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Lemma 6.11. (Gauss) If \( \mathrm{D} \) is a unique factorization domain and \( \mathrm{f},\mathrm{g} \in \mathrm{D}\left\lbrack \mathrm{x}\right\rbrack \), then \( \mathrm{C}\left( \mathrm{{fg}}\right) = \mathrm{C}\left( \mathrm{f}\right) \mathrm{C}\left( \mathrm{g}\right) \) . In particular, the product of primitive ... | PROOF. \( f = C\left( f\right) {f}_{1} \) and \( g = C\left( g\right) {g}_{1} \) with \( {f}_{1},{g}_{1} \) primitive. Consequently, \( C\left( {fg}\right) = C\left( {C\left( f\right) {f}_{1}C\left( g\right) {g}_{1}}\right) = C\left( f\right) C\left( g\right) C\left( {{f}_{1}{g}_{1}}\right) \) . Hence it suffices to pr... | Yes |
Lemma 6.12. Let \( \mathrm{D} \) be a unique factorization domain with quotient field \( \mathrm{F} \) and let \( \mathrm{f} \) and \( \mathrm{g} \) be primitive polynomials in \( \mathrm{D}\left\lbrack \mathrm{x}\right\rbrack \) . Then \( \mathrm{f} \) and \( \mathrm{g} \) are associates in \( \mathrm{D}\left\lbrack \... | PROOF. If \( f \) and \( g \) are associates in the integral domain \( F\left\lbrack x\right\rbrack \), then \( f = {gu} \) for some unit \( u \in F\left\lbrack x\right\rbrack \) (Theorem 3.2 (vi)). By Corollary \( {6.4u} \in F \), whence \( u = b/c \) with \( b, c \in D \) and \( c \neq 0 \) . Therefore, \( {cf} = {bg... | Yes |
Lemma 6.13. Let \( \mathrm{D} \) be a unique factorization domain with quotient field \( \mathrm{F} \) and \( \mathrm{f} \) a primitive polynomial of positive degree in \( \mathrm{D}\left\lbrack \mathrm{x}\right\rbrack \) . Then \( \mathrm{f} \) is irreducible in \( \mathrm{D}\left\lbrack \mathrm{x}\right\rbrack \) if ... | SKETCH OF PROOF. Suppose \( f \) is irreducible in \( D\left\lbrack x\right\rbrack \) and \( f = {gh} \) with \( g, h \in F\left\lbrack x\right\rbrack \) and \( \deg g \geq 1 \), deg \( h \geq 1 \) . Then \( g = \mathop{\sum }\limits_{{i = 0}}^{n}\left( {{a}_{i}/{b}_{i}}\right) {x}^{i} \) and \( h = \mathop{\sum }\limi... | No |
Theorem 6.14. If \( \mathbf{D} \) is a unique factorization domain, then so is the polynomial ring \( D\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . | SKETCH OF PROOF OF 6.14. We shall prove only that \( D\left\lbrack x\right\rbrack \) is a unique factorization domain. Since \( D\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack = D\left\lbrack {{x}_{1},\ldots ,{x}_{n - 1}}\right\rbrack \left\lbrack {x}_{n}\right\rbrack \) by Corollary 5.7, a routine inductive argum... | No |
Theorem 6.15. (Eisenstein's Criterion). Let \( \mathbf{D} \) be a unique factorization domain with quotient field \( \mathrm{F} \) . If \( \mathrm{f} = \mathop{\sum }\limits_{{i = 0}}^{n}{\mathrm{a}}_{\mathrm{i}}{\mathrm{x}}^{\mathrm{i}}\varepsilon \mathrm{D}\left\lbrack \mathrm{x}\right\rbrack \), deg \( \mathrm{f} \g... | PROOF. \( f = C\left( f\right) {f}_{1} \) with \( {f}_{1} \) primitive in \( D\left\lbrack x\right\rbrack \) and \( C\left( f\right) {\varepsilon D} \) ; (in particular \( {f}_{1} = f \) if \( f \) is primitive). Since \( C\left( f\right) \) is a unit in \( F \) (Corollary 6.4), it suffices to show that \( {f}_{1} \) i... | Yes |
Theorem 1.5. Let \( \mathrm{R} \) be a ring, \( \mathrm{A} \) an \( \mathrm{R} \) -module, \( \mathrm{X} \) a subset of \( \mathrm{A},\left\{ {{\mathrm{B}}_{\mathrm{i}} \mid \mathrm{i} \in \mathrm{I}}\right\} \) a family of submodules of \( \mathrm{A} \) and \( \mathrm{a}\varepsilon \mathrm{A} \) . Let \( \mathrm{{Ra}}... | PROOF. Exercise; note that if \( R \) has an identity \( {1}_{R} \) and \( A \) is unitary, then \( n{1}_{R} \in R \) for all \( {n\varepsilon }\mathbf{Z} \) and \( {na} = \left( {n{1}_{R}}\right) a \) for all \( {a\varepsilon A} \) . | No |
Theorem 1.6. Let \( \mathrm{B} \) be a submodule of a module \( \mathrm{A} \) over a ring \( \mathrm{R} \) . Then the quotient group \( \mathrm{A}/\mathrm{B} \) is an \( \mathrm{R} \) -module with the action of \( \mathrm{R} \) on \( \mathrm{A}/\mathrm{B} \) given by:\n\n\[ \mathrm{r}\left( {\mathrm{a} + \mathrm{B}}\ri... | SKETCH OF PROOF OF 1.6. Since \( A \) is an additive abelian group, \( B \) is a normal subgroup, and \( A/B \) is a well-defined abelian group. If \( a + B = {a}^{\prime } + B \) , then \( a - {a}^{\prime }{\varepsilon B} \) . Since \( B \) is a submodule \( {ra} - r{a}^{\prime } = r\left( {a - {a}^{\prime }}\right) {... | No |
Theorem 1.7. If \( \mathrm{R} \) is a ring and \( \mathrm{f} : \mathrm{A} \rightarrow \mathrm{B} \) is an \( \mathrm{R} \) -module homomorphism and \( \mathrm{C} \) is a submodule of Ker \( \mathrm{f} \), then there is a unique \( \mathrm{R} \) -module homomorphism \( \bar{\mathrm{f}} : \mathrm{A}/\mathrm{C} \rightarro... | PROOF. See Theorem I.5.6 and Corollary I.5.7. | No |
Corollary 1.8. If \( \mathrm{R} \) is a ring and \( {\mathrm{A}}^{\prime } \) is a submodule of the \( \mathrm{R} \) -module \( \mathrm{A} \) and \( {\mathrm{B}}^{\prime } \) a submodule of the R-module B and \( \mathrm{f} : \mathrm{A} \rightarrow \mathrm{B} \) is an R-module homomorphism such that \( \mathrm{f}\left( ... | \( \bar{\mathrm{A}} \) is an \( \mathrm{R} \) -module isomorphism if and only if \( \operatorname{Im}\mathrm{f} + {\mathrm{B}}^{\prime } = \mathrm{B} \) and \( {\mathrm{f}}^{-1}\left( {\mathrm{\;B}}^{\prime }\right) \subset {\mathrm{A}}^{\prime } \) . In particular if \( \mathrm{f} \) is an epimorphism such that \( \m... | Yes |
Theorem 1.9. Let \( \mathrm{B} \) and \( \mathrm{C} \) be submodules of a module \( \mathrm{A} \) over a ring \( \mathrm{R} \) .\n\n(i) There is an R-module isomorphism \( \mathrm{B}/\left( {\mathrm{B} \cap \mathrm{C}}\right) \cong \left( {\mathrm{B} + \mathrm{C}}\right) /\mathrm{C} \) ;\n\n(ii) if \( \mathrm{C} \subse... | PROOF. See Corollaries I.5.9 and I.5.10. | No |
Theorem 1.10. If \( \mathrm{R} \) is a ring and \( \mathrm{B} \) is a submodule of an \( \mathrm{R} \) -module \( \mathrm{A} \), then there is a one-to-one correspondence between the set of all submodules of \( \mathrm{A} \) containing \( \mathrm{B} \) and the set of all submodules of \( \mathrm{A}/\mathrm{B} \), given... | PROOF. See Theorem I.5.11 and Corollary I.5.12. | No |
Theorem 1.11. Let \( \mathrm{R} \) be a ring and \( \left\{ {{\mathrm{A}}_{\mathrm{i}} \mid \mathrm{i}\varepsilon \mathrm{I}}\right\} \) a nonempty family of \( \mathrm{R} \) -modules, \( \mathop{\prod }\limits_{{i \in I}}{\mathrm{\;A}}_{\mathrm{i}} \) the direct product of the abelian groups \( {\mathrm{A}}_{\mathrm{i... | PROOF. Exercise. - | No |
Theorem 1.12. If \( \mathrm{R} \) is a ring, \( \left\{ {{\mathrm{A}}_{\mathrm{i}} \mid \mathrm{i} \in \mathrm{I}}\right\} \) a family of \( \mathrm{R} \) -modules, \( \mathrm{C} \) an \( \mathrm{R} \) -module, and \( \left\{ {{\varphi }_{\mathrm{i}} : \mathrm{C} \rightarrow {\mathrm{A}}_{\mathrm{i}} \mid \mathrm{i} \i... | PROOF. By Theorem I.8.2 there is a unique group homomorphism \( \varphi : C \rightarrow \prod {A}_{i} \) which has the desired property, given by \( \varphi \left( c\right) = {\left\{ {\varphi }_{i}\left( c\right) \right\} }_{i : I} \) . Since each \( {\varphi }_{i} \) is an \( R \) - module homomorphism, \( \varphi \l... | Yes |
Theorem 1.13. If \( \mathrm{R} \) is a ring, \( \left\{ {{\mathrm{A}}_{\mathrm{i}} \mid \mathrm{i} \in \mathrm{I}}\right\} \) a family of \( \mathrm{R} \) -modules, \( \mathrm{D} \) an \( \mathrm{R} \) -module, and \( \left\{ {{\psi }_{\mathrm{i}} : {\mathrm{A}}_{\mathrm{i}} \rightarrow \mathrm{D} \mid \mathrm{i} \in \... | PROOF. By Theorem I.8.5 there is a unique abelian group homomorphism \( \psi : \sum {A}_{i} \rightarrow D \) with the desired property, given by \( \psi \left( \left\{ {a}_{i}\right\} \right) = \mathop{\sum }\limits_{i}{\psi }_{i}\left( {a}_{i}\right) \), where the sum\n\nis taken over the finite set of indices \( i \)... | Yes |
Theorem 1.14. Let \( \mathrm{R} \) be a ring and \( \mathrm{A},{\mathrm{A}}_{1},{\mathrm{\;A}}_{2},\ldots ,{\mathrm{A}}_{\mathrm{n}}\mathrm{R} \) -modules. Then \( \mathrm{A} \cong {\mathrm{A}}_{1} \oplus \) \( {\mathrm{A}}_{2} \oplus \cdots \oplus {\mathrm{A}}_{\mathrm{n}} \) if and only if for each \( \mathrm{i} = 1,... | PROOF. ( \( \Rightarrow \) ) If \( A \) is the module \( {A}_{1}\bigoplus {A}_{2}\bigoplus \cdots \bigoplus {A}_{n} \), then the canonical injections \( {\iota }_{i} \) and projections \( {\pi }_{i} \) satisfy (i)-(iii) as the reader may easily verify. Likewise if \( A \cong {A}_{1} \oplus \cdots \oplus {A}_{n} \), und... | Yes |
Theorem 1.15. Let \( \mathrm{R} \) be a ring and \( \left\{ {{\mathrm{A}}_{\mathrm{i}} \mid \mathrm{i} \in \mathrm{I}}\right\} \) a family of submodules of an \( \mathrm{R} \) -module A such that\n\n(i) \( \mathbf{A} \) is the sum of the family \( \left\{ {{\mathbf{A}}_{\mathrm{i}} \mid \mathrm{i}\varepsilon \mathbf{I}... | ## PROOF. Exercise; see Theorem I.8.6. | No |
Lemma 1.17. (The Short Five Lemma) Let \( \mathrm{R} \) be a ring and a commutative diagram of \( \mathbf{R} \) -modules and \( \mathbf{R} \) -module homomorphisms such that each row is a short exact sequence. Then (i) \( \alpha ,\gamma \) monomorphisms \( \Rightarrow \beta \) is a monomorphism; (ii) \( \alpha ,\gamma ... | PROOF. (i) Let \( b \in B \) and suppose \( \beta \left( b\right) = 0 \) ; we must show that \( b = 0 \) . By commutativity we have \[ {\gamma g}\left( b\right) = {g}^{\prime }\beta \left( b\right) = {g}^{\prime }\left( 0\right) = 0. \] This implies \( g\left( b\right) = 0 \), since \( \gamma \) is a monomorphism. By e... | Yes |
Theorem 1.18. Let \( \mathrm{R} \) be a ring and \( 0 \rightarrow {\mathrm{A}}_{1}\overset{\mathrm{f}}{ \rightarrow }\mathrm{B}\overset{\mathrm{g}}{ \rightarrow }{\mathrm{A}}_{2} \rightarrow 0 \) a short exact sequence of R-module homomorphisms. Then the following conditions are equivalent.\n\n(i) There is an R-module ... | SKETCH OF PROOF OF 1.18. (i) \( \Rightarrow \) (iii) By Theorem 1.13 the homomorphisms \( f \) and \( h \) induce a module homomorphism \( \varphi : {A}_{1} \oplus {A}_{2} \rightarrow B \), given by \( \left( {{a}_{1},{a}_{2}}\right) \mapsto f\left( {a}_{1}\right) + h\left( {a}_{2}\right) \) . Verify that the diagram\n... | Yes |
Theorem 2.1. Let \( \mathrm{R} \) be a ring with identity. The following conditions on a unitary \( \mathrm{R} \) -module \( \mathrm{F} \) are equivalent:\n\n(i) \( \mathrm{F} \) has a nonempty basis;\n\n(ii) \( \mathrm{F} \) is the internal direct sum of a family of cyclic \( \mathrm{R} \) -modules, each of which is i... | SKETCH OF PROOF OF 2.1. (i) \( \Rightarrow \) (ii) Let \( X \) be a basis of \( F \) and \( x \in X \) . The map \( R \rightarrow {Rx} \), given by \( r \mapsto {rx} \), is an \( R \) -module epimorphism by Theorem 1.5. If \( {rx} = 0 \), then \( r = 0 \) by linear independence, whence the map is a monomorphism and \( ... | Yes |
Every (unitary) module A over a ring \( \mathrm{R} \) (with identity) is the homomorphic image of a free R-module F. If A is finitely generated, then \( \mathrm{F} \) may be chosen to be finitely generated. | SKETCH OF PROOF OF 2.2. Let \( X \) be a set of generators of \( A \) and \( F \) the free \( R \) -module on the set \( X \) . Then the inclusion map \( X \rightarrow A \) induces an \( R \) -module homomorphism \( \bar{f} : F \rightarrow A \) such that \( X \subset \operatorname{Im}\bar{f} \) (Theorem 2.1 (iv)). Sinc... | Yes |
Lemma 2.3. A maximal linearly independent subset \( \mathrm{X} \) of a vector space \( \mathrm{V} \) over a division ring \( \mathrm{D} \) is a basis of \( \mathrm{V} \) . | PROOF. Let \( W \) be the subspace of \( V \) spanned by the set \( X \) . Since \( X \) is linearly independent and spans \( W, X \) is a basis of \( W \) . If \( W = V \), we are done. If not, then there exists a nonzero \( {a\varepsilon V} \) with \( {a\varepsilon W} \) . Consider the set \( X \cup \{ a\} \) . If \(... | Yes |
Theorem 2.5. If \( \mathrm{V} \) is a vector space over a division ring \( \mathrm{D} \) and \( \mathrm{X} \) is a subset that spans \( \mathrm{V} \), then \( \mathrm{X} \) contains a basis of \( \mathrm{V} \) . | SKETCH OF PROOF. Partially order the set \( \mathcal{S} \) of all linearly independent subsets of \( X \) by inclusion. Zorn’s Lemma implies the existence of a maximal linearly independent subset \( Y \) of \( X \) . Every element of \( X \) is a linear combination of elements of \( Y \) (otherwise, as in Lemma 2.3, we... | No |
Theorem 2.7. If \( \mathrm{V} \) is a vector space over a division ring \( \mathrm{D} \), then any two bases of \( \mathrm{V} \) have the same cardinality. | PROOF. Let \( X \) and \( Y \) be bases of \( V \) . If either \( X \) or \( Y \) is infinite, then \( \left| X\right| = \left| Y\right| \) by Theorem 2.6. Hence we assume \( X \) and \( Y \) are finite, say \( X = \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \), and \( Y = \left\{ {{y}_{1},\ldots ,{y}_{m}}\right\} \) . S... | Yes |
Proposition 2.9. Let \( \mathrm{E} \) and \( \mathrm{F} \) be free modules over a ring \( \mathrm{R} \) that has the invariant dimension property. Then \( \mathrm{E} \cong \mathrm{F} \) if and only if \( \mathrm{E} \) and \( \mathrm{F} \) have the same rank. | PROOF. Exercise; see Proposition II.1.3. | No |
Lemma 2.10. Let \( \mathrm{R} \) be a ring with identity, \( \mathrm{I}\left( { \neq \mathrm{R}}\right) \) an ideal of \( \mathrm{R},\mathrm{F} \) a free \( \mathrm{R} \) -module with basis \( \mathrm{X} \) and \( \pi : \mathrm{F} \rightarrow \mathrm{F}/\mathrm{{IF}} \) the canonical epimorphism. Then \( \mathrm{F}/\ma... | PROOF OF 2.10. If \( u + {IF} \in F/{IF} \), then \( u = \mathop{\sum }\limits_{{j = 1}}^{n}{r}_{j}{x}_{j} \) with \( {r}_{j} \in R,{x}_{j} \in X \) since \( {u\varepsilon F} \) and \( X \) is a basis of \( F \) . Consequently, \( u + {IF} = \left( {\mathop{\sum }\limits_{j}^{{j = 1}}{r}_{j}{x}_{j}}\right) + {IF} = \ma... | Yes |
Proposition 2.11. Let \( \mathrm{f} : \mathrm{R} \rightarrow \mathrm{S} \) be a nonzero epimorphism of rings with identity. If \( \mathrm{S} \) has the invariant dimension property, then so does \( \mathbf{R} \) . | PROOF. Let \( I = \operatorname{Ker}f \) ; then \( S \cong R/I \) (Corollary III.2.10). Let \( X \) and \( Y \) be bases of the free \( R \) -module \( F \) and \( \pi : F \rightarrow F/{IF} \) the canonical epimorphism. By Lemma 2.10 \( F/{IF} \) is a free \( R/I \) -module (and hence a free \( S \) -module) with base... | Yes |
Corollary 2.12. If \( \mathrm{R} \) is a ring with identity that has a homomorphic image which is a division ring, then \( \mathrm{R} \) has the invariant dimension property. In particular, every commutative ring with identity has the invariant dimension property. | PROOF. The first statement follows from Theorem 2.7 and Proposition 2.11. If \( R \) is commutative with identity, then \( R \) contains a maximal ideal \( M \) (Theorem III.2.18) and \( R/M \) is a field (Theorem III.2.20). Thus the second statement is a special case of the first. | Yes |
Theorem 2.13. Let \( \mathrm{W} \) be a subspace of a vector space \( \mathrm{V} \) over a division ring \( \mathrm{D} \) . (i) \( {\dim }_{\mathrm{D}}\mathrm{W} \leq {\dim }_{\mathrm{D}}\mathrm{V} \) ; (ii) if \( {\dim }_{\mathrm{D}}\mathrm{W} = {\dim }_{\mathrm{D}}\mathrm{V} \) and \( {\dim }_{\mathrm{D}}\mathrm{V} \... | SKETCH OF PROOF. (i) Let \( Y \) be a basis of \( W \) . By Theorem 2.4 there is a basis \( X \) of \( V \) containing \( Y \) . Therefore, \( {\dim }_{D}W = \left| Y\right| \leq \left| X\right| = {\dim }_{D}V \) . (ii) If \( \left| Y\right| = \left| X\right| \) and \( \left| X\right| \) is finite, then since \( Y \sub... | Yes |
Corollary 2.14. If \( \mathrm{f} : \mathrm{V} \rightarrow {\mathrm{V}}^{\prime } \) is a linear transformation of vector spaces over a division ring \( \mathrm{D} \), then there exists a basis \( \mathrm{X} \) of \( \mathrm{V} \) such that \( \mathrm{X} \cap \operatorname{Ker}\mathrm{f} \) is a basis of \( \operatornam... | SKETCH OF PROOF. To prove the first statement let \( W = \operatorname{Ker}f \) and let \( Y, X \) be as in the proof of Theorem 2.13. The second statement follows from Theorem 2.13 (iii) since \( V/W \cong \operatorname{Im}f \) by Theorem 1.7. | No |
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\[ \n{\\dim }_{\\mathrm{R}}\\mathrm{T} = \\left( {{\\dim }_{\\mathrm{S}}\\mathrm{T}}\\right) \\left( {{\\dim }_{\\mathrm{R}}\\mathrm{S}}\\right) .\n\]\n\nFurthermo... | 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, w... | 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}\overset{\mathr... | 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 \rightarro... | 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 \( \mathrm{D} \)\n\n\[ 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). | 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 \( \left( {a, b}\right) \mapsto g\left( {a, b}\right) \in C \) determines a unique homomorphism \( {g}_{1} : F \rightarrow C \) by Theore... | Yes |
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 |
Theorem 5.5. Let \( \mathrm{R} \) and \( \mathrm{S} \) be rings and \( {}_{\mathrm{S}}{\mathrm{A}}_{\mathrm{R}},{}_{\mathrm{R}}\mathrm{B},{\mathrm{C}}_{\mathrm{R}},{}_{\mathrm{R}}{\mathrm{D}}_{\mathrm{S}} \) (bi)modules as indicated.\n\n(i) \( \mathrm{A}{ \otimes }_{\mathrm{R}}\mathrm{B} \) is a left \( \mathrm{S} \) -... | 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 \( \bar{g} : A{ \otimes }_{R}B \rightarrow C \) 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 \( A \times B \) (defined... | 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\[ \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( {\matho... | Yes |
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... | No |
Theorem 6.12. Let \( \mathrm{A} \) be a finitely generated module over a principal ideal domain \( \mathrm{R} \) .\n\n(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 ,... | 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... | Yes |
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 \[ \pi : A{\bigotimes }_{K}A \rightarrow A \] by Theorem 5.6. Verify that \( \pi \) has the required proper... | 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 \( \left\lbrack {\mathrm{F} : \mathrm{K}}\right\rbrack = \left\lbrack {\mathrm{F} : \mathrm{E}}\right\rbrack \left\lbrack {\mathrm{E} : \mathrm{K}}\right\rbrack \). Furthermore \... | 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: (i) \( \mathrm{u}\varepsilon \mathrm... | 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 \( \mathbf{K} \), then \( \mathbf{F} \) 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}}^{\pri... | 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 |
Lemma 2.10. Let \( \mathrm{F} \) be an extension field of \( \mathrm{K},\mathrm{L} \) and \( \mathrm{M} \) intermediate fields with \( \mathrm{L} \subset \mathrm{M} \), and \( \mathrm{H},\mathrm{J} \) subgroups of the Galois group Aut \( {}_{\mathrm{K}}\mathrm{F} \) with \( \mathrm{H} < \mathrm{J} \) .\n\n(i) If \( \ma... | 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... | No |
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 |
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