let \( {G}_{f} \) denote the group of permutations of the set \( Z\left( n\right) \) of roots induced by \( G \) . Since the elements of \( {G}_{f} \) are restrictions of automorphisms, it is clear that they are automor-phisms of the multiplicative group of \( Z\left( n\right) \) . Hence \( {G}_{f... | Yes |
Theorem 4. If \( \Phi \) contains \( n \) distinct \( n \) -th roots of 1 then the Galois group of the equation \( {x}^{n} = \alpha \) over \( \Phi \) is cyclic of order a divisor of \( n \) . | Proof. Let \( \mathrm{P}/\Phi \) be a splitting field over \( \Phi \) of \( {x}^{n} - \alpha, G \) its Galois group. We have to show that \( G \) is cyclic. If \( \alpha = 0 \), we have \( \mathrm{P} = \Phi, G = 1 \) . Hence we assume \( \alpha \neq 0 \) . Let \( \rho \) be one of the roots of \( {x}^{n} - \alpha \) in... | Yes |
Theorem 5. Assume \( \Phi \) has \( n \) distinct \( n \) -th roots of 1 and let \( \mathrm{P}/\Phi \) be a cyclic \( n \) dimensional extension field. Then \( \mathrm{P} = \Phi \left( \xi \right) \) where \( {\xi }^{n} = {\alpha \varepsilon \Phi } \) | Proof. The hypothesis on \( \mathrm{P} \) is that \( \mathrm{P}/\Phi \) is Galois with Galois group \( G \) which is cyclic of order \( n \) . Since \( \mathrm{P} \) is separable over \( \Phi \) it has a primitive element so \( \mathrm{P} = \Phi \left( \theta \right) \) . Let \( s \) be a generator of \( G \) and let \... | No |
Theorem 6. Let \( \mathrm{P}/\Phi \) be finite dimensional Galois over \( \Phi \) and let \( {\mathrm{P}}^{\prime } \) be an extension field of \( \mathrm{P} \) such that \( {\mathrm{P}}^{\prime } \) is generated by \( \mathrm{P} \) and a second subfield \( {\Phi }^{\prime } \supseteq \Phi \) . Then \( {\mathrm{P}}^{\p... | Proof. We know that \( \mathrm{P} = \Phi \left( {{\xi }_{1},\cdots ,{\xi }_{n}}\right) \) where the \( {\xi }_{i} \) are the roots of a separable polynomial \( f\left( x\right) \) e \( \Phi \left\lbrack x\right\rbrack \) . Since \( {\mathrm{P}}^{\prime } \) is generated by \( {\Phi }^{\prime } \supseteq \Phi \) and \( ... | Yes |
Theorem 7. The general equation of the n-th degree (13) is irreducible in \( \sum = \Phi \left( {{t}_{1},{t}_{2},\cdots ,{t}_{n}}\right) \) and has distinct roots. The Galois group of \( f\left( x\right) = 0 \) is the symmetric group \( {S}_{n} \) . | Since \( {S}_{n} \) is not solvable if \( n > 4 \) this implies the\n\nTheorem of Abel-Ruffini. The general equation of the n-th degree is not solvable by radicals if \( n > 4 \) (characteristic 0 ). | Yes |
Theorem 8. Let \( f\left( x\right) \) be a polynomial of prime degree with rational coefficients which is irreducible in the rational field. Suppose \( f\left( x\right) = 0 \) has exactly two non-real roots in the field \( C \) of complex numbers. Then the group \( {G}_{f} \) of \( f\left( x\right) = 0 \) over the rati... | Proof. The fundamental theorem of algebra asserts that \( f\left( x\right) = \left( {x - {\rho }_{1}}\right) \left( {x - {\rho }_{2}}\right) \cdots \left( {x - {\rho }_{p}}\right) \) in \( C\left\lbrack x\right\rbrack \) . Then the subfield \( \mathrm{P} = {R}_{0}\left( {{\rho }_{1},{\rho }_{2},\cdots ,{\rho }_{p}}\rig... | Yes |
Theorem 4. \( U\left( 2\right) \) and \( U\left( 4\right) \) are cyclic and, if \( e \geq 3 \), then \( U\left( {2}^{e}\right) \) is a direct product of a cyclic group of order 2 and one of order \( {2}^{e - 2} \) . | Proof. The order of \( U\left( {2}^{e}\right) \) is \( \varphi \left( {2}^{e}\right) = {2}^{e - 1} \) . If \( e = 1,\left( {U\left( 2\right) : 1}\right) \) \( = 1 \) and if \( e = 2, U\left( {2}^{e}\right) = U\left( 4\right) \) has only two elements and so is cyclic. Suppose \( e \geq 3 \) . We show first that there ar... | Yes |
Corollary 1. If \( a \neq 1 \) in \( A \), then there exists a character \( {\chi \varepsilon } \) Hom \( \left( {A, Z}\right) \) such that \( {a}^{x} \neq 1 \) . | Proof. Let \( B \) be the subgroup of \( A \) of elements \( b \) such that \( {b}^{x} = \) 1 for all \( {\chi \varepsilon } \) Hom \( \left( {A, Z}\right) \) . Then we see immediately that our assertion will follow if we can show that \( B = 1 \) . Now let \( \chi \mathbf{e} \) Hom \( \left( {A, Z}\right) \) . Since \... | Yes |
For a \( \mathrm{e}A \) define a mapping \( {\eta }_{a} \) of Hom \( \left( {A, Z}\right) \) into \( Z \) by \( {\chi }^{{\eta }_{a}} = {a}^{\chi } \) . Then \( {\eta }_{a}\varepsilon \operatorname{Hom}\left( {\operatorname{Hom}\left( {A, Z}\right), Z}\right) \) and the mapping \( a \rightarrow {\eta }_{a} \) is an iso... | Proof. Observe first that \( a \rightarrow {\eta }_{a} \) is a homomorphism since \( {\chi }^{{\eta }_{ab}} = {\left( ab\right) }^{\chi } = {a}^{\chi }{b}^{\chi } = {\chi }^{{\eta }_{a}}{\chi }^{{\eta }_{b}} = {\chi }^{{\eta }_{a}{\eta }_{b}} \) (the last equation by the definition of the product in a character group).... | Yes |
Corollary 3. A set \( \left\{ {{\chi }_{1},{\chi }_{2},\cdots ,{\chi }_{r}}\right\} \) of characters generate the character group Hom \( \left( {A, Z}\right) \) if and only if the only a \( {\varepsilon A} \) satisfying \( {a}^{{\chi }_{i}} = 1, i = 1,2,\cdots, r \) is \( a = 1 \) . | Proof. This is equivalent to the dual statement \( \left\{ {{a}_{1},{a}_{2},\cdots ,{a}_{r}}\right\} \) generate \( A \) if and only if \( {a}_{i}x = 1 \), for \( i = 1,2,\cdots, r \) holds only for the character 1. This is easy; for, if \( {a}_{1},\cdots ,{a}_{r} \) generate \( A \) and \( {a}_{i}{}^{x} = 1 \) holds f... | Yes |
Theorem 7. Let \( \Phi \) be a field containing \( m \) distinct \( m \) -th roots of 1 and let \( \mathrm{P}/\Phi \) be a Kummer \( {m}^{\prime } \) -extension where \( {m}^{\prime } \mid m \) . Let \( M\left( \mathrm{P}\right) \) be defined by (7) where \( {\mathrm{P}}^{ * } \) is the multiplicative group of \( \math... | Proof. The first statement on the exactness of the displayed sequence means that \( {\Phi }^{ * } \) is the kernel of the mapping \( \rho \rightarrow {\chi }_{\rho } \) and this mapping is surjective on Hom \( \left( {G, Z}\right) \) . Both of these facts were established above. Consequently, we have Hom \( \left( {G, ... | Yes |
Theorem 8. Let \( \Phi \) be a field containing \( m \) distinct \( m \) -th roots of 1 and let \( N \) be a subgroup of \( {\Phi }^{ * } \) containing \( {\Phi }^{*m} \) such that \( N/{\Phi }^{*m} \) is finite. Then there exists a Kummer \( {m}^{\prime } \) -extension \( \mathrm{P}/\Phi \) with \( {m}^{\prime } \mid ... | Proof. The foregoing analysis of Kummer extensions gives the clue to the definition of \( \mathrm{P}/\Phi \) . In view of this, we are led to choose \( {\alpha }_{1},{\alpha }_{2},\cdots ,{\alpha }_{r} \) in the given group \( N \) so that the cosets \( {\alpha }_{i}{\Phi }^{*m} \) generate \( N/{\Phi }^{*m} \) . Let \... | Yes |
Lemma 1. Let \( \mu \geq 1,0 \leq k \leq m - 1, a = \left( {a}_{\nu }\right), b = \left( {b}_{\nu }\right) \) , \( 0 \leq \nu \leq m - 1,{a}_{\nu },{b}_{\nu }{\varepsilon I}\left\lbrack {{x}_{i},{y}_{j}}\right\rbrack \) . Write \( {a}^{\varphi } = \left( {a}^{\left( \nu \right) }\right) ,{b}^{\varphi } = \left( {b}^{\l... | Proof. We have \( {a}^{\left( 0\right) } = {a}_{0},{b}^{\left( 0\right) } = {b}_{0} \), so the result is clear for \( k = 0 \) . To prove the result by induction on \( k \) we may assume that both sets (16) and (17) hold for \( 0 \leq \nu \leq k - 1 \) and prove that under these conditions \( {a}_{k} \equiv {b}_{k}\lef... | Yes |
Theorem 9. If \( x \circ y \) denotes \( x + y,{xy} \) or \( x - y \), then \( {\left( x \circ y\right) }_{v} \) is a polynomial in \( {x}_{0},{y}_{0},{x}_{1},{y}_{1},\cdots ,{x}_{\nu },{y}_{\nu } \) with integer coefficients. | Proof. Since \( {\left( x \circ y\right) }_{\nu } \) is a polynomial in \( {x}_{0},{y}_{0},\cdots ,{x}_{\nu },{y}_{\nu } \) with rational coefficients, it suffices to prove that \( {\left( x \circ y\right) }_{v} \) e \( I\left\lbrack {{x}_{i},{y}_{j}}\right\rbrack \) . This is clear for \( {\left( x \circ y\right) }_{0... | Yes |
Theorem 10. \( {\mathfrak{W}}_{m}\left( \mathfrak{A}\right) \) is a commutative ring. | Proof. Let \( a = \left( {a}_{\nu }\right), b = \left( {b}_{\nu }\right), c = \left( {c}_{\nu }\right) \) be any three elements of \( {\mathfrak{W}}_{m}\left( \mathfrak{A}\right) \) . Then we have just seen that we have a homomorphism of \( {\mathfrak{J}}_{m} \) into \( {\mathfrak{W}}_{m}\left( \mathfrak{A}\right) \) s... | Yes |
\[ {p1} = \overset{p}{\overbrace{1 + 1 + 1\cdots + 1}} = {1}^{VR} \] | Proof. Consider the subrings \( {\Im }_{m - 1},{\Im }_{m},{\Im }_{m + 1} \) of \( {\mathfrak{X}}_{m - 1},{\mathfrak{X}}_{m} \) , \( {\mathfrak{X}}_{m + 1} \) of elements with components in \( I\left\lbrack {{x}_{i},{y}_{j},{z}_{k}}\right\rbrack \) and define the mappings \( R \) and \( V \) for these in the same way as... | No |
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