Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Theorem 31.12. (Jordan Decomposition) Let \( f : E \rightarrow E \) be a linear map on the finite-dimensional vector space \( E \) over the field \( K \). If all the eigenvalues \( {\lambda }_{1},\ldots ,{\lambda }_{k} \) of \( f \) belong to \( K \), then there exist a diagonalizable linear map \( D \) and a nilpotent... | Proof. We already proved the existence part. Suppose we also have \( f = {D}^{\prime } + {N}^{\prime } \), with \( {D}^{\prime }{N}^{\prime } = {N}^{\prime }{D}^{\prime } \), where \( {D}^{\prime } \) is diagonalizable, \( {N}^{\prime } \) is nilpotent, and both are polynomials in \( f \). We need to prove that \( D = ... | Yes |
Proposition 31.13. Given a nilpotent linear map \( f \) with \( {f}^{r} \neq 0 \) and \( {f}^{r + 1} = 0 \) as above, the inclusions in the following sequence are strict:\n\n\[ \left( 0\right) = {N}_{0} \subset {N}_{1} \subset \cdots \subset {N}_{r} \subset {N}_{r + 1} = E. \] | Proof. We proceed by contradiction. Assume that \( {N}_{i} = {N}_{i + 1} \) for some \( i \) with \( 0 \leq i \leq r \) . Since \( {f}^{r + 1} = 0 \), for every \( u \in E \), we have\n\n\[ 0 = {f}^{r + 1}\left( u\right) = {f}^{i + 1}\left( {{f}^{r - i}\left( u\right) }\right) \]\n\nwhich shows that \( {f}^{r - i}\left... | Yes |
Proposition 31.14. Given a nilpotent linear map \( f \) with \( {f}^{r} \neq 0 \) and \( {f}^{r + 1} = 0 \), for any integer \( i \) with \( 1 \leq i \leq r \), for any subspace \( U \) of \( E \), if \( U \cap {N}_{i} = \left( 0\right) \), then \( f\left( U\right) \cap {N}_{i - 1} = \left( 0\right) \) , and the restri... | Proof. Pick \( v \in f\left( U\right) \cap {N}_{i - 1} \) . We have \( v = f\left( u\right) \) for some \( u \in U \) and \( {f}^{i - 1}\left( v\right) = 0 \), which means that \( {f}^{i}\left( u\right) = 0 \) . Then \( u \in U \cap {N}_{i} \), so \( u = 0 \) since \( U \cap {N}_{i} = \left( 0\right) \), and \( v = f\l... | Yes |
Proposition 31.15. Given a nilpotent linear map \( f \) with \( {f}^{r} \neq 0 \) and \( {f}^{r + 1} = 0 \), there exists a sequence of subspace \( {U}_{1},\ldots ,{U}_{r + 1} \) of \( E \) with the following properties:\n\n(1) \( {N}_{i} = {N}_{i - 1} \oplus {U}_{i} \), for \( i = 1,\ldots, r + 1 \) .\n\n(2) We have \... | Proof. We proceed inductively, by defining the sequence \( {U}_{r + 1},{U}_{r},\ldots ,{U}_{1} \) . We pick \( {U}_{r + 1} \) to be any supplement of \( {N}_{r} \) in \( {N}_{r + 1} = E \), so that\n\n\[ E = {N}_{r + 1} = {N}_{r} \oplus {U}_{r + 1} \]\n\nSince \( {f}^{r + 1} = 0 \) and \( {N}_{r} = \operatorname{Ker}\l... | Yes |
Theorem 31.16. For any nilpotent linear map \( f : E \rightarrow E \) on a finite-dimensional vector space \( E \) of dimension \( n \) over a field \( K \), there is a basis of \( E \) such that the matrix \( N \) of \( f \) is of the form\n\n\[ N = \left( \begin{matrix} 0 & {\nu }_{1} & 0 & \cdots & 0 & 0 \\ 0 & 0 & ... | Proof. First apply Proposition 31.15 to obtain a direct sum \( E = {\bigoplus }_{i = 1}^{r + 1}{U}_{i} \) . Then we define a basis of \( E \) inductively as follows. First we choose a basis\n\n\[ {e}_{1}^{r + 1},\ldots ,{e}_{{n}_{r + 1}}^{r + 1} \]\n\nof \( {U}_{r + 1} \) . Next, for \( i = r + 1,\ldots ,2 \), given th... | Yes |
Theorem 31.17. (Jordan form) Let \( E \) be a vector space of dimension \( n \) over a field \( K \) and let \( f : E \rightarrow E \) be a linear map. The following properties are equivalent:\n\n(1) The eigenvalues of \( f \) all belong to \( K \) (i.e. the roots of the characteristic polynomial \( {\chi }_{f} \) all ... | Proof. Assume (1). First we apply Theorem 31.11, and we get a direct sum \( E = {\bigoplus }_{j = 1}^{k}{W}_{k} \) , such that the restriction of \( {g}_{i} = f - {\lambda }_{j} \) id to \( {W}_{i} \) is nilpotent. By Theorem 31.16, there is a basis of \( {W}_{i} \) such that the matrix of the restriction of \( {g}_{i}... | Yes |
Proposition 32.1. Let \( A \) be an integral domain. For any \( a \in A \) with \( a \neq 0 \), if the principal ideal (a) is a prime ideal, then a is irreducible. | Proof. If \( \left( a\right) \) is prime, then \( \left( a\right) \neq A \) and \( a \) is not a unit. Assume that \( a = {bc} \) . Then, \( {bc} \in \left( a\right) \) , and since \( \left( a\right) \) is prime, either \( b \in \left( a\right) \) or \( c \in \left( a\right) \) . Consider the case where \( b \in \left(... | Yes |
Proposition 32.2. Let \( A \) be an integral domain satisfying condition (1) in Definition 32.2. Then, condition (2) in Definition 32.2 is equivalent to the following condition:\n\n\( \left( {2}^{\prime }\right) \) If \( a \in A \) is irreducible and a divides the product \( {bc} \), where \( b, c \in A \) and \( b, c ... | Proof. First, assume that (2) holds. Let \( {bc} = {ad} \), where \( d \in A, d \neq 0 \) . If \( b \) is a unit, then\n\n\[ c = {ad}{b}^{-1}, \]\n\nand \( c \) is divisible by \( a \) . A similar argument applies to \( c \) . Thus, we may assume that \( b \) and \( c \) are not units. In view of (1), we can write\n\n\... | Yes |
Proposition 32.3. Let \( A \) be a factorial ring. For any \( a \in A \) with \( a \neq 0 \), the principal ideal (a) is a prime ideal iff \( a \) is irreducible. | Proof. In view of Proposition 32.1, we just have to prove that if \( a \in A \) is irreducible, then the principal ideal \( \left( a\right) \) is a prime ideal. Indeed, if \( {bc} \in \left( a\right) \), then \( a \) divides \( {bc} \), and by Proposition 32.2, property \( \left( {2}^{\prime }\right) \) implies that ei... | Yes |
Proposition 32.4. Let \( A \) be an integral domain. For any \( a \in A, a \neq 0 \), if a divides a nonnull polynomial \( f\left( X\right) \in A\left\lbrack X\right\rbrack \), then a divides every coefficient of \( f\left( X\right) \) . | Proof. Assume that \( f\left( X\right) = {ag}\left( X\right) \), for some \( g\left( X\right) \in A\left\lbrack X\right\rbrack \) . Since \( a \neq 0 \) and \( A \) is an integral ring, \( f\left( X\right) \) and \( g\left( X\right) \) have the same degree \( m \), and since for every \( i\left( {0 \leq i \leq m}\right... | Yes |
Proposition 32.6. Let \( A \) be a UFD. For any \( a \in A, a \neq 0 \), if a divides the product \( f\left( X\right) g\left( X\right) \) of two polynomials \( f\left( X\right), g\left( X\right) \in A\left\lbrack X\right\rbrack \) and \( f\left( X\right) \) is irreducible and of degree at least 1, then a divides \( g\l... | Proof. The Proposition is trivial is \( a \) is a unit. Otherwise, \( a = {a}_{1}\cdots {a}_{m} \) where \( {a}_{i} \in A \) is irreducible. Using induction and applying Lemma 32.5, we conclude that \( a \) divides \( g\left( X\right) \) . | No |
Proposition 32.7. Let \( A \) be an integral domain.\n\n(1) There is a field \( F \) and an injective ring homomorphism \( i : A \rightarrow F \) such that every element of \( F \) is of the form \( i\left( a\right) i{\left( b\right) }^{-1} \), where \( a, b \in A, b \neq 0 \) . | Proof. (1) Consider the binary relation \( \simeq \) on \( A \times \left( {A-\{ 0\} }\right) \) defined as follows:\n\n\[ \left( {a, b}\right) \simeq \left( {{a}^{\prime },{b}^{\prime }}\right) \;\text{ iff }\;a{b}^{\prime } = {a}^{\prime }b. \]\n\nIt is easily seen that \( \simeq \) is an equivalence relation. Note t... | Yes |
Lemma 32.9. Let \( A \) be a UFD. Given any three nonnull polynomials \( f\left( X\right), g\left( X\right), h\left( X\right) \in \) \( A\left\lbrack X\right\rbrack \), if \( f\left( X\right) \) is irreducible and \( f\left( X\right) \) divides the product \( g\left( X\right) h\left( X\right) \), then either \( f\left(... | Proof. If \( f\left( X\right) \) has degree 0, then the result follows from Lemma 32.5. Thus, we may assume that the degree of \( f\left( X\right) \) is \( m \geq 1 \) . Let \( F \) be the fraction field of \( A \) . By Lemma 32.8, \( f\left( X\right) \) is also irreducible in \( F\left\lbrack X\right\rbrack \) . Since... | Yes |
Proposition 32.11. Let \( A \) be a PID.\n\n(1) For any \( a, b, d \in A\left( {a, b, d \neq 0}\right), d \) is a gcd of \( a \) and \( b \) iff\n\n\[ \left( d\right) = \left( {a, b}\right) = \left( a\right) + \left( b\right) \]\n\ni.e., \( d \) generates the principal ideal generated by \( a \) and \( b \) . | Proof. (1) Recall that the ideal generated by \( a \) and \( b \) is the set\n\n\[ \left( a\right) + \left( b\right) = {aA} + {bA} = \{ {ax} + {by} \mid x, y \in A\} .\n\nFirst, assume that \( d \) is a gcd of \( a \) and \( b \) . If so, \( a \in {Ad}, b \in {Ad} \), and thus, \( \left( a\right) \subseteq \left( d\rig... | Yes |
Let \( A \) be a ring that is a UFD, and not a field. Then, \( A \) is a PID iff every nonzero prime ideal is maximal. | Proof. Assume that \( A \) is a PID that is not a field. Consider any nonzero prime ideal, \( \left( p\right) \) , and pick any proper ideal \( \mathfrak{A} \) in \( A \) such that\n\n\[ \left( p\right) \subseteq \mathfrak{A} \]\n\nSince \( A \) is a PID, the ideal \( \mathfrak{A} \) is a principal ideal, so \( \mathfr... | Yes |
Theorem 32.14. Given a commutative ring \( A \) , let \( \mathfrak{a} \) and \( \mathfrak{b} \) be any two ideals of \( A \) such that \( \mathfrak{a} + \mathfrak{b} = A \) . Then, the homomorphism \( \theta : A/\mathfrak{{ab}} \rightarrow A/\mathfrak{a} \times A/\mathfrak{b} \) is an isomorphism. | Proof. We already showed that \( \theta \) is injective, so we need to prove that \( \theta \) is surjective. We need to prove that for any \( y, z \in A \), there is some \( x \in A \) such that\n\n\[ x \equiv y\;\left( {\;\operatorname{mod}\;\mathfrak{a}}\right) \]\n\n\[ x \equiv z\;\left( {\;\operatorname{mod}\;\mat... | Yes |
Proposition 32.17. A ring A satisfies the a.c.c if and only if it satisfies the maximum condition. | Proof. Suppose that \( A \) does not satisfy the a.c.c. Then, there is an infinite strictly ascending sequence of ideals\n\n\[{\mathfrak{A}}_{1} \subset {\mathfrak{A}}_{2} \subset \cdots \subset {\mathfrak{A}}_{i} \subset \cdots ,\]\n\nand the collection \( C = \left\{ {\mathfrak{A}}_{i}\right\} \) has no maximal eleme... | Yes |
Proposition 32.18. A ring A satisfies the a.c.c if and only if every ideal is finitely generated. | Proof. Assume that every ideal is finitely generated. Consider an ascending sequence of ideals\n\n\[ \n{\\mathfrak{A}}_{1} \\subseteq {\\mathfrak{A}}_{2} \\subseteq \\cdots \\subseteq {\\mathfrak{A}}_{i} \\subseteq \\cdots \n\]\n\nObserve that \( \\mathfrak{A} = \\mathop{\\bigcup }\\limits_{i}{\\mathfrak{A}}_{i} \) is ... | Yes |
Lemma 32.19. Let \( A \) be a (commutative) ring. For every ideal \( \mathfrak{A} \) in \( A\left\lbrack X\right\rbrack \), for every \( i \geq 0 \) , let \( {L}_{i}\left( \mathfrak{A}\right) \) denote the set of elements of \( A \) consisting of 0 and of the coefficients of \( {X}^{i} \) in all the polynomials \( f\le... | Proof. That \( {L}_{i}\left( \mathfrak{A}\right) \) is an ideal and that \( {L}_{i}\left( \mathfrak{A}\right) \subseteq {L}_{i + 1}\left( \mathfrak{A}\right) \) follows from the fact that if \( f\left( X\right) \in \) \( \mathfrak{A} \) and \( g\left( X\right) \in \mathfrak{A} \), then \( f\left( X\right) + g\left( X\r... | Yes |
Proposition 33.1. For every nondegenerate pairing \( \langle - , - \rangle : E \times F \rightarrow K \), the induced maps \( \varphi : E \rightarrow {F}^{ * } \) and \( \psi : F \rightarrow {E}^{ * } \) are linear and injective. Furthermore, if \( E \) and \( F \) are finite dimensional, then \( \varphi : E \rightarro... | Proof. The maps \( \varphi : E \rightarrow {F}^{ * } \) and \( \psi : F \rightarrow {E}^{ * } \) are linear because \( u, v \mapsto \langle u, v\rangle \) is bilinear. Assume that \( \varphi \left( u\right) = 0 \) . This means that \( \varphi \left( u\right) \left( y\right) = \langle u, y\rangle = 0 \) for all \( y \in... | Yes |
For any vector space \( E \), given a basis \( \left( {{e}_{1},\ldots ,{e}_{n}}\right) \) for \( E \) and its dual basis \( \left( {{e}_{1}^{ * },\ldots ,{e}_{n}^{ * }}\right) \) for \( {E}^{ * } \), for any inner product \( \langle - , - \rangle \) on \( E \), if \( \left( {g}_{ij}\right) \) is its Gram matrix, with \... | Proof. For every \( u = \mathop{\sum }\limits_{{j = 1}}^{n}{u}^{j}{e}_{j} \), since \( {u}^{b}\left( v\right) = \langle u, v\rangle \) for all \( v \in E \), we have \( {u}^{b}\left( {e}_{i}\right) = \left\langle {u,{e}_{i}}\right\rangle = \left\langle {\mathop{\sum }\limits_{{j = 1}}^{n}{u}^{j}{e}_{j},{e}_{i}}\right\r... | Yes |
Proposition 33.3. If \( \langle - , - \rangle \) is an inner product on a finite vector space \( E \) (over a field, \( K \) ), then for every bilinear form \( f : E \times E \rightarrow K \), there is a unique linear map \( {f}^{\natural } : E \rightarrow E \) such that\n\n\[ f\left( {u, v}\right) = \left\langle {{f}^... | Proof. For every \( g \in \operatorname{Hom}\left( {E, E}\right) \), the map given by\n\n\[ f\left( {u, v}\right) = \langle g\left( u\right), v\rangle ,\;u, v \in E, \]\n\n is clearly bilinear. It is also clear that the above defines a linear map from \( \operatorname{Hom}\left( {E, E}\right) \) to \( \operatorname{Hom... | Yes |
Proposition 33.4. Let \( E \) and \( F \) be two nontrivial vector spaces and let \( {\left( {u}_{i}\right) }_{i \in I} \) be any family of vectors \( {u}_{i} \in E \) . The family \( {\left( {u}_{i}\right) }_{i \in I} \) is linearly independent iff for every family \( {\left( {v}_{i}\right) }_{i \in I} \) of vectors \... | Proof. Left as an exercise. | No |
Proposition 33.8. Given a tensor product \( {E}_{1} \otimes \cdots \otimes {E}_{n} \) ,, there is a canonical isomorphism\n\n\[ \mathrm{L}\left( {{E}_{1},\ldots ,{E}_{n};K}\right) \cong {\left( {E}_{1} \otimes \cdots \otimes {E}_{n}\right) }^{ * } \]\n\nbetween the vector space of multilinear maps \( \mathcal{L}\left( ... | The fact that the map \( \varphi : {E}_{1} \times \cdots \times {E}_{n} \rightarrow {E}_{1} \otimes \cdots \otimes {E}_{n} \) is multilinear, can also be expressed as follows:\n\n\[ {u}_{1} \otimes \cdots \otimes \left( {{v}_{i} + {w}_{i}}\right) \otimes \cdots \otimes {u}_{n} = \left( {{u}_{1} \otimes \cdots \otimes {... | No |
Proposition 33.9. Given two linear maps \( f : E \rightarrow {E}^{\prime } \) and \( g : F \rightarrow {F}^{\prime } \), there is a unique linear map\n\n\[ f \otimes g : E \otimes F \rightarrow {E}^{\prime } \otimes {F}^{\prime } \]\n\nsuch that\n\n\[ \left( {f \otimes g}\right) \left( {u \otimes v}\right) = f\left( u\... | Proof. We can define \( h : E \times F \rightarrow {E}^{\prime } \otimes {F}^{\prime } \) by\n\n\[ h\left( {u, v}\right) = f\left( u\right) \otimes g\left( v\right) \]\n\nIt is immediately verified that \( h \) is bilinear, and thus it induces a unique linear map\n\n\[ f \otimes g : E \otimes F \rightarrow {E}^{\prime ... | Yes |
Proposition 33.10. Suppose we have linear maps \( f : E \rightarrow {E}^{\prime }, g : F \rightarrow {F}^{\prime },{f}^{\prime } : {E}^{\prime } \rightarrow {E}^{\prime \prime } \) and \( {g}^{\prime } : {F}^{\prime } \rightarrow {F}^{\prime \prime } \) . Then the following identity holds:\n\n\[ \left( {{f}^{\prime } \... | Proof. We have the commutative diagram \n\nand thus the commutative diagram. \n\nWe also have the commutativ... | Yes |
Proposition 33.11. If \( f : E \rightarrow {E}^{\prime } \) and \( g : F \rightarrow {F}^{\prime } \) are isomorphims, then \( f \otimes g : E \otimes F \rightarrow {E}^{\prime } \otimes {F}^{\prime } \) is also an isomorphism. | Proof. If \( {f}^{-1} : {E}^{\prime } \rightarrow E \) is the inverse of \( f : E \rightarrow {E}^{\prime } \) and \( {g}^{-1} : {F}^{\prime } \rightarrow F \) is the inverse of \( g : F \rightarrow {F}^{\prime } \), then \( {f}^{-1} \otimes {g}^{-1} : {E}^{\prime } \otimes {F}^{\prime } \rightarrow E \otimes F \) is t... | Yes |
Proposition 33.12. Given \( n \geq 2 \) vector spaces \( {E}_{1},\ldots ,{E}_{n} \), if \( {\left( {u}_{i}^{k}\right) }_{i \in {I}_{k}} \) is a basis for \( {E}_{k} \) , \( 1 \leq k \leq n \), then the family of vectors\n\n\[ \n{\left( {u}_{{i}_{1}}^{1} \otimes \cdots \otimes {u}_{{i}_{n}}^{n}\right) }_{\left( {{i}_{1}... | Proof. For each \( k,1 \leq k \leq n \), every \( {v}^{k} \in {E}_{k} \) can be written uniquely as\n\n\[ \n{v}^{k} = \mathop{\sum }\limits_{{j \in {I}_{k}}}{v}_{j}^{k}{u}_{j}^{k} \n\]\n\nfor some family of scalars \( {\left( {v}_{j}^{k}\right) }_{j \in {I}_{k}} \) . Let \( F \) be any nontrivial vector space. We show ... | Yes |
Proposition 33.14. Given any three vector spaces \( E, F, G \), we have the canonical isomorphism\n\n\[ \operatorname{Hom}\left( {E, F;G}\right) \cong \operatorname{Hom}\left( {E,\operatorname{Hom}\left( {F, G}\right) }\right) . \] | Proof. Any bilinear map \( f : E \times F \rightarrow G \) gives the linear map \( \varphi \left( f\right) \in \operatorname{Hom}\left( {E,\operatorname{Hom}\left( {F, G}\right) }\right) \) , where \( \varphi \left( f\right) \left( u\right) \) is the linear map in \( \operatorname{Hom}\left( {F, G}\right) \) given by\n... | Yes |
Proposition 33.16. We have canonical isomorphisms\n\n\[ \n{\left( {E}_{1} \otimes \cdots \otimes {E}_{n}\right) }^{ * } \cong {E}_{1}^{ * } \otimes \cdots \otimes {E}_{n}^{ * } \]\n\nand\n\n\[ \n\mu : {E}_{1}^{ * } \otimes \cdots \otimes {E}_{n}^{ * } \cong \operatorname{Hom}\left( {{E}_{1},\ldots ,{E}_{n};K}\right) .\... | Proof. The second isomorphism follows from the isomorphism \( {\left( {E}_{1} \otimes \cdots \otimes {E}_{n}\right) }^{ * } \cong {E}_{1}^{ * } \otimes \cdots \otimes {E}_{n}^{ * } \) together with the isomorphism \( \operatorname{Hom}\left( {{E}_{1},\ldots ,{E}_{n};K}\right) \cong {\left( {E}_{1} \otimes \cdots \otime... | Yes |
Proposition 33.17. If \( E \) and \( F \) are vector spaces (not necessarily finite dimensional), then the following properties hold:\n\n(1) The linear map \( {\alpha }_{ \otimes } : {E}^{ * } \otimes F \rightarrow \operatorname{Hom}\left( {E, F}\right) \) is injective. | Proof. (1) Let \( {\left( {e}_{i}^{ * }\right) }_{i \in I} \) be a basis of \( {E}^{ * } \) and let \( {\left( {f}_{j}\right) }_{j \in J} \) be a basis of \( F \) . Then we know that \( {\left( {e}_{i}^{ * } \otimes {f}_{j}\right) }_{i \in I, j \in J} \) is a basis of \( {E}^{ * } \otimes F \) . To prove that \( {\alph... | Yes |
Proposition 33.18. If the \( {E}_{1},\ldots {E}_{n} \) are finite-dimensional vector spaces and \( F \) is any vector space, then we have the canonical isomorphism\n\n\[ \n\operatorname{Hom}\left( {{E}_{1},\ldots ,{E}_{n};F}\right) \cong {E}_{1}^{ * } \otimes \cdots \otimes {E}_{n}^{ * } \otimes F.\n\] | Proof. In view of the canonical isomorphism\n\n\[ \n\operatorname{Hom}\left( {{E}_{1},\ldots ,{E}_{n};F}\right) \cong \operatorname{Hom}\left( {{E}_{1} \otimes \cdots \otimes {E}_{n}, F}\right)\n\]\n\ngiven by Proposition 33.7 and the canonical isomorphism \( {\left( {E}_{1} \otimes \cdots \otimes {E}_{n}\right) }^{ * ... | Yes |
Proposition 33.19. Given any \( K \) -algebra \( A \), for any linear map \( f : V \rightarrow A \), there is a unique \( K \) -algebra homomorphism \( \bar{f} : T\left( V\right) \rightarrow A \) so that\n\n\[ f = \bar{f} \circ i \] | Proof. Left an an exercise (use Theorem 33.6). A proof can be found in Knapp [103] (Appendix A, Proposition A.14) or Bertin [15] (Chapter 4, Theorem 2.4). | No |
Proposition 33.23. Given any two symmetric \( n \) -th tensor powers \( \left( {{S}_{1},{\varphi }_{1}}\right) \) and \( \left( {{S}_{2},{\varphi }_{2}}\right) \) for \( E \), there is an isomorphism \( h : {S}_{1} \rightarrow {S}_{2} \) such that\n\n\[{\varphi }_{2} = h \circ {\varphi }_{1}\] | Proof. Replace tensor product by \( n \) -th symmetric tensor power in the proof of Proposition 33.5. | No |
Proposition 33.25. There is a canonical isomorphism\n\n\[ \operatorname{Hom}\left( {{\mathrm{S}}^{n}\left( E\right), F}\right) \cong {\operatorname{Sym}}^{n}\left( {E;F}\right) \]\n\nbetween the vector space of linear maps \( \operatorname{Hom}\left( {{\mathrm{S}}^{n}\left( E\right), F}\right) \) and the vector space o... | Proof. The map \( h \circ \varphi \) is clearly symmetric multilinear. By Theorem 33.24, for every symmetric multilinear map \( f \in {\operatorname{Sym}}^{n}\left( {E;F}\right) \) there is a unique linear map \( {f}_{ \odot } \in \operatorname{Hom}\left( {{\mathrm{S}}^{n}\left( E\right), F}\right) \) such that \( f = ... | Yes |
Proposition 33.30. Assume the field \( K \) has characteristic zero. We have the canonical isomorphisms\n\n\[{\left( {\mathrm{S}}^{n}\left( E\right) \right) }^{ * } \cong {\mathrm{S}}^{n}\left( {E}^{ * }\right)\]\n\nand\n\n\[{\mathrm{S}}^{n}\left( {E}^{ * }\right) \cong {\operatorname{Sym}}^{n}\left( {E;K}\right) = {\o... | Proof. The isomorphism\n\n\[ \mu : {\mathrm{S}}^{n}\left( {E}^{ * }\right) \cong {\operatorname{Sym}}^{n}\left( {E;K}\right) \]\n\nfollows from the isomorphisms \( {\left( {\mathrm{S}}^{n}\left( E\right) \right) }^{ * } \cong {\mathrm{S}}^{n}\left( {E}^{ * }\right) \) and \( {\left( {\mathrm{S}}^{n}\left( E\right) \rig... | Yes |
Proposition 33.32. For any two vector spaces \( E \) and \( F \), there is a canonical isomorphism (of \( K \) -algebras) | \[ \mathrm{S}\left( {E \oplus F}\right) \cong \mathrm{S}\left( E\right) \otimes \mathrm{S}\left( F\right) \] | Yes |
Problem 33.5. Induct on \( m \geq 2 \) to prove the canonical isomorphism\n\n\[ \n{V}^{\otimes m} \otimes {V}^{\otimes n} \cong {V}^{\otimes \left( {m + n}\right) }.\n\] | Use this isomorphism to show that \( \cdot : {V}^{\otimes m} \times {V}^{\otimes n} \rightarrow {V}^{\otimes \left( {m + n}\right) } \) defined as\n\n\[ \n\left( {{v}_{1} \otimes \cdots \otimes {v}_{m}}\right) \cdot \left( {{w}_{1} \otimes \cdots \otimes {w}_{n}}\right) = {v}_{1} \otimes \cdots \otimes {v}_{m} \otimes ... | No |
Problem 33.6. Prove Proposition 33.19. | Hint. See Knapp [103] (Appendix A, Proposition A.14) or Bertin [15] (Chapter 4, Theorem 2.4). | No |
Proposition 34.1. Let \( f : {E}^{n} \rightarrow F \) be a multilinear map. If \( f \) is alternating, then the following properties hold:\n\n(1) For all \( i \), with \( 1 \leq i \leq n - 1 \),\n\n\[ f\left( {\ldots ,{u}_{i},{u}_{i + 1},\ldots }\right) = - f\left( {\ldots ,{u}_{i + 1},{u}_{i},\ldots }\right) . \]\n\n(... | Proof. (1) By multilinearity applied twice, we have\n\n\[ f\left( {\ldots ,{u}_{i} + {u}_{i + 1},{u}_{i} + {u}_{i + 1},\ldots }\right) = f\left( {\ldots ,{u}_{i},{u}_{i},\ldots }\right) + f\left( {\ldots ,{u}_{i},{u}_{i + 1},\ldots }\right) \]\n\n\[ + f\left( {\ldots ,{u}_{i + 1},{u}_{i},\ldots }\right) + f\left( {\ldo... | Yes |
Proposition 34.2. Let \( f : {E}^{n} \rightarrow F \) be an alternating multilinear map. For any families of vectors, \( \left( {{u}_{1},\ldots ,{u}_{n}}\right) \) and \( \left( {{v}_{1},\ldots ,{v}_{n}}\right) \), with \( {u}_{i},{v}_{i} \in E \), if\n\n\[ \n{v}_{j} = \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{ij}{u}_{i}... | Proof. Use Property (ii) of Proposition 34.1. | No |
Proposition 34.3. Given any two n-th exterior tensor powers \( \left( {{A}_{1},{\varphi }_{1}}\right) \) and \( \left( {{A}_{2},{\varphi }_{2}}\right) \) for \( E \) , there is an isomorphism \( h : {A}_{1} \rightarrow {A}_{2} \) such that | Proof. Replace tensor product by \( n \) -th exterior tensor power in the proof of Proposition 33.5. | No |
Theorem 34.4. Given a vector space \( E \), an \( n \) -th exterior tensor power \( \left( {\mathop{\bigwedge }\limits^{n}\left( E\right) ,\varphi }\right) \) for \( E \) can be constructed \( \left( {n \geq 1}\right) \) . Furthermore, denoting \( \varphi \left( {{u}_{1},\ldots ,{u}_{n}}\right) \) as \( {u}_{1} \land \... | Proof sketch. We can give a quick proof using the tensor algebra \( T\left( E\right) \) . Let \( {\mathfrak{I}}_{a} \) be the two-sided ideal of \( T\left( E\right) \) generated by all tensors of the form \( u \otimes u \in {E}^{\otimes 2} \) . Then let\n\n\[ \mathop{\bigwedge }\limits^{n}\left( E\right) = {E}^{\otimes... | Yes |
Given any vector space \( E \), if \( E \) has finite dimension \( d = \dim \left( E\right) \), then for all \( n > d \), the exterior power \( \mathop{\bigwedge }\limits^{n}\left( E\right) \) is trivial; that is \( \mathop{\bigwedge }\limits^{n}\left( E\right) = \left( 0\right) \). | First assume that \( E \) has finite dimension \( d = \dim \left( E\right) \) and that \( n > d \) . We know that \( \mathop{\bigwedge }\limits^{n}\left( E\right) \) is generated by the tensors of the form \( {v}_{1} \land \cdots \land {v}_{n} \), with \( {v}_{i} \in E \) . If \( {u}_{1},\ldots ,{u}_{d} \) is a basis o... | Yes |
Proposition 34.8. For any vector space \( E \), the vectors \( {u}_{1},\ldots ,{u}_{n} \in E \) are linearly independent iff \( {u}_{1} \land \cdots \land {u}_{n} \neq 0 \) . | Proof. If \( {u}_{1} \land \cdots \land {u}_{n} \neq 0 \), then \( {u}_{1},\ldots ,{u}_{n} \) must be linearly independent. Otherwise, some \( {u}_{i} \) would be a linear combination of the other \( {u}_{j} \) ’s (with \( j \neq i \) ), and then, as in the proof of Proposition 34.7, \( {u}_{1} \land \cdots \land {u}_{... | Yes |
Proposition 34.9. We have the following isomorphism: | \[ \mathop{\bigwedge }\limits^{n}\left( {E \oplus F}\right) \cong {\bigoplus }_{k = 0}^{n}\mathop{\bigwedge }\limits^{k}\left( E\right) \otimes \mathop{\bigwedge }\limits^{{n - k}}\left( F\right) \] | Yes |
Proposition 34.10. There is a canonical isomorphism\n\n\\[ \n{\\left( \\mathop{\\bigwedge }\\limits^{n}\\left( E\\right) \\right) }^{ * } \\cong \\mathop{\\bigwedge }\\limits^{n}\\left( {E}^{ * }\\right) \n\\] | The second isomorphism follows from the canonical isomorphism \\( {\\left( \\mathop{\\bigwedge }\\limits^{n}\\left( E\\right) \\right) }^{ * } \\cong \\mathop{\\bigwedge }\\limits^{n}\\left( {E}^{ * }\\right) \\) and the canonical isomorphism \\( {\\left( \\mathop{\\bigwedge }\\limits^{n}\\left( E\\right) \\right) }^{ ... | No |
Proposition 34.11. If \( f : E \rightarrow F \) is any linear map between two finite-dimensional vector spaces \( E \) and \( F \), then\n\n\[ \mu \left( {\left( {\mathop{\bigwedge }\limits^{p}{f}^{\top }}\right) \left( \omega \right) }\right) \left( {{u}_{1},\ldots ,{u}_{p}}\right) = \mu \left( \omega \right) \left( {... | Proof. It is enough to prove the formula on generators. By definition of \( \mu \), we have\n\n\[ \mu \left( {\left( {\mathop{\bigwedge }\limits^{p}{f}^{\top }}\right) \left( {{v}_{1}^{ * } \land \cdots \land {v}_{p}^{ * }}\right) }\right) \left( {{u}_{1},\ldots ,{u}_{p}}\right) = \mu \left( {{f}^{\top }\left( {v}_{1}^... | Yes |
For all \( \alpha \in \mathop{\bigwedge }\limits^{m}\left( V\right) \) and all \( \beta \in \mathop{\bigwedge }\limits^{n}\left( V\right) \), we have\n\n\[ \beta \land \alpha = {\left( -1\right) }^{mn}\alpha \land \beta . \] | Proof. Since \( v \land u = - u \land v \) for all \( u, v \in V \), Proposition 34.12 follows by induction. | No |
Proposition 34.15. Let \( V \) be any oriented Euclidean vector space whose orientation is given by some chosen orthonormal basis \( \left( {{e}_{1},\ldots ,{e}_{n}}\right) \) . For any alternating tensor \( \alpha \in \mathop{\bigwedge }\limits^{k}V \), there is a unique alternating tensor \( * \alpha \in \mathop{\big... | Proof. Since \( \mathop{\bigwedge }\limits^{n}V \) has dimension 1, the alternating tensor \( {e}_{1} \land \cdots \land {e}_{n} \) is a basis of \( \mathop{\bigwedge }\limits^{n}V \) . It follows that for any fixed \( \alpha \in \mathop{\bigwedge }\limits^{k}V \), the linear map \( {\lambda }_{\alpha } \) from \( \mat... | Yes |
Proposition 34.16. If \( V \) is any oriented vector space of dimension \( n \), for every \( k \) with \( 0 \leq k \leq n \), we have\n\n(i) \( * * = {\left( -\mathrm{{id}}\right) }^{k\left( {n - k}\right) } \).\n\n(ii) \( \langle x, y{\rangle }_{ \land } = * \left( {x \land * y}\right) = * \left( {y \land * x}\right)... | Proof. (1) Let \( {\left( {e}_{i}\right) }_{i = 1}^{n} \) is an orthonormal basis of \( V \) . It is enough to check the identity on basis elements. We have\n\n\[ * \left( {{e}_{{i}_{1}} \land \cdots \land {e}_{{i}_{k}}}\right) = \operatorname{sign}\left( {{i}_{1},\ldots {i}_{k},{j}_{1},\ldots ,{j}_{n - k}}\right) {e}_... | Yes |
Proposition 34.17. If \( V \) is any finite-dimensional oriented vector space, for any basis \( \left( {{v}_{!},\ldots ,{v}_{n}}\right) \) of \( V \), we have\n\n\[ \n* \left( 1\right) = \frac{1}{\sqrt{\det \left( \left\langle {{v}_{i},{v}_{j}}\right\rangle \right) }}{v}_{1} \land \cdots \land {v}_{n}.\n\] | Proof. If \( \left( {{e}_{1},\ldots ,{e}_{n}}\right) \) is an orthonormal basis of \( V \) and \( \left( {{v}_{1},\ldots ,{v}_{n}}\right) \) is any other basis of \( V \) ,\nthen\n\n\[ \n{\left\langle {v}_{1} \land \cdots \land {v}_{n},{v}_{1} \land \cdots \land {v}_{n}\right\rangle }_{ \land } = \det \left( \left\lang... | Yes |
Proposition 34.18. For any basis \( \left( {{e}_{1},\ldots ,{e}_{n}}\right) \) of \( E \) the following properties hold:\n\n(1) If \( H \cap L = \varnothing ,\left| H\right| = p \), and \( \left| L\right| = q \), then\n\n\[ \n{\rho }_{H, L}{\rho }_{L, H} = {\left( -1\right) }^{\nu }{\left( -1\right) }^{{pq} - \nu } = {... | Proof. These are proved in Bourbaki [25] (Chapter III, §11, Section 11), but the proofs of (3) and (4) are very concise. We elaborate on the proofs of (2) and (4), the proof of (3) being similar.\n\nIn (2) if \( H \cap L \neq \varnothing \), then \( {e}_{H} \land {e}_{L} \) contains some vector twice and so \( {e}_{H} ... | Yes |
For the left hook\n\n\[ \lrcorner : E \times \mathop{\bigwedge }\limits^{{q + 1}}{E}^{ * } \rightarrow \mathop{\bigwedge }\limits^{q}{E}^{ * } \]\n\nfor every \( u \in E,{x}^{ * } \in \mathop{\bigwedge }\limits^{{q + 1 - s}}{E}^{ * } \), and \( {y}^{ * } \in \mathop{\bigwedge }\limits^{s}{E}^{ * } \), we have\n\n\[ u\l... | Proof. We can prove the above identity assuming that \( {x}^{ * } \) and \( {y}^{ * } \) are of the form \( {e}_{I}^{ * } \) and \( {e}_{J}^{ * } \) using Proposition 34.18 and leave the details as an exercise for the reader.\n\nThus, \( \lrcorner : E \times \mathop{\bigwedge }\limits^{{q + 1}}{E}^{ * } \rightarrow \ma... | No |
Proposition 34.20. The following identities hold:\n\n\[ \n{z}^{ * }\llcorner u = {\left( -1\right) }^{pq}u\lrcorner {z}^{ * }\;\text{for all}u \in \mathop{\bigwedge }\limits^{p}E\text{and all}{z}^{ * } \in \mathop{\bigwedge }\limits^{{p + q}}{E}^{ * }\n\]\n\n\[ \nz\llcorner {u}^{ * } = {\left( -1\right) }^{pq}{u}^{ * }... | Therefore the left and right hooks are not independent, and in fact each one determines the other. As a consequence, we can restrict our attention to only one of the hooks, for example the left hook, but there are a few situations where it is nice to use both, for example in Proposition 34.23. | No |
Proposition 34.22. For the right hook\n\n\\[ \n\\mu : \\mathop{\\bigwedge }\\limits^{{q + 1}}{E}^{ * } \\times E \\rightarrow \\mathop{\\bigwedge }\\limits^{q}{E}^{ * }\n\\]\n\nfor every \\( u \\in E,{x}^{ * } \\in \\mathop{\\bigwedge }\\limits^{r}{E}^{ * } \\), and \\( {y}^{ * } \\in \\mathop{\\bigwedge }\\limits^{{q ... | Proof. A proof involving determinants can be found in Warner [184], Chapter 2. | No |
Proposition 34.23. The linear maps \( \gamma : \mathop{\bigwedge }\limits^{p}E \rightarrow \mathop{\bigwedge }\limits^{{n - p}}{E}^{ * } \) and \( \delta : \mathop{\bigwedge }\limits^{p}{E}^{ * } \rightarrow \mathop{\bigwedge }\limits^{{n - p}}E \) are isomorphims, and \( {\gamma }^{-1} = \delta \) . The isomorphisms \... | Proof. Using Propositions 34.18 and 34.21, for any subset \( J \subseteq \{ 1,\ldots, n\} = M \) such that \( \left| J\right| = p \), we have\n\n\[ \gamma \left( {e}_{J}\right) = {e}_{J}\lrcorner {e}^{ * } = {\rho }_{M - J, J}{e}_{M - J}^{ * }\;\text{ and }\;\delta \left( {e}_{M - J}^{ * }\right) = {e}_{▱}{e}_{M - J}^{... | Yes |
Any nonzero \( z \in \mathop{\bigwedge }\limits^{p}E \) is decomposable iff the smallest subspace \( W \) of \( E \) such that \( z \in \mathop{\bigwedge }\limits^{p}W \) has dimension \( p \) . Furthermore, if \( z = {u}_{1} \land \cdots \land {u}_{p} \) is decomposable, then \( \left( {{u}_{1},\ldots ,{u}_{p}}\right)... | Proof. If \( \dim \left( W\right) = p \), then for any basis \( \left( {{e}_{1},\ldots ,{e}_{p}}\right) \) of \( W \) we know that \( \mathop{\bigwedge }\limits^{p}W \) has \( {e}_{1} \land \cdots \land {e}_{p} \) has a basis, and thus has dimension 1. Since \( z \in \mathop{\bigwedge }\limits^{p}W \), we have \( z = \... | Yes |
Proposition 34.27. Any nonzero \( z \in \mathop{\bigwedge }\limits^{p}E \) is decomposable iff\n\n\[ \left( {{u}^{ * }\lrcorner z}\right) \land z = 0,\;\text{ for all }{u}^{ * } \in \mathop{\bigwedge }\limits^{{p - 1}}{E}^{ * }.\] | Proof. First assume that \( z \in \mathop{\bigwedge }\limits^{p}E \) is decomposable. If so, by Corollary 34.26, the smallest subspace \( W \) of \( E \) such that \( z \in \mathop{\bigwedge }\limits^{p}W \) has dimension \( p \), so we have \( z = {e}_{1} \land \cdots \land {e}_{p} \) where \( {e}_{1},\ldots ,{e}_{p} ... | Yes |
Proposition 34.28. Given any vector space \( E \) of dimension \( n \), a vector \( x \in \mathop{\bigwedge }\limits^{2}E \) is decomposable iff \( x \land x = 0 \) . | Proof. Recall that as an application of Proposition 34.19 we proved the formula \( \left( \dagger \right) \), namely\n\n\[ \n{u}^{ * }\lrcorner \left( {x \land x}\right) = 2\left( {\left( {{u}^{ * }\lrcorner x}\right) \land x}\right) \n\]\n\nfor all \( x \in \mathop{\bigwedge }\limits^{2}E \) and all \( {u}^{ * } \in {... | Yes |
Proposition 34.30. The map \( {i}_{G} : G\left( {p, n}\right) \rightarrow {\mathbb{{RP}}}^{\left( \begin{array}{l} n \\ p \end{array}\right) - 1} \) is injective. | Proof. Let \( U \) and \( V \) be any two \( p \) -planes and assume that \( {i}_{G}\left( U\right) = {i}_{G}\left( V\right) \) . This means that there is a basis \( \left( {{u}_{1},\ldots ,{u}_{p}}\right) \) of \( U \) and a basis \( \left( {{v}_{1},\ldots ,{v}_{p}}\right) \) of \( V \) such that\n\n\[ \n{v}_{1} \land... | Yes |
Proposition 34.31. Let \( E \) be any \( n \) -dimensional vector space over a field \( K \), and let \( U \) and \( V \) be the smallest subspaces of \( E \) associated with two nonzero decomposable vectors \( u = {u}_{1} \land \cdots \land {u}_{p} \in \mathop{\bigwedge }\limits^{p}U \) and \( v = {v}_{1} \land \cdots... | Proof. Assume \( U \cap V = \left( 0\right) \) . We know by Corollary 34.26 that \( \left( {{u}_{1},\ldots ,{u}_{p}}\right) \) is a basis of \( U \) and \( \left( {{v}_{1},\ldots ,{v}_{q}}\right) \) is a basis of \( V \) . Since \( U \cap V = \left( 0\right) ,\left( {{u}_{1},\ldots ,{u}_{p},{v}_{1},\ldots ,{v}_{q}}\rig... | Yes |
Proposition 34.33. The map \[ {\mu }_{F} : \left( {\mathop{\bigwedge }\limits^{n}\left( {E}^{ * }\right) }\right) \otimes F \rightarrow {\operatorname{Alt}}^{n}\left( {E;F}\right) \] defined as above is a canonical isomorphism for every \( n \geq 0 \) . Furthermore, given any three vector spaces, \( F, G, H \), and any... | Proof. Since we already know that \( \left( {\mathop{\bigwedge }\limits^{n}\left( {E}^{ * }\right) }\right) \otimes F \) and \( {\operatorname{Alt}}^{n}\left( {E;F}\right) \) are isomorphic, it is enough to show that \( {\mu }_{F} \) maps some basis of \( \left( {\mathop{\bigwedge }\limits^{n}\left( {E}^{ * }\right) }\... | Yes |
Proposition 34.34. If \( \left( {{e}_{1},\ldots ,{e}_{p}}\right) \) is any basis of \( E \), then every element \( \omega \in \left( {\mathop{\bigwedge }\limits^{n}\left( {E}^{ * }\right) }\right) \otimes F \) can be written in a unique way as\n\n\[ \omega = \mathop{\sum }\limits_{I}{e}_{I}^{ * } \otimes {f}_{I},\;{f}_... | Proof. Since, by Proposition 34.7, the \( {e}_{I}^{ * } \) form a basis of \( \mathop{\bigwedge }\limits^{n}\left( {E}^{ * }\right) \), elements of the form \( {e}_{I}^{ * } \otimes f \) span \( \left( {\mathop{\bigwedge }\limits^{n}\left( {E}^{ * }\right) }\right) \otimes F \) . Now if we apply \( {\mu }_{F}\left( \om... | Yes |
Theorem 35.1. For any free module \( M \), any two bases of \( M \) have the same cardinality. | Proof sketch. We give the argument for finite bases, but it also holds for infinite bases. The trick is to pick any maximal ideal \( \mathfrak{m} \) in \( A \) (whose existence is guaranteed by Theorem B.3). Then, \( A/\mathfrak{m} \) is a field, and \( M/\mathfrak{m}M \) can be made into a vector space over \( A/\math... | No |
Proposition 35.2. Let \( f : E \rightarrow F \) be a surjective linear map between two \( A \) -modules with \( F \) a free module. Given any basis \( \left( {{v}_{1},\ldots ,{v}_{r}}\right) \) of \( F \), for any \( r \) vectors \( {u}_{1},\ldots ,{u}_{r} \in E \) such that \( f\left( {u}_{i}\right) = {v}_{i} \) for \... | Proof. Pick any \( w \in E \), write \( f\left( w\right) \) over the basis \( \left( {{v}_{1},\ldots ,{v}_{r}}\right) \) as \( f\left( w\right) = {a}_{1}{v}_{1} + \cdots + {a}_{r}{v}_{r} \) , and let \( u = {a}_{1}{u}_{1} + \cdots + {a}_{r}{u}_{r} \) . Observe that\n\n\[ f\left( {w - u}\right) = f\left( w\right) - f\le... | Yes |
Proposition 35.3. If \( A \) is an integral domain, then for any \( A \)-module \( M \), the set \( {M}_{\text{tor }} \) of torsion elements in \( M \) is a submodule of \( M \). | Proof. If \( x, y \in M \) are torsion elements \( \left( {x, y \neq 0}\right) \), then there exist some nonzero elements \( a, b \in A \) such that \( {ax} = 0 \) and \( {by} = 0 \). Since \( A \) is an integral domain, \( {ab} \neq 0 \), and then for all \( \lambda ,\mu \in A \), we have\n\n\[ {ab}\left( {{\lambda x}... | Yes |
Proposition 35.4. If \( A \) is an integral domain, then for any \( A \) -module \( M \), the quotient module \( M/{M}_{\text{tor }} \) is torsion free. | Proof. Let \( \bar{x} \) be an element of \( M/{M}_{\text{tor }} \) and assume that \( a\bar{x} = 0 \) for some \( a \neq 0 \) in \( A \) . This means that \( {ax} \in {M}_{\text{tor }} \), so there is some \( b \neq 0 \) in \( A \) such that \( {bax} = 0 \) . Since \( a, b \neq 0 \) and \( A \) is an integral domain, ... | Yes |
Proposition 35.5. If \( A \) is a PID and if \( F \) is a free \( A \) -module of dimension \( n \), then every submodule \( M \) of \( F \) is a free module of dimension at most \( n \) . | Proof. Let \( \left( {{u}_{1},\ldots ,{u}_{n}}\right) \) be a basis of \( F \), and let \( {M}_{r} = M \cap \left( {A{u}_{1} \oplus \cdots \oplus A{u}_{r}}\right) \), the intersection of \( M \) with the free module generated by \( \left( {{u}_{1},\ldots ,{u}_{r}}\right) \), for \( r = 1,\ldots, n \) . We prove by indu... | Yes |
Proposition 35.6. If \( A \) is a PID and if \( M \) is a finitely generated module which is torsion-free, then \( M \) is free. | Proof. Let \( \left( {{y}_{1},\ldots ,{y}_{n}}\right) \) be some generators for \( M \), and let \( \left( {{u}_{1},\ldots ,{u}_{m}}\right) \) be a maximal subsequence of \( \left( {{y}_{1},\ldots ,{y}_{n}}\right) \) which is linearly independent. If \( m = n \), we are done. Otherwise, due to the maximality of \( m \)... | Yes |
Theorem 35.7. Let \( M \) be a finitely generated module over a PID. Then \( M/{M}_{\text{tor }} \) is free, and there exit a free submodule \( F \) of \( M \) such that \( M \) is the direct sum\n\n\[ M = {M}_{\text{tor }} \oplus F \]\n\nThe dimension of \( F \) is uniquely determined. | Proof. By Proposition 35.4 \( M/{M}_{\text{tor }} \) is torsion-free, and since \( M \) is finitely generated, it is also finitely generated. By Proposition 35.6, \( M/{M}_{\text{tor }} \) is free. We have the quotient linear map \( \pi : M \rightarrow M/{M}_{\text{tor }} \), which is surjective, and \( M/{M}_{\text{to... | Yes |
Proposition 35.8. If \( R \) is an \( m \times n \) matrix presenting an \( A \) -module \( M \), then the matrices \( S \) of the form listed below present the same module (a module isomorphic to \( M \) ):\n\n(1) \( S = {QR}{P}^{-1} \), where \( Q \) is a \( m \times m \) invertible matrix and \( P \) a \( n \times n... | Proof. (1) By definition, we have an isomorphism \( M \approx {A}^{m}/R{A}^{n} \), where we denote by \( R{A}^{n} \) the image of \( {A}^{n} \) by the linear map defined by \( R \) . Going from \( R \) to \( {QR}{P}^{-1} \) corresponds to making a change of basis in \( {A}^{m} \) and a change of basis in \( {A}^{n} \),... | Yes |
Proposition 35.9. Let \( f : E \rightarrow F \) be a linear map between two \( A \) -modules \( E \) and \( F \). (1) Given any set of generators \( \left( {{v}_{1},\ldots ,{v}_{r}}\right) \) of \( \operatorname{Im}\left( f\right) \), for any \( r \) vectors \( {u}_{1},\ldots ,{u}_{r} \in E \) such that \( f\left( {u}_... | Proof. (1) Pick any \( w \in E \), write \( f\left( w\right) \) over the generators \( \left( {{v}_{1},\ldots ,{v}_{r}}\right) \) of \( \operatorname{Im}\left( f\right) \) as \( f\left( w\right) = \) \( {a}_{1}{v}_{1} + \cdots + {a}_{r}{v}_{r} \), and let \( u = {a}_{1}{u}_{1} + \cdots + {a}_{r}{u}_{r} \) . Observe tha... | Yes |
A ring A is Noetherian iff every submodule \( N \) of a finitely generated A-module \( M \) is itself finitely generated. | First, assume that every submodule \( N \) of a finitely generated \( A \) -module \( M \) is itself finitely generated. The ring \( A \) is a module over itself and it is generated by the single element 1. Furthermore, every submodule of \( A \) is an ideal, so the hypothesis implies that every ideal in \( A \) is fin... | Yes |
Proposition 35.11. For any finitely-generated projective A-modules, P, and any A-module, \( Q \), we have the isomorphisms:\n\n\[ \n{P}^{* * } \cong P \]\n\n\[ \n{\operatorname{Hom}}_{A}\left( {P, Q}\right) \cong {P}^{ * }{ \otimes }_{A}Q. \]\n | Proof sketch. We only consider the second isomorphism. Since \( P \) is projective, we have some \( A \) -modules, \( {P}_{1}, F \), with\n\n\[ \nP \oplus {P}_{1} = F \]\n\nwhere \( F \) is some free module. Now, we know that for any \( A \) -modules, \( U, V, W \), we have\n\n\[ \n{\operatorname{Hom}}_{A}\left( {U \op... | Yes |
Proposition 35.13. Let \( M \) and \( N \) be two A-module with \( N \) a free module, and pick any basis \( \left( {{v}_{1},\ldots ,{v}_{n}}\right) \) for \( N \) . Then, every element of \( M \otimes N \) can expressed in a unique way as a sum of the form\n\n\[{u}_{1} \otimes {v}_{1} + \cdots + {u}_{n} \otimes {v}_{n... | Proof. Since \( N \) is free with basis \( \left( {{v}_{1},\ldots ,{v}_{n}}\right) \), we have an isomorphism\n\n\[N \approx A{v}_{1} \oplus \cdots \oplus A{v}_{n}\]\n\nBy Proposition 35.12, we obtain an isomorphism\n\n\[M \otimes N \approx M \otimes \left( {A{v}_{1} \oplus \cdots \oplus A{v}_{n}}\right) \approx \left(... | Yes |
Proposition 35.14. Given any A-module \( M \) and any ideal \( \mathfrak{a} \) in \( A \), there is an isomorphism\n\n\[ \left( {A/\mathfrak{a}}\right) { \otimes }_{A}M \approx M/\mathfrak{a}M \]\n\ngiven by the map \( \left( {\bar{a} \otimes u}\right) \mapsto {au}\left( {{\;\operatorname{mod}\;\mathfrak{a}}M}\right) \... | Sketch of proof. Consider the map \( \varphi : \left( {A/\mathfrak{a}}\right) \times M \rightarrow M/\mathfrak{a}M \) given by\n\n\[ \varphi \left( {\bar{a}, u}\right) = {au}\;\left( {{\;\operatorname{mod}\;\mathfrak{a}}M}\right) \]\n\nfor all \( \bar{a} \in A/\mathfrak{a} \) and all \( u \in M \) . It is immediately c... | Yes |
Proposition 35.15. Given a ring \( A \approx A/{\mathfrak{b}}_{1} \times \cdots \times A/{\mathfrak{b}}_{n} \) as above, the \( A \) -module \( M \) is the direct sum\n\n\[ M = {M}_{1} \oplus \cdots \oplus {M}_{n} \]\n\nwhere \( {M}_{i} \) is the submodule of \( M \) annihilated by \( {\mathfrak{b}}_{i} \) . | Proof. For \( i = 1,\ldots, n \), let \( {p}_{i} : M \rightarrow M \) be the map given by\n\n\[ {p}_{i}\left( x\right) = {e}_{i}x,\;x \in M. \]\n\nThe map \( {p}_{i} \) is clearly linear, and because of the properties satisfied by the \( {e}_{i}\mathrm{\;s} \), we have\n\n\[ {p}_{i}^{2} = {p}_{i} \]\n\n\[ {p}_{i}{p}_{j... | Yes |
Proposition 35.16. Let \( M \) be module over a PID \( A \) . For every nonzero \( \alpha \in A \), if\n\n\[ \n\alpha = u{p}_{1}^{{n}_{1}}\cdots {p}_{r}^{{n}_{r}}\n\]\n\nis a factorization of \( \alpha \) into prime factors (where \( u \) is a unit), then the module \( M\left( \alpha \right) \) annihilated by \( \alpha... | Proof. First observe that since \( M\left( \alpha \right) \) is annihilated by \( \alpha \), we can view \( M\left( \alpha \right) \) as a \( A/\left( \alpha \right) \) - module. By the Chinese remainder theorem (Theorem 32.15) applied to the ideals \( \left( {u{p}_{1}^{{n}_{1}}}\right) = \) \( \left( {p}_{1}^{{n}_{1}}... | Yes |
Theorem 35.17. (Primary Decomposition Theorem) Let \( M \) be a torsion-module over a PID. For every irreducible element \( p \in P \), let \( {M}_{p} \) be the submodule of \( M \) annihilated by some power of \( p \). Then, \( M \) is the (possibly infinite) direct sum\n\n\[ M = {\bigoplus }_{p \in P}{M}_{p} \] | Proof. Since \( M \) is a torsion-module, for every \( x \in M \), there is some \( \alpha \in A \) such that \( x \in M\left( \alpha \right) \). By Proposition 35.16, if \( \alpha = u{p}_{1}^{{n}_{1}}\cdots {p}_{r}^{{n}_{r}} \) is a factorization of \( \alpha \) into prime factors (where \( u \) is a unit), then the m... | Yes |
Proposition 35.18. Two torsion modules \( M \) and \( N \) over a PID are isomorphic iff for every every irreducible element \( p \in P \), the \( p \) -primary components \( {M}_{p} \) and \( {N}_{p} \) of \( M \) and \( N \) are isomorphic. | Proof. Let \( f : M \rightarrow N \) be an isomorphism. For any \( p \in P \), we have \( x \in {M}_{p} \) iff \( {p}^{k}x = 0 \) for some \( k \geq 1 \), so\n\n\[ 0 = f\left( {{p}^{k}x}\right) = {p}^{k}f\left( x\right) \]\n\nwhich shows that \( f\left( x\right) \in {N}_{p} \) . Therefore, \( f \) restricts to a linear... | Yes |
Theorem 35.19. (Primary Decomposition Theorem for finitely generated torsion modules) Let \( M \) be a finitely generated torsion-module over a PID \( A \) . If \( \operatorname{Ann}\left( M\right) = \left( a\right) \) and if \( a = \) \( u{p}_{1}^{{n}_{1}}\cdots {p}_{r}^{{n}_{r}} \) is a factorization of a into prime ... | Proof. This is an immediate consequence of Proposition 35.16. | No |
Proposition 35.20. If \( M \) is a torsion module over a PID, for every submodule \( N \) of \( M \) , we have a direct sum | Proof. It is easily verified that \( N \cap {M}_{p} \) is the \( p \) -primary component of \( N \) . | No |
Proposition 35.21. If \( M \) is a torsion module over a PID, a submodule \( N \) of \( M \) is a direct factor of \( M \) iff \( {N}_{p} \) is a direct factor of \( {M}_{p} \) for every irreducible element \( p \in A \) . | Proof. This is because if \( N \) and \( {N}^{\prime } \) are two submodules of \( M \), we have \( M = N \oplus {N}^{\prime } \) iff, by Proposition 35.20, \( {M}_{p} = {N}_{p} \oplus {N}_{p}^{\prime } \) for every irreducible elements \( p \in A \) . | Yes |
Proposition 35.22. Let \( A \) be a PID which is not a field, and let \( M \) be any \( A \)-module. Then \( M \) is semi-simple iff it is a torsion module and if \( {M}_{p} = M\left( p\right) \) for every irreducible element \( p \in A \) (in other words, if \( x \in M \) is annihilated by a power of \( p \), then it ... | Proof. Assume that \( M \) is semi-simple. Let \( x \in M \) and pick any irreducible element \( p \in A \). Then, the submodule \( {pAx} \) has a supplement \( N \) such that\n\n\[ M = {pAx} \oplus N \]\n\nso we can write \( x = {pax} + y \), for some \( y \in N \) and some \( a \in A \). But then,\n\n\[ y = \left( {1... | Yes |
Proposition 35.24. For any commutative ring \( A \), if \( F \) is a free \( A \) -module and if \( \left( {{e}_{1},\ldots ,{e}_{n}}\right) \) is a basis of \( F \), for any elements \( {a}_{1},\ldots ,{a}_{n} \in A \), there is an isomorphism\n\n\[ F/\left( {A{a}_{1}{e}_{1} \oplus \cdots \oplus A{a}_{n}{e}_{n}}\right)... | Proof. Let \( \sigma : F \rightarrow A/\left( {{a}_{1}A}\right) \oplus \cdots \oplus A/\left( {{a}_{n}A}\right) \) be the linear map given by\n\n\[ \sigma \left( {{x}_{1}{e}_{1} + \cdots + {x}_{n}{e}_{n}}\right) = \left( {{\bar{x}}_{1},\ldots ,{\bar{x}}_{n}}\right) ,\]\n\nwhere \( {\bar{x}}_{i} \) is the equivalence cl... | Yes |
Theorem 35.25. Let \( M \) be a finitely generated nontrivial A-module, where A a PID. Then, \( M \) is isomorphic to a direct sum of cyclic modules\n\n\[ M \approx A/{\mathfrak{a}}_{1} \oplus \cdots \oplus A/{\mathfrak{a}}_{m} \]\n\nwhere the \( {\mathfrak{a}}_{i} \) are proper ideals of \( A \) (possibly zero) such t... | Proof. Since \( M \) is finitely generated and nontrivial, there is a surjective homomorphism \( \varphi : {A}^{n} \rightarrow M \) for some \( n \geq 1 \), and \( M \) is isomorphic to \( {A}^{n}/\operatorname{Ker}\left( \varphi \right) \) . Since \( \operatorname{Ker}\left( \varphi \right) \) is a submodule of the fr... | Yes |
Proposition 35.26. If \( A \) is a commutative ring and if \( {\mathfrak{a}}_{1},\ldots ,{\mathfrak{a}}_{m} \) are ideals of \( A \), then there is an isomorphism\n\n\[ A/{\mathfrak{a}}_{1} \otimes \cdots \otimes A/{\mathfrak{a}}_{m} \approx A/\left( {{\mathfrak{a}}_{1} + \cdots + {\mathfrak{a}}_{m}}\right) . | Sketch of proof. We proceed by induction on \( m \) . For \( m = 2 \), we define the map \( \varphi : A/{\mathfrak{a}}_{1} \times A/{\mathfrak{a}}_{2} \rightarrow A/\left( {{\mathfrak{a}}_{1} + {\mathfrak{a}}_{2}}\right) \) by\n\n\[ \varphi \left( {\bar{a},\bar{b}}\right) = {ab}\;\left( {{\;\operatorname{mod}\;{\mathfr... | Yes |
Proposition 35.27. If \( A \) is a commutative ring, then for any \( n \) modules \( {M}_{i} \), there is an isomorphism\n\n\[ \bigwedge \left( {{\bigoplus }_{i = 1}^{n}{M}_{i}}\right) \approx {\bigotimes }_{i = 1}^{n}\bigwedge {M}_{i} \] | A proof can be found in Bourbaki [25] (Chapter III, Section 7, No 7, Proposition 10). | No |
Proposition 35.28. Let \( A \) be a commutative ring and let \( {\mathfrak{a}}_{1},\ldots ,{\mathfrak{a}}_{n} \) be \( n \) ideals of \( A \). If the module \( M \) is the direct sum of \( n \) cyclic modules\n\n\[ M = A/{\mathfrak{a}}_{1} \oplus \cdots \oplus A/{\mathfrak{a}}_{n} \]\n\nthen for every \( p > 0 \), the ... | Proof. If \( {u}_{i} \) is the image of 1 in \( A/{\mathfrak{a}}_{i} \), then \( A/{\mathfrak{a}}_{i} \) is equal to \( A{u}_{i} \). By Proposition 35.27, we have\n\n\[ \bigwedge M \approx {\bigotimes }_{i = 1}^{n}\bigwedge \left( {A{u}_{i}}\right) \]\n\nWe also have\n\n\[ \bigwedge \left( {A{u}_{i}}\right) = {\bigoplu... | Yes |
Proposition 35.29. Let \( A \) be a commutative ring and let \( {\mathfrak{a}}_{1},\ldots ,{\mathfrak{a}}_{n} \) be \( n \) ideals of \( A \) such that \( {\mathfrak{a}}_{1} \subseteq {\mathfrak{a}}_{2} \subseteq \cdots \subseteq {\mathfrak{a}}_{n} \) . If the module \( M \) is the direct sum of \( n \) cyclic modules\... | Proof. With the notation of Proposition 35.28, we have \( {\mathfrak{a}}_{H} = {\mathfrak{a}}_{\max \left( H\right) } \), where \( \max \left( H\right) \) is the greatest element in the set \( H \) . Since \( \max \left( H\right) \geq p \) for any subset with \( p \) elements and since \( \max \left( H\right) = p \) wh... | Yes |
Theorem 35.31. (Invariant Factors Decomposition) Let \( M \) be a finitely generated nontrivial A-module, where \( A \) a PID. Then, \( M \) is isomorphic to a direct sum of cyclic modules\n\n\[ M \approx A/{\mathfrak{a}}_{1} \oplus \cdots \oplus A/{\mathfrak{a}}_{m} \]\n\nwhere the \( {\mathfrak{a}}_{i} \) are proper ... | Proof. By Theorem 35.7, since \( {M}_{\text{tor }} = A/{\mathfrak{a}}_{r + 1} \oplus \cdots \oplus A/{\mathfrak{a}}_{m} \), we know that the dimension \( r \) of the free summand only depends on \( M \) . The uniqueness of the sequence of ideals follows from Proposition 35.30. | Yes |
Let \( F \) be a finitely generated free module over a PID \( A \), and let \( M \) be any submodule of \( F \). Then, \( M \) is a free module and there is a basis \( \left( {{e}_{1},\ldots ,{e}_{n}}\right) \) of \( F \), some \( q \leq n \), and some nonzero elements \( {a}_{1},\ldots ,{a}_{q} \in A \), such that \( ... | Proof. Since \( {a}_{i} \neq 0 \) for \( i = 1,\ldots, q \), observe that\n\n\[ {M}^{\prime } = \{ x \in F \mid \left( {\exists \beta \in A,\beta \neq 0}\right) \left( {{\beta x} \in M}\right) \} ,\]\n\nwhich shows that \( {M}^{\prime }/M \) is the torsion module of \( F/M \). Therefore, \( {M}^{\prime } \) is uniquely... | Yes |
Proposition 35.33. Let \( F \) be a free module of finite dimension over a PID, \( \left( {{u}_{1},\ldots ,{u}_{n}}\right) \) be a basis of \( F, M \) be a submodule of \( F \), and \( \left( {{x}_{1},\ldots ,{x}_{m}}\right) \) be a set of generators of \( M \) . If \( {a}_{1}A,\ldots ,{a}_{q}A \) are the invariant fac... | Proof. Proposition 35.23 shows that \( M \subseteq {a}_{1}F \) . Consequently, the coordinates of any element of \( M \) are multiples of \( {a}_{1} \) . On the other hand, we know that there is a linear form \( f \) for which \( {a}_{1}A \) is a maximal ideal and some \( {e}^{\prime } \in M \) such that \( f\left( {e}... | Yes |
Proposition 35.34. Let \( A \) be a PID, let \( F \) be a free module of dimension \( n \), \( {F}^{\prime } \) be a free module of dimension \( m \), and \( f : F \rightarrow {F}^{\prime } \) be a linear map from \( F \) to \( {F}^{\prime } \). Then, there exist a basis \( \left( {{e}_{1},\ldots ,{e}_{n}}\right) \) of... | Proof. Let \( {F}_{0} \) be the kernel of \( f \). Since \( {M}^{\prime } = f\left( F\right) \) is a submodule of the free module \( {F}^{\prime } \), it is free, and similarly \( {F}_{0} \) is free as a submodule of the free module \( F \) (by Proposition 35.23). By Proposition 35.2, we have \[ F = {F}_{0} \oplus {F}_... | Yes |
Proposition 35.36. Two \( m \times n \) matrices \( X \) and \( Y \) are equivalent iff they have the same invariant factors. | If \( X \) is the matrix of a linear map \( f : F \rightarrow {F}^{\prime } \) with respect to some basis \( \left( {{u}_{1},\ldots ,{u}_{n}}\right) \) of \( F \) and some basis \( \left( {{u}_{1}^{\prime },\ldots ,{u}_{m}^{\prime }}\right) \) of \( {F}^{\prime } \), then the columns of \( X \) are the coordinates of t... | No |
Theorem 35.38. (Elementary Divisors Decomposition) Let \( M \) be a finitely generated nontrivial A-module, where A a PID. Then, \( M \) is isomorphic to the direct sum \( {A}^{r} \oplus {M}_{\text{tor }} \), where \( {A}^{r} \) is a free module and where the torsion module \( {M}_{\text{tor }} \) is a direct sum of cy... | Proof. By Theorem 35.31, we already know that \( M \approx {A}^{r} \oplus {M}_{\text{tor }} \), where \( r \) is uniquely determined, and where\n\n\[ {M}_{\text{tor }} \approx A/{\mathfrak{a}}_{r + 1} \oplus \cdots \oplus A/{\mathfrak{a}}_{m} \]\n\na direct sum of cyclic modules, with \( \left( 0\right) \neq {\mathfrak... | Yes |
Proposition 35.39. Given a ring homomomorphism \( \rho : A \rightarrow B \) and given any \( A \) -module \( M \), the map \( \varphi : M \rightarrow {\rho }_{ * }\left( {{\rho }^{ * }\left( M\right) }\right) \) given by \( \varphi \left( x\right) = 1{ \otimes }_{A}x \) is \( A \) -linear and \( \varphi \left( M\right)... | \[ \bar{f} : {\rho }^{ * }\left( M\right) \rightarrow N \] such that \[ {\rho }_{ * }\left( \bar{f}\right) \circ \varphi = f \] as in the following commutative diagram  or equivalently, \[ \bar{f}\left( {1{ \otimes... | Yes |
Proposition 35.40. Given a ring homomomorphism \( \rho : A \rightarrow B \), for any two \( A \) -modules \( M \) an \( N \), for every \( A \) -linear map \( f : M \rightarrow N \), there is a unique \( B \) -linear map \( {\rho }^{ * }\left( f\right) : {\rho }^{ * }\left( M\right) \rightarrow \) \( {\rho }^{ * }\left... | Proof. Apply Proposition 35.40 to the \( A \) -linear map \( {\varphi }_{N} \circ f \) . | No |
Given a ring homomomorphism \( \rho : A \rightarrow B \), for any \( A \) -module \( M \), if \( \left( {{u}_{1},\ldots ,{u}_{n}}\right) \) is a basis of \( M \), then \( \left( {\varphi \left( {u}_{1}\right) ,\ldots ,\varphi \left( {u}_{n}\right) }\right) \) is a basis of \( {\rho }^{ * }\left( M\right) \), where \( \... | The first assertion follows immediately from Proposition 35.13, since it asserts that every element \( z \) of \( {\rho }^{ * }\left( M\right) = {\rho }_{ * }\left( B\right) { \otimes }_{A}M \) can be written in a unique way as\n\n\[ z = {b}_{1} \otimes {u}_{1} + \cdots + {b}_{n} \otimes {u}_{n} = {b}_{1}\left( {1 \oti... | Yes |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.