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Corollary 2.15. If \( \\mathrm{V} \) and \( \\mathrm{W} \) are finite dimensional subspaces of a vector space over a division ring \( \\mathrm{D} \), then\n\n\[ \n{\\dim }_{\\mathrm{D}}\\mathrm{V} + {\\dim }_{\\mathrm{D}}\\mathrm{W} = {\\dim }_{\\mathrm{D}}\\left( {\\mathrm{V} \\cap \\mathrm{W}}\\right) + {\\dim }_{\\m... | SKETCH OF PROOF. Let \( X \) be a basis of \( V \\cap W, Y \) a (finite) basis of \( V \) that contains \( X \), and \( Z \) a (finite) basis of \( W \) that contains \( X \) (Theorem 2.4). Show that \( X \\cup \\left( {Y - X}\\right) \\cup \\left( {Z - X}\\right) \) is a basis of \( V + W \), whence\n\n\[ \n{\\dim }_{... | No |
Theorem 2.16. Let \( \mathrm{R},\mathrm{S},\mathrm{T} \) be division rings such that \( \mathrm{R} \subset \mathrm{S} \subset \mathrm{T} \) . Then\n\n\[{\dim }_{\mathrm{R}}\mathrm{T} = \left( {{\dim }_{\mathrm{S}}\mathrm{T}}\right) \left( {{\dim }_{\mathrm{R}}\mathrm{S}}\right) .\n\]\n\nFurthermore, \( {\dim }_{\mathrm... | PROOF. Let \( U \) be a basis of \( T \) over \( S \), and let \( V \) a basis of \( S \) over \( R \) . It suffices to show that \( \{ {vu} \mid v \in V, u \in U\} \) is a basis of \( T \) over \( R \) . For the elements \( {vu} \) are all distinct by the linear independence of \( U \) over \( S \) . Consequently, we ... | Yes |
Theorem 3.2. Every free module \( \mathrm{F} \) over a ring \( \mathrm{R} \) with identity is projective. | PROOF OF 3.2. In view of the remarks preceding the theorem we may assume that we are given a diagram of homomorphisms of unitary \( R \) -modules:\n\n\n\nwith \( g \) an epimorphism and \( F \) a free \( R \) -module... | Yes |
Corollary 3.3. Every module A over a ring \( \mathrm{R} \) is the homomorphic image of a projective \( \mathrm{R} \) -module. | PROOF. Immediate from Theorem 3.2 and Corollary 2.2. | No |
Theorem 3.4. Let \( \mathrm{R} \) be a ring. The following conditions on an \( \mathrm{R} \) -module \( \mathrm{P} \) are equivalent.\n\n(i) \( \mathrm{P} \) is projective;\n\n(ii) every short exact sequence \( 0 \rightarrow \mathrm{A}\overset{\mathrm{f}}{ \rightarrow }\mathrm{B}\overset{\mathrm{g}}{ \rightarrow }\math... | PROOF OF 3.4. (i) \( \Rightarrow \) (ii) Consider the diagram \n\nwith bottom row exact by hypothesis. Since \( P \) is projective there is an \( R \) -module homomorphism \( h : P \rightarrow B \) such that \( {gh} ... | Yes |
Proposition 3.5. Let \( \mathrm{R} \) be a ring. A direct sum of \( \mathrm{R} \) -modules \( \mathop{\sum }\limits_{{i \in I}}{\mathrm{P}}_{\mathrm{i}} \) is projective if and only if each \( {\mathrm{P}}_{\mathrm{i}} \) is projective. | SKETCH OF PROOF. Suppose \( \sum {P}_{i} \) is projective. Since the proof of (iii) \( \Rightarrow \) (i) in Theorem 3.4 uses only the fact that \( F \) is projective, it remains valid with \( \mathop{\sum }\limits_{{i \in I}}{P}_{i} \) , \( \mathop{\sum }\limits_{{i \neq j}}{P}_{i} \) and \( {P}_{j} \) in place of \( ... | No |
Proposition 3.7. A direct product of \( \mathrm{R} \) -modules \( \mathop{\prod }\limits_{{i \in I}}{\mathrm{\;J}}_{\mathrm{i}} \) is injective if and only if \( {\mathrm{J}}_{\mathrm{i}} \) is injective for every \( \mathrm{i} \in \mathbf{I} \) . | ## PROOF. Exercise; see Proposition 3.5. | No |
Lemma 3.11. If \( \mathrm{J} \) is a divisible abelian group and \( \mathrm{R} \) is a ring with identity, then \( {\operatorname{Hom}}_{\mathbf{Z}}\left( {\mathrm{R},\mathrm{J}}\right) \) is an injective left \( \mathrm{R} \) -module. | SKETCH OF PROOF. By Lemma 3.8 it suffices to show that for each left ideal \( L \) of \( R \), every \( R \) -module homomorphism \( f : L \rightarrow {\operatorname{Hom}}_{\mathbf{Z}}\left( {R, J}\right) \) may be extended to an \( R \) -module homomorphism \( h : R \rightarrow {\operatorname{Hom}}_{\mathbf{Z}}\left( ... | No |
Proposition 3.12. Every unitary module A over a ring \( \mathbf{R} \) with identity may be embedded in an injective \( \mathrm{R} \) -module. | SKETCH OF PROOF. Since \( A \) is an abelian group, there is a divisible group \( J \) and a group monomorphism \( f : A \rightarrow J \) by Lemma 3.10. The map \( \bar{f} : {\operatorname{Hom}}_{\mathbf{Z}}\left( {R, A}\right) \) \( \rightarrow {\operatorname{Hom}}_{\mathbf{Z}}\left( {R, J}\right) \) given on \( {g\va... | No |
Proposition 3.13. Let \( \\mathrm{R} \) be a ring with identity. The following conditions on a unitary \( \\mathrm{R} \) -module \( \\mathrm{J} \) are equivalent.\n\n(i) \( \\mathrm{J} \) is injective;\n\n(ii) every short exact sequence \( 0 \\rightarrow \\mathrm{J}\\overset{\\mathrm{f}}{ \\rightarrow }\\mathrm{B}\\ove... | SKETCH OF PROOF. (i) \\(\\Rightarrow\\) (ii) Dualize the proof of (i) \\(\\Rightarrow\\) (ii) of Theorem 3.4. (ii) \\(\\Rightarrow\\) (iii) since the sequence \( 0 \\rightarrow J\\overset{ \\subset }{ \\rightarrow }B\\overset{\\pi }{ \\rightarrow }B/J \\rightarrow 0 \) is split exact, there is a homomorphism \( g : B/J... | No |
Theorem 4.1. Let A, B, C, D be modules over a ring \( \mathrm{R} \) and \( \varphi : \mathrm{C} \rightarrow \mathrm{A} \) and \( \psi : \mathrm{B} \rightarrow \mathrm{D} \) \( \mathrm{R} \) -module homomorphisms. Then the map \( \theta : {\operatorname{Hom}}_{\mathrm{R}}\left( {\mathrm{A},\mathrm{B}}\right) \rightarrow... | SKETCH OF PROOF. \( \theta \) is well defined since composition of \( R \) -module homomorphisms is an \( R \) -module homomorphism. \( \theta \) is a homomorphism since such composition of homomorphisms is distributive with respect to addition. | No |
Proposition 4.3. Let \( \mathrm{R} \) be a ring. \( \mathrm{A}\overset{\theta }{ \rightarrow }\mathrm{B}\overset{\zeta }{ \rightarrow }\mathrm{C} \rightarrow 0 \) is an exact sequence of \( \mathrm{R} \) -modules if and only if for every \( \mathrm{R} \) -module \( \mathrm{D} \) \[ 0 \rightarrow {\operatorname{Hom}}_{\... | SKETCH OF PROOF. If \( A\overset{\theta }{ \rightarrow }B\overset{\xi }{ \rightarrow }C \rightarrow 0 \) is exact, we shall show that \( \operatorname{Ker}\bar{\theta } \subset \operatorname{Im}\bar{\zeta } \) . If \( {f\varepsilon }\operatorname{Ker}\bar{\theta } \), then \( 0 = \bar{\theta }\left( f\right) = {f\theta... | No |
Proposition 4.4. The following conditions on modules over a ring \( \mathbf{R} \) are equivalent.\n\n(i) \( 0 \rightarrow \mathrm{A}\overset{\varphi }{ \rightarrow }\mathrm{B}\overset{\psi }{ \rightarrow }\mathrm{C} \rightarrow 0 \) is a split exact sequence of \( \mathrm{R} \) -modules;\n\n(ii) \( 0 \rightarrow {\oper... | SKETCH OF PROOF. (i) \( \Rightarrow \) (iii) By Theorem 1.18 there is a homomorphism \( \alpha : B \rightarrow A \) such that \( {\alpha \varphi } = {1}_{A} \) . Verify that the induced-homomorphism\n\n\[ \bar{\alpha } : {\operatorname{Hom}}_{R}\left( {A, D}\right) \rightarrow {\operatorname{Hom}}_{R}\left( {B, D}\righ... | No |
Theorem 4.5. The following conditions on a module \( \mathrm{P} \) over a ring \( \mathrm{R} \) are equivalent\n\n(i) \( \mathrm{P} \) is projective;\n\n(ii) if \( \psi : \mathrm{B} \rightarrow \mathrm{C} \) is any \( \mathrm{R} \) -module epimorphism then \( \bar{\psi } : {\operatorname{Hom}}_{\mathrm{R}}\left( {\math... | SKETCH OF PROOF. (i) \( \Leftrightarrow \) (ii) The map \( \bar{\psi } : {\operatorname{Hom}}_{R}\left( {P, B}\right) \rightarrow {\operatorname{Hom}}_{R}\left( {P, C}\right) \) (given by \( g \mapsto {\psi g} \) ) is an epimorphism if and only if for every \( R \) -module homomorphism \( f : P \rightarrow C \), there ... | No |
Proposition 4.6. The following conditions on a module \( \mathbf{J} \) over a ring \( \mathbf{R} \) are equivalent.\n\n(i) \( \mathrm{J} \) is injective;\n\n(ii) if \( \theta : \mathrm{A} \rightarrow \mathrm{B} \) is any \( \mathrm{R} \) -module monomorphism, then \( \bar{\theta } : {\operatorname{Hom}}_{\mathrm{R}}\le... | PROOF. The proof is dual to that of Theorem 4.5 and is left as an exercise. | No |
Theorem 4.7. Let \( A, B,\left\{ {{A}_{i} \mid {i\varepsilon I}}\right\} \) and \( \left\{ {{B}_{j} \mid {j\varepsilon J}}\right\} \) be modules over a ring \( R \) . Then there are isomorphisms of abelian groups:\n\n(i) \( {\operatorname{Hom}}_{\mathrm{R}}\left( {\mathop{\sum }\limits_{{i \in I}}{\mathrm{\;A}}_{\mathr... | SKETCH OF PROOF OF 4.7. (i) For each \( {i\varepsilon I} \) let \( {\iota }_{i} : {A}_{i} \rightarrow \mathop{\sum }\limits_{{i\varepsilon I}}{A}_{i} \) be the canonical injection (Theorem 1.11). Given \( \left\{ {g}_{i}\right\} \varepsilon \mathop{\prod }\limits_{{i \in I}}{\operatorname{Hom}}_{R}\left( {{A}_{i}, B}\r... | No |
Theorem 4.8. Let \( \mathrm{R} \) and \( \mathrm{S} \) be rings and let \( {}_{\mathrm{R}}\mathrm{A},{}_{\mathrm{R}}{\mathrm{B}}_{\mathrm{S}},{}_{\mathrm{R}}{\mathrm{C}}_{\mathrm{S}},{}_{\mathrm{R}}\mathrm{D} \) be (bi)modules as indicated.\n\n(i) \( {\operatorname{Hom}}_{\mathrm{R}}\left( {\mathrm{A},\mathrm{B}}\right... | SKETCH OF PROOF. (i) The verification that \( {fs} \) is a well-defined module homomorphism and that \( {\operatorname{Hom}}_{R}\left( {A, B}\right) \) is actually a right \( S \) -module is tedious but straight-forward; similarly for (iii). (ii) \( \bar{\varphi } \) is an abelian group homomorphism by Theorem 4.1. If ... | No |
Theorem 4.9. If \( \mathrm{A} \) is a unitary left module over a ring \( \mathrm{R} \) with identity then there is an isomorphism of left \( \mathrm{R} \) -modules \( \mathrm{A} \cong {\operatorname{Hom}}_{\mathrm{R}}\left( {\mathrm{R},\mathrm{A}}\right) \) . | SKETCH OF PROOF. Since \( R \) is an \( R - R \) bimodule, the left module structure of \( {\operatorname{Hom}}_{R}\left( {R, A}\right) \) is given by Theorem 4.8(iii). Verify that the map \( \varphi : {\operatorname{Hom}}_{R}\left( {R, A}\right) \) \( \rightarrow A \) given by \( f \mapsto f\left( {1}_{R}\right) \) is... | No |
Theorem 4.11. Let \( \mathrm{F} \) be a free left module over a ring \( \mathrm{R} \) with identity. Let \( \mathrm{X} \) be a basis of \( \mathrm{F} \) and for each \( \mathrm{x}\varepsilon \mathrm{X} \) let \( {\mathrm{f}}_{\mathrm{x}} : \mathrm{F} \rightarrow \mathrm{R} \) be given by \( {\mathrm{f}}_{\mathrm{x}}\le... | PROOF OF 4.11. (i) If \( {f}_{{x}_{1}}{r}_{1} + {f}_{{x}_{2}}{r}_{2} + \cdots + {f}_{{x}_{n}}{r}_{n} = 0\left( {{r}_{i}{\varepsilon R};{x}_{i}{\varepsilon X}}\right) \), then for each \( j = 0,1,2,\ldots, n \) ,\n\n\[ 0 = \left\langle {{x}_{j},0}\right\rangle = \left\langle {{x}_{j},\mathop{\sum }\limits_{{i = 1}}^{n}{... | Yes |
Theorem 4.12. Let \( \mathrm{A} \) be a left module over a ring \( \mathrm{R} \) . (i) There is an R-module homomorphism \( \theta : \mathrm{A} \rightarrow {\mathrm{A}}^{* * } \) . (ii) If \( \mathrm{R} \) has an identity and \( \mathrm{A} \) is free, then \( \theta \) is a monomorphism. (iii) If \( \mathrm{R} \) has a... | PROOF OF 4.12. (i) For each \( {a\varepsilon A} \) let \( \theta \left( a\right) : {A}^{ * } \rightarrow R \) be the map defined by \( \left\lbrack {\theta \left( a\right) }\right\rbrack \left( f\right) = \langle a, f\rangle {\varepsilon R} \) . Statement (2) after Theorem 4.10 shows that \( \theta \left( a\right) \) i... | Yes |
Theorem 5.2. Let \( {\mathrm{A}}_{\mathrm{R}} \) and \( {}_{\mathrm{R}}\mathrm{B} \) be modules over a ring \( \mathrm{R} \), and let \( \mathrm{C} \) be an abelian group. If \( \mathrm{g} : \mathrm{A} \times \mathrm{B} \rightarrow \mathrm{C} \) is a middle linear map, then there exists a unique group homomorphism \( \... | SKETCH OF PROOF. Let \( F \) be the free abelian group on the set \( A \times B \), and let \( K \) be the subgroup described in Definition 5.1. Since \( F \) is free, the assignment \( \left( {a, b}\right) \mapsto g\left( {a, b}\right) {\varepsilon C} \) determines a unique homomorphism \( {g}_{1} : F \rightarrow C \)... | No |
Corollary 5.3. If \( {\mathrm{A}}_{\mathrm{R}},{\mathrm{A}}_{\mathrm{R}}{}^{\prime },{}_{\mathrm{R}}\mathrm{B} \) and \( {}_{\mathrm{R}}{\mathrm{B}}^{\prime } \) are modules over a ring \( \mathrm{R} \) and \( \mathrm{f} : \mathrm{A} \rightarrow {\mathrm{A}}^{\prime } \) , \( \mathrm{g} : \mathrm{B} \rightarrow {\mathr... | SKETCH OF PROOF. Verify that the assignment \( \left( {a, b}\right) \mapsto f\left( a\right) \otimes g\left( b\right) \) defines a middle linear map \( h : A \times B \rightarrow C = {A}^{\prime }{\bigotimes }_{R}{B}^{\prime } \) . By Theorem 5.2 there is a unique homomorphism \( \bar{h} : A{\bigotimes }_{R}B \rightarr... | No |
(i) \( \mathrm{A}{ \otimes }_{\mathrm{R}}\mathrm{B} \) is a left \( \mathrm{S} \) -module such that \( \mathrm{s}\left( {\mathrm{a} \otimes \mathrm{b}}\right) = \mathrm{{sa}} \otimes \mathrm{b} \) for all \( \mathrm{s} \in \mathrm{S},\mathrm{a} \in \mathrm{A} \) , \( \mathrm{b}\varepsilon \mathrm{B} \) . | SKETCH OF PROOF. (i) For each \( s \in S \) the map \( A \times B \rightarrow A{\bigotimes }_{R}B \) given by \( \left( {a, b}\right) \mapsto {sa} \otimes b \) is \( R \) -middle linear, and therefore induces a unique group homomorphism \( {\alpha }_{s} : A{ \otimes }_{R}B \rightarrow A{ \otimes }_{R}B \) such that \( ... | No |
Theorem 5.6. If \( \mathrm{A},\mathrm{B},\mathrm{C} \) are modules over a commutative ring \( \mathrm{R} \) and \( \mathrm{g} : \mathrm{A} \times \mathrm{B} \rightarrow \mathrm{C} \) is a bilinear map, then there is a unique \( \mathrm{R} \) -module homomorphism \( \overline{\mathrm{g}} : \mathrm{A}{\bigotimes }_{\math... | SKETCH OF PROOF. Verify that the unique homomorphism of abelian groups \( \bar{g} : A{\bigotimes }_{R}B \rightarrow C \) given by Theorem 5.2 is actually an \( R \) -module homomorphism. To prove the last statement let \( \mathcal{B}\left( {A, B}\right) \) be the category of all bilinear maps on \( A \times B \) (defin... | No |
Theorem 5.7. If \( \mathrm{R} \) is a ring with identity and \( {\mathrm{A}}_{\mathrm{R}},{}_{\mathrm{R}}\mathrm{B} \) are unitary \( \mathrm{R} \) -modules, then there are \( \mathrm{R} \) -module isomorphisms\n\n\[ \mathrm{A}{\bigotimes }_{\mathrm{R}}\mathrm{R} \cong \mathrm{A}\;\text{ and }\;\mathrm{R}{\bigotimes }_... | SKETCH OF PROOF. Since \( R \) is an \( R - R \) bimodule \( R{\bigotimes }_{R}B \) is a left \( R \) - module by Theorem 5.5. The assignment \( \left( {r, b}\right) \mapsto {rb} \) defines a middle linear map \( R \times B \rightarrow B \) . By Theorem 5.2 there is a group homomorphism \( \alpha : R{\bigotimes }_{R}B ... | No |
Theorem 5.8. If \( \mathrm{R} \) and \( \mathrm{S} \) are rings and \( {\mathrm{A}}_{\mathrm{R}},{}_{\mathrm{R}}{\mathrm{B}}_{\mathrm{S}},{}_{\mathrm{S}}\mathrm{C} \) are (bi)modules, then there is an isomorphism\n\n\[ \left( {\mathrm{A}{ \otimes }_{\mathrm{R}}\mathrm{B}}\right) { \otimes }_{\mathrm{S}}\mathrm{C} \cong... | PROOF. By definition every element \( v \) of \( \left( {A{\bigotimes }_{R}B}\right) {\bigotimes }_{S}C \) is a finite sum \( \mathop{\sum }\limits_{{i = 1}}^{n}{u}_{i} \otimes {c}_{i}\left( {{u}_{i} \in A{ \otimes }_{R}B,{c}_{i} \in C}\right) \) . Since each \( {u}_{i} \in A{ \otimes }_{R}B \) is a finite sum \( \math... | Yes |
Theorem 5.10. (Adjoint Associativity) Let \( \\mathrm{R} \) and \( \\mathrm{S} \) be rings and \( {\\mathrm{A}}_{\\mathrm{R}},{}_{\\mathrm{R}}{\\mathrm{B}}_{\\mathrm{S}},{\\mathrm{C}}_{\\mathrm{S}} \) (bi)- modules. Then there is an isomorphism of abelian groups\n\n\[ \n\\alpha : {\\operatorname{Hom}}_{\\mathrm{S}}\\le... | SKETCH OF PROOF OF 5.10. The proof is a straightforward exercise in the use of the appropriate definitions. The following items must be checked.\n\n(i) For each \( a \\in A \), and \( f \\in {\\operatorname{Hom}}_{S}\\left( {A{\\bigotimes }_{R}B, C}\\right) ,\\left( {\\alpha f}\\right) \\left( a\\right) : B \\rightarro... | No |
Theorem 5.11. Let \( \mathrm{R} \) be a ring with identity. If \( \mathrm{A} \) is a unitary right \( \mathrm{R} \) -module and \( \mathrm{F} \) is a free left R-module with basis \( \mathrm{Y} \), then every element \( \mathrm{u} \) of \( \mathrm{A}{\bigotimes }_{\mathrm{R}}\mathrm{F} \) may be written uniquely in the... | PROOF OF 5.11. For each \( y \in Y \), let \( {A}_{y} \) be a copy of \( A \) and consider the direct \( \operatorname{sum}\mathop{\sum }\limits_{{y \in Y}}{A}_{y} \) . We first construct an isomorphism \( \theta : A{ \otimes }_{R}F \cong \mathop{\sum }\limits_{{y \in Y}}{A}_{y} \) as follows. Since \( Y \) is a basis,... | Yes |
Corollary 5.12. If \( \mathrm{R} \) is a ring with identity and \( {\mathrm{A}}_{\mathrm{R}} \) and \( {}_{\mathrm{R}}\mathrm{B} \) are free \( \mathrm{R} \) -modules with bases \( \mathrm{X} \) and \( \mathrm{Y} \) respectively, then \( \mathrm{A}{ \otimes }_{\mathrm{R}}\mathrm{B} \) is a free (right) \( \mathrm{R} \)... | SKETCH OF PROOF OF 5.12. By the proof of Theorem 5.11 and by Theorem 2.1 (for right \( R \) -modules) there is a group isomorphism\n\n\[ \n\theta : A{\bigotimes }_{R}B \cong \mathop{\sum }\limits_{{y\varepsilon Y}}{A}_{y} = \mathop{\sum }\limits_{{y\varepsilon Y}}A = \mathop{\sum }\limits_{{y\varepsilon Y}}\left( {\mat... | No |
Corollary 5.13. Let \( \mathrm{S} \) be a ring with identity and \( \mathrm{R} \) a subring of \( \mathrm{S} \) that contains \( {1}_{\mathrm{S}} \) . If \( \mathrm{F} \) is a free left \( \mathrm{R} \) -module with basis \( \mathrm{X} \), then \( \mathrm{S}{\bigotimes }_{\mathrm{R}}\mathrm{F} \) is a free left \( \mat... | SKETCH OF PROOF. Since \( S \) is clearly an \( S - R \) bimodule, \( S{\bigotimes }_{R}F \) is a left \( S \) -module by Theorem 5.5. The proof of Theorem 5.11 shows that there is a group isomorphism \( \theta : S{ \otimes }_{R}F \cong \mathop{\sum }\limits_{{x \in X}}{S}_{x} \), with each \( {S}_{x} = S \) . Furtherm... | No |
Corollary 6.2. Let \( \mathrm{R} \) be a principal ideal domain. If \( \mathrm{A} \) is a finitely generated \( \mathrm{R} \) -module generated by \( \mathrm{n} \) elements, then every submodule of \( \mathrm{A} \) may be generated by \( \mathrm{m} \) elements with \( \mathrm{m} \leq \mathrm{n} \) . | PROOF. Exercise; see Corollary II.1.7 and Corollary 2.2. | No |
Corollary 6.3. A unitary module A over a principal ideal domain is free if and only if A is projective. | PROOF. ( \( \Rightarrow \) ) Theorem 3.2. ( \( \Leftarrow \) ) There is a short exact sequence \( 0 \rightarrow K\overset{ \subset }{ \rightarrow }F\overset{f}{ \rightarrow } \) \( A \rightarrow 0 \) with \( F \) free, \( f \) an epimorphism and \( K = \ker f \) by Corollary 2.2. If \( A \) is projective, then \( F \co... | No |
Theorem 6.4. Let \( \mathrm{A} \) be a left module over an integral domain \( \mathrm{R} \) and for each \( \mathrm{a}\varepsilon \mathrm{A} \) let \( {\mathcal{O}}_{\mathrm{a}} = \{ \mathrm{r}\varepsilon \mathrm{R} \mid \mathrm{{ra}} = 0\} \) . (iii) For each \( \mathrm{a}\varepsilon \mathrm{A} \) there is an isomorph... | SKETCH OF PROOF OF 6.4. (iii) Use Theorems 1.5(i) and 1.7. | No |
Theorem 6.5. A finitely generated torsion-free module A over a principal ideal domain \( \mathrm{R} \) is free. | REMARK. The hypothesis that \( A \) is finitely generated is essential (Exercise II.1.10).\n\nPROOF OF 6.5. We may assume \( A \neq 0 \) . Let \( X \) be a finite set of nonzero generators of \( A \) . If \( {x\varepsilon X} \), then \( {rx} = 0\left( {r\varepsilon R}\right) \) if and only if \( r = 0 \) since \( A \) ... | No |
Theorem 6.5. If \( \mathrm{A} \) is a finitely generated module over a principal ideal domain \( \mathrm{R} \) , then \( \mathrm{A} = {\mathrm{A}}_{\mathrm{t}} \oplus \mathrm{F} \), where \( \mathrm{F} \) is a free \( \mathrm{R} \) -module of finite rank and \( \mathrm{F} \cong \mathrm{A}/{\mathrm{A}}_{\mathrm{t}} \) . | SKETCH OF PROOF. The quotient module \( A/{A}_{t} \) is torsion-free since for each \( r \neq 0 \) ,\n\n\[ r\left( {a + {A}_{t}}\right) = {A}_{t} \Rightarrow {ra}\varepsilon {A}_{t} \Rightarrow {r}_{1}\left( {ra}\right) = 0\text{ for some }{r}_{1} \neq 0 \Rightarrow a\varepsilon {A}_{t} \]\n\nFurthermore, \( A/{A}_{t} ... | No |
Theorem 6.7. Let \( \mathrm{A} \) be a torsion module over a principal ideal domain \( \mathrm{R} \) and for each prime \( \mathrm{p}\varepsilon \mathrm{R} \) let \( \mathrm{A}\left( \mathrm{p}\right) = \{ \mathrm{a}\varepsilon \mathrm{A} \mid \mathrm{a} \) has order a power of \( \mathrm{p}\} \) .\n\n(i) \( \mathrm{A}... | PROOF. (i) Let \( a, b \in A\left( p\right) \) . If \( {\mathcal{O}}_{a} = \left( {p}^{r}\right) \) and \( {\mathcal{O}}_{b} = \left( {p}^{s}\right) \) let \( k = \max \left( {r, s}\right) \) . Then \( {p}^{k}\left( {a + b}\right) = 0 \), whence \( {\mathcal{O}}_{a + b} = \left( {p}^{i}\right) \) with \( 0 \leq i \leq ... | Yes |
Theorem 6.9. Let \( \mathrm{A} \) be a finitely generated module over a principal ideal domain \( \mathrm{R} \) such that every element of \( \mathrm{A} \) has order a power of some prime \( \mathrm{p}\varepsilon \mathrm{R} \) . Then \( \mathrm{A} \) is a direct sum of cyclic \( \mathrm{R} \) -modules of orders \( {\ma... | PROOF. The proof proceeds by induction on the number \( r \) of generators of \( A \) , with the case \( r = 1 \) being trivial. If \( r > 1 \), then \( A \) is generated by elements \( {a}_{1},\ldots ,{a}_{r} \) whose orders are respectively \( {p}^{{n}_{1}},{p}^{{m}_{2}},{p}^{{m}_{3}},\ldots ,{p}^{{m}_{r}} \) . We ma... | Yes |
Lemma 6.11. Let \( \mathrm{R} \) be a principal ideal domain. If \( \mathrm{r} \in \mathrm{R} \) factors as \( \mathrm{r} = {\mathrm{p}}_{1}{}^{{\mathrm{n}}_{1}}\cdots {\mathrm{p}}_{\mathrm{k}}{}^{{\mathrm{n}}_{\mathrm{k}}} \) with \( {\mathrm{p}}_{\mathrm{i}},\ldots ,{\mathrm{p}}_{\mathrm{k}} \in \mathrm{R} \) distinc... | SKETCH OF PROOF. We shall prove that if \( s, t \in R \) are relatively prime, then \( R/\left( {st}\right) \cong R/\left( s\right) \oplus R/\left( t\right) \) . The first part of the lemma then follows by induction on the number of distinct primes in the prime decomposition of \( r \) . The last statement of the lemma... | Yes |
Theorem 6.12. Let \( \mathrm{A} \) be a finitely generated module over a principal ideal domain \( \mathrm{R} \). (i) A is the direct sum of a free submodule \( \mathrm{F} \) of finite rank and a finite number of cyclic torsion modules. The cyclic torsion summands (if any) are of orders \( {\mathrm{r}}_{1},\ldots ,{\ma... | SKETCH OF PROOF OF 6.12. The existence of a direct sum decomposition of the type described in (ii) is an immediate consequence of Theorems 6.6, 6.7, and 6.9. Thus \( A \) is the direct sum of a free module and a finite family of cyclic \( R \) -modules, each of which has order a power of a prime. In the case of abelian... | No |
Corollary 6.13. Two finitely generated modules over a principal ideal domain, A and \( \mathrm{B} \), are isomorphic if and only if \( \mathrm{A}/{\mathrm{A}}_{\mathrm{t}} \) and \( \mathrm{B}/{\mathrm{B}}_{\mathrm{t}} \) have the same rank and \( \mathrm{A} \) and \( \mathrm{B} \) have the same invariant factors [resp... | PROOF. Exercise. | No |
Theorem 7.2. Let \( \mathrm{K} \) be a commutative ring with identity and \( \mathrm{A} \) a unitary left \( \mathrm{K} \) -module. Then \( \mathrm{A} \) is a \( \mathrm{K} \) -algebra if and only if there exists a \( \mathrm{K} \) -module homomorphism \( \pi : \mathrm{A}{ \otimes }_{\mathrm{K}}\mathrm{A} \rightarrow \... | SKETCH OF PROOF. If \( A \) is a \( K \) -algebra, then the map \( A \times A \rightarrow A \) given by \( \left( {a, b}\right) \mapsto {ab} \) is \( K \) -bilinear, whence there is a \( K \) -module homomorphism\n\n\[ \pi : A{\bigotimes }_{K}A \rightarrow A \]\n\nby Theorem 5.6. Verify that \( \pi \) has the required ... | No |
Theorem 7.4. Let \( \mathrm{A} \) and \( \mathrm{B} \) be algebras [with identity] over a commutative ring \( \mathrm{K} \) with identity. Let \( \pi \) be the composition\n\n\[ \n\\left( {\\mathrm{A}{\\bigotimes }_{\\mathrm{K}}\\mathrm{B}}\\right) {\\bigotimes }_{\\mathrm{K}}\\left( {\\mathrm{A}{\\bigotimes }_{\\mathr... | PROOF. Exercise; note that for generators \( a \\otimes b \) and \( {a}_{1} \\otimes {b}_{1} \) of \( A{ \\otimes }_{K}B \) the product is defined to be\n\n\[ \n\\left( {a \\otimes b}\\right) \\left( {{a}_{1} \\otimes {b}_{1}}\\right) = \\pi \\left( {a \\otimes b \\otimes {a}_{1} \\otimes {b}_{1}}\\right) = a{a}_{1} \\... | No |
Theorem 1.2. Let \( \\mathrm{F} \) be an extension field of \( \\mathrm{E} \) and \( \\mathrm{E} \) an extension field of \( \\mathrm{K} \). Then \( \\left\\lbrack {\\mathrm{F} : \\mathrm{K}}\\right\\rbrack = \\left\\lbrack {\\mathrm{F} : \\mathrm{E}}\\right\\rbrack \\left\\lbrack {\\mathrm{E} : \\mathrm{K}}\\right\\rb... | PROOF. This is a restatement of Theorem IV.2.16. | Yes |
Theorem 1.3. If \( \mathrm{F} \) is an extension field of a field \( \mathrm{K},\mathrm{u},{\mathrm{u}}_{\mathrm{i}}\varepsilon \mathrm{F} \), and \( \mathrm{X} \subset \mathrm{F} \), then\n\n(i) the subring \( \mathrm{K}\left\lbrack \mathrm{u}\right\rbrack \) consists of all elements of the form \( \mathrm{f}\left( \m... | SKETCH OF PROOF. (vi) Every field that contains \( K \) and \( X \) must contain the \( \mathrm{{set}}E = \left\{ \begin{matrix} f\left( {{u}_{1},\ldots ,{u}_{n}}\right) /g\left( {{u}_{1},\ldots ,{u}_{n}}\right) & \;|\;n \in {\mathbf{N}}^{ * };\;f, \\ g \in K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack ;\;{u}_{i... | Yes |
Theorem 1.5. If \( \mathrm{F} \) is an extension field of \( \mathrm{K} \) and \( \mathrm{u}\varepsilon \mathrm{F} \) is transcendental over \( \mathrm{K} \), then there is an isomorphism of fields \( \mathrm{K}\left( \mathrm{u}\right) \cong \mathrm{K}\left( \mathrm{x}\right) \) which is the identity on \( \mathrm{K} \... | SKETCH OF PROOF. Since \( u \) is transcendental \( f\left( u\right) \neq 0, g\left( u\right) \neq 0 \) for all nonzero \( f, g \in K\left\lbrack x\right\rbrack \) . Consequently, the map \( \varphi : K\left( x\right) \rightarrow F \) given by \( f/g \mapsto f\left( u\right) /g\left( u\right) \) \( = f\left( u\right) g... | No |
Theorem 1.6. If \( \mathrm{F} \) is an extension field of \( \mathrm{K} \) and \( \mathrm{u} \in \mathrm{F} \) is algebraic over \( \mathrm{K} \), then\n\n(i) \( \mathrm{K}\left( \mathrm{u}\right) = \mathrm{K}\left\lbrack \mathrm{u}\right\rbrack \) ;\n\n(ii) \( \mathrm{K}\left( \mathrm{u}\right) \cong \mathrm{K}\left\l... | PROOF. (i) and (ii) The map \( \varphi : K\left\lbrack x\right\rbrack \rightarrow K\left\lbrack u\right\rbrack \) given by \( g \mapsto g\left( u\right) \) is a nonzero ring epimorphism by Theorems III.5.5. and 1.3. Since \( K\left\lbrack x\right\rbrack \) is a principal ideal domain (Corollary III.6.4), Ker \( \varphi... | Yes |
Theorem 1.8. Let \( \sigma : \mathrm{K} \rightarrow \mathrm{L} \) be an isomorphism of fields, \( \mathrm{u} \) an element of some extension field of \( \mathrm{K} \) and \( \mathrm{v} \) an element of some extension field of \( \mathrm{L} \) . Assume either\n\n(i) \( \mathrm{u} \) is transcendental over \( \mathrm{K} ... | SKETCH OF PROOF. (i) By the remarks preceding the theorem \( \sigma \) extends to an isomorphism \( K\left\lbrack x\right\rbrack \cong L\left\lbrack x\right\rbrack \) . Verify that this map in turn extends to an isomorphism \( K\left( x\right) \rightarrow L\left( x\right) \) given by \( h/g \mapsto {\sigma h}/{\sigma g... | No |
Corollary 1.9. Let \( \mathrm{E} \) and \( \mathrm{F} \) each be extension fields of \( \mathrm{K} \) and let \( \mathrm{u}\varepsilon \mathrm{E} \) and \( \mathrm{v}\varepsilon \mathrm{F} \) be algebraic over \( \mathbf{K} \) . Then \( \mathbf{u} \) and \( \mathbf{v} \) are roots of the same irreducible polynomial \( ... | PROOF. ( \( \Rightarrow \) ) Apply Theorem 1.8 with \( \sigma = {1}_{K} \) (so that \( {\sigma f} = f \) for all \( {f\varepsilon K}\left\lbrack x\right\rbrack \) ).\n\n\( \left( \Leftarrow \right) \) Suppose \( \sigma : K\left( u\right) \cong K\left( v\right) \) with \( \sigma \left( u\right) = v \) and \( \sigma \lef... | Yes |
Theorem 1.10. If \( \mathrm{K} \) is a field and \( \mathrm{f} \in \mathrm{K}\left\lbrack \mathrm{x}\right\rbrack \) polynomial of degree \( \mathrm{n} \), then there exists a simple extension field \( \mathrm{F} = \mathrm{K}\left( \mathrm{u}\right) \) of \( \mathrm{K} \) such that:\n\n(i) \( \mathrm{u}\varepsilon \mat... | SKETCH OF PROOF OF 1.10. We may assume that \( f \) is irreducible (if not, replace \( f \) by one of its irreducible factors). Then the ideal \( \left( f\right) \) is maximal in \( K\left\lbrack x\right\rbrack \) (Theorem III.3.4 and Corollary III.6.4) and the quotient ring \( F = K\left\lbrack x\right\rbrack /\left( ... | Yes |
Theorem 1.11. If \( \mathrm{F} \) is a finite dimensional extension field of \( \mathrm{K} \), then \( \mathrm{F} \) is finitely generated and algebraic over \( \mathrm{K} \) . | PROOF. If \( \left\lbrack {F : K}\right\rbrack = n \) and \( u \in F \), then the set of \( n + 1 \) elements \( \left\{ {{1}_{K}, u,{u}^{2},\ldots ,{u}^{n}}\right\} \) must be linearly dependent. Hence there are \( {a}_{i}{\varepsilon K} \), not all zero, such that \( {a}_{0} + {a}_{1}u + \) \( {a}_{2}{u}^{2} + \cdots... | Yes |
Theorem 1.12. If \( \mathrm{F} \) is an extension field of \( \mathrm{K} \) and \( \mathrm{X} \) is a subset of \( \mathrm{F} \) such that \( \mathrm{F} = \mathrm{K}\left( \mathrm{X}\right) \) and every element of \( \mathrm{X} \) is algebraic over \( \mathrm{K} \), then \( \mathrm{F} \) is an algebraic extension of \(... | PROOF. If \( v \in F \), then \( v \in K\left( {{u}_{1},\ldots ,{u}_{n}}\right) \) for some \( {u}_{i} \in X \) (Theorem 1.3) and there is a tower of subfields:\n\n\[ K \subset K\left( {u}_{1}\right) \subset K\left( {{u}_{1},{u}_{2}}\right) \subset \cdots \subset K\left( {{u}_{1},\ldots ,{u}_{n - 1}}\right) \subset K\l... | Yes |
Theorem 1.13. If \( \mathrm{F} \) is an algebraic extension field of \( \mathrm{E} \) and \( \mathrm{E} \) is an algebraic extension field of \( \mathbf{K} \), then \( \mathbf{F} \) is an algebraic extension of \( \mathbf{K} \) . | PROOF. Let \( u \in F \) ; since \( u \) is algebraic over \( E,{b}_{n}{u}^{n} + \cdots + {b}_{1}u + {b}_{0} = 0 \) for some \( {b}_{i} \in E\left( {{b}_{n} \neq 0}\right) \) . Therefore, \( u \) is algebraic over the subfield \( K\left( {{b}_{0},\ldots ,{b}_{n}}\right) \) . Consequently, there is a tower of fields\n\n... | Yes |
Theorem 1.14. Let \( \\mathrm{F} \) be an extension field of \( \\mathrm{K} \) and \( \\mathrm{E} \) the set of all elements of \( \\mathrm{F} \) which are algebraic over \( \\mathrm{K} \) . Then \( \\mathrm{E} \) is a subfield of \( \\mathrm{F} \) (which is, of course, algebraic over \( \\mathbf{K} \) ). | PROOF OF 1.14. If \( u, v \\in E \), then \( K\\left( {u, v}\\right) \) is an algebraic extension field of \( K \) by Theorem 1.12. Therefore, since \( u - v \) and \( u{v}^{-1}\\left( {v \\neq 0}\\right) \) are in \( K\\left( {u, v}\\right), u - v \) and \( u{v}^{-1}\\varepsilon E \) . This implies that \( E \) is a f... | Yes |
Lemma 1.15. Let \( \mathrm{F} \) be a subfield of the field \( \mathbf{R} \) of real numbers and let \( {\mathrm{L}}_{1},{\mathrm{\;L}}_{2} \) be nonparallel lines in \( \mathrm{F} \) and \( {\mathrm{C}}_{1},{\mathrm{C}}_{2} \) distinct circles in \( \mathrm{F} \) . Then\n\n(i) \( {\mathrm{L}}_{1} \cap {\mathrm{L}}_{2}... | SKETCH OF PROOF. (i) Exercise. (iii) If the circles are \( {C}_{1} : {x}^{2} + {y}^{2} + {a}_{1}x + \) \( {b}_{1}y + {c}_{1} = 0 \) and \( {C}_{2} : {x}^{2} + {y}^{2} + {a}_{2}x + {b}_{2}y + {c}_{2} = 0\left( {{a}_{i},{b}_{i},{c}_{i} \in F}\right. \) by the remarks preceding the lemma), show that \( {C}_{1} \cap {C}_{2... | No |
Corollary 1.17. An angle of \( {60}^{ \circ } \) cannot be trisected by ruler and compass constructions. | PROOF. If it were possible to trisect a \( {60}^{ \circ } \) angle, we would then be able to construct a right triangle with one acute angle of \( {20}^{ \circ } \) . It would then be possible to construct the real number (ratio) \( \cos {20}^{ \circ } \) (Exercise 25). However for any angle \( \alpha \) , elementary t... | Yes |
Corollary 1.18. It is impossible by ruler and compass constructions to duplicate a cube of side length 1 (that is, to construct the side of a cube of volume 2). | PROOF. If \( s \) is the side length of a cube of volume 2, then \( s \) is a root of \( {x}^{3} - 2 \) , which is irreducible in \( \mathbf{Q}\left\lbrack x\right\rbrack \) by Eisenstein’s Criterion (Theorem III.6.15). Therefore \( s \) is not constructible by Proposition 1.16. | Yes |
Theorem 2.2. Let \( \mathrm{F} \) be an extension field of \( \mathrm{K} \) and \( \mathrm{f} \in \mathrm{K}\left\lbrack \mathrm{x}\right\rbrack \) . If \( \mathrm{u} \in \mathrm{F} \) is a root of \( \mathrm{f} \) and \( {\sigma \varepsilon Au}{t}_{\mathrm{K}}\mathrm{F} \), then \( \sigma \left( \mathrm{u}\right) \in ... | PROOF. If \( f = \mathop{\sum }\limits_{{i = 1}}^{n}{k}_{i}{x}^{i} \), then \( f\left( u\right) = 0 \) implies \( 0 = \sigma \left( {f\left( u\right) }\right) = \sigma \left( {\sum {k}_{i}{u}^{i}}\right) \n\n\[ \n= \sum \sigma \left( {k}_{i}\right) \sigma \left( {u}^{i}\right) = \mathop{\sum }\limits_{i}{k}_{i}\sigma {... | Yes |
Theorem 2.3. Let \( \mathrm{F} \) be an extension field of \( \mathrm{K},\mathrm{E} \) an intermediate field and \( \mathrm{H} \) a subgroup of \( {Au}{t}_{\mathrm{K}}\mathrm{F} \) . Then\n\n(i) \( {\mathrm{H}}^{\prime } = \{ \mathrm{v}\varepsilon \mathrm{F} \mid \sigma \left( \mathrm{v}\right) = \mathrm{v} \) for all ... | PROOF. Exercise. | No |
Lemma 2.6. Let \( \mathrm{F} \) be an extension field of \( \mathrm{K} \) with intermediate fields \( \mathrm{L} \) and \( \mathrm{M} \). Let \( \mathrm{H} \) and \( \mathrm{J} \) be subgroups of \( \mathrm{G} = {Au}{t}_{\mathrm{K}}\mathrm{F} \). Then:\n\n(i) \( {\mathrm{F}}^{\prime } = 1 \) and \( {\mathrm{K}}^{\prime... | SKETCH OF PROOF. (i)-(iii) follow directly from the appropriate definitions. To prove the first part of (iv) observe that (iii) and (ii) imply \( {L}^{\prime \prime \prime } < {L}^{\prime } \) and that (iii) applied with \( {L}^{\prime } \) in place of \( H \) implies \( {L}^{\prime } < {L}^{\prime \prime \prime } \). ... | No |
Theorem 2.7. If \( \mathrm{F} \) is an extension field of \( \mathrm{K} \), then there is a one-to-one correspondence between the closed intermediate fields of the extension and the closed subgroups of the Galois group, given by \( \mathrm{E} \mapsto {\mathrm{E}}^{\prime } = {Au}{t}_{\mathrm{E}}\mathrm{F} \) . | PROOF. Exercise; the inverse of the correspondence is given by assigning to each closed subgroup \( H \) its fixed field \( {H}^{\prime } \). Note that by Lemma 2.6(iv) all primed objects are closed. | No |
(ii) if \( \mathrm{H} \) is closed and \( \left\lbrack {\mathrm{J} : \mathrm{H}}\right\rbrack \) finite, then \( \mathrm{J} \) is closed and \( \left\lbrack {{\mathrm{H}}^{\prime } : {\mathrm{J}}^{\prime }}\right\rbrack = \left\lbrack {\mathrm{J} : \mathrm{H}}\right\rbrack \); | SKETCH OF PROOF OF 2.10. (ii) Applying successively the facts that \( J \subset {J}^{\prime \prime } \) and \( H = {H}^{\prime \prime } \) and Lemmas 2.8 and 2.9 yields\n\n\[ \left\lbrack {J : H}\right\rbrack \leq \left\lbrack {{J}^{\prime \prime } : H}\right\rbrack = \left\lbrack {{J}^{\prime \prime } : {H}^{\prime \p... | Yes |
Lemma 2.12. If \( \mathrm{F} \) is a Galois extension field of \( \mathrm{K} \) and \( \mathrm{E} \) is a stable intermediate field of the extension, then \( \mathrm{E} \) is Galois over \( \mathrm{K} \) . | PROOF. If \( u : E - K \), then there exists \( \sigma \in {\operatorname{Aut}}_{K}F \) such that \( \sigma \left( u\right) \neq u \) since \( F \) is Galois over \( K \) . But \( \sigma \mid E \) e Aut \( {}_{K}E \) by stability. Therefore, \( E \) is Galois over \( K \) by the Remarks after Definition 2.4. | No |
Lemma 2.13. If \( \mathrm{F} \) is an extension field of \( \mathrm{K} \) and \( \mathrm{E} \) is an intermediate field of the extension such that \( \mathrm{E} \) is algebraic and Galois over \( \mathrm{K} \), then \( \mathrm{E} \) is stable (relative to \( \mathrm{F} \) and \( \mathrm{K} \) ). | PROOF OF 2.13. If \( u \in E \), let \( f \in K\left\lbrack x\right\rbrack \) be the irreducible polynomial of \( u \) and let \( u = {u}_{1},{u}_{2},\ldots ,{u}_{r} \) be the distinct roots of \( f \) that lie in \( E \) . Then \( r \leq n = \deg f \) by Theorem III.6.7. If \( \tau \) e Aut \( {}_{K}E \), then it foll... | Yes |
Lemma 2.14. Let \( \mathrm{F} \) be an extension field of \( \mathrm{K} \) and \( \mathrm{E} \) a stable intermediate field of the extension. Then the quotient group \( {Au}{t}_{\mathrm{K}}\mathrm{F}/{Au}{t}_{\mathrm{E}}\mathrm{F} \) is isomorphic to the group of all \( \mathrm{K} \) -automorphisms of \( \mathrm{E} \) ... | SKETCH OF PROOF. Since \( E \) is stable, the assignment \( \sigma \left| { \rightarrow \sigma }\right| E \) defines a group homomorphism \( {\operatorname{Aut}}_{K}F \rightarrow {\operatorname{Aut}}_{K}E \) whose image is clearly the subgroup of all \( K \) -automorphisms of \( E \) that are extendible to \( F \) . Ob... | No |
Theorem 2.15. (Artin) Let \( \mathrm{F} \) be a field, \( \mathrm{G} \) a group of automorphisms of \( \mathrm{F} \) and \( \mathrm{K} \) the fixed field of \( \mathrm{G} \) in \( \mathrm{F} \) . Then \( \mathrm{F} \) is Galois over \( \mathrm{K} \) . If \( \mathrm{G} \) is finite, then \( \mathrm{F} \) is a finite dim... | PROOF. In any case \( G \) is a subgroup of \( {\operatorname{Aut}}_{K}F \) . If \( u \in F - K \), then there must be a \( \sigma \in G \) such that \( \sigma \left( u\right) \neq u \) . Therefore, the fixed field of \( {\operatorname{Aut}}_{K}F \) is \( K \), whence \( F \) is Galois over \( K \) . If \( G \) is fini... | Yes |
Proposition 2.16. If \( \mathrm{G} \) is a finite group, then there exists a Galois field extension with Galois group isomorphic to \( \mathbf{G} \) . | PROOF. Cayley’s Theorem II.4.6 states that for \( n = \left| G\right|, G \) is isomorphic to a subgroup of \( {S}_{n} \) (also denoted \( G \) ). Let \( K \) be any field and \( E \) the subfield of symmetric rational functions in \( K\left( {{x}_{1},\ldots ,{x}_{n}}\right) \) . The discussion preceding the theorem sho... | Yes |
Lemma 2.17. Let \( \mathrm{K} \) be a field, \( {\mathrm{f}}_{1},\ldots ,{\mathrm{f}}_{\mathrm{n}} \) the elementary symmetric functions in \( {\mathrm{x}}_{1},\ldots ,{\mathrm{x}}_{\mathrm{n}} \) over \( \mathrm{K} \) and \( \mathrm{k} \) an integer with \( 1 \leq \mathrm{k} \leq \mathrm{n} - 1 \) . If \( {\mathrm{h}}... | SKETCH OF PROOF. The theorem is true when \( k = n - 1 \) since in that case \( {h}_{1} = {f}_{1} - {x}_{n} \) and \( {h}_{j} = {f}_{j} - {h}_{j - 1}{x}_{n}\left( {2 \leq j \leq n}\right) \) . Complete the proof by induction on \( k \) in reverse order: assume that the theorem is true when \( k = r + 1 \) and \( r + 1 ... | No |
Theorem 2.18. If \( \mathrm{K} \) is a field, \( \mathrm{E} \) the subfield of all symmetric rational functions in \( \mathrm{K}\left( {{\mathrm{x}}_{1},\ldots ,{\mathrm{x}}_{\mathrm{n}}}\right) \) and \( {\mathrm{f}}_{1},\ldots ,{\mathrm{f}}_{\mathrm{n}} \) the elementary symmetric functions, then \( \mathrm{E} = \mat... | SKETCH OF PROOF. Since \( \left\lbrack {K\left( {{x}_{1},\ldots ,{x}_{n}}\right) : E}\right\rbrack = n \) ! and \( K\left( {{f}_{1},\ldots ,{f}_{n}}\right) \subset E \subset \) \( K\left( {{x}_{1},\ldots ,{x}_{n}}\right) \), it suffices by Theorem 1.2 to show that \( \left\lbrack {K\left( {{x}_{1},\ldots ,{x}_{n}}\righ... | Yes |
Lemma 2.19. Let \( \mathrm{K} \) be a field and \( \mathrm{E} \) the subfield of all symmetric rational functions in \( \mathrm{K}\left( {{\mathrm{x}}_{\mathrm{l}},\ldots ,{\mathrm{x}}_{\mathrm{n}}}\right) \) . Then the set \( \mathrm{X} = \left\{ {{\mathrm{x}}_{\mathrm{l}}{}^{{\mathrm{i}}_{1}}{\mathrm{x}}_{2}{}^{{\mat... | SKETCH OF PROOF. Since \( \left\lbrack {K\left( {{x}_{1},\ldots ,{x}_{n}}\right) : E}\right\rbrack = n \) ! and \( \left| X\right| = n \) !, it suffices to show that \( X \) spans \( K\left( {{x}_{1},\ldots ,{x}_{n}}\right) \) (see Theorem IV.2.5). Consider the tower of fields \( E \subset E\left( {x}_{n}\right) \subse... | No |
Proposition 2.20. Let \( \mathrm{K} \) be a field and let \( {\mathrm{f}}_{1},\ldots ,{\mathrm{f}}_{\mathrm{n}} \) be the elementary symmetric functions in \( \mathrm{K}\left( {{\mathrm{x}}_{1},\ldots ,{\mathrm{x}}_{\mathrm{n}}}\right) \) .\n\n(i) Every polynomial in \( \mathrm{K}\left\lbrack {{\mathrm{x}}_{1},\ldots ,... | PROOF. Let \( {g}_{k}\left( y\right) \left( {k = 1,\ldots, n}\right) \) be as in the proof of Theorem 2.18. As noted there the coefficients of \( {g}_{k}\left( y\right) \) are polynomials (over \( K \) ) in \( {f}_{1},\ldots ,{f}_{n} \) and \( {x}_{k + 1},\ldots ,{x}_{n} \) . Since \( {g}_{k} \) is monic of degree \( k... | Yes |
Theorem 3.2. If \( \mathrm{K} \) is a field and \( \mathrm{f} \in \mathrm{K}\left\lbrack \mathrm{x}\right\rbrack \) has degree \( \mathrm{n} \geq 1 \), then there exists a splitting field \( \mathrm{F} \) of \( \mathrm{f} \) with \( \left\lbrack {\mathrm{F} : \mathrm{K}}\right\rbrack \leq \mathrm{n} \) ! | SKETCH OF PROOF. Use induction on \( n = \deg f \) . If \( n = 1 \) or if \( f \) splits over \( K \), then \( F = K \) is a splitting field. If \( n > 1 \) and \( f \) does not split over \( K \), let \( g \in K\left\lbrack x\right\rbrack \) be an irreducible factor of \( f \) of degree greater than one. By Theorem 1.... | No |
Theorem 3.3. The following conditions on a field \( \mathrm{F} \) are equivalent.\n\n(i) Every nonconstant polynomial \( \mathrm{f} \in \mathrm{F}\left\lbrack \mathrm{x}\right\rbrack \) has a root in \( \mathrm{F} \) ;\n\n(ii) every nonconstant polynomial \( \mathrm{f}\varepsilon \mathrm{F}\left\lbrack \mathrm{x}\right... | PROOF. Exercise; see Section III. 6 and Theorems 1.6, 1.10, 1.12 and 1.13. | No |
Theorem 3.4. If \( \mathrm{F} \) is an extension field of \( \mathrm{K} \), then the following conditions are equivalent.\n\n(i) \( \mathrm{F} \) is algebraic over \( \mathrm{K} \) and \( \mathrm{F} \) is algebraically closed;\n\n(ii) \( \mathrm{F} \) is a splitting field over \( \mathrm{K} \) of the set of all [irredu... | PROOF. Exercise; also see Exercises 9, 10. | No |
Corollary 3.7. If \( \mathrm{K} \) is a field and \( \mathrm{S} \) a set of polynomials (of positive degree) in \( \mathrm{K}\left\lbrack \mathrm{x}\right\rbrack \) , then there exists a splitting field of \( \mathbf{S} \) over \( \mathbf{K} \) . | PROOF. Exercise. | No |
Corollary 3.9. Let \( \mathrm{K} \) be a field and \( \mathrm{S} \) a set of polynomials (of positive degree) in \( \mathrm{K}\left\lbrack \mathrm{x}\right\rbrack \) . Then any two splitting fields of \( \mathrm{S} \) over \( \mathrm{K} \) are \( \mathrm{K} \) -isomorphic. In particular, any two algebraic closures of \... | SKETCH OF PROOF. Apply Theorem 3.8 with \( \sigma = {1}_{K} \) . The last statement is then an immediate consequence of Theorem 3.4(ii). | No |
Theorem 3.12. (Generalized Fundamental Theorem) If \( \mathrm{F} \) is an algebraic Galois extension field of \( \mathbf{K} \), then there is a one-to-one correspondence between the set of all intermediate fields of the extension and the set of all closed subgroups of the Galois group \( {\operatorname{Aut}}_{\mathrm{K... | PROOF OF 3.12. In view of Theorem 2.7 we need only show that every intermediate field \( E \) is closed in order to establish the one-to-one correspondence. By Theorem 3.11 \( F \) is the splitting field over \( K \) of a set \( T \) of separable polynomials. Therefore, \( F \) is also a splitting field of \( T \) over... | Yes |
Theorem 3.14. If \( \mathrm{F} \) is an algebraic extension field of \( \mathrm{K} \), then the following statements are equivalent.\n\n(i) \( \mathrm{F} \) is normal over \( \mathrm{K} \) ;\n\n(ii) \( \mathrm{F} \) is a splitting field over \( \mathrm{K} \) of some set of polynomials in \( \mathrm{K}\left\lbrack \math... | PROOF OF 3.14. (i) \( \Rightarrow \) (ii) \( F \) is a splitting field over \( K \) of \( \left\{ {{f}_{i} \in K\left\lbrack x\right\rbrack \mid i \in I}\right\} \) , where \( \left\{ {{u}_{i} \mid i \in I}\right\} \) is a basis of \( F \) over \( K \) and \( {f}_{i} \) is the irreducible polynomial of \( {u}_{i} \) .\... | Yes |
Corollary 3.15. Let \( \mathrm{F} \) be an algebraic extension field of \( \mathrm{K} \) . Then \( \mathrm{F} \) is Galois over \( \mathrm{K} \) if and only if \( \mathrm{F} \) is normal and separable over \( \mathrm{K} \) . If char \( \mathrm{K} = 0 \), then \( \mathrm{F} \) is Galois over \( \mathrm{K} \) if and only... | PROOF. Exercise; use Theorems 3.11 and 3.14. | No |
Theorem 3.16. If \( \\mathrm{E} \) is an algebraic extension field of \( \\mathbf{K} \), then there exists an extension field \( \\mathrm{F} \) of \( \\mathrm{E} \) such that\n\n(i) \( \\mathrm{F} \) is normal over \( \\mathrm{K} \) ;\n\n(ii) no proper subfield of \( \\mathrm{F} \) containing \( \\mathrm{E} \) is norma... | PROOF OF 3.16. (i) Let \( X = \\left\\{ {{u}_{i} \\mid {i\\varepsilon I}}\\right\\} \) be a basis of \( E \) over \( K \) and let \( {f}_{i} \\in K\\left\\lbrack x\\right\\rbrack \) be the irreducible polynomial of \( {u}_{i} \). If \( F \) is a splitting field of \( S = \\left\\{ {{f}_{i} \\mid i \\in I}\\right\\} \) ... | Yes |
Lemma 3.17. If \( \mathrm{F} \) is a finite dimensional separable extension of an infinite field \( \mathbf{K} \) , then \( \mathrm{F} = \mathrm{K}\left( \mathrm{u}\right) \) for some \( \mathrm{u}\varepsilon \mathrm{F} \) . | SKETCH OF PROOF. By Theorem 3.16 there is a finite dimensional Galois extension field \( {F}_{1} \) of \( K \) that contains \( F \) . The Fundamental Theorem 2.5 implies that \( {\text{Aut}}_{K}{F}_{1} \) is finite and that the extension of \( K \) by \( {F}_{1} \) has only finitely many intermediate fields. Therefore... | No |
Lemma 3.18. There are no extension fields of dimension 2 over the field of complex numbers. | SKETCH OF PROOF. It is easy to see that any extension field \( F \) of dimension 2 over \( \mathbf{C} \) would necessarily be of the form \( F = \mathbf{C}\left( u\right) \) for any \( {u\varepsilon F} - \mathbf{C} \) . By Theorem \( {1.6u} \) would be the root of an irreducible monic polynomial \( {f\varepsilon }\math... | No |
Corollary 3.20. Every proper algebraic extension field of the field of real numbers is isomorphic to the field of complex numbers. | PROOF. If \( F \) is an algebraic extension of \( \mathbf{R} \) and \( u \in F - \mathbf{R} \) has irreducible polynomial \( {f\varepsilon R}\left\lbrack x\right\rbrack \) of degree greater than one, then \( f \) splits over \( \mathbf{C} \) by Theorem 3.19. If \( v \in \mathbf{C} \) is a root of \( f \), then by Corol... | Yes |
Theorem 4.2. Let \( \mathrm{K} \) be a field and \( \mathrm{f} \in \mathrm{K}\left\lbrack \mathrm{x}\right\rbrack \) a polynomial with Galois group \( \mathrm{G} \). (i) \( \mathrm{G} \) is isomorphic to a subgroup of some symmetric group \( {\mathrm{S}}_{\mathrm{n}} \). (ii) If \( \mathrm{f} \) is (irreducible) separa... | SKETCH OF PROOF. (i) If \( {u}_{1},\ldots ,{u}_{n} \) are the distinct roots of \( f \) in some splitting field \( F\left( {1 \leq n \leq \deg f}\right) \), then Theorem 2.2 implies that every \( \sigma \) e \( {\operatorname{Aut}}_{K}F \) induces a unique permutation of \( \left\{ {{u}_{1},\ldots ,{u}_{n}}\right\} \) ... | No |
Corollary 4.3. Let \( \mathrm{K} \) be a field and \( \mathrm{f} \in \mathrm{K}\left\lbrack \mathrm{x}\right\rbrack \) an irreducible polynomial of degree 2 with Galois group \( \mathrm{G} \) . If \( \mathrm{f} \) is separable (as is always the case when char \( \mathrm{K} \neq 2 \) ), then \( \mathrm{G} \cong {\mathrm... | SKETCH OF PROOF. Note that \( {S}_{2} = {Z}_{2} \) . Use Remark (ii) after Definition 3.10 and Theorem 4.2. | No |
Proposition 4.5. Let \( \mathrm{K},\mathrm{f},\mathrm{F} \) and \( \Delta \) be as in Definition 4.4.\n\n(i) The discriminant \( {\Delta }^{2} \) of \( \mathrm{f} \) actually lies in \( \mathrm{K} \) . | SKETCH OF PROOF. For (ii) see the proof of Theorem I.6.7. Assuming (ii) note that for every \( \sigma \in {\operatorname{Aut}}_{K}F,\sigma \left( {\Delta }^{2}\right) = \sigma {\left( \Delta \right) }^{2} = {\left( \pm \Delta \right) }^{2} = {\Delta }^{2} \) . Therefore, \( {\Delta }^{2} \in K \) since \( F \) is Galoi... | No |
Corollary 4.6. Let \( \mathrm{K},\mathrm{f},\mathrm{F},\Delta \) be as in Definition 4.4 (so that \( \mathrm{F} \) is Galois over \( \mathrm{K} \) ) and consider \( \mathrm{G} = {Au}{t}_{\mathrm{K}}\mathrm{F} \) as a subgroup of \( {\mathrm{S}}_{\mathrm{n}} \) . In the Galois correspondence (Theorem 2.5) the subfield \... | PROOF. Exercise. | No |
Corollary 4.7. Let \( \mathrm{K} \) be a field and \( \mathrm{f}\varepsilon \mathrm{K}\left\lbrack \mathrm{x}\right\rbrack \) an (irreducible) separable polynomial of degree 3 . The Galois group of \( \mathrm{f} \) is either \( {\mathrm{S}}_{3} \) or \( {\mathrm{A}}_{3} \) . If char \( \mathrm{K} \neq 2 \), it is \( {\... | PROOF. Exercise; use Theorem 4.2 and Corollary 4.6. | No |
Proposition 4.8. Let \( \mathrm{K} \) be a field with char \( \mathrm{K} \neq 2,3 \) . If \( \mathrm{f}\left( \mathrm{x}\right) = {\mathrm{x}}^{3} + \mathrm{b}{\mathrm{x}}^{2} + \mathrm{{cx}} + \) \( \mathrm{d} \in \mathrm{K}\left\lbrack \mathrm{x}\right\rbrack \) has three distinct roots in some splitting field, then ... | SKETCH OF PROOF. Let \( F \) be a splitting field of \( f \) over \( K \) and verify that \( u : F \) is a root of \( f \) if and only if \( u + b/3 \) is a root of \( g = f\left( {x - b/3}\right) \) . This implies that \( g \) has the same discriminant as \( f \) . Verify that \( g \) has the form \( {x}^{3} + {px} + ... | No |
Lemma 4.9. Let \( \mathrm{K},\mathrm{f},\mathrm{F},{\mathrm{u}}_{\mathrm{i}},\mathrm{V} \), and \( \mathrm{G} = {Au}{t}_{\mathrm{K}}\mathrm{F} < {\mathrm{S}}_{4} \) be as in the preceding paragraph. If \( \alpha = {\mathrm{u}}_{1}{\mathrm{u}}_{2} + {\mathrm{u}}_{3}{\mathrm{u}}_{4},\beta = {\mathrm{u}}_{1}{\mathrm{u}}_{... | SKETCH OF PROOF. Clearly every element in \( G \cap V \) fixes \( \alpha ,\beta ,\gamma \) and hence \( K\left( {\alpha ,\beta ,\gamma }\right) \) . In order to complete the proof it suffices, in view of the Fundamental Theorem, to show that every element of \( G \) not in \( V \) moves at least one of \( \alpha ,\beta... | No |
Lemma 4.10. If \( \mathrm{K} \) is a field and \( \mathrm{f} = {\mathrm{x}}^{4} + {\mathrm{{bx}}}^{3} + {\mathrm{{cx}}}^{2} + \mathrm{{dx}} + \mathrm{e} \in \mathrm{K}\left\lbrack \mathrm{x}\right\rbrack \), then the resolvant cubic of \( \mathrm{f} \) is the polynomial \( {\mathrm{x}}^{3} - {\mathrm{{cx}}}^{2} + \left... | SKETCH OF PROOF. Let \( f \) have roots \( {u}_{1},\ldots ,{u}_{4} \) in some splitting field \( F \) . Then use the fact that \( f = \left( {x - {u}_{1}}\right) \left( {x - {u}_{2}}\right) \left( {x - {u}_{3}}\right) \left( {x - {u}_{4}}\right) \) to express \( b, c, d, e \) in terms of the \( {u}_{i} \) . Expand the ... | No |
Theorem 4.12. If \( \mathrm{p} \) is prime and \( \mathrm{f} \) is an irreducible polynomial of degree \( \mathrm{p} \) over the field of rational numbers which has precisely two nonreal roots in the field of complex numbers, then the Galois group of \( \mathrm{f} \) is (isomorphic to) \( {\mathrm{S}}_{\mathrm{p}} \) . | SKETCH OF PROOF. Let \( G \) be the Galois group of \( f \) considered as a subgroup of \( {S}_{p} \) . Since \( p\left| \right| G \mid \) (Theorem 4.2), \( G \) contains an element \( \sigma \) of order \( p \) by Cauchy’s Theorem II.5.2. \( \sigma \) is a \( p \) -cycle by Corollary I.6.4. Now complex conjugation \( ... | No |
Theorem 5.1. Let \( \mathrm{F} \) be a field and let \( \mathrm{P} \) be the intersection of all subfields of \( \mathrm{F} \) . Then \( \mathrm{P} \) is a field with no proper subfields. If char \( \mathrm{F} = \mathrm{p} \) (prime), then \( \mathrm{P} \cong {\mathbf{Z}}_{\mathrm{p}} \) . If char \( \mathrm{F} = 0 \) ... | SKETCH OF PROOF OF 5.1. Note that every subfield of \( F \) must contain 0 and \( {1}_{F} \) . It follows readily that \( P \) is a field that has no proper subfields. Clearly \( P \) contains all elements of the form \( m{1}_{F}\left( {m \in \mathbf{Z}}\right) \) . To complete the proof one may either show directly th... | No |
Corollary 5.2. If \( \mathrm{F} \) is a finite field, then char \( \mathrm{F} = \mathrm{p} \neq 0 \) for some prime \( \mathrm{p} \) and \( \left| \mathrm{F}\right| = {\mathrm{p}}^{\mathrm{n}} \) for some integer \( \mathrm{n} \geq 1 \) . | PROOF. Theorem III.1.9 and Theorem 5.1 imply that \( F \) has prime characteristic \( p \neq 0 \) . Since \( F \) is a finite dimensional vector space over its prime subfield \( {Z}_{p}, F \cong {Z}_{p} \oplus \cdots \oplus {Z}_{p} \) ( \( n \) summands) by Theorem IV.2.4 and hence \( \left| F\right| = {p}^{n} \) . | Yes |
Theorem 5.3. If \( \mathrm{F} \) is a field and \( \mathrm{G} \) is a finite subgroup of the multiplicative group of nonzero elements of \( \mathrm{F} \), then \( \mathrm{G} \) is a cyclic group. In particular, the multiplicative group of all nonzero elements of a finite field is cyclic. | PROOF. If \( G\left( { \neq 1}\right) \) is a finite abelian group, \( G \cong {Z}_{{m}_{1}}\bigoplus {Z}_{{m}_{2}}\bigoplus \cdots \bigoplus {Z}_{{m}_{k}} \) where \( {m}_{1} > 1 \) and \( {m}_{1}\left| {m}_{2}\right| \cdots \mid {m}_{k} \) by Theorem II.2.1. Since \( {m}_{k}\left( {\sum {Z}_{{m}_{i}}}\right) = 0 \), ... | Yes |
Corollary 5.4. If \( \mathrm{F} \) is a finite field, then \( \mathrm{F} \) is a simple extension of its prime subfield \( {\mathrm{Z}}_{\mathrm{p}} \) ; that is, \( \mathrm{F} = {\mathrm{Z}}_{\mathrm{p}}\left( \mathrm{u}\right) \) for some \( \mathrm{u} \in \mathrm{F} \) . | SKETCH OF PROOF. Let \( u \) be a generator of the multiplicative group of nonzero elements of \( F \) . | No |
Lemma 5.5. If \( \mathrm{F} \) is a field of characteristic \( \mathrm{p} \) and \( \mathrm{r} \geq 1 \) is an integer, then the map \( \varphi : \mathrm{F} \rightarrow \mathrm{F} \) given by \( \mathrm{u} \mapsto {\mathrm{u}}^{\mathrm{{pr}}} \) is a \( {\mathrm{Z}}_{\mathrm{p}} \) -monomorphism of fields. If \( \mathr... | SKETCH OF PROOF. The key fact is that for characteristic \( p,{\left( u \pm v\right) }^{{p}^{r}} \) \( = {u}^{{p}^{r}} \pm {v}^{{p}^{r}} \) for all \( u, v \in F \) (Exercise III.1.11). Since \( {1}_{F} \mapsto {1}_{F},\varphi \) fixes each element in the prime subfield \( {Z}_{p} \) of \( F \) . | No |
Proposition 5.6. Let \( \mathrm{p} \) be a prime and \( \mathrm{n} \geq 1 \) an integer. Then \( \mathrm{F} \) is a finite field with \( {\mathrm{p}}^{\mathrm{n}} \) elements if and only if \( \mathrm{F} \) is a splitting field of \( {\mathrm{x}}^{\mathrm{{pn}}} - \mathrm{x} \) over \( {\mathrm{Z}}_{\mathrm{p}} \) . | PROOF. If \( \left| F\right| = {p}^{n} \), then the multiplicative group of nonzero elements of \( F \) has order \( {p}^{n} - 1 \) and hence every nonzero \( u \in F \) satisfies \( {u}^{{p}^{n} - 1} = {1}_{F} \) . Thus every nonzero \( u \in F \) is a root of \( {x}^{p - 1} - {1}_{F} \) and therefore a root of \( x\l... | Yes |
Corollary 5.7. If \( \mathrm{p} \) is a prime and \( \mathrm{n} \geq 1 \) an integer, then there exists a field with \( {\mathrm{p}}^{\mathrm{n}} \) elements. Any two finite fields with the same number of elements are isomorphic. | PROOF. Given \( p \) and \( n \), a splitting field \( F \) of \( {x}^{{p}^{n}} - x \) over \( {Z}_{p} \) exists by Theorem 3.2 and has order \( {p}^{n} \) by Proposition 5.6. Since every finite field of order \( {p}^{n} \) is a splitting field of \( {x}^{{p}^{n}} - x \) over \( {Z}_{p} \) by Proposition 5.6, any two s... | Yes |
Corollary 5.8. If \( \mathrm{K} \) is a finite field and \( \mathrm{n} \geq 1 \) is an integer, then there exists a simple extension field \( \mathrm{F} = \mathrm{K}\left( \mathrm{u}\right) \) of \( \mathrm{K} \) such that \( \mathrm{F} \) is finite and \( \left\lbrack {\mathrm{F} : \mathrm{K}}\right\rbrack = \mathrm{n... | SKETCH OF PROOF. Given \( K \) of order \( {p}^{r} \) let \( F \) be a splitting field of \( f = {x}^{p\prime n} - x \) over \( K \) . By Proposition 5.6 every \( {u\varepsilon K} \) satisfies \( {u}^{p\prime } = u \) and it follows inductively that \( {u}^{{p}^{rn}} = u \) for all \( u \in K \) . Therefore, \( F \) is... | No |
Corollary 5.9. If \( \mathrm{K} \) is a finite field and \( \mathrm{n} \geq 1 \) an integer, then there exists an irreducible polynomial of degree \( \mathrm{n} \) in \( \mathrm{K}\left\lbrack \mathrm{x}\right\rbrack \) . | PROOF. Exercise; use Corollary 5.8 and Theorem 1.6. | No |
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