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Lemma 40.6. Suppose that \( R \) is a UFD and \( A \) an \( m \times n \) matrix in \( {R}^{m \times n} \) . Let \( P \) and \( Q \) be invertible matrices in \( {\mathrm{{GL}}}_{m}\left( R\right) \) and \( {\mathrm{{GL}}}_{n}\left( R\right) \) respectively. Set \( B = {PAQ} \) . If \( 1 \leq l \leq \min \{ m, n\} \) a... | Proof. Let \( P = \left( {p}_{ij}\right), A = \left( {a}_{ij}\right), Q = \left( {q}_{ij}\right) \) and \( c \) in \( R \) a gcd of all the \( l \) -order minors of \( {PA} \) . Then the \( {ki} \) th entry of \( {PA} \) is \( \mathop{\sum }\limits_{j}{p}_{kj}{a}_{ji} \), so the \( k \) th row of \( {PA} \) is \( \math... | Yes |
Corollary 41.2. Let \( R \) be a PID and \( M \) a finitely generated \( R \) module. Then there exists an exact sequence of \( R \) -modules\n\n\[ 0 \rightarrow {R}^{m}\overset{g}{ \rightarrow }{R}^{n}\overset{f}{ \rightarrow }M \rightarrow 0. \]\n\nwith \( m \leq n \) . | Proof. As \( M \) is generated by \( n \) elements, for some positive integer \( n \), we have an exact sequence\n\n\[ 0 \rightarrow \ker f\overset{inc}{ \rightarrow }{R}^{n}\overset{f}{ \rightarrow }M \rightarrow 0. \]\n\nAs \( \ker f \subset {R}^{n} \), it is \( R \) -free of rank at most \( n \) by the proposition. | No |
Corollary 41.3. Let \( R \) be a PID and \( M \) a finitely generated \( R \) -module. Suppose that \( M \) can be generated by \( n \) elements. Then any submodule of \( M \) can be generated by \( n \) elements. | Proof. Let\n\n\[ 0 \rightarrow {R}^{m}\overset{g}{ \rightarrow }{R}^{n}\overset{f}{ \rightarrow }M \rightarrow 0 \]\n\nbe a free resolution of \( M \) and \( N \) a submodule of \( M \) . By the Correspondence Principle, there exists a submodule \( B \) of \( {R}^{n} \) satisfying \( \ker f \subset B \subset {R}^{n} \)... | Yes |
Theorem 41.6. (Fundamental Theorem of fg Modules over a PID, Form I) Let \( R \) be a PID and \( M \) a nontrivial finitely generated \( R \) -module. Then \( M \) is a direct sum of cyclic \( R \) -modules. More precisely, there exist a nonnegative integer \( r \) and\n\n(*)\n\n\[{m}_{1},\ldots ,{m}_{r} \in M\\text{sa... | Proof. Existence: We have shown that there exists an exact sequence\n\n(41.7)\n\n\[0 \\rightarrow {R}^{m}\\overset{g}{ \\rightarrow }{R}^{n} \\rightarrow M \\rightarrow 0\]\n\nof \( R \) -modules with \( m \\leq n \), i.e, \( M \\cong \\operatorname{coker}g = {R}^{n}/\\operatorname{im}g \) . Let \( {\\mathcal{S}}_{m} \... | Yes |
Let \( R \) be a PID and \( {d}_{1}\left| \cdots \right| {d}_{r} \) nonzero nonunits in \( R \) . Set\n\n\[ N = R/\left( {d}_{1}\right) \coprod \cdots \coprod R/\left( {d}_{r}\right) \coprod {R}^{s}\text{and} \]\n\n\[ {N}_{0} = R/\left( {d}_{1}\right) \coprod \cdots \coprod R/\left( {d}_{r}\right) \]\n\nso \( N = {N}_{... | As \( {R}^{s} \) is \( R \) -free, it is \( R \) -torsion-free, so \( v = 0 \) and we conclude that \( {N}_{0} = {N}_{t} \) . As the\n\ndiagram\n\n\n\nhas exact rows and is commutative, the vertical right hand arrow ... | No |
Let \( R \) be a PID and \( M \) a nontrivial cyclic \( R \) -module. Then \( M = {Rm} \cong R/\left( d\right) \), for some \( m \in M \) and unique \( \left( d\right) = {\operatorname{ann}}_{R}m < \) \( R \) . Assume that \( M \) is not \( R \) -free, then \( \left( d\right) > 0 \) and \( d \) is not a unit. Let \( d ... | By the Chinese Remainder Theorem, \( R/\left( d\right) \cong R/\left( {p}_{1}^{{e}_{1}}\right) \times \cdots \times R/\left( {p}_{r}^{{e}_{r}}\right) \) as rings. It follows that \( R/\left( d\right) \cong R/\left( {p}_{1}^{{e}_{1}}\right) \coprod \cdots \coprod R/\left( {p}_{r}^{{e}_{r}}\right) \) as \( R \) -modules,... | Yes |
Theorem 41.20. (Primary Decomposition Theorem) Let \( R \) be a PID and \( M \) a torsion \( R \) -module. If \( p \in \mathcal{P} \), let \[ {M}_{p} \mathrel{\text{:=}} \left\{ {x \in M \mid {p}^{r}x = 0\text{ for some positive integer }r}\right\} . \] Then \( {M}_{p} \) is a p-primary submodule of \( M \) and \( M = ... | Proof. Clearly, \( {M}_{p} \) is a \( p \) -primary submodule of \( M \) for all \( p \in \mathcal{P} \) . We may assume that \( M \) is not trivial. \( M = \mathop{\sum }\limits_{\mathcal{P}}{M}_{p} : \) Let \( x \) be a nonzero element of \( M \) and \( \left( d\right) = {\operatorname{ann}}_{R}x \) . Then \( \bar{d}... | Yes |
Theorem 41.22. (Fundamental Theorem of fg Modules over a PID, Form II) Let \( R \) be a PID and \( M \) a nontrivial finitely generated \( R \) - module. Then there exists a finitely generated free submodule \( P \) of \( M \) of unique rank and \( {p}_{i} \in \mathcal{P}, i = 1,\ldots, r \) some \( r \) and \( {x}_{ij... | Moreover, this decomposition is unique relative the ordering of the \( {p}_{i} \) subject to the order of the \( {e}_{ij} \) above and the uniqueness of the rank of \( P \) . Generators of the ideals \( \left( {p}_{i}^{{e}_{ij}}\right) \) in the theorem are called elementary \( {di} \) - visors of \( M \) . They are un... | Yes |
Find all abelian groups up to isomorphism of order \( {2}^{3} \cdot {3}^{2} \cdot 5 \) . | We write these groups in both forms. Given the factorization, it is easier to write down the representatives of isomorphism classes in Form II. To get Form I, we take the biggest \( p \) -primary pieces for each prime \( p \) and combine them to get one term, then continue with what is left. This is illustrated in our ... | Yes |
Lemma 42.2. Suppose \( V \) is a finite dimensional vector space over \( F \) and \( T : V \rightarrow V \) a linear operator. Let \( V \) be an \( F\left\lbrack t\right\rbrack \) -module by evaluation at \( T \), then\n\n\[ \left( {q}_{T}\right) = {\operatorname{ann}}_{F\left\lbrack t\right\rbrack }V \mathrel{\text{:=... | Proof. We know that \( {q}_{T}v \) is zero for all \( v \) in \( V \), so \( {q}_{T} \) lies in \( {\operatorname{ann}}_{F\left\lbrack t\right\rbrack }V \) . Let \( \varphi : F\left\lbrack t\right\rbrack \rightarrow {\operatorname{End}}_{F}\left( V\right) \) be evaluation at \( T \) . Suppose that \( f \in {\operatorna... | Yes |
Proposition 42.5. Suppose \( V \) is a finite dimensional vector space over \( F \) and \( T : V \rightarrow V \) a linear operator. Let \( V \) be a cyclic \( F\left\lbrack t\right\rbrack \) -module by evaluation at \( T \), i.e., there exists a vector \( v \) in \( V \) satisfying \( V = \) \( F\left\lbrack t\right\r... | Proof. \( \mathcal{B} \) spans \( \mathrm{V} \) as a vector space over \( F \) : Let \( w \) be a vector in \( V = F\left\lbrack t\right\rbrack v \), so there exists a polynomial \( h \) in \( F\left\lbrack t\right\rbrack \) satisfying \( w = {hv} = h\left( T\right) \left( v\right) \), As \( F\left\lbrack t\right\rbrac... | Yes |
Theorem 42.6. (Rational Canonical Form) Suppose \( V \) is a nontrivial finite dimensional vector space over \( F \) and \( T : V \rightarrow V \) a linear operator. Let \( V \) be an \( F\left\lbrack t\right\rbrack \) -module by evaluation at \( T \) . Then there exists a direct sum decomposition of \( F\left\lbrack t... | Proof. Apply the Fundamental Theorem 41.6 and Proposition 42.5 after noting that \( {V}_{i} \) being \( F\left\lbrack t\right\rbrack \) -cyclic implies that \( {\operatorname{ann}}_{F\left\lbrack t\right\rbrack }{V}_{i} = \) \( \left( {q}_{i}\right) = {\left. \left( {\left. q\right| }_{{V}_{i}}\right) \text{ and }T\rig... | Yes |
Corollary 42.11. Let \( F \) be a field and \( A \in {\mathbb{M}}_{n}\left( F\right) \) . Then \( A \) is similar to its transpose \( {A}^{t} \) . | Proof. It follows Theorem 40.8 (or by the algorithm given in the proof of Theorem 104.2 in Appendix §104) that a matrix and its transpose in \( {\mathbb{M}}_{n}\left( {F\left\lbrack t\right\rbrack }\right) \) have equivalent Smith Normal Forms. In particular, the matrices \( {tI} - A \) and \( {tI} - {A}^{t} \) have th... | Yes |
Lemma 42.14. Suppose \( V \) is a finite dimensional vector space over \( F \) and \( T : V \rightarrow V \) a linear operator. Let \( V \) be a cyclic \( F\left\lbrack t\right\rbrack \) -module by evaluation at \( T \), i.e., there exists a vector \( v \) in \( V \) satisfying \( V = F\left\lbrack t\right\rbrack v \) ... | Proof. We know that \( \mathcal{C} = \left\{ {v, T\left( v\right) ,\ldots ,{T}^{d - 1}\left( v\right) }\right\} \) is an \( F \) -basis for \( V \) by our previous work. If \( \mathcal{B} \) is linearly dependent, then there exists an equation \[ \text{(*)}\;0 = \mathop{\sum }\limits_{{i = 0}}^{{d - 1}}{a}_{i}{\left( T... | Yes |
Theorem 43.3. Let \( V \) be a finite dimensional vector space over the field \( F \) and \( \mathcal{S} \) a set of commuting semisimple linear operators in \( {\operatorname{End}}_{F}\left( V\right) \) , i.e., if \( T \) and \( S \) lie in \( \mathcal{S} \), they are semisimple and \( {ST} = {TS} \) . Then there exis... | Proof. If \( T = \rho {1}_{V} \) for some \( \rho \in F \) for every \( T \in \mathcal{S} \), the result is trivial. So we may assume that there exists a \( T \) in \( \mathcal{S} \) so that this is not true. As \( T \) can have only finitely many eigenvalues, say \( {\lambda }_{1},\ldots ,{\lambda }_{r} \) , we have \... | "No" |
Theorem 43.5. Let \( V \) a finite dimensional vector space over \( F \), and \( T \) a linear operator on \( V \). Suppose that \( {f}_{T} \) splits over \( F \). Then there exist unique linear operators \( {T}_{s} \) and \( {T}_{n} \) on \( V \) semisimple and nilpotent respectively and if \( T \) is a linear automor... | Proof. Let \( V \) be an \( F\left\lbrack t\right\rbrack \) -module by evaluation at \( T \) and suppose that the characteristic polynomial of \( T \) is \( {f}_{T} = \mathop{\prod }\limits_{{i = 1}}^{r}{\left( t - {\lambda }_{i}\right) }^{{e}_{i}} \) in \( F\left\lbrack t\right\rbrack \) with \( {\lambda }_{i} \neq {\... | Yes |
Theorem 44.1. Let \( R \) be a commutative ring and \( M \) a finitely generated \( R \) -module. Suppose that \( f \) is an \( R \) -endomorphism of \( M \) . Then, in the notation above, we have an exact sequence\n\n\[ \n0 \rightarrow M\left\lbrack t\right\rbrack \xrightarrow[]{t{1}_{M\left\lbrack t\right\rbrack } - ... | Proof. Exactness at the middle \( M\left\lbrack t\right\rbrack \) : As \n\n\[ \n\left( {{\varphi }_{f} \circ \left( {t{1}_{M\left\lbrack t\right\rbrack } - f\left\lbrack t\right\rbrack }\right) }\right) \left( {\sum {t}^{i}{x}_{i}}\right) = {\varphi }_{f}\left( {\sum {t}^{i + 1}{x}_{i} - {t}^{i}f\left( {x}_{i}\right) }... | No |
Proposition 45.3. Let \( L/K/F \) be an extension of fields. Then \( L/F \) is a finite extension if and only if both \( L/K \) and \( K/F \) are finite field extensions. Moreover, if \( L/F \) is a finite extension, then \[ \left\lbrack {L : F}\right\rbrack = \left\lbrack {L : K}\right\rbrack \left\lbrack {K : F}\righ... | Proof. We prove a stronger statement. Let \[ \mathcal{B} = {\left\{ {x}_{i}\right\} }_{I}\text{be an}F\text{-basis for}K\text{,} \] \[ \mathcal{C} = {\left\{ {y}_{j}\right\} }_{J}\text{be an}K\text{-basis for}L\text{, and} \] \[ \mathcal{D} = {\left\{ {x}_{i}{y}_{j}\right\} }_{I \times J} \] Claim. \( \mathcal{D} \) is... | Yes |
Lemma 45.6. Let \( K \) be a field and \( S \) a non-empty subset of \( K \) . Then there exists a unique minimal subfield \( {F}_{0} \) of \( K \) containing \( S \), i.e., if \( S \subset F \subset K \) with \( F \) a field, then \( {F}_{0} \subset F \) . | Proof. Let\n\n\[ \mathcal{F} = \{ F \mid F\text{ a field satisfying }S \subset F \subset K\} .\n\]\n\nAs \( K \in \mathcal{F} \), the set \( \mathcal{F} \) is not empty. Set \( {F}_{0} = \mathop{\bigcap }\limits_{\mathcal{F}}F \) . Certainly, \( S \subset {F}_{0} \) . We show that \( {F}_{0} \) works. We first prove th... | Yes |
Theorem 45.12. Let \( K/F \) be an extension of fields and \( \alpha \) an element in \( K \) that is algebraic over \( F \) . Then there exists a unique monic irreducible polynomial \( {m}_{F}\left( \alpha \right) \) in \( F\left\lbrack t\right\rbrack \) satisfying all of the following:\n\n(1) \( \alpha \) is a root o... | Proof. Let \( {e}_{\alpha } : F\left\lbrack t\right\rbrack \rightarrow F\left\lbrack \alpha \right\rbrack \) be given by \( f \mapsto f\left( \alpha \right) \), evaluation at \( \alpha \), a ring epimorphism and \( F \) -linear transformation of vector spaces over \( F \) . As \( \alpha \) is algebraic over \( F,\alpha... | Yes |
Proposition 45.14. (Characterization of algebraic elements) Let \( K/F \) be a field extension and \( \alpha \in K \) . Then \( \alpha \) is algebraic over \( F \) if and only if \( \left\lbrack {F\left( \alpha \right) : F}\right\rbrack \) is finite. | Proof. \( \left( \Rightarrow \right) \) : This follows from the Theorem.\n\n\( \left( \Leftarrow \right) \) : Suppose that \( n = \left\lbrack {F\left( \alpha \right) : F}\right\rbrack \) . Then \( \left\{ {1,\alpha ,\ldots ,{\alpha }^{n}}\right\} \) is \( F \) -linearly dependent. Therefore, there exist \( {a}_{i} \in... | Yes |
Proposition 45.16. (Characterization of transcendental elements) Let \( K/F \) be a field extension and \( u \in K \) . Then the following are equivalent:\n\n(1) \( u \) is transcendental over \( F \) .\n\n(2) \( F\left\lbrack u\right\rbrack \cong F\left\lbrack t\right\rbrack \) .\n\n(3) \( F\left\lbrack u\right\rbrack... | Proof. By Example 45.11(8), we know that (1) implies (2), (3), (4), and (5), and if (2),(3),(4), or (5) holds, then \( u \) cannot be algebraic and (1) must also hold. | Yes |
Corollary 45.17. Let \( K/F \) be a field extension. If \( {a}_{1},\ldots ,{a}_{r} \) are elements of \( K \) algebraic over \( F \), then \( F\left\lbrack {{a}_{1},\ldots ,{a}_{r}}\right\rbrack = F\left( {{a}_{1},\ldots ,{a}_{r}}\right) \) and \( \left\lbrack {F\left( {{a}_{1},\ldots ,{a}_{r}}\right) : F}\right\rbrack... | Proof. As \( {a}_{r} \) is algebraic over \( F \), it is algebraic over \( F\left( {{a}_{1},\ldots ,{a}_{r - 1}}\right) \) . By induction and the \( r = 1 \) case (that we have done), \( F\left\lbrack {{a}_{1},\ldots ,{a}_{r}}\right\rbrack = \n\n\[ F\left\lbrack {{a}_{1},\ldots ,{a}_{r - 1}}\right\rbrack \left\lbrack {... | Yes |
Corollary 45.18. Let \( A \) be a domain containing a field \( F \) . If \( A \) is a finite dimensional vector space over \( F \), then \( A \) is a field. | Proof. If \( A \) has an \( F \) -basis \( \mathcal{B} \), then \( A = F\left\lbrack \mathcal{B}\right\rbrack \) and its quotient field \( = K = F\left( \mathcal{B}\right) \) is a finite dimensional vector space over \( F \) on basis \( \mathcal{B} \) . In particular, \( A = K \) . | Yes |
Theorem 45.19. Let \( K/F \) be a field extension with \( a, b \) elements in \( K \) algebraic over \( F \) . Then \( a \pm b \) , \( {ab} \) and \( {b}^{-1} \), if \( b \neq 0 \), are algebraic over \( F \) . In particular, the set of elements in \( K \) algebraic over \( F \) forms an intermediate field of \( K/F \)... | Proof. Let\\gamma = a \\pm b\\text{,}{ab}\\text{or}{b}^{-1}\\text{, if}b \\neq 0\\text{. Then}F \\subset F\\left( \\gamma \\right) \\subset F\\left( {a, b}\\right) \\text{,}\n\nso\n\n\\left\\lbrack {F\\left( \\gamma \\right) : F}\\right\\rbrack \\leq \\left\\lbrack {F\\left( {a, b}\\right) : F}\\right\\rbrack \\leq \\l... | Yes |
Theorem 45.21. Let \( L/K/F \) be field extensions. Then \( L/F \) is algebraic if and only if both \( L/K \) and \( K/F \) are algebraic. | Proof. \( \left( \Rightarrow \right) \) : Any element in \( L \) algebraic over \( F \) is clearly algebraic over \( K \) and any element in \( K \) lies in \( L \) so is algebraic over \( F \) .\n\n\( \left( \Leftarrow \right) \) : Let \( \alpha \) be an element of \( L \) . We must show that \( \alpha \) is algebraic... | Yes |
Proposition 46.3. Let \( K/F \) be an extension of fields. Then any two transcendence bases of \( K \) over \( F \) have the same cardinality. | Proof. Although this follows from the fact that the cardinality of any basis for a fixed vector space is well-defined, we present a proof in the case that \( K/F \) is a field extension and \( K \) has a finite transcendence basis \( \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \) over \( F \) . Suppose that \( \left\{ {{... | Yes |
Theorem 47.2. (Kronecker’s Theorem) Let \( F \) be a field and \( f \) a nonconstant polynomial in \( F\left\lbrack t\right\rbrack \) . Then there exists an algebraic extension \( K/F \) such that \( f \) has a root in \( K \) and \( \left\lbrack {K : F}\right\rbrack \leq \deg f \) . | Proof. Let \( {f}_{1} \mid f \) in the UFD \( F\left\lbrack t\right\rbrack \) with \( {f}_{1} \) irreducible. Then \( f \) has a root in the field \( K = F\left\lbrack t\right\rbrack /\left( {f}_{1}\right) \), as \( {f}_{1} \) does, with \( \left\lbrack {K : F}\right\rbrack = \deg {f}_{1} \leq \) \( \deg f \) . | Yes |
Theorem 47.4. Let \( F \) be a field and \( f \) a non-constant polynomial in \( F\left\lbrack t\right\rbrack \) . Then there exists a splitting field \( K \) of \( f \) over \( F \) satisfying \( \left\lbrack {K : F}\right\rbrack \leq \left( {\deg f}\right) ! \) . | Proof. Let \( n = \deg f \) . By Kronecker’s Theorem, there exists a field extension \( {K}_{1}/F \) such that \( f = \left( {t - {\alpha }_{1}}\right) {f}_{1} \) in \( {K}_{1}\left\lbrack t\right\rbrack \) with \( {f}_{1} \in {K}_{1}\left\lbrack t\right\rbrack \) and \( \left\lbrack {{K}_{1} : F}\right\rbrack \leq n \... | Yes |
Lemma 47.8. Let \( \varphi : F \rightarrow {F}^{\prime } \) be a field isomorphism, \( f \) an irreducible polynomial in \( F\left\lbrack t\right\rbrack \) and \( {f}^{\prime } = \widetilde{\varphi }\left( f\right) \) . Then\n\n(1) \( {f}^{\prime } \) is an irreducible polynomial in \( {F}^{\prime }\left\lbrack t\right... | Proof. (1) is clear, so both \( F\left\lbrack t\right\rbrack /\left( f\right) \) and \( {F}^{\prime }\left\lbrack t\right\rbrack /\left( {f}^{\prime }\right) \) are fields. Let\n\n- : \( {F}^{\prime }\left\lbrack t\right\rbrack \rightarrow {F}^{\prime }\left\lbrack t\right\rbrack /\left( {f}^{\prime }\right) \) be the ... | Yes |
Lemma 47.9. Let \( K/F \) be an extension of fields and \( f \) an irreducible polynomial in \( F\left\lbrack t\right\rbrack \) . If \( \alpha \) in \( K \) is a root of \( f \), then there exists an \( F \) - isomorphism\n\n\[ \theta : F\left\lbrack t\right\rbrack /\left( f\right) \rightarrow F\left( \alpha \right) \t... | Proof. Let \( {e}_{\alpha } : F\left\lbrack t\right\rbrack \rightarrow F\left\lbrack \alpha \right\rbrack \) by \( g \mapsto g\left( \alpha \right) \) be the evaluation map at \( \alpha \), a ring epimorphism, and, in fact, an \( F \) -epimorphism. As \( f\left( \alpha \right) = 0 \) , the element \( \alpha \) is algeb... | Yes |
Proposition 47.10. Let \( \varphi : F \rightarrow {F}^{\prime } \) be a field isomorphism, \( f \) an irreducible polynomial in \( F\left\lbrack t\right\rbrack \) and \( {f}^{\prime } = \widetilde{\varphi }\left( f\right) \in {F}^{\prime }\left\lbrack t\right\rbrack \) . Suppose that \( K/F \) is a field extension with... | Proof. The two lemmas determine isomorphisms\n\n\[ F\left( \alpha \right) \underset{\text{ lifts }{1}_{F}}{\overset{\theta }{ \leftarrow }}F\left\lbrack t\right\rbrack /\left( f\right) \xrightarrow[{\text{ lifts }\varphi }]{\tau }{F}^{\prime }\left\lbrack t\right\rbrack /\left( {f}^{\prime }\right) \underset{\text{ lif... | Yes |
Theorem 47.12. Let \( \varphi : F \rightarrow {F}^{\prime } \) be an isomorphism of fields, \( f \) a non-constant polynomial in \( F\left\lbrack t\right\rbrack \) and \( {f}^{\prime } = \widetilde{\varphi }\left( f\right) \in {F}^{\prime }\left\lbrack t\right\rbrack \) . Suppose that \( E \) is a splitting field of \(... | An immediate consequence of this theorem is the result that we had wished to prove. | No |
Corollary 47.13. (Uniqueness of Splitting Fields) Let \( F \) be a field and \( f \) a non-constant polynomial in \( F\left\lbrack t\right\rbrack \) . If \( E \) and \( {E}^{\prime } \) are splitting fields of \( f \) over \( F \), then there exists an \( F \) -isomorphism \( E \rightarrow {E}^{\prime } \) taking the s... | Proof. (of the theorem.) We induct on \( n = \left\lbrack {E : F}\right\rbrack \) .\n\n\( n = 1 \) : We have \( E = F \), so \( f \) splits over \( F \), hence \( {f}^{\prime } \) splits over \( {F}^{\prime } \) . In particular, if \( f = a\prod \left( {t - {\alpha }_{i}}\right) \) in \( F\left\lbrack t\right\rbrack \)... | Yes |
Theorem 48.3. Let \( K/F \) be an algebraic extension of fields and \( L \) an algebraically closed field. Suppose that there exists a homomorphism \( \sigma : F \rightarrow L \) . Then there exists a homomorphism \( \tau : K \rightarrow L \) that lifts \( \sigma \) . In particular, if \( K \) is also algebraically clo... | Proof. We use a Zorn Lemma argument that is often repeated in algebra to define maps. Let\n\n\[ S \mathrel{\text{:=}} \{ \left( {E,\eta }\right) \mid K/E/F\text{an intermediate field} \]\n\n\[ \eta : E \rightarrow L\text{a homomorphism lifting}\sigma \} \text{.} \]\n\nPartially order the set \( S \) by \( \leq \) defin... | Yes |
Theorem 49.2. Let \( n \geq 2 \) and \( {z}_{1}\left( { = 0}\right) ,{z}_{2}\left( { = 1}\right) ,\ldots ,{z}_{n} \) be complex numbers. Then \( C\left( {{z}_{1},\ldots ,{z}_{n}}\right) \) is a field satisfying \( \mathbb{Q} \subset C\left( {{z}_{1},\ldots ,{z}_{n}}\right) \subset \) C. | Proof. As \( C\left( {{z}_{1},\ldots ,{z}_{n}}\right) \subset \mathbb{C} \) with \( \operatorname{char}\mathbb{C} = 0 \), it suffices to show that \( C\left( {{z}_{1},\ldots ,{z}_{n}}\right) \) is a field. Let \( z,{z}^{\prime } \) lie in \( C\left( {{z}_{1},\ldots ,{z}_{n}}\right) \), with \( z \) nonzero. As we can d... | Yes |
Theorem 49.4. Let \( {z}_{1}\left( { = 0}\right) ,{z}_{2}\left( { = 1}\right) ,\ldots ,{z}_{n} \) be points in \( \mathbb{C} \) with \( n \geq \) 2. Then the field \( C\left( {{z}_{1},\ldots ,{z}_{n}}\right) \) has the following properties:\n\n(1) The point \( {z}_{i} \) lies in \( C\left( {{z}_{1},\ldots ,{z}_{n}}\rig... | Proof. (1) is clear.\n\n(2). If \( z = r{e}^{\sqrt{-1}\theta } \), then \( \bar{z} = r{e}^{-\sqrt{-1}\theta } \), and (2) follows.\n\n(3). Since we can construct rays of angle \( \theta /2 \) from rays of angle \( \theta \), it suffices to show that we can construct \( \sqrt{r} \) given a segment of length \( r \) . Dr... | Yes |
Theorem 49.9. (Constructibility Criterion) Given complex numbers \( {z}_{1}\left( { = 0}\right) ,{z}_{2}\left( { = 1}\right) ,\ldots ,{z}_{n} \) with \( n \geq 2 \), set \( F = \mathbb{Q}\left( {{z}_{1},{z}_{2},\ldots ,{z}_{n},\overline{{z}_{1}},\ldots ,\overline{{z}_{n}}}\right) \) . If \( z \) is a complex number, th... | Proof. Let\n\n\[ \n{C}^{\prime } = \{ z \in \mathbb{C} \mid \text{There exists a square root tower}K/F\text{with}z \in K\} \text{.} \]\n\nBy definition, \( F \subset {C}^{\prime } \) . We must show that \( {C}^{\prime } \subset C\left( {{z}_{1},\ldots ,{z}_{n}}\right) \) . We know that \( C\left( {{z}_{1},\ldots ,{z}_{... | Yes |
Corollary 49.10. Let \( {z}_{1}\left( { = 0}\right) ,{z}_{2}\left( { = 1}\right) ,\ldots ,{z}_{n} \) be complex numbers with \( n \geq 2 \) and \( z \) a complex number in \( C\left( {{z}_{1},\ldots ,{z}_{n}}\right) \) . Then \( z \) is algebraic over \( F = \mathbb{Q}\left( {{z}_{1},{z}_{2},\ldots ,{z}_{n},\overline{{... | Proof. As \( z \) lies in \( C\left( {{z}_{1},\ldots ,{z}_{n}}\right) \), there exists a square root tower \( K/F \) with \( z \in K \), hence \( \left\lbrack {F\left( z\right) : F}\right\rbrack \mid \left\lbrack {K : F}\right\rbrack \), a power of two. | Yes |
Lemma 49.11. Let \( \alpha \) be a ray with the \( X \) -axis as one side. Then it is constructible form \( {z}_{1}\left( { = 0}\right) ,{z}_{2}\left( { = 1}\right) ,\ldots ,{z}_{n} \) if and only if \( {e}^{\sqrt{-1}\alpha } \) is constructible from \( {z}_{1}\left( { = 0}\right) ,{z}_{2}\left( { = 1}\right) ,\ldots ,... | Proof. As \( z = {e}^{\sqrt{-1}\alpha } = \cos \alpha + \sqrt{-1}\sin \alpha \) and we can erect and drop perpendiculars, this is immediate. | No |
Lemma 49.14. Any \( {2}^{n} \) -gon, \( n > 1 \), is constructible from two points. | Proof. As we can erect perpendiculars, i.e., angles \( {2\pi }/4 \) radians, we can construct a square. As we can bisect any angle, we can construct a ray of \( \frac{1}{{2}^{m}}\left( {\pi /2}\right) \) for any \( m \) by induction. | No |
Lemma 49.15. Let \( m \) and \( n \) be relatively prime integers each at least three. Then\n\n(1) A regular \( \left( {mn}\right) \) -gon is constructible from two points if and only if both a regular \( m \) -gon and a regular \( n \) -gon can be constructible from two points.\n\n(2) a regular \( n \) -gon can be con... | Proof. (2): Angles can be bisected or doubled using straight-edge and compass.\n\n(1): Suppose that a regular \( \left( {mn}\right) \) -gon can be constructed, i.e., a ray of angle \( \theta = {2\pi }/{mn} \) radians. Then rays of angle \( {m\theta } \) and \( {n\theta } \) can be constructed. Conversely, suppose that ... | Yes |
Theorem 49.17. Let \( n \) be an integer at least three. Then a regular \( n \) -gon can be constructed from two points if and only if \( n = {2}^{k}{p}_{1}\cdots {p}_{r} \) with \( k \) non-negative, and \( {p}_{1},\ldots ,{p}_{r} \) distinct Fermat primes for some \( r \) or \( n = {2}^{k} \) for some integer \( k \)... | Proof. By the lemmas, it suffices to determine when a regular \( {p}^{r} \) - gon can be constructed with \( p \) an odd prime, \( r \geq 1 \), i.e., to determine when \( \zeta = \cos \frac{2\pi }{{p}^{r}} + \sqrt{-1}\sin \frac{2\pi }{{p}^{r}} \) is constructible. By Example 47.15(6), we know that \( \left\lbrack {\mat... | Yes |
Lemma 50.3. Let \( K/F \) be an extension of fields, \( f \) a non-constant polynomial with a root \( \alpha \) in \( K \) . Then \( \alpha \) is a simple root of \( f \) if and only if \( \alpha \) is not a root of \( {f}^{\prime } \) . | Proof. We can write \( f = {\left( t - \alpha \right) }^{r}g \) for some \( g \) in \( K\left\lbrack t\right\rbrack \) with \( g\left( \alpha \right) \neq 0 \) for some integer \( r \) . Therefore, \( {f}^{\prime } = r{\left( t - \alpha \right) }^{r - 1}g + {\left( t - \alpha \right) }^{r}{g}^{\prime } \) in \( K\left\... | Yes |
Corollary 50.4. Let \( f \) be a non-constant polynomial in \( F\left\lbrack t\right\rbrack \) . Then \( f \) has a multiple root in some extension field \( K \) of \( F \) if and only if the ideal \( \left( {f,{f}^{\prime }}\right) \) in \( F\left\lbrack t\right\rbrack \) is not the unit ideal. | Proof. ( \( \Rightarrow \) ): Let \( \alpha \) be a multiple root of \( f \) in an extension field \( K \) of \( F \) . Then \( {m}_{F}\left( \alpha \right) \mid f \) and \( {m}_{F}\left( \alpha \right) \mid {f}^{\prime } \) in \( F\left\lbrack t\right\rbrack \), so \( \left( {f,{f}^{\prime }}\right) \subset \left( {{m... | Yes |
Corollary 50.5. Let \( f \) be an irreducible polynomial in \( F\left\lbrack t\right\rbrack \) . (1) If \( \operatorname{char}F = 0 \), then \( f \) has no multiple roots in any field extension of \( F \) . In particular, \( f \) can have only simple roots, if any, in a field extension of \( F \) . (2) If \( \operatorn... | Proof. We begin the proof, leaving the rest of the proof as exercises. Let \( K/F \) be a field extension such that \( f \) has a root \( \alpha \) in \( K \) , so \( f = a{m}_{F}\left( \alpha \right) \) for some nonzero \( a \) in \( F \) . In particular, if \( \alpha \) is a multiple root, we must have \( f \mid {f}^... | No |
Lemma 51.3. (Dedekind’s Lemma) Let \( {\sigma }_{1},\ldots ,{\sigma }_{n} : G \rightarrow {F}^{ \times } \) be distinct characters. Then \( {\sigma }_{1},\ldots ,{\sigma }_{n} : G \rightarrow {F}^{ \times } \) are independent. | Proof. We prove this by induction on \( n \) .\n\n\( n = 1 \) : If \( a{\sigma }_{1}\left( g\right) = 0 \) for all \( g \) in \( G \), then \( a = 0 \) as \( {\sigma }_{1}\left( g\right) \neq 0 \) for all \( g \) in \( G \).\n\n\( n > 1 \) : Suppose that we have an equation\n\n(*) \( \;\mathop{\sum }\limits_{{i = 1}}^{... | Yes |
Lemma 51.8. (Artin’s Lemma) Suppose that \( S \) is a finite non-empty set of field homomorphisms \( F \rightarrow K \), then \( \left\lbrack {F : {F}^{S}}\right\rbrack \geq \left| S\right| \) . | Proof. Let \( S = \left\{ {{\sigma }_{1},\ldots ,{\sigma }_{n}}\right\} \) with \( n \geq 1 \) . Suppose that \( r = \) \( \left\lbrack {F : {F}^{S}}\right\rbrack < \left| S\right| = n \) and \( \left\{ {{\omega }_{1},\ldots ,{\omega }_{r}}\right\} \) is an \( {F}^{S} \) -basis for \( F \) . Consider\nthe following sys... | Yes |
Corollary 51.11. Let \( K/F \) be a finite extension of fields. Then the Galois group \( G\left( {K/F}\right) \) is a finite group and satisfies \( \left\lbrack {K : F}\right\rbrack \geq \left| {G\left( {K/F}\right) }\right| \) . | Proof. As \( F \) is a subfield of \( {K}^{S} \) for any non-empty subset \( S \) of \( G\left( {K/F}\right) \), we have\n\n\[ \left\lbrack {K : F}\right\rbrack \geq \left\lbrack {K : {K}^{S}}\right\rbrack \geq \left| S\right| \text{ for any finite }S. \]\n\nIt follows that \( G\left( {K/F}\right) \) is finite and sati... | Yes |
Corollary 51.16. Let \( K \) be a field, \( {G}_{1},{G}_{2} \) two finite subgroups of \( \operatorname{Aut}\left( K\right) \), the group of field automorphisms of \( K \), and \( {F}_{i} = {K}^{{\widehat{G}}_{i}} \) for \( i = 1,2 \) . Then \( {F}_{1} = {F}_{2} \) if and only if \( {G}_{1} = {G}_{2} \). | Proof. \( \left( \Leftarrow \right) \) is clear.\n\n\( \left( \Rightarrow \right) \) : By Artin’s Theorem, \( {G}_{i} = G\left( {K/{F}_{i}}\right) \), so this also follows. | No |
Corollary 51.17. Suppose that \( K/F \) is a finite extension of fields. Then \( K/F \) is Galois if and only if \( \left\lbrack {K : F}\right\rbrack = \left| {G\left( {K/F}\right) }\right| \) . | Proof. \( \left( \Rightarrow \right) \) follows from Artin’s Theorem and the special case of Artin's Lemma, Corollary 51.9.\n\n\( \left( \Leftarrow \right) \) : Let \( E = {K}^{G\left( {K/F}\right) } \), then we know that \( G\left( {K/E}\right) = G\left( {K/F}\right) \) . As \( K/E \) is Galois, by the proven sufficie... | Yes |
Lemma 52.2. Let \( G \) be a finite cyclic group of order \( n \) and \( \operatorname{Aut}\left( G\right) \) the automorphism group of \( G \) . Then \( \operatorname{Aut}\left( G\right) \cong {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ \times } \) . In particular, \( \operatorname{Aut}\left( G\right) \) is abelian (and... | Proof. Let \( G = \langle a\rangle \) and \( \sigma \in \operatorname{Aut}\left( G\right) \) . Then \( \sigma \left( a\right) = {a}^{i} \), for some \( 1 \leq i \leq n \) and \( \langle a\rangle = \left\langle {a}^{i}\right\rangle \) . It follows that \( i \) and \( n \) are relatively prime. Conversely, if \( i \) and... | Yes |
Proposition 52.4. Let \( K \) be a splitting field of \( {t}^{n} - 1 \) in \( F\left\lbrack t\right\rbrack \) over \( F \) and\n\n\[ U = \left\{ {z \in K \mid {z}^{n} = 1}\right\} = \left\{ {z \mid z\text{ a root of }{t}^{n} - 1\text{ in }K}\right\} .\n\]\n\nThen \( U \) is a cyclic subgroup of \( {K}^{ \times } \) . S... | Proof. As \( {\left( {z}_{i}{z}_{j}\right) }^{n} = 1 = {\left( {z}_{i}^{-1}\right) }^{n} \), for all \( {z}_{i},{z}_{j} \in U \), we know that \( U \) is a group. Since it is a finite subgroup of \( {K}^{ \times } \), it is cyclic by Theorem 31.15. As \( K = F\left( U\right) \), every \( \sigma \) in \( G\left( {K/F}\r... | Yes |
Corollary 52.7. Let \( F \) be a field of characteristic zero or \( \operatorname{char}F//n \) satisfying the polynomial \( {t}^{n} - 1 \) in \( F\left\lbrack t\right\rbrack \) splits over \( F \) with \( U \) the set of \( n \) th roots of units in \( F \) . Suppose that \( K \) is a splitting field of the polynomial ... | Proof. Let \( U = \left\{ {{z}_{1},\ldots ,{z}_{n}}\right\} \), a cyclic subgroup of \( {F}^{ \times } \) of order \( n \) . As \( {\left( {t}^{n} - a\right) }^{\prime } = n{t}^{n - 1} \) has only zero as a root, \( {t}^{n} - a \) has \( n \) distinct roots in \( K \) . Let \( r \) in \( K \) be a root of \( {t}^{n} - ... | Yes |
Proposition 53.2. Let \( K/F \) be a finite extension of fields. Then \( K/F \) is normal if and only if any irreducible polynomial in \( F\left\lbrack t\right\rbrack \) having a root in \( K \) splits over \( K \) . | Proof. \( \left( \Rightarrow \right) \) : Let \( K \) be a splitting field the non-constant polynomial \( g \) in \( F\left\lbrack t\right\rbrack \) over \( F \) and \( f \in F\left\lbrack t\right\rbrack \) an irreducible polynomial having a root \( \alpha \) in \( K \) . Let \( L/K \) with \( \beta \in L \) a root of ... | Yes |
Proposition 53.4. Let \( K/F \) be a finite, normal extension and \( K/E/F \) an intermediate field. Suppose that \( \varphi : E \rightarrow K \) is an \( F \)-homomorphism. Then there exists an \( F \)-automorphism \( \sigma \in G\left( {K/F}\right) \) satisfying \( {\left. \sigma \right| }_{E} = \varphi \), i.e., eve... | Proof. Let \( K \) be a splitting field of the non-constant polynomial \( f \) in \( F\left\lbrack t\right\rbrack \), hence a splitting field of \( f \) over \( E \). In addition, \( K \) is a splitting field of \( f = \widetilde{\varphi }\left( f\right) \) over \( \varphi \left( E\right) \), so there exists an automor... | Yes |
Corollary 53.7. Let \( K/F \) be a finite normal field extension and \( K/E/F \) an intermediate field extension. Then \( K/E \) is normal. Moreover, the following are equivalent:\n\n(1) \( E/F \) is normal.\n\n(2) \( \sigma \left( E\right) = E \) for every \( \sigma \in G\left( {K/F}\right) \) .\n\n(3) \( {\left. \sig... | Proof. If \( K \) is the splitting field of the non-constant polynomial \( f \in F\left\lbrack t\right\rbrack \) over \( F \), then it is also the splitting field of \( f \) over \( E \), so the first statement follows.\n\nWe turn to the equivalences. By Proposition 53.4, any \( \varphi \in G\left( {E/F}\right) \) lift... | Yes |
Proposition 53.10. Let \( K/F \) be a finite field extension. Then a normal closure of \( K/F \) exists and is unique up to a \( K \) -isomorphism. | Proof. Let \( K = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) with each \( {\alpha }_{i} \) algebraic over \( F \) , \( f = \prod {m}_{F}\left( {\alpha }_{i}\right) \) in \( F\left\lbrack t\right\rbrack \), and \( L \) a splitting field of \( f \) over \( K \), hence also a splitting field of \( f \) over \... | Yes |
Corollary 53.13. Let \( K/F \) be a extension of fields with \( \alpha \) an element in \( K \) algebraic over \( F \) . Then \( \alpha \) is separable over \( F \) if and only if \( F\left( \alpha \right) /F \) is separable. | Proof. Let \( L/F\left( \alpha \right) \) be a field extension such that \( L/F \) is finite and normal. We know that \( {m}_{F}\left( \alpha \right) \) splits over \( L \) . Let \( \alpha = {\alpha }_{1},\ldots ,{\alpha }_{m} \) be the distinct roots of \( {m}_{F}\left( \alpha \right) \) in \( L \) . We know that ther... | Yes |
Proposition 53.14. Let \( K \) be a splitting field of a separable polynomial in \( F\left\lbrack t\right\rbrack \) . Then \( K/F \) is separable and Galois. | Proof. Suppose that \( K \) is a splitting field of the separable polynomial \( f \in F\left\lbrack t\right\rbrack \) . We may assume that \( F < K \) . Let \( \alpha \in K \) be a root of \( f \) not in \( F \) . As \( \alpha \) is separable over \( F \), the extension \( F\left( \alpha \right) /F \) is separable by t... | Yes |
Corollary 53.15. Let \( K/F \) be a finite separable field extension and \( L/K \) a normal closure of \( K/F \) . Then \( L/F \) is separable and Galois. | Proof. Let \( K = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \), then \( L \) is a splitting field of the separable polynomial \( \prod {m}_{F}\left( {\alpha }_{i}\right) \) in \( F\left\lbrack t\right\rbrack \), hence \( L/F \) is separable and Galois. | Yes |
Corollary 53.16. Let \( K = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) with \( {\alpha }_{i} \) separable over \( F \) for each \( i = 1,\ldots, n \) . Then \( K/F \) is separable. | Proof. Let \( L/K \) be a normal closure of \( K/F \), hence the splitting field of the separable polynomial \( \prod {m}_{F}\left( {\alpha }_{i}\right) \) . Therefore, \( L/F \) is separable hence so is \( K/F \) . | Yes |
Corollary 53.17. If \( K/F \) is finite field extension and \( K/E/F \), then \( K/F \) is separable if and only if \( K/E \) and \( E/F \) are separable. | Proof. \( \left( \Rightarrow \right) \) has already been done.\n\n\( \left( \Leftarrow \right) \) : Let \( L/K \) be a (finite) normal extension and\n\n\( {\varphi }_{1},\ldots ,{\varphi }_{m} : E \rightarrow L \) be all the distinct \( F \) -homomorphisms\n\n\( {\psi }_{1},\ldots ,{\psi }_{n} : K \rightarrow L \) be a... | Yes |
Theorem 53.18. Let \( K/F \) be a finite extension of fields. Then \( K/F \) is Galois if and only if \( K/F \) is normal and separable. | Proof. \( \left( \Leftarrow \right) \) has already been done.\n\n\( \left( \Rightarrow \right) \) : Let \( L/K \) be a normal closure of \( K/F \) . As \( K/F \) is Galois, the number of \( F \) -homomorphisms \( K \rightarrow L \) is at least \( \left| {G\left( {K/F}\right) }\right| = \left\lbrack {K : F}\right\rbrack... | Yes |
Corollary 53.19. Let \( K/F \) be a finite extension and \( K/E/F \) an intermediate field. Suppose that \( K/F \) is Galois. Then\n\n(1) \( K/E \) is Galois.\n\n(2) \( E/F \) is Galois if and only if \( E/F \) is normal. | Proof. As \( K/F \) is Galois, it is separable and normal, so \( K/E \) and \( E/F \) are separable and \( K/E \) is normal. Therefore, \( K/E \) is Galois and \( E/F \) is normal if and only if \( E/F \) is Galois. | Yes |
Lemma 54.1. Let \( K/F \) be a finite Galois extension of fields, \( {H}_{i} \subset \) \( G\left( {K/F}\right) \) a subgroup and \( {E}_{i} = {K}^{{H}_{i}} \) for \( i = 1,2 \) . If \( \sigma \) is an element of the Galois group \( G\left( {K/F}\right) \), then\n\n\[ \n{\left. \sigma \right| }_{{E}_{1}} : {E}_{1} \rig... | Proof. We know that \( K/{E}_{i} \) is Galois and \( {H}_{i} = G\left( {K/{E}_{i}}\right) \) for \( i = 1,2 \) . In particular, \( \left\lbrack {K : {E}_{i}}\right\rbrack = \left| {H}_{i}\right| \) for \( i = 1,2 \), so under either condition, we have\n\n\[ \n\left| {H}_{1}\right| = \left| {H}_{2}\right| = \left\lbrack... | Yes |
Theorem 54.3. (The Fundamental Theorem of Galois Theory) Suppose that \( K/F \) is a finite Galois extension. Then\n\n\[ i : \mathcal{G}\left( {K/F}\right) \rightarrow \mathcal{F}\left( {K/F}\right) \text{given by}H \mapsto {K}^{H} \]\n\nis a bijection of sets with inverse\n\n\[ j : \mathcal{F}\left( {K/F}\right) \righ... | Proof. We already know that \( {H}_{1} = {H}_{2} \) in \( \mathcal{G}\left( {K/F}\right) \) if and only if \( {K}^{{H}_{1}} = {K}^{{H}_{2}} \) in \( \mathcal{F}\left( {K/F}\right) \), so \( i : \mathcal{G}\left( {K/F}\right) \rightarrow \mathcal{F}\left( {K/F}\right) \) is injective. Let \( E \in \mathcal{F}\left( {K/F... | Yes |
Corollary 54.4. Let \( K/F \) be a finite Galois extension of fields, then \( \mathcal{F}\left( {K/F}\right) \) is finite, i.e., there exist only finitely many intermediate fields \( E \) with \( K/E/F \) . | Proof. By the Fundamental Theorem of Galois Theory, we have \( \left| {\mathcal{F}\left( {K/F}\right) }\right| = \left| {\mathcal{G}\left( {K/F}\right) }\right| \), hence \( \mathcal{F}\left( {K/F}\right) \) is a finite set. | Yes |
Corollary 54.5. Let \( K/F \) be a finite separable extension. Then \( \mathcal{F}\left( {K/F}\right) \) is finite. | Proof. Let \( L/K \) be a normal closure of \( K/F \) . Then \( L/F \) is finite, separable, and normal, so Galois. Thus \( \left| {\mathcal{F}\left( {K/F}\right) }\right| \leq \left| {\mathcal{F}\left( {L/F}\right) }\right| \) is finite. | Yes |
We know that \( K = \mathbb{Q}\left( {\sqrt{2},\sqrt{3}}\right) \) is the splitting field of \( \left( {{t}^{2} - 2}\right) \left( {{t}^{2} - 3}\right) \) in \( \mathbb{C} \) over \( \mathbb{Q} \). | Then we have seen that \( G\left( {K/\mathbb{Q}}\right) = \) \( \{ 1,\sigma ,\tau ,{\sigma \tau }\} \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \) with\n\n\( \sigma : K \rightarrow K \) the \( \mathbb{Q}\left( \sqrt{3}\right) \) -automorphism given by \( \sqrt{2} \mapsto - \sqrt{2} \),\n\n\( \tau : K \ri... | Yes |
Corollary 54.7. Let \( F \) be a field and \( f \) be a symmetric polynomial in \( F\left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \), then \( f \) is a rational function in the elementary symmetric functions in \( {t}_{1},\ldots ,{t}_{n} \), i.e., \( f \in F\left( {{s}_{1},\ldots ,{s}_{n}}\right) \) . | The Fundamental Theorem of Symmetric Functions implies that a symmetric polynomial in \( F\left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \) actually lies in \( F\left\lbrack {{s}_{1},\ldots ,{s}_{n}}\right\rbrack \), even with \( F \) replaced by a commutative ring. (cf. Theorem 66.4 for the proof.) | Yes |
Theorem 54.9. (Primitive Element Theorem) Let \( K/F \) be a finite extension of fields. If \( \mathcal{F}\left( {K/F}\right) \) is finite, then there exists an element \( \alpha \) in \( K \) such that \( K = F\left( \alpha \right) \) . In particular, this is true if \( K/F \) is separable. | Proof. Case 1. \( F \) is finite:\n\n\( K \) must also be a finite field, so \( {K}^{ \times } = \langle \alpha \rangle \) for some \( \alpha \in K \) . Therefore, \( K = F\left( \alpha \right) \) and \( \mathcal{F}\left( {K/F}\right) \) is finite.\n\nCase 2. \( F \) is infinite:\n\nChoose \( \alpha \in K \) with \( \l... | Yes |
Theorem 54.11. (Square Tower Theorem) Let \( F \) be a field of characteristic different from two and \( K/F \) a finite normal field extension. Then \( K/F \) is a square root tower if and only if \( \left\lbrack {K : F}\right\rbrack = {2}^{e} \) for some positive integer \( e \) . | Proof. \( \left( \Rightarrow \right) \) has previously been established.\n\n\( \left( \Leftarrow \right) \) : Suppose that \( \left\lbrack {K : F}\right\rbrack = {2}^{e} \), for some integer \( e \) . Let \( K = \) \( F\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) . We know that \( \deg {m}_{F}\left( {\alpha }... | Yes |
Proposition 55.1. Let \( K/F \) be an algebraic extension. If \( K/F \) is normal, and \( \sigma : K \rightarrow K \) an \( F \)-homomorphism, then \( \sigma \) lies in \( G\left( {K/F}\right) \) , i.e., \( \sigma \) is onto. | Proof. Let \( \alpha \) lie in \( K \), then the minimal polynomial \( {m}_{F}\left( \alpha \right) \) splits over \( K \) (just as in the finite normal case by Exercise 53.22(2)), so there exists an intermediate field \( K/E/F \) with \( E \) a splitting field of \( {m}_{F}\left( \alpha \right) \) over \( F \) . Thus ... | Yes |
Proposition 55.2. Let \( K/F \) be an algebraic extension. If \( K/F \) is normal and \( K/E/F \) is an intermediate field, then any \( F \) -homomorphism \( \sigma : E \rightarrow K \) lifts to an element of \( G\left( {K/F}\right) \) . | Proof. This is Exercise 53.22(3), which follows by a Zorn Lemma argument and the finite extension case. | No |
Theorem 55.3. Let \( K/F \) be an algebraic extension. If \( K/F \) is normal, then \( K/F \) is Galois if and only if \( K/F \) is separable and normal. | Proof. \( \left( \Leftarrow \right) \) : Let \( \alpha \) lie in \( {K}^{G\left( {K/F}\right) } \) . As above there exists an intermediate field \( K/E/F \) with \( E \) a splitting field of \( {m}_{F}\left( \alpha \right) \) over \( F \) . As \( K/F \) is separable, so is \( E/F \), hence \( E/F \) is finite normal an... | Yes |
Lemma 55.5. Let \( K/F \) be an algebraic and Galois extension of fields and \( K/E/F \) an intermediate field. If \( \left\lbrack {E : F}\right\rbrack \) finite, then the index \( \left\lbrack {G\left( {K/F}\right) : G\left( {K/E}\right) }\right\rbrack \) is finite and \( \left\lbrack {G\left( {K/F}\right) : G\left( {... | Proof. Suppose that \( {\sigma }_{1},\ldots ,{\sigma }_{n} : E \rightarrow K \) are all the distinct \( F \) - homomorphisms. These lift to \( {\widehat{\sigma }}_{1},\ldots ,{\widehat{\sigma }}_{n} \) in \( G\left( {K/F}\right) \), i.e., \( {\left. {\widehat{\sigma }}_{i}\right| }_{E} = {\sigma }_{i} \) for each \( i ... | Yes |
Theorem 55.7. (Krull) Suppose that \( K/F \) is an algebraic Galois extension of fields. Let\n\n\[ \n\mathcal{N}\left( {K/F}\right) \mathrel{\text{:=}} \{ G\left( {K/E}\right) \mid K/E/F\text{with}E/F\text{finite Galois}\} \text{.} \]\n\nThen there exists a topology on \( G \) compatible with the group structure of \( ... | Let \( X \) be a topological space. It is called Hausdorff if given any two distinct points in \( X \), there exist disjoint neighborhoods of these two points and compact if every collection of open sets that cover \( X \) , i.e., whose union is \( X \), contains a finite subcollection that covers \( X \) . Finally, \(... | Yes |
Lemma 55.8. Let \( K/F \) be an algebraic Galois extension of fields, \( H \) a subgroup of \( G\left( {K/F}\right) \), and \( E = {K}^{H} \) . Then \( G\left( {K/E}\right) = \bar{H} \), the closure of \( H \) in \( G\left( {K/F}\right) \) . | Proof. \( G\left( {K/E}\right) \subset \bar{H} \) : Let \( \sigma \) lie in \( G\left( {K/E}\right) \) . We must show that \( \sigma \in \bar{H} \) . As remarked above, since \( \sigma \mathcal{N}\left( {K/F}\right) \) is a fundamental system of neighborhoods for \( \sigma \), it suffices to show that \( H \cap \sigma ... | Yes |
Theorem 55.9. Let \( K/F \) be an algebraic Galois extension of fields and \[ {\mathcal{G}}_{c}\left( {K/F}\right) \mathrel{\text{:=}} \{ H \mid H\text{is a closed subgroup of}G\left( {K/F}\right) \} \text{,} \] then \[ i : {\mathcal{G}}_{c}\left( {K/F}\right) \rightarrow \mathcal{F}\left( {K/F}\right) \text{given by}H... | Proof. By the lemma, we know that \[ {\mathcal{G}}_{c}\left( {K/F}\right) = \{ G\left( {K/E}\right) \mid E \in \mathcal{F}\left( {K/F}\right) \} . \] \( i \) is injective: If \( {K}^{{H}_{1}} = {K}^{{H}_{2}} \), then by the lemma, \[ {H}_{1} = \overline{{H}_{1}} = G\left( {K/{K}^{{H}_{1}}}\right) = G\left( {K/{}^{{H}_{... | Yes |
Example 55.11. Let \( S = \{ \sqrt{a} \mid a > 0 \) a prime or \( a = - 1\} \subset \mathbb{C} \) and \( K \) the splitting field of \( \left\{ {t + {a}^{2} \mid a \in S}\right\} \) in \( \mathbb{C} \) . Then \( K/\mathbb{Q} \) is Galois, and it is easy to see that \( K \) is the compositum of all quadratic extensions ... | If \( \sigma \) and \( \tau \) lie in \( G\left( {K/\mathbb{Q}}\right) \), then we have \( {\sigma }^{2} = {1}_{K} \) and \( {\sigma \tau } = {\tau \sigma } \), i.e., \( G\left( {K/\mathbb{Q}}\right) \) is abelian and all non-identity elements are of order two. It follows that \( G\left( {K/\mathbb{Q}}\right) \) is a v... | No |
Lemma 56.2. The Möbius function \( \mu \) is multiplicative and satisfies:\n\n(1) \( I = \mu \star U \) i.e.,\n\n\[ \mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) = \left\{ \begin{array}{ll} 1 & \text{ if }n = 1 \\ 0 & \text{ if }n > 1 \end{array}\right. \]\n\n(2) If \( n = {p}_{1}^{{e}_{1}}\cdots {p}_{r}^{{e}_{... | Proof. Let \( m \) and \( n \) be relatively prime positive integers. Clearly, either \( m \) or \( n \) is not square free if and only if \( {mn} \) is not square free. It follows easily that \( \mu \) is multiplicative. Let \( \epsilon = \mu \) or \( \left| \mu \right| \) . If \( n > 1 \) is not square-free, then \( ... | Yes |
Proposition 56.3. (Möbius Inversion Formula) Let \( f \) and \( g \) be arithmetic functions not zero at 1 , then\n\n\[ f\left( n\right) = \mathop{\sum }\limits_{{d \mid n}}g\left( d\right) \text{ if and only if }g\left( n\right) = \mathop{\sum }\limits_{{d \mid n}}f\left( d\right) \mu \left( \frac{n}{d}\right) . \] | Proof. The proposition is equivalent to \( f = g \star U \) if and only if \( g = f \star \mu \) . By Exercise 8.5(15), the Dirichlet product is associative with \( I \) a unity, so \( f = g \star U \) if and only if \( f \star \mu = \left( {g \star U}\right) \star \mu = g \star \left( {U \star \mu }\right) = \) \( g \... | Yes |
Using the formula above, we have\n\n\[ \n{\Phi }_{12} = \mathop{\prod }\limits_{{d \mid {12}}}{\left( {t}^{\frac{12}{d}} - 1\right) }^{\mu \left( d\right) } \n\] | \[ \n= {\left( {t}^{12} - 1\right) }^{\mu \left( 1\right) }{\left( {t}^{6} - 1\right) }^{\mu \left( 2\right) }{\left( {t}^{4} - 1\right) }^{\mu \left( 3\right) }{\left( {t}^{3} - 1\right) }^{\mu \left( 4\right) } \n\]\n\n\[ \n{\left( {t}^{2} - 1\right) }^{\mu \left( 6\right) }{\left( t - 1\right) }^{\mu \left( {12}\rig... | Yes |
Theorem 56.6. Let \( \zeta \) be a primitive \( n \) root of unity in \( \mathbb{C} \) . Then \( \mathbb{Q}\left( \zeta \right) /\mathbb{Q} \) is an abelian extension of degree \( \varphi \left( n\right) \) . | Proof. We must show that \( {\Phi }_{n} \) is irreducible in \( \mathbb{Q}\left\lbrack t\right\rbrack \), equivalently in \( \mathbb{Z}\left\lbrack t\right\rbrack \) . Let \( \zeta \) be a primitive \( n \) th root of unity. As \( \mathbb{Z}\left\lbrack t\right\rbrack \) is a UFD, we see that there exist polynomials \(... | Yes |
Corollary 56.8. Let \( {\zeta }_{m},{\zeta }_{n} \), and \( {\zeta }_{mn} \) be primitive mth, nth, and (mn)th roots of unity over \( \mathbb{Q} \) with \( m \) and \( n \) relatively prime positive integers. Then \( \mathbb{Q}\left( {\zeta }_{mn}\right) = \mathbb{Q}\left( {{\zeta }_{m},{\zeta }_{n}}\right) \) and \( \... | Proof. As \( m \) and \( n \) are relatively prime, \( {\zeta }_{m}{\zeta }_{n} \) is a primitive \( \left( {mn}\right) \) th root of unity. Since the Euler phi-function is multiplicative, the result follows from the theorem. | No |
Proposition 56.10. Let \( n > 1 \) be an integer. Then there exist infinitely many primes \( p \) satisfying \( p \equiv 1{\;\operatorname{mod}\;n} \) . | To prove this, we first establish the following lemma:\n\nLemma 56.11. Let a be a | No |
Lemma 56.11. Let a be a positive integer and \( p \) a (positive) prime not dividing \( n \) . If \( p \mid {\Phi }_{n}\left( a\right) \) in \( \mathbb{Z} \), then \( p \equiv 1{\;\operatorname{mod}\;n} \) . | Proof. Let \( - : \mathbb{Z} \rightarrow \mathbb{Z}/p\mathbb{Z} \) be the canonical epimorphism. We know that \( {t}^{n} - 1 = \mathop{\prod }\limits_{{d \mid n}}{\Phi }_{d} \) is a factorization of \( {t}^{n} - 1 \) into irreducible polynomials in \( \mathbb{Z}\left\lbrack t\right\rbrack \) . In particular, \( {a}^{n}... | Yes |
Theorem 56.12. (Kronecker-Weber Theorem) Let \( K/\mathbb{Q} \) be a abelian extension of fields. Then there exists a root of unity \( \zeta \) in \( \mathbb{C} \) such that \( K \) is a subfield of \( \mathbb{Q}\left( \zeta \right) \) | We shall prove this in the case that \( K \) is a quadratic extension of \( \mathbb{Q} \) . The general theorem shows that abelian extensions of \( \mathbb{Q} \) are determined by the unit circle. A similar geometric interpretation is true when \( \mathbb{Q} \) is replaced by an imaginary quadratic extension of \( \mat... | No |
Lemma 56.13. Let \( p \) be an odd prime. Then for all integers \( a \) and \( b \) not divisible by \( p \), we have the following:\n\n(1) \( \left( \frac{a}{p}\right) = \left( \frac{b}{p}\right) \) if \( a \equiv b{\;\operatorname{mod}\;p} \) .\n\n(2) \( \left( \frac{ab}{p}\right) = \left( \frac{a}{p}\right) \left( \... | Proof. We have already observed (1), and (4) is the fact that half of \( {\left( \mathbb{Z}/p\mathbb{Z}\right) }^{ \times } \) are squares and half not.\n\n(3): We know that \( {x}^{p - 1} \equiv 1{\;\operatorname{mod}\;p} \) if \( p \nmid x \), so if \( a \) is a square modulo \( p \) , then \( {a}^{\frac{p - 1}{2}} \... | No |
Proposition 56.14. Let \( p \) be an odd prime and \( \zeta \) a primitive pth root of unity. Set\n\n\[ S \mathrel{\text{:=}} \mathop{\sum }\limits_{{a = 1}}^{{p - 1}}\left( \frac{a}{p}\right) {\zeta }^{a} \]\n\nin \( \mathbb{Q}\left( \zeta \right) \) (even in \( \mathbb{Z}\left\lbrack \zeta \right\rbrack \) ). Then\n\... | Proof. We have\n\n\[ {S}^{2} = \mathop{\sum }\limits_{{a = 1, b = 1}}^{{p - 1}}\left( \frac{a}{p}\right) \left( \frac{b}{p}\right) {\zeta }^{a + b} = \mathop{\sum }\limits_{{a = 1, b = 1}}^{{p - 1}}\left( \frac{ab}{p}\right) {\zeta }^{a + b}. \]\n\nAs \( a \) ranges over \( 1,\ldots, p - 1{\;\operatorname{mod}\;p} \), ... | Yes |
Corollary 56.16. Let \( p \) be a (positive) prime, \( {\zeta }_{p} \) a primitive pth root of unity, and \( {\zeta }_{4p} \) a primitive \( {4p} \) th root of unity, then \( \sqrt{p} \) and \( \sqrt{-p} \) lie in \( \mathbb{Q}\left( {\zeta }_{4p}\right) \) . | Proof. Let \( {\zeta }_{n} \) be a fixed primitive \( n \) th root of unity in \( \mathbb{C} \) for each \( n \) .\n\nCase 1. \( p \) an odd prime:\n\nIf \( p \equiv 1{\;\operatorname{mod}\;4} \), then by the proposition, we know that \( \sqrt{p} \) lies in \( \mathbb{Q}\left( {\zeta }_{p}\right) \subset \mathbb{Q}\lef... | Yes |
Theorem 56.17. Let \( n \) be a nonzero integer. Then \( \sqrt{n} \) lies in \( \mathbb{Q}\left( {\zeta }_{4n}\right) \) , where \( {\zeta }_{4n} \) is a primitive \( {4n} \) th root of unity. In particular, any quadratic extension of \( \mathbb{Q} \) lies in \( \mathbb{Q}\left( \zeta \right) \) for some root of unity ... | Proof. Let \( {\zeta }_{n} \) be a fixed primitive \( n \) th root of unity in \( \mathbb{C} \) for each \( n \) . We may assume that \( n \) is square-free. We know the result if \( n = - 1 \), so we may assume that \( n = \pm {p}_{1}\cdots {p}_{r} \) with \( {p}_{1} < \cdots < {p}_{r} \) positive primes with \( r \ge... | Yes |
Theorem 57.3. (Constructibility Criterion (Refined Form)) Let \( z \) be a complex number and \( {z}_{1}\left( { = 0}\right) ,{z}_{2}\left( { = 1}\right) ,\ldots ,{z}_{n} \) other complex numbers with \( n \geq 2 \) . Set \( F = \mathbb{Q}\left( {{z}_{1},{z}_{2},\ldots ,{z}_{n},\overline{{z}_{1}},\ldots ,\overline{{z}_... | Proof. (2) if and only if (3) follows from the Square Root Tower Theorem 54.11.\n\n\( \left( 2\right) \Rightarrow \left( 1\right) \) follows from the original Constructibility Criterion 49.9.\n\n\( \left( 1\right) \Rightarrow \left( 2\right) \) : By the original Constructibility Criterion 49.9, there exists a square ro... | Yes |
Lemma 57.7. Let \( n \) be a positive integer and \( F \) a field satisfying \( \operatorname{char}F = 0 \) or \( \operatorname{char}F//n \) . Suppose that \( K/F \) is a splitting field of a separable polynomial \( f \) over \( F \) and \( L/F \) is a a splitting field of \( {t}^{n} - 1 \) over \( K \) . Then \( G\lef... | Proof. We know that \( L \) is a splitting field of the separable polynomial \( \left( {{t}^{n} - 1}\right) f \) over \( F \), so \( L/F \) is Galois. As \( L/K \) is abelian, by Remark 57.6, the group \( G\left( {L/F}\right) \) is solvable if and only if the group \( G\left( {K/F}\right) \) is solvable. | Yes |
Lemma 57.10. Let \( p \) be a prime and \( G \) a subgroup of the symmetric group \( {S}_{p} \) containing a p-cycle and a transposition. Then \( G = {S}_{p} \). | Proof. (of the lemma) By changing notation, we may assume that \( \left( {12}\right) \in G \) . Let \( \sigma = \left( {{a}_{1}\cdots {a}_{p}}\right) \) lie in \( G \) . As \( p \) is a prime, \( {\sigma }^{i},1 \leq i < p \) , is also a \( p \) -cycle (cf. Exercise \( {22.22}\left( 1\right) \), so we may assume that \... | No |
Proposition 57.11. Let \( f \) be an irreducible polynomial in \( \mathbb{Q}\left\lbrack t\right\rbrack \) of degree \( p \) with \( p \) a prime. Suppose that \( f \) has precisely two nonreal roots in the splitting field \( K \) in \( \mathbb{C} \) of \( f \) over \( \mathbb{Q} \) . Then \( G\left( {K/\mathbb{Q}}\rig... | Proof. Let \( \alpha \) in \( K \) be a root of \( f \), so \( \left\lbrack {\mathbb{Q}\left( \alpha \right) : \mathbb{Q}}\right\rbrack = \deg f = p \) . As char \( K = 0 \), we know that \( K/\mathbb{Q} \) is Galois, hence \( \mid G\left( {K/\mathbb{Q}}\right) = \left\lbrack {K : \mathbb{Q}}\right\rbrack \leq \) \( p ... | Yes |
Theorem 57.12. (Abel-Ruffini) In general, there is no formula to determine the roots of a general fifth degree polynomial in \( \mathbb{Q}\left\lbrack t\right\rbrack \) that involves only addition, multiplication, and extraction of nth roots for various \( n \) . | Proof. It suffices to give an example of a polynomial \( f \) in \( \mathbb{Q}\left\lbrack t\right\rbrack \) that has a nonsolvable Galois group. Let \( f = {t}^{5} - {6t} + 3 \) . We know that \( f \) is irreducible by Eisenstein’s Criterion, and by calculus it has three real roots in \( \mathbb{R} \), since \( {f}^{\... | Yes |
Corollary 57.13. In general, there is no formula to determine the roots of a general nth degree polynomial in \( \mathbb{Q}\left\lbrack t\right\rbrack \) that involves only addition, multiplication, and extraction of nth roots for various n. | Proof. Apply the theorem to any \( n \) th degree polynomial with \( n \geq \) 5 divisible by a fifth degree polynomial not solvable by radicals. | Yes |
Lemma 57.19. (Hilbert Theorem 90) Let \( K/F \) be a Galois extension of fields of degree \( n \) . Suppose that \( K/F \) is cyclic with \( G\left( {K/F}\right) = \langle \sigma \rangle \) and \( x \in K \) . Then\n\n\( {\mathrm{N}}_{K/F}\left( x\right) = 1 \) if and only if there exists a \( y \in {K}^{ \times } \) s... | Proof. ( \( \Leftarrow \) ): If \( x = y/\sigma \left( y\right) \), then\n\n\[ {\mathrm{N}}_{K/F}\left( x\right) = \frac{{\mathrm{N}}_{K/F}\left( y\right) }{{\mathrm{N}}_{K/F}\left( {\sigma \left( y\right) }\right) } = 1. \]\n\n\( \left( \Rightarrow \right) \) : Suppose that\n\n\[ 1 = {\mathrm{N}}_{K/F}\left( x\right) ... | Yes |
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