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Corollary 20. Let \( n \geq 2 \) be an integer with factorization \( n = {p}_{1}^{{\alpha }_{1}}{p}_{2}^{{\alpha }_{2}}\cdots {p}_{r}^{{\alpha }_{r}} \) in \( \mathbb{Z} \), where \( {p}_{1},\ldots ,{p}_{r} \) are distinct primes. We have the following isomorphisms of (multiplicative) groups:\n\n(1) \( {\left( \mathbb{... | Proof: This is mainly a matter of collecting previous results. The isomorphism in (1) follows from the Chinese Remainder Theorem (see Corollary 18, Section 7.6). | Yes |
Corollary 22. Every ideal in the polynomial ring \( F\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack \) with coefficients from a field \( F \) is finitely generated. | If \( I \) is an ideal in \( F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) generated by a (possibly infinite) set \( \mathcal{S} \) of polynomials, Corollary 22 shows that \( I \) is finitely generated, and in fact \( I \) is generated by a finite number of the polynomials from the set \( \mathcal{S} \) (cf. ... | No |
Theorem 23. Fix a monomial ordering on \( R = F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) and suppose \( \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\} \) is a Gröbner basis for the nonzero ideal \( I \) in \( R \) . Then\n\n(1) Every polynomial \( f \in R \) can be written uniquely in the form\n\n\[ f = {f}_{I}... | Proof: Letting \( {f}_{I} = \mathop{\sum }\limits_{{i = 1}}^{m}{q}_{i}{g}_{i} \in I \) in the general polynomial division of \( f \) by \( {g}_{1},\ldots ,{g}_{m} \) immediately gives a decomposition \( f = {f}_{l} + r \) for any generators \( {g}_{1},\ldots ,{g}_{m} \) . Suppose now that \( \left\{ {{g}_{1},\ldots ,{g... | Yes |
(1) If \( {g}_{1},\ldots ,{g}_{m} \) are any elements of \( I \) such that \( {LT}\left( I\right) = \left( {{LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right) }\right) \) , then \( \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\} \) is a Gröbner basis for \( I \) . | Proof: Suppose \( {g}_{1},\ldots ,{g}_{m} \in I \) with \( {LT}\left( I\right) = \left( {{LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right) }\right) \) . We need to see that \( {g}_{1},\ldots ,{g}_{m} \) generate the ideal \( I \) . If \( f \in I \), use general polynomial division to write \( f = \mathop{\su... | Yes |
Lemma 25. Suppose \( {f}_{1},\ldots ,{f}_{m} \in F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) are polynomials with the same multidegree \( \alpha \) and that the linear combination \( h = {a}_{1}{f}_{1} + \cdots + {a}_{m}{f}_{m} \) with constants \( {a}_{i} \in F \) has strictly smaller multidegree. Then\n\n... | Proof: Write \( {f}_{i} = {c}_{i}{f}_{i}^{\prime } \) where \( {c}_{i} \in F \) and \( {f}_{i}^{\prime } \) is a monic polynomial of multidegree \( \alpha \) . We have\n\n\[ h = \sum {a}_{i}{c}_{i}{f}_{i}^{\prime } = {a}_{1}{c}_{1}\left( {{f}_{1}^{\prime } - {f}_{2}^{\prime }}\right) + \left( {{a}_{1}{c}_{1} + {a}_{2}{... | Yes |
Theorem 27. Fix a monomial ordering on \( R = F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . Then there is a unique reduced Gröbner basis for every nonzero ideal \( I \) in \( R \) . | Proof: By Exercise 15, two reduced bases have the same number of elements and the same leading terms since reduced bases are also minimal bases. If \( G = \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\} \) and \( {G}^{\prime } = \left\{ {{g}_{1}^{\prime },\ldots ,{g}_{m}^{\prime }}\right\} \) are two reduced bases for the sa... | No |
Proposition 29. (Elimination) Suppose \( G = \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\} \) is a Gröbner basis for the nonzero ideal \( I \) in \( F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) with respect to the lexicographic monomial ordering \( {x}_{1} > \cdots > {x}_{n} \) . Then \( G \cap F\left\lbrack {{x... | Proof: Denote \( {G}_{i} = G \cap F\left\lbrack {{x}_{i + 1},\ldots ,{x}_{n}}\right\rbrack \) . Then \( {G}_{i} \subseteq {I}_{i} \), so by Proposition 24, to see that \( {G}_{i} \) is a Gröbner basis of \( {I}_{i} \) it suffices to see that \( {LT}\left( {G}_{i}\right) \), the leading terms of the elements in \( {G}_{... | Yes |
Proposition 30. If \( I \) and \( J \) are any two ideals in \( F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) then \( {tI} + \left( {1 - t}\right) J \) is an ideal in \( F\left\lbrack {t,{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) and \( I \cap J = \left( {{tI} + \left( {1 - t}\right) J}\right) \cap F\left\lbrac... | Proof: First, \( {tI} \) and \( \left( {1 - t}\right) J \) are clearly ideals in \( F\left\lbrack {{x}_{1},\ldots ,{x}_{n}, t}\right\rbrack \), so also their sum \( {tI} + \left( {1 - t}\right) J \) is an ideal in \( F\left\lbrack {{x}_{1},\ldots ,{x}_{n}, t}\right\rbrack \) . If \( f \in I \cap J \), then \( f = {tf} ... | Yes |
Proposition 1 . (The Submodule Criterion) Let \( R \) be a ring and let \( M \) be an \( R \) -module. A subset \( N \) of \( M \) is a submodule of \( M \) if and only if\n\n(1) \( N \neq \varnothing \), and\n\n(2) \( x + {ry} \in N \) for all \( r \in R \) and for all \( x, y \in N \) . | Proof: If \( N \) is a submodule, then \( 0 \in N \) so \( N \neq \varnothing \) . Also \( N \) is closed under addition and is sent to itself under the action of elements of \( R \) . Conversely, suppose (1) and (2) hold. Let \( r = - 1 \) and apply the subgroup criterion (in additive form) to see that \( N \) is a su... | Yes |
A map \( \varphi : M \rightarrow N \) is an \( R \) -module homomorphism if and only if \( \varphi \left( {{rx} + y}\right) = {r\varphi }\left( x\right) + \varphi \left( y\right) \) for all \( x, y \in M \) and all \( r \in R. \) | \( \textit{Proof:}\;\left( 1\right) \;\textit{Certainly}\;\varphi \left( {{rx} + y}\right) = {r\varphi }\left( x\right) + \varphi \left( y\right) \;\textit{if}\;\varphi \;\textit{is an}\;R\textit{-module homomorphism}. \) Conversely, if \( \varphi \left( {{rx} + y}\right) = {r\varphi }\left( x\right) + \varphi \left( y... | Yes |
Proposition 3 . Let \( R \) be a ring, let \( M \) be an \( R \) -module and let \( N \) be a submodule of \( M \) . The (additive, abelian) quotient group \( M/N \) can be made into an \( R \) -module by defining an action of elements of \( R \) by\n\n\[ r\left( {x + N}\right) = \left( {rx}\right) + N,\;\text{ for all... | Proof: Since \( M \) is an abelian group under + the quotient group \( M/N \) is defined and is an abelian group. To see that the action of the ring element \( r \) on the coset \( x + N \) is well defined, suppose \( x + N = y + N \), i.e., \( x - y \in N \) . Since \( N \) is a (left) \( R \) -submodule, \( r\left( {... | Yes |
(1) (The First Isomorphism Theorem for Modules) Let \( M, N \) be \( R \) -modules and let \( \varphi : M \rightarrow N \) be an \( R \) -module homomorphism. Then \( \ker \varphi \) is a submodule of \( M \) and \( M/\ker \varphi \cong \varphi \left( M\right) \) . | Proof: Exercise. | No |
Proposition 5. Let \( {N}_{1},{N}_{2},\ldots ,{N}_{k} \) be submodules of the \( R \) -module \( M \) . Then the following are equivalent:\n\n(1) The map \( \pi : {N}_{1} \times {N}_{2} \times \cdots \times {N}_{k} \rightarrow {N}_{1} + {N}_{2} + \cdots + {N}_{k} \) defined by\n\n\[ \pi \left( {{a}_{1},{a}_{2},\ldots ,... | Proof: To prove (1) implies (2), suppose for some \( j \) that (2) fails to hold and let \( {a}_{j} \in \left( {{N}_{1} + \cdots + {N}_{j - 1} + {N}_{j + 1} + \cdots + {N}_{k}}\right) \cap {N}_{j} \), with \( {a}_{j} \neq 0 \) . Then\n\n\[ {a}_{j} = {a}_{1} + \cdots + {a}_{j - 1} + {a}_{j + 1} + \cdots + {a}_{k} \]\n\n... | Yes |
Theorem 6. For any set \( A \) there is a free \( R \) -module \( F\left( A\right) \) on the set \( A \) and \( F\left( A\right) \) satisfies the following universal property: if \( M \) is any \( R \) -module and \( \varphi : A \rightarrow M \) is any map of sets, then there is a unique \( R \) -module homomorphism \(... | Proof: Let \( F\left( A\right) = \{ 0\} \) if \( A = \varnothing \) . If \( A \) is nonempty let \( F\left( A\right) \) be the collection of all set functions \( f : A \rightarrow R \) such that \( f\left( a\right) = 0 \) for all but finitely many \( a \in A \) . Make\n\n\( F\left( A\right) \) into an \( R \) -module b... | No |
Theorem 8. Let \( R \) be a subring of \( S \), let \( N \) be a left \( R \) -module and let \( \iota : N \rightarrow S{ \otimes }_{R}N \) be the \( R \) -module homomorphism defined by \( \iota \left( n\right) = 1 \otimes n \) . Suppose that \( L \) is any left \( S \) - module (hence also an \( R \) -module) and tha... | Proof: Suppose \( \varphi : N \rightarrow L \) is an \( R \) -module homomorphism to the \( S \) -module \( L \) . By the universal property of free modules (Theorem 6 in Section 3) there is a \( \mathbb{Z} \) -module homomorphism from the free \( \mathbb{Z} \) -module \( F \) on the set \( S \times N \) to \( L \) tha... | Yes |
Let \( \iota : N \rightarrow S{ \otimes }_{R}N \) be the \( R \) -module homomorphism in Theorem 8. Then \( N/\ker \iota \) is the unique largest quotient of \( N \) that can be embedded in any \( S \) -module. In particular, \( N \) can be embedded as an \( R \) -submodule of some left \( S \) -module if and only if \... | The quotient \( N/\ker \iota \) is mapped injectively (by \( \iota \) ) into the \( S \) -module \( S{ \otimes }_{R}N \) . Suppose now that \( \varphi \) is an \( R \) -module homomorphism injecting the quotient \( N/\ker \varphi \) of \( N \) into an \( S \) -module \( L \) . Then, by Theorem 8, \( \ker \iota \) is ma... | Yes |
Suppose \( R \) is a ring with \( 1, M \) is a right \( R \) -module, and \( N \) is a left \( R \) -module. Let \( M{ \otimes }_{R}N \) be the tensor product of \( M \) and \( N \) over \( R \) and let \( \iota : M \times N \rightarrow \) \( M{ \otimes }_{R}N \) be the \( R \) -balanced map defined above.\n\n(1) If \(... | The proof of (1) is immediate from the properties of \( \iota \) above. | No |
Corollary 11. Suppose \( D \) is an abelian group and \( {\iota }^{\prime } : M \times N \rightarrow D \) is an \( R \) -balanced map such that\n\n(i) the image of \( {\iota }^{\prime } \) generates \( D \) as an abelian group, and\n\n(ii) every \( R \) -balanced map defined on \( M \times N \) factors through \( {\iot... | Proof: Since \( {\iota }^{\prime } : M \times N \rightarrow D \) is a balanced map, the universal property in (2) of Theorem 10 implies there is a (unique) homomorphism \( f : M{ \otimes }_{R}N \rightarrow D \) with \( {\iota }^{\prime } = f \circ \iota \) . In particular \( {\iota }^{\prime }\left( {m, n}\right) = f\l... | Yes |
Suppose \( R \) is a commutative ring. Let \( M \) and \( N \) be two left \( R \) -modules and let \( M{ \otimes }_{R}N \) be the tensor product of \( M \) and \( N \) over \( R \), where \( M \) is given the standard \( R \) -module structure. Then \( M{ \otimes }_{R}N \) is a left \( R \) -module with\n\n\[ r\left( ... | Proof: We have shown \( M{ \otimes }_{R}N \) is an \( R \) -module and that \( \iota \) is bilinear. It remains only to check that in the bijective correspondence in Theorem 10 the bilinear maps correspond with the \( R \) -module homomorphisms. If \( \varphi : M \times N \rightarrow L \) is bilinear then it is an \( R... | Yes |
Theorem 14. (Associativity of the Tensor Product) Suppose \( M \) is a right \( R \) -module, \( N \) is an \( \left( {R, T}\right) \) -bimodule, and \( L \) is a left \( T \) -module. Then there is a unique isomorphism\n\n\[ \left( {M{ \otimes }_{R}N}\right) { \otimes }_{T}L \cong M{ \otimes }_{R}\left( {N{ \otimes }_... | Proof: Note first that the \( \left( {R, T}\right) \) -bimodule structure on \( N \) makes \( M{ \otimes }_{R}N \) into a right \( T \) -module and \( N{ \otimes }_{T}L \) into a left \( R \) -module, so both sides of the isomorphism are well defined. For each fixed \( l \in L \), the mapping \( \left( {m, n}\right) \m... | Yes |
Corollary 16. Let \( R \) be a commutative ring and let \( {M}_{1},\ldots ,{M}_{n}, L \) be \( R \) -modules. Let \( {M}_{1} \otimes {M}_{2} \otimes \cdots \otimes {M}_{n} \) denote any bracketing of the tensor product of these modules and let\n\n\[ \iota : {M}_{1} \times \cdots \times {M}_{n} \rightarrow {M}_{1} \otim... | Hence there is a bijection\n\n\[ \left\{ \begin{matrix} n\text{-multilinear maps} \\ \varphi : {M}_{1} \times \cdots \times {M}_{n} \rightarrow L \end{matrix}\right\} \leftrightarrow \left\{ \begin{matrix} R\text{-module homomorphisms} \\ \Phi : {M}_{1} \otimes \cdots \otimes {M}_{n} \rightarrow L \end{matrix}\right\} ... | Yes |
Theorem 17. (Tensor Products of Direct Sums) Let \( M,{M}^{\prime } \) be right \( R \) -modules and let \( N,{N}^{\prime } \) be left \( R \) -modules. Then there are unique group isomorphisms\n\n\[ \left( {M \oplus {M}^{\prime }}\right) { \otimes }_{R}N \cong \left( {M{ \otimes }_{R}N}\right) \oplus \left( {{M}^{\pri... | Proof: The map \( \left( {M \oplus {M}^{\prime }}\right) \times N \rightarrow \left( {M{ \otimes }_{R}N}\right) \oplus \left( {{M}^{\prime }{ \otimes }_{R}N}\right) \) defined by \( \left( {\left( {m,{m}^{\prime }}\right), n}\right) \mapsto \) \( \left( {m \otimes n,{m}^{\prime } \otimes n}\right) \) is well defined si... | Yes |
Corollary 18. (Extension of Scalars for Free Modules) The module obtained from the free \( R \) -module \( N \cong {R}^{n} \) by extension of scalars from \( R \) to \( S \) is the free \( S \) -module \( {S}^{n} \) , i.e., \[ S{ \otimes }_{R}{R}^{n} \cong {S}^{n} \] as left \( S \) -modules. | Proof: This follows immediately from Theorem 17 and the isomorphism \( S{ \otimes }_{R}R \cong \) \( S \) proved in Example 7 previously. | Yes |
Corollary 19. Let \( R \) be a commutative ring and let \( M \cong {R}^{s} \) and \( N \cong {R}^{t} \) be free \( R \) -modules with bases \( {m}_{1},\ldots ,{m}_{s} \) and \( {n}_{1},\ldots ,{n}_{t} \), respectively. Then \( M{ \otimes }_{R}N \) is a free \( R \) -module of rank \( {st} \), with basis \( {m}_{i} \oti... | Proof: This follows easily from Theorem 17 and the first example following Corollary 9. | No |
Proposition 20. Suppose \( R \) is a commutative ring and \( M, N \) are left \( R \) -modules, considered with the standard \( R \) -module structures. Then there is a unique \( R \) -module isomorphism\n\n\[ M{ \otimes }_{R}N \cong N{ \otimes }_{R}M \]\n\nmapping \( m \otimes n \) to \( n \otimes m \) . | Proof: The map \( M \times N \rightarrow N \otimes M \) defined by \( \left( {m, n}\right) \mapsto n \otimes m \) is \( R \) -balanced. Hence it induces a unique homomorphism \( f \) from \( M \otimes N \) to \( N \otimes M \) with \( f\left( {m \otimes n}\right) = \) \( n \otimes m \) . Similarly, we have a unique hom... | Yes |
Proposition 21. Let \( R \) be a commutative ring and let \( A \) and \( B \) be \( R \) -algebras. Then the multiplication \( \left( {a \otimes b}\right) \left( {{a}^{\prime } \otimes {b}^{\prime }}\right) = a{a}^{\prime } \otimes b{b}^{\prime } \) is well defined and makes \( A{ \otimes }_{R}B \) into an \( R \) -alg... | Proof: Note first that the definition of an \( R \) -algebra shows that\n\n\[ \begin{matrix} r\left( {a \otimes b}\right) = {ra} \otimes b = {ar} \otimes b = a \otimes {rb} = a \otimes {br} = \left( {a \otimes b}\right) r \end{matrix} \]\n\nfor every \( r \in R, a \in A \) and \( b \in B \) . To show that \( A \otimes ... | No |
Proposition 22. Let \( A, B \) and \( C \) be \( R \) -modules over some ring \( R \) . Then\n\n(1) The sequence \( 0 \rightarrow A\overset{\psi }{ \rightarrow }B \) is exact (at \( A \) ) if and only if \( \psi \) is injective.\n\n(2) The sequence \( B\overset{\varphi }{ \rightarrow }C \rightarrow 0 \) is exact (at \(... | Proof: The (uniquely defined) homomorphism \( 0 \rightarrow A \) has image \( 0 \) in A. This will be the kernel of \( \psi \) if and only if \( \psi \) is injective. Similarly, the kernel of the (uniquely defined) zero homomorphism \( C \rightarrow 0 \) is all of \( C \), which is the image of \( \varphi \) if and onl... | Yes |
Proposition 24. (The Short Five Lemma) Let \( \alpha ,\beta ,\gamma \) be a homomorphism of short exact sequences\n\n\n\n(1) If \( \alpha \) and \( \gamma \) are injective then so is \( \beta \) .\n\n(2) If \( \alpha... | Proof: We shall prove (1), leaving the proof of (2) as an exercise (and (3) follows immediately from (1) and (2)). Suppose then that \( \alpha \) and \( \gamma \) are injective and suppose \( b \in B \) with \( \beta \left( b\right) = 0 \) . Let \( \psi : A \rightarrow B \) and \( \varphi : B \rightarrow C \) denote th... | No |
Proposition 25. The short exact sequence \( 0 \rightarrow A\overset{\psi }{ \rightarrow }B\overset{\varphi }{ \rightarrow }C \rightarrow 0 \) of \( R \) -modules is split if and only if there is an \( R \) -module homomorphism \( \mu : C \rightarrow B \) such that \( \varphi \circ \mu \) is the identity map on \( C \) ... | Proof: This follows directly from the definitions: if \( \mu \) is given define \( {C}^{\prime } = \mu \left( C\right) \subseteq \) \( B \) and if \( {C}^{\prime } \) is given define \( \mu = {\varphi }^{-1} : C \cong {C}^{\prime } \subseteq B \) . | No |
Proposition 26. Let \( 0 \rightarrow A\overset{\psi }{ \rightarrow }B\overset{\varphi }{ \rightarrow }C \rightarrow 0 \) be a short exact sequence of modules (respectively, \( 1 \rightarrow A\overset{\psi }{ \rightarrow }B\overset{\varphi }{ \rightarrow }C \rightarrow 1 \) a short exact sequence of groups). Then \( B =... | Proof: This is similar to the proof of Proposition 25. If \( \lambda \) is given, define \( {C}^{\prime } = \) \( \ker \lambda \subseteq B \) and if \( {C}^{\prime } \) is given define \( \lambda : B = \psi \left( A\right) \oplus {C}^{\prime } \rightarrow A \) by \( \lambda \left( {\left( {\psi \left( a\right) ,{c}^{\p... | Yes |
Proposition 27. Let \( D, L \) and \( M \) be \( R \) -modules and let \( \psi : L \rightarrow M \) be an \( R \) -module homomorphism. Then the map\n\n\[ \n{\psi }^{\prime } : {\operatorname{Hom}}_{R}\left( {D, L}\right) \rightarrow {\operatorname{Hom}}_{R}\left( {D, M}\right)\n\]\n\n\[ \nf \mapsto {f}^{\prime } = \ps... | Proof: The fact that \( {\psi }^{\prime } \) is a homomorphism is immediate. If \( \psi \) is injective, then distinct homomorphisms \( f \) and \( g \) from \( D \) into \( L \) give distinct homomorphisms \( \psi \circ f \) and \( \psi \circ g \) from \( D \) into \( M \), which is to say that \( {\psi }^{\prime } \)... | Yes |
Theorem 28. Let \( D, L, M \), and \( N \) be \( R \) -modules. If\n\n\[ 0 \rightarrow L\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }N \rightarrow 0\;\text{ is exact,}\]\n\nthen the associated sequence\n\n\[ 0 \rightarrow {\operatorname{Hom}}_{R}\left( {D, L}\right) \overset{{\psi }^{\prime }}{ \righ... | Proof: The only item in the first statement that has not already been proved is the exactness of (10) at \( {\operatorname{Hom}}_{R}\left( {D, M}\right) \), i.e., \( \ker {\varphi }^{\prime } = \operatorname{image}{\psi }^{\prime } \) . Suppose \( F : D \rightarrow M \) is an element of \( {\operatorname{Hom}}_{R}\left... | Yes |
Proposition 29. Let \( D, L \) and \( N \) be \( R \) -modules. Then\n\n(1) \( {\operatorname{Hom}}_{R}\left( {D, L \oplus N}\right) \cong {\operatorname{Hom}}_{R}\left( {D, L}\right) \oplus {\operatorname{Hom}}_{R}\left( {D, N}\right) \), and\n\n(2) \( {\operatorname{Hom}}_{R}\left( {L \oplus N, D}\right) \cong {\oper... | Proof: Let \( {\pi }_{1} : L \oplus N \rightarrow L \) be the natural projection from \( L \oplus N \) to \( L \) and similarly let \( {\pi }_{2} \) be the natural projection to \( N \) . If \( f \in {\operatorname{Hom}}_{R}\left( {D, L \oplus N}\right) \) then the compositions \( {\pi }_{1} \circ f \) and \( {\pi }_{2... | No |
Proposition 30. Let \( P \) be an \( R \) -module. Then the following are equivalent:\n\n(1) For any \( R \) -modules \( L, M \), and \( N \), if\n\n\[ 0 \rightarrow L\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }N \rightarrow 0 \]\n\nis a short exact sequence, then\n\n\[ 0 \rightarrow {\operatorname{... | Proof: The equivalence of (1) and (2) is a restatement of a result in Theorem 28. Suppose now that (2) is satisfied, and let \( 0 \rightarrow L\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }P \rightarrow 0 \) be exact. By (2), the identity map from \( P \) to \( P \) lifts to a homomorphism \( \mu \) m... | Yes |
If \( D \) is an \( R \) -module, then the functor \( {\operatorname{Hom}}_{R}\left( {D, \bot }\right) \) from the category of \( R \) -modules to the category of abelian groups is left exact. It is exact if and only if \( D \) is a projective \( R \) -module. | Note that if \( {\operatorname{Hom}}_{R}\left( {D, \bot }\right) \) takes short exact sequences to short exact sequences, then it takes exact sequences of any length to exact sequences since any exact sequence can be broken up into a succession of short exact sequences.\n\nAs we have seen, the functor \( {\operatorname... | No |
Theorem 33. Let \( D, L, M \), and \( N \) be \( R \) -modules. If\n\n\[ 0 \rightarrow L\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }N \rightarrow 0\;\text{ is exact,}\]\n\nthen the associated sequence\n\n\[ 0 \rightarrow {\operatorname{Hom}}_{R}\left( {N, D}\right) \overset{{\varphi }^{\prime }}{ \r... | Proof: The only item remaining to be proved in the first statement is the exactness of (12) at \( {\operatorname{Hom}}_{R}\left( {M, D}\right) \) . The proof of this statement is very similar to the proof of the corresponding result in Theorem 28 and is left as an exercise. Note also that the injectivity of \( \psi \) ... | No |
Proposition 34. Let \( Q \) be an \( R \) -module. Then the following are equivalent:\n\n(1) For any \( R \) -modules \( L, M \), and \( N \), if\n\n\[ 0 \rightarrow L\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }N \rightarrow 0 \]\n\nis a short exact sequence, then\n\n\[ 0 \rightarrow {\operatorname{... | Proof: The equivalence of (1) and (2) is part of Theorem 33. Suppose now that (2) is satisfied and let \( 0 \rightarrow Q\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }N \rightarrow 0 \) be exact. Taking \( L = Q \) and \( f \) the identity map from \( Q \) to itself, it follows by (2) that there is a ... | No |
Proposition 36. Let \( Q \) be an \( R \) -module.\n\n(1) (Baer’s Criterion) The module \( Q \) is injective if and only if for every left ideal \( I \) of \( R \) any \( R \) -module homomorphism \( g : I \rightarrow Q \) can be extended to an \( R \) -module homomorphism \( G : R \rightarrow Q \) . | Proof: If \( Q \) is injective and \( g : I \rightarrow Q \) is an \( R \) -module homomorphism from the nonzero ideal \( I \) of \( R \) into \( Q \), then \( g \) can be extended to an \( R \) -module homomorphism from \( R \) into \( Q \) by Proposition 34(2) applied to the exact sequence \( 0 \rightarrow I \rightar... | No |
Theorem 38. Let \( R \) be a ring with 1 and let \( M \) be an \( R \) -module. Then \( M \) is contained in an injective \( R \) -module. | ## Proof: A proof is outlined in Exercises 15 to 17. | No |
Theorem 39. Suppose that \( D \) is a right \( R \) -module and that \( L, M \) and \( N \) are left \( R \) -modules. If\n\n\[ 0 \rightarrow L\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }N \rightarrow 0\;\text{ is exact,}\]\n\nthen the associated sequence of abelian groups\n\n\[ D{ \otimes }_{R}L\ov... | Proof: For the first statement it remains to prove the exactness of (13) at \( D{ \otimes }_{R}M \) . Since \( \varphi \circ \psi = 0 \), we have\n\n\[ \left( {1 \otimes \varphi }\right) \left( {\sum {d}_{i} \otimes \psi \left( {l}_{i}\right) }\right) = \sum {d}_{i} \otimes \left( {\varphi \circ \psi \left( {l}_{i}\rig... | Yes |
Corollary 42. Free modules are flat; more generally, projective modules are flat. | Proof: To show that the free \( R \) -module \( F \) is flat it suffices to show that for any injective map \( \psi : L \rightarrow M \) of \( R \) -modules \( L \) and \( M \) the induced map \( 1 \otimes \psi : F{ \otimes }_{R}L \rightarrow \) \( F{ \otimes }_{R}M \) is also injective. Suppose first that \( F \cong {... | Yes |
Theorem 43. (Adjoint Associativity) Let \( R \) and \( S \) be rings, let \( A \) be a right \( R \) -module, let \( B \) be an \( \left( {R, S}\right) \) -bimodule and let \( C \) be a right \( S \) -module. Then there is an isomorphism of abelian groups:\n\n\[{\operatorname{Hom}}_{S}\left( {A{ \otimes }_{R}B, C}\righ... | Proof: Suppose \( \varphi : A{ \otimes }_{R}B \rightarrow C \) is a homomorphism. For any fixed \( a \in A \) define the map \( \Phi \left( a\right) \) from \( B \) to \( C \) by \( \Phi \left( a\right) \left( b\right) = \varphi \left( {a \otimes b}\right) \) . It is easy to check that \( \Phi \left( a\right) \) is a h... | Yes |
Corollary 44. If \( R \) is commutative then the tensor product of two projective \( R \) -modules is projective. | Proof: Let \( {P}_{1} \) and \( {P}_{2} \) be projective modules. Then by Corollary 32, \( {\operatorname{Hom}}_{R}\left( {{P}_{2}, \bot }\right) \) is an exact functor from the category of \( R \) -modules to the category of \( R \) -modules. Then the composition \( {\operatorname{Hom}}_{R}\left( {{P}_{1},{\operatorna... | Yes |
Proposition 1. Assume the set \( \mathcal{A} = \left\{ {{v}_{1},{v}_{2},\ldots ,{v}_{n}}\right\} \) spans the vector space \( V \) but no proper subset of \( \mathcal{A} \) spans \( V \) . Then \( \mathcal{A} \) is a basis of \( V \) . In particular, any finitely generated (i.e., finitely spanned) vector space over \( ... | Proof: It is only necessary to prove that \( {v}_{1},{v}_{2},\ldots ,{v}_{n} \) are linearly independent. Suppose \( {\alpha }_{1}{v}_{1} + {\alpha }_{2}{v}_{2} + \cdots + {\alpha }_{n}{v}_{n} = 0 \) where not all of the \( {\alpha }_{i} \) are 0 . By reordering, we may assume that \( {\alpha }_{1} \neq 0 \) and then\n... | Yes |
Corollary 2. Assume the finite set \( \mathcal{A} \) spans the vector space \( V \) . Then \( \mathcal{A} \) contains a basis of \( V \) . | Proof: Any subset \( \mathcal{B} \) of \( \mathcal{A} \) spanning \( V \) such that no proper subset of \( \mathcal{B} \) also spans \( V \) (there clearly exist such subsets) is a basis for \( V \) by Proposition 1. | Yes |
Theorem 3. (A Replacement Theorem) Assume \( \mathcal{A} = \left\{ {{a}_{1},{a}_{2},\ldots ,{a}_{n}}\right\} \) is a basis for \( V \) containing \( n \) elements and \( \left\{ {{b}_{1},{b}_{2},\ldots ,{b}_{m}}\right\} \) is a set of linearly independent vectors in \( V \) . Then there is an ordering \( {a}_{1},{a}_{2... | Proof: Proceed by induction on \( k \) . If \( k = 0 \) there is nothing to prove, since \( \mathcal{A} \) is given as a basis for \( V \) . Suppose now that \( \left\{ {{b}_{1},{b}_{2},\ldots ,{b}_{k},{a}_{k + 1},{a}_{k + 2},\ldots ,{a}_{n}}\right\} \) is a basis for \( V \) . Then in particular this is a spanning set... | Yes |
Corollary 5. (Building-Up Lemma) If \( A \) is a set of linearly independent vectors in the finite dimensional space \( V \) then there exists a basis of \( V \) containing \( A \) . | Proof: This is also immediate from Theorem 3, since we can use the elements of \( A \) to successively replace the elements of any given basis for \( V \) (which exists by the assumption that \( V \) is finite dimensional). | No |
Theorem 6. If \( V \) is an \( n \) dimensional vector space over \( F \) , then \( V \cong {F}^{n} \) . In particular, any two finite dimensional vector spaces over \( F \) of the same dimension are isomorphic. | Proof: Let \( {v}_{1},{v}_{2},\ldots ,{v}_{n} \) be a basis for \( V \) . Define the map\n\n\[ \varphi : {F}^{n} \rightarrow V\;\text{by}\;\varphi \left( {{\alpha }_{1},{\alpha }_{2},\ldots ,{\alpha }_{n}}\right) = {\alpha }_{1}{v}_{1} + {\alpha }_{2}{v}_{2} + \cdots + {\alpha }_{n}{v}_{n}. \]\n\nThe map \( \varphi \) ... | Yes |
Theorem 7. Let \( V \) be a vector space over \( F \) and let \( W \) be a subspace of \( V \) . Then \( V/W \) is a vector space with \( \dim V = \dim W + \dim V/W \) (where if one side is infinite then both are). | Proof: Suppose \( W \) has dimension \( m \) and \( V \) has dimension \( n \) over \( F \) and let \( {w}_{1},{w}_{2},\ldots ,{w}_{m} \) be a basis for \( W \) . By Corollary 5, these linearly independent elements of \( V \) can be extended to a basis \( {w}_{1},{w}_{2},\ldots ,{w}_{m},{v}_{m + 1},\ldots ,{v}_{n} \) o... | Yes |
Corollary 8. Let \( \varphi : V \rightarrow U \) be a linear transformation of vector spaces over \( F \) . Then \( \ker \varphi \) is a subspace of \( V,\varphi \left( V\right) \) is a subspace of \( U \) and \( \dim V = \dim \ker \varphi + \dim \varphi \left( V\right) \) . | Proof: This follows immediately from Theorem 7. Note that the proof of Theorem 7 is in fact the special case of Corollary 8 where \( U \) is the quotient \( V/W \) and \( \varphi \) is the natural projection homomorphism. | No |
Corollary 9. Let \( \varphi : V \rightarrow W \) be a linear transformation of vector spaces of the same finite dimension. Then the following are equivalent:\n\n(1) \( \varphi \) is an isomorphism\n\n(2) \( \varphi \) is injective, i.e., \( \ker \varphi = 0 \)\n\n(3) \( \varphi \) is surjective, i.e., \( \varphi \left(... | Proof: The equivalence of these conditions follows from Corollary 8 by counting dimensions. | No |
Theorem 10. Let \( V \) be a vector space over \( F \) of dimension \( n \) and let \( W \) be a vector space over \( F \) of dimension \( m \), with bases \( \mathcal{B},\mathcal{E} \) respectively. Then the map \( {\operatorname{Hom}}_{F}\left( {V, W}\right) \rightarrow \) \( {M}_{m \times n}\left( F\right) \) from t... | Proof: The columns of the matrix \( {M}_{\mathcal{B}}^{\mathcal{E}}\left( \varphi \right) \) are determined by the action of \( \varphi \) on the basis \( \mathcal{B} \) as in equation (3). This shows in particular that the map \( \varphi \mapsto {M}_{\mathcal{B}}^{\mathcal{E}}\left( \varphi \right) \) is an \( F \) -l... | Yes |
Corollary 11. The dimension of \( {\operatorname{Hom}}_{F}\left( {V, W}\right) \) is \( \left( {\dim V}\right) \left( {\dim W}\right) \) . | Proof: The dimension of \( {M}_{m \times n}\left( F\right) \) is \( {mn} \) . | No |
Corollary 13. Matrix multiplication is associative and distributive (whenever the dimensions are such as to make products defined). An \( n \times n \) matrix \( A \) is nonsingular if and only if it is invertible. | Proof: Let \( A, B \) and \( C \) be matrices such that the products \( \left( {AB}\right) C \) and \( A\left( {BC}\right) \) are defined, and let \( S, T \) and \( R \) denote the associated linear transformations. By Theorem 12, the linear transformation corresponding to \( {AB} \) is the composite \( S \circ T \) so... | Yes |
Proposition 17. Let \( \varphi : V \rightarrow X \) and \( \psi : W \rightarrow Y \) be linear transformations of finite dimensional vector spaces. Then the Kronecker product of matrices representing \( \varphi \) and \( \psi \) is a matrix representation of \( \varphi \otimes \psi \) . | ## Example\n\nLet \( V = X = {\mathbb{R}}^{3} \), both with basis \( {v}_{1},{v}_{2},{v}_{3} \), and \( W = Y = {\mathbb{R}}^{2} \), both with basis \( {w}_{1},{w}_{2} \) . Suppose \( \varphi : {\mathbb{R}}^{3} \rightarrow {\mathbb{R}}^{3} \) is the linear transformation given by \( \varphi \left( {a{v}_{1} + b{v}_{2} ... | No |
Proposition 18. With notations as above, \( \left\{ {{v}_{1}^{ * },{v}_{2}^{ * },\ldots ,{v}_{n}^{ * }}\right\} \) is a basis of \( {V}^{ * } \) . In particular, if \( V \) is finite dimensional then \( {V}^{ * } \) has the same dimension as \( V \) . | Proof: Observe that since \( V \) is finite dimensional, \( \dim {V}^{ * } = \dim {\operatorname{Hom}}_{F}\left( {V, F}\right) = \) \( \dim V = n \) (Corollary 11), so since there are \( n \) of the \( {v}_{i}^{ * } \) ’s it suffices to prove that they are linearly independent. If\n\n\[ \n{\alpha }_{1}{v}_{1}^{ * } + {... | Yes |
Theorem 19. There is a natural injective linear transformation from \( V \) to \( {V}^{* * } \) . If \( V \) is finite dimensional then this linear transformation is an isomorphism. | Proof: Let \( v \in V \) . Define the map (evaluation at \( v \) ) \[ {E}_{v} : {V}^{ * } \rightarrow F\;\text{ by }\;{E}_{v}\left( f\right) = f\left( v\right) . \] Then \( {E}_{v}\left( {f + {\alpha g}}\right) = \left( {f + {\alpha g}}\right) \left( v\right) = f\left( v\right) + {\alpha g}\left( v\right) = {E}_{v}\lef... | Yes |
Theorem 20. With notations as above, \( {\varphi }^{ * } \) is a linear transformation from \( {W}^{ * } \) to \( {V}^{ * } \) and \( {M}_{{\mathcal{E}}^{ * }}^{{\mathcal{B}}^{ * }}\left( {\varphi }^{ * }\right) \) is the transpose of the matrix \( {M}_{\mathcal{B}}^{\mathcal{E}}\left( \varphi \right) \) (recall that t... | Proof: The map \( {\varphi }^{ * } \) is linear because \( \left( {f + {\alpha g}}\right) \circ \varphi = \left( {f \circ \varphi }\right) + \alpha \left( {g \circ \varphi }\right) \) . The equations which define \( \varphi \) are (from its matrix)\n\n\[ \varphi \left( {v}_{j}\right) = \mathop{\sum }\limits_{{i = 1}}^{... | Yes |
For any matrix \( A \) , the row rank of \( A \) equals the column rank of \( A \) . | Let \( \varphi : V \rightarrow W \) be a linear transformation whose matrix with respect to some fixed bases of \( V \) and \( W \) is \( A \) . By Theorem 20 the matrix of \( {\varphi }^{ * } : {W}^{ * } \rightarrow {V}^{ * } \) with respect to the dual bases is the transpose of \( A \) . The column rank of \( A \) is... | Yes |
Proposition 22. Let \( \varphi \) be an \( n \) -multilinear alternating function on \( V \) . Then\n\n(1) \( \varphi \left( {{v}_{1},\ldots ,{v}_{i - 1},{v}_{i + 1},{v}_{i},{v}_{i + 2},\ldots ,{v}_{n}}\right) = - \varphi \left( {{v}_{1},{v}_{2},\ldots ,{v}_{n}}\right) \) for any \( i \in \) \( \{ 1,2,\ldots, n - 1\} \... | Proof: (1) Let \( \psi \left( {x, y}\right) \) be the function \( \varphi \) with variable entries \( x \) and \( y \) in positions \( i \) and \( i + 1 \) respectively and fixed entries \( {v}_{j} \) in position \( j \), for all other \( j \) . Thus (1) is the same as showing \( \psi \left( {y, x}\right) = - \psi \lef... | Yes |
Theorem 24. There is a unique \( n \times n \) determinant function on \( R \) and it can be computed for any \( n \times n \) matrix \( \left( {\alpha }_{ij}\right) \) by the formula:\n\n\[ \det \left( {\alpha }_{ij}\right) = \mathop{\sum }\limits_{{\sigma \in {S}_{n}}}\epsilon \left( \sigma \right) {\alpha }_{\sigma ... | Proof: Let \( {A}_{1},{A}_{2},\ldots ,{A}_{n} \) be the column vectors in a general \( n \times n \) matrix \( \left( {\alpha }_{ij}\right) \) . We leave it as an exercise to check that the formula given in the statement of the theorem does satisfy the axioms of a determinant function - this gives existence of a determ... | No |
Corollary 25. The determinant is an \( n \) -multilinear function of the rows of \( {M}_{n \times n}\left( R\right) \) and for any \( n \times n \) matrix \( A \) , \( \det A = \det \left( {A}^{t}\right) \), where \( {A}^{t} \) is the transpose of \( A \) . | Proof: The first statement is an immediate consequence of the second, so it suffices to prove that a matrix and its transpose have the same determinant. For \( A = \left( {\alpha }_{ij}\right) \) one calculates that\n\n\[ \det {A}^{t} = \mathop{\sum }\limits_{{\sigma \in {S}_{n}}}\epsilon \left( \sigma \right) {\alpha ... | Yes |
Theorem 26. (Cramer’s Rule) If \( {A}_{1},{A}_{2},\ldots ,{A}_{n} \) are the columns of an \( n \times n \) matrix \( A \) and \( B = {\beta }_{1}{A}_{1} + {\beta }_{2}{A}_{2} + \cdots + {\beta }_{n}{A}_{n} \), for some \( {\beta }_{1},\ldots ,{\beta }_{n} \in R \), then\n\n\[{\beta }_{i}\det A = \det \left( {{A}_{1},\... | Proof: This follows immediately from Proposition 22(3) on replacing the given expression for \( B \) in the \( {i}^{\text{th }} \) position and expanding by multilinearity in that position. | No |
Corollary 27. If \( R \) is an integral domain, then \( \det A = 0 \) for \( A \in {M}_{n}\left( R\right) \) if and only if the columns of \( A \) are \( R \) -linearly dependent as elements of the free \( R \) -module of rank \( n \) . Also, det \( A = 0 \) if and only if the rows of \( A \) are \( R \) -linearly depe... | Proof: Since \( \det A = \det {A}^{\prime } \) the first sentence implies the second.\n\nAssume first that the columns of \( A \) are linearly dependent and\n\n\[ 0 = {\beta }_{1}{A}_{1} + {\beta }_{2}{A}_{2} + \cdots + {\beta }_{n}{A}_{n} \] is a dependence relation on the columns of \( A \) with, say, \( {\beta }_{i}... | Yes |
Theorem 28. For matrices \( A, B \in {M}_{n \times n}\left( R\right) \), det \( {AB} = \left( {\det A}\right) \left( {\det B}\right) \) . | Proof: Let \( B = \left( {\beta }_{ij}\right) \) and let \( {A}_{1},{A}_{2},\ldots ,{A}_{n} \) be the columns of \( A \) . Then \( C = {AB} \) is the \( n \times n \) matrix whose \( {j}^{\text{th }} \) column is \( {C}_{j} = {\beta }_{1j}{A}_{1} + {\beta }_{2j}{A}_{2} + \cdots + {\beta }_{nj}{A}_{n} \) . By Propositio... | Yes |
Theorem 29. (The Cofactor Expansion Formula along the \( {i}^{\text{th }} \) row) If \( A = \left( {\alpha }_{ij}\right) \) is an \( n \times n \) matrix, then for each fixed \( i \in \{ 1,2,\ldots, n\} \) the determinant of \( A \) can be computed from the formula\n\n\[ \det A = {\left( -1\right) }^{i + 1}{\alpha }_{i... | Proof: For each \( A \) let \( D\left( A\right) \) be the element of \( R \) obtained from the cofactor expansion formula described above. We prove that \( D \) satisfies the axioms of a determinant function, hence is the determinant function. Proceed by induction on \( n \) . If \( n = 1 \) , \( D\left( \left( \alpha ... | Yes |
Theorem 30. (Cofactor Formula for the Inverse of a Matrix) Let \( A = \left( {\alpha }_{ij}\right) \) be an \( n \times n \) matrix and let \( B \) be the transpose of its matrix of cofactors, i.e., \( B = \left( {\beta }_{ij}\right) \), where \( {\beta }_{ij} = {\left( -1\right) }^{i + j}\det {A}_{ji},1 \leq i, j \leq... | Proof: The \( i, j \) entry of \( {AB} \) is \( {\alpha }_{i1}{\beta }_{1j} + {\alpha }_{i2}{\beta }_{2j} + \cdots + {\alpha }_{in}{\beta }_{nj} \) . By definition of the entries of \( B \) this equals\n\n\[ \n{\alpha }_{i1}{\left( -1\right) }^{j + 1}D\left( {A}_{j1}\right) + {\alpha }_{i2}{\left( -1\right) }^{j + 2}D\... | Yes |
Theorem 31. If \( M \) is any \( R \) -module over the commutative ring \( R \) then\n\n(1) \( \mathcal{T}\left( M\right) \) is an \( R \) -algebra containing \( M \) with multiplication defined by mapping\n\n\[ \left( {{m}_{1} \otimes \cdots \otimes {m}_{i}}\right) \left( {{m}_{1}^{\prime } \otimes \cdots \otimes {m}_... | Proof: The map\n\n\[ \underset{i\text{ factors }}{\underbrace{M \times M \times \cdots \times M}} \times \underset{j\text{ factors }}{\underbrace{M \times M \times \cdots \times M}} \rightarrow {\mathcal{T}}^{i + j}\left( M\right) \]\n\ndefined by\n\n\[ \left( {{m}_{1},\ldots ,{m}_{i},{m}_{1}^{\prime },\ldots ,{m}_{j}^... | Yes |
Proposition 32. Let \( V \) be a finite dimensional vector space over the field \( F \) with basis \( \mathcal{B} = \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) . Then the \( k \) -tensors\n\n\[ \n{v}_{{i}_{1}} \otimes {v}_{{i}_{2}} \otimes \cdots \otimes {v}_{{i}_{k}}\;\text{ with }{v}_{{i}_{j}} \in \mathcal{B} \n\]\n... | Proof: This follows immediately from Proposition 16 of Section 2. | No |
Proposition 33. Let \( S \) be a graded ring, let \( I \) be a graded ideal in \( S \) and let \( {I}_{k} = I \cap {S}_{k} \) for all \( k \geq 0 \) . Then \( S/I \) is naturally a graded ring whose homogeneous component of degree \( k \) is isomorphic to \( {S}_{k}/{I}_{k} \) . | Proof: The map\n\n\[ S = { \oplus }_{k = 0}^{\infty }{S}_{k} \rightarrow { \oplus }_{k = 0}^{\infty }\left( {{S}_{k}/{I}_{k}}\right) \]\n\n\[ \left( {\ldots ,{s}_{k},\ldots }\right) \mapsto \left( {\ldots ,{s}_{k}{\;\operatorname{mod}\;{I}_{k}},\ldots }\right) \]\n\n is surjective with kernel \( I = { \oplus }_{k = 0}^... | No |
Theorem 34. Let \( M \) be an \( R \) -module over the commutative ring \( R \) and let \( \mathcal{S}\left( M\right) \) be its symmetric algebra.\n\n(1) The \( {k}^{\text{th }} \) symmetric power, \( {\mathcal{S}}^{k}\left( M\right) \), of \( M \) is equal to \( M \otimes \cdots \otimes M \) ( \( k \) factors) modulo ... | Proof: The \( k \) -tensors \( {\mathcal{C}}^{k}\left( M\right) \) in the ideal \( \mathcal{C}\left( M\right) \) are finite sums of elements of the form\n\n\[ {m}_{1} \otimes \ldots \otimes {m}_{i - 1} \otimes \left( {{m}_{i} \otimes {m}_{i + 1} - {m}_{i + 1} \otimes {m}_{i}}\right) \otimes {m}_{i + 2} \otimes \ldots \... | Yes |
Corollary 35. Let \( V \) be an \( n \) -dimensional vector space over the field \( F \) . Then \( \mathcal{S}\left( V\right) \) is isomorphic as a graded \( F \) -algebra to the ring of polynomials in \( n \) variables over \( F \) (i.e., the isomorphism is also a vector space isomorphism from \( {\mathcal{S}}^{k}\lef... | Proof: Let \( \mathcal{B} = \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) be a basis of \( V \) . By Proposition 32 there is a bijection between a basis of \( {\mathcal{T}}^{k}\left( V\right) \) and the set \( {\mathcal{B}}^{k} \) of ordered \( k \) -tuples of elements from \( \mathcal{B} \) . Define two \( k \) -tuples... | No |
Corollary 37. Let \( V \) be a finite dimensional vector space over the field \( F \) with basis \( \mathcal{B} = \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) . Then the vectors\n\n\[ \n{v}_{{i}_{1}} \land {v}_{{i}_{2}} \land \cdots \land {v}_{{i}_{k}}\;\text{for }1 \leq {i}_{1} < {i}_{2} < \cdots < {i}_{k} \leq n \n\]... | Proof: As the proof of Theorem 36 shows, modulo \( {\mathcal{A}}^{k}\left( M\right) \), the order of the terms in any simple \( k \) -tensor can be rearranged up to introducing a sign change. It follows that the \( k \) -tensors in the corollary (which have been arranged with increasing subscripts on the \( {v}_{i} \) ... | No |
Proposition 39. Let \( \sigma \) be an element in the symmetric group \( {S}_{k} \) and let \( \epsilon \left( \sigma \right) \) be the sign of the permutation \( \sigma \) . Then\n\n(1) for every \( w \in {\mathcal{S}}^{k}\left( M\right) \) we have \( {\sigma w} = w \), and\n\n(2) for every \( w \in \mathop{\bigwedge ... | Proof: The first statement is immediate from (1) in Theorem 34. We showed in the course of the proof of Theorem 36 that\n\n\[ \n{m}_{1} \land \cdots \land {m}_{i} \land {m}_{i + 1} \land \cdots \land {m}_{k} = - {m}_{1} \land \cdots \land {m}_{i + 1} \land {m}_{i} \land \cdots \land {m}_{k}, \n\]\n\nwhich shows that th... | Yes |
Proposition 40. Suppose \( k \) ! is a unit in the ring \( R \) and \( M \) is an \( R \) -module. Then\n\n(1) The map \( \left( {1/k!}\right) {Sym} \) induces an \( R \) -module isomorphism between the \( {k}^{\text{th }} \) symmetric power of \( M \) and the \( R \) -submodule of symmetric \( k \) -tensors:\n\n\[ \fr... | Proof: We have seen that the respective maps are surjective \( R \) -homomorphisms from \( {\mathcal{T}}^{k}\left( M\right) \) so to prove the proposition it suffices to check that their kernels are \( {\mathcal{C}}^{k}\left( M\right) \) and \( {\mathcal{A}}^{k}\left( M\right) \), respectively. We show the first and le... | No |
Theorem 1. Let \( R \) be a ring and let \( M \) be a left \( R \) -module. Then the following are equivalent:\n\n(1) \( M \) is a Noetherian \( R \) -module.\n\n(2) Every nonempty set of submodules of \( M \) contains a maximal element under inclusion.\n\n(3) Every submodule of \( M \) is finitely generated. | Proof: [(1) implies (2)] Assume \( M \) is Noetherian and let \( \sum \) be any nonempty collection of submodules of \( M \) . Choose any \( {M}_{1} \in \sum \) . If \( {M}_{1} \) is a maximal element of \( \sum \) ,(2) holds, so assume \( {M}_{1} \) is not maximal. Then there is some \( {M}_{2} \in \sum \) such that \... | Yes |
Proposition 3. Let \( R \) be an integral domain and let \( M \) be a free \( R \) -module of rank \( n < \infty \) . Then any \( n + 1 \) elements of \( M \) are \( R \) -linearly dependent, i.e., for any \( {y}_{1},{y}_{2},\ldots ,{y}_{n + 1} \in M \) there are elements \( {r}_{1},{r}_{2},\ldots ,{r}_{n + 1} \in R \)... | Proof: The quickest way of proving this is to embed \( R \) in its quotient field \( F \) (since \( R \) is an integral domain) and observe that since \( M \cong R \oplus R \oplus \cdots \oplus R \) ( \( n \) times) we obtain \( M \subseteq F \oplus F \oplus \cdots \oplus F \) . The latter is an \( n \) -dimensional ve... | Yes |
Theorem 5. (Fundamental Theorem, Existence: Invariant Factor Form) Let \( R \) be a P.I.D. and let \( M \) be a finitely generated \( R \)-module.\n\n(1) Then \( M \) is isomorphic to the direct sum of finitely many cyclic modules. More precisely,\n\n\[ M \cong {R}^{r} \oplus R/\left( {a}_{1}\right) \oplus R/\left( {a}... | Proof: The module \( M \) can be generated by a finite set of elements by assumption so let \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \) be a set of generators of \( M \) of minimal cardinality. Let \( {R}^{n} \) be the free \( R \)-module of rank \( n \) with basis \( {b}_{1},{b}_{2},\ldots ,{b}_{n} \) and define the homomor... | Yes |
Theorem 6. (Fundamental Theorem, Existence: Elementary Divisor Form) Let \( R \) be a P.I.D. and let \( M \) be a finitely generated \( R \) -module. Then \( M \) is the direct sum of a finite number of cyclic modules whose annihilators are either (0) or generated by powers of primes in \( R \), i.e., \[ M \cong {R}^{r... | We proved Theorem 6 by using the prime power factors of the invariant factors for \( M \) . In fact we shall see that the decomposition of \( M \) into a direct sum of cyclic modules whose annihilators are (0) or prime powers as in Theorem 6 is unique, i.e., the integer \( r \) and the ideals \( \left( {p}_{1}^{{\alpha... | Yes |
Theorem 7. (The Primary Decomposition Theorem) Let \( R \) be a P.I.D. and let \( M \) be a nonzero torsion \( R \) -module (not necessarily finitely generated) with nonzero annihilator \( a \) . Suppose the factorization of \( a \) into distinct prime powers in \( R \) is\n\n\[ a = u{p}_{1}^{{\alpha }_{1}}{p}_{2}^{{\a... | Proof: We have already proved these results in the case where \( M \) is finitely generated over \( R \) . In the general case it is clear that \( {N}_{i} \) is a submodule of \( M \) with annihilator dividing \( {p}_{i}^{{\alpha }_{i}} \) . Since \( R \) is a P.I.D. the ideals \( \left( {p}_{i}^{{\alpha }_{i}}\right) ... | Yes |
Lemma 8. Let \( R \) be a P.I.D. and let \( p \) be a prime in \( R \) . Let \( F \) denote the field \( R/\left( p\right) \) . (1) Let \( M = {R}^{r} \) . Then \( M/{pM} \cong {F}^{r} \) . | Proof: (1) There is a natural map from \( {R}^{r} \) to \( {\left( R/\left( p\right) \right) }^{r} \) defined by mapping \( \left( {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right) \) to \( \left( {{\alpha }_{1}{\;\operatorname{mod}\;\left( p\right) },\ldots ,{\alpha }_{r}{\;\operatorname{mod}\;\left( p\right) }}\right) \) ... | Yes |
Corollary 10. Let \( R \) be a P.I.D. and let \( M \) be a finitely generated \( R \) -module.\n\n(1) The elementary divisors of \( M \) are the prime power factors of the invariant factors of \( M \).\n\n(2) The largest invariant factor of \( M \) is the product of the largest of the distinct prime powers among the el... | Proof: The procedure in (1) gives \( a \) set of elementary divisors and since the elementary divisors for \( M \) are unique by the theorem, it follows that the procedure in (1) gives the set of elementary divisors. Similarly for (2). | No |
Corollary 11. (The Fundamental Theorem of Finitely Generated Abelian Groups) \( \mathrm{{See}} \) Theorem 5.3 and Theorem 5.5. | Proof: Take \( R = \mathbb{Z} \) in Theorems 5, 6 and 9 (note however that the invariant factors are listed in reverse order in Chapter 5 for computational convenience). | No |
Proposition 12. The following are equivalent:\n\n(1) \( \lambda \) is an eigenvalue of \( T \)\n\n(2) \( {\lambda I} - T \) is a singular linear transformation of \( V \)\n\n(3) \( \det \left( {{\lambda I} - T}\right) = 0 \). | Proof: Since \( \lambda \) is an eigenvalue of \( T \) with corresponding eigenvector \( v \) if and only if \( v \) is a nonzero vector in the kernel of \( {\lambda I} - T \), it follows that (1) and (2) are equivalent.\n\n(2) and (3) are equivalent by our results on determinants. | Yes |
Theorem 15. Let \( S \) and \( T \) be linear transformations of \( V \) . Then the following are equivalent:\n\n(1) \( S \) and \( T \) are similar linear transformations\n\n(2) the \( F\left\lbrack x\right\rbrack \) -modules obtained from \( V \) via \( S \) and via \( T \) are isomorphic \( F\left\lbrack x\right\rbr... | Proof: [(1) implies (2)] Assume there is a nonsingular linear transformation \( U \) such that \( S = {UT}{U}^{-1} \) . The vector space isomorphism \( U : V \rightarrow V \) is also an \( F\left\lbrack x\right\rbrack \) -module homomorphism, where \( x \) acts on the first \( V \) via \( T \) and on the second via \( ... | Yes |
Corollary 18. Let \( A \) and \( B \) be two \( n \times n \) matrices over a field \( F \) and suppose \( F \) is a subfield of the field \( K \). (1) The rational canonical form of \( A \) is the same whether it is computed over \( K \) or over \( F \). The minimal and characteristic polynomials and the invariant fac... | Proof: (1) Let \( M \) be the rational canonical form of \( A \) when computed over the smaller field \( F \). Since \( M \) satisfies the conditions in the definition of the rational canonical form over \( K \), the uniqueness of the rational canonical form implies that \( M \) is also the rational canonical form of \... | Yes |
Lemma 19. Let \( a\left( x\right) \in F\left\lbrack x\right\rbrack \) be any monic polynomial.\n\n(1) The characteristic polynomial of the companion matrix of \( a\left( x\right) \) is \( a\left( x\right) \).\n\n(2) If \( M \) is the block diagonal matrix\n\n\[ M = \left( \begin{matrix} {A}_{1} & 0 & \ldots & 0 \\ 0 & ... | Proof: These are both straightforward exercises. | No |
Proposition 20. Let \( A \) be an \( n \times n \) matrix over the field \( F \) .\n\n(1) The characteristic polynomial of \( A \) is the product of all the invariant factors of A.\n\n(2) (The Cayley-Hamilton Theorem) The minimal polynomial of \( A \) divides the characteristic polynomial of \( A \) .\n\n(3) The charac... | Proof: Let \( B \) be the rational canonical form of \( A \) . By the previous lemma the block diagonal form of \( B \) shows that the characteristic polynomial of \( B \) is the product of the characteristic polynomials of the companion matrices of the invariant factors of \( A \) . By the first part of the lemma abov... | Yes |
Theorem 21. Let \( A \) be an \( n \times n \) matrix over the field \( F \) . Using the three elementary row and column operations above, the \( n \times n \) matrix \( {xI} - A \) with entries from \( F\left\lbrack x\right\rbrack \) can be put into the diagonal form (called the Smith Normal Form for \( A \) ) | Proof: cf. the exercises. | No |
(1) If a matrix \( A \) is similar to a diagonal matrix \( D \), then \( D \) is the Jordan canonical form of \( A \) . | Proof: The first assertion is immediate from the uniqueness of Jordan canonical forms because a diagonal matrix is itself in Jordan form (with Jordan blocks of size 1). | No |
Corollary 25. If \( A \) is an \( n \times n \) matrix with entries from \( F \) and \( F \) contains all the eigenvalues of \( A \), then \( A \) is similar to a diagonal matrix over \( F \) if and only if the minimal polynomial of \( A \) has no repeated roots. | Proof: Suppose \( A \) is similar to a diagonal matrix. The minimal polynomial of a diagonal matrix has no repeated roots (its roots are precisely the distinct elements along the diagonal). Since similar matrices have the same minimal polynomial it follows that the minimal polynomial for \( A \) has no repeated roots.\... | Yes |
Proposition 1. The characteristic of a field \( F,\operatorname{ch}\left( F\right) \), is either 0 or a prime \( p \) . If \( \operatorname{ch}\left( F\right) = p \) then for any \( \alpha \in F \) , | \[ p \cdot \alpha = \underset{p\text{ times }}{\underbrace{\alpha + \alpha + \cdots + \alpha }} = 0. \] Proof: Only the second statement has not been proved, and this follows immediately from the evident equality \( p \cdot \alpha = p \cdot \left( {{1}_{F}\alpha }\right) = \left( {p \cdot {1}_{F}}\right) \left( \alpha ... | No |
Theorem 3. Let \( F \) be a field and let \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) be an irreducible polynomial. Then there exists a field \( K \) containing an isomorphic copy of \( F \) in which \( p\left( x\right) \) has a root. Identifying \( F \) with this isomorphic copy shows that there exists an ... | Proof: Consider the quotient\n\n\[ K = F\left\lbrack x\right\rbrack /\left( {p\left( x\right) }\right) \]\n\nof the polynomial ring \( F\left\lbrack x\right\rbrack \) by the ideal generated by \( p\left( x\right) \). Since by assumption \( p\left( x\right) \) is an irreducible polynomial in the P.I.D. \( F\left\lbrack ... | Yes |
Theorem 4. Let \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) be an irreducible polynomial of degree \( n \) over the field \( F \) and let \( K \) be the field \( F\left\lbrack x\right\rbrack /\left( {p\left( x\right) }\right) \) . Let \( \theta = x{\;\operatorname{mod}\;\left( {p\left( x\right) }\right) } \i... | Proof: Let \( a\left( x\right) \in F\left\lbrack x\right\rbrack \) be any polynomial with coefficients in \( F \) . Since \( F\left\lbrack x\right\rbrack \) is a Euclidean Domain (this is Theorem 3 of Chapter 9), we may divide \( a\left( x\right) \) by \( p\left( x\right) \) :\n\n\[ a\left( x\right) = q\left( x\right) ... | Yes |
Corollary 5. Let \( K \) be as in Theorem 4, and let \( a\left( \theta \right), b\left( \theta \right) \in K \) be two polynomials of degree \( < n \) in \( \theta \) . Then addition in \( K \) is defined simply by usual polynomial addition and multiplication in \( K \) is defined by\n\n\[ a\left( \theta \right) b\left... | By the results proved above, this definition of addition and multiplication on the polynomials of degree \( < n \) in \( \theta \) make \( K \) into a field, so that one can also divide by nonzero elements as well, which is not so immediately obvious from the definitions of the operations. | Yes |
Theorem 6. Let \( F \) be a field and let \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) be an irreducible polynomial. Suppose \( K \) is an extension field of \( F \) containing a root \( \alpha \) of \( p\left( x\right) : p\left( \alpha \right) = 0 \) . Let \( F\left( \alpha \right) \) denote the subfield of... | Proof: There is a natural homomorphism\n\n\[ \varphi : F\left\lbrack x\right\rbrack \rightarrow F\left( \alpha \right) \subseteq K \]\n\n\[ a\left( x\right) \mapsto a\left( \alpha \right) \]\n\nobtained by mapping \( F \) to \( F \) by the identity map and sending \( x \) to \( \alpha \) and then extending so that the ... | Yes |
Theorem 8. Let \( \varphi : F\overset{ \sim }{ \rightarrow }{F}^{\prime } \) be an isomorphism of fields. Let \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) be an irreducible polynomial and let \( {p}^{\prime }\left( x\right) \in {F}^{\prime }\left\lbrack x\right\rbrack \) be the irreducible polynomial obtaine... | Proof: As noted above, the isomorphism \( \varphi \) induces a natural isomorphism from \( F\left\lbrack x\right\rbrack \) to \( {F}^{\prime }\left\lbrack x\right\rbrack \) which maps the maximal ideal \( \left( {p\left( x\right) }\right) \) to the maximal ideal \( \left( {{p}^{\prime }\left( x\right) }\right) \) . Tak... | Yes |
Proposition 9. Let \( \alpha \) be algebraic over \( F \) . Then there is a unique monic irreducible polynomial \( {m}_{\alpha, F}\left( x\right) \in F\left\lbrack x\right\rbrack \) which has \( \alpha \) as a root. A polynomial \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) has \( \alpha \) as a root if and o... | Proof: Let \( g\left( x\right) \in F\left\lbrack x\right\rbrack \) be a polynomial of minimal degree having \( \alpha \) as a root. Multiplying \( g\left( x\right) \) by a constant, we may assume \( g\left( x\right) \) is monic. Suppose \( g\left( x\right) \) were reducible in \( F\left\lbrack x\right\rbrack \), say \(... | Yes |
Corollary 10. If \( L/F \) is an extension of fields and \( \alpha \) is algebraic over both \( F \) and \( L \) , then \( {m}_{\alpha, L}\left( x\right) \) divides \( {m}_{\alpha, F}\left( x\right) \) in \( L\left\lbrack x\right\rbrack \) . | Proof: This is immediate from the second statement in Proposition 9 applied to \( L \) , since \( {m}_{\alpha, F}\left( x\right) \) is a polynomial in \( L\left\lbrack x\right\rbrack \) having \( \alpha \) as a root. | Yes |
Proposition 11. Let \( \alpha \) be algebraic over the field \( F \) and let \( F\left( \alpha \right) \) be the field generated by \( \alpha \) over \( F \) . Then\n\n\[ F\left( \alpha \right) \cong F\left\lbrack x\right\rbrack /\left( {{m}_{\alpha }\left( x\right) }\right) \]\n\nso that in particular\n\n\[ \left\lbra... | Proof: This follows immediately from Theorem 6. | No |
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