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Corollary 9.5. Let \( G \) be a product of a p-group and a cyclic group, and let \( k \) be algebraically closed. If \( E \) is a simple \( \left( {G, k}\right) \) -space and is not 1-dimensional, then \( E \) is induced by a 1-dimensional representation of some subgroup. | Proof. We apply the theorem step by step using the transitivity of induced representations until we get a 1-dimensional representation of a subgroup. | No |
Lemma 10.1. \( {V}_{R}\left( G\right) \) is an ideal in \( {X}_{R}\left( G\right) \) . | Proof. This is immediate from Theorem 6.1. | No |
Proposition 10.2. Every subgroup and every factor group of a p-elementary group is p-elementary. If \( S \) is a subgroup of the p-elementary group \( P \times C \) , where \( P \) is a p-group, and \( C \) is cyclic, of order prime to \( p \), then | Proof. Clear. | No |
Lemma 10.3. If \( d \in \mathbf{Z} \) and the constant function \( d.{1}_{G} \) belongs to \( {V}_{R} \) then \( d.{1}_{G} \) belongs to \( {V}_{\mathbf{z}} \) . | Proof. We contend that \( 1,\zeta ,\ldots ,{\zeta }^{N - 1} \) are linearly independent over \( {X}_{\mathbf{Z}}\left( G\right) \) . Indeed, a relation of linear dependence would yield\n\n\[ \mathop{\sum }\limits_{{v = 1}}^{s}\mathop{\sum }\limits_{{j = 0}}^{{N - 1}}{c}_{vj}{\chi }_{v}{\zeta }^{j} = 0 \]\n\nwith intege... | Yes |
Lemma 10.4. Let \( f \in {X}_{R}\left( G\right) \), and assume that \( f\left( \sigma \right) \in \mathbf{Z} \) for all \( \sigma \in G \) . Then \( f \) is constant \( {\;\operatorname{mod}\;p} \) on every \( p \) -class. | Proof. Let \( x = {\sigma \tau } \), where \( \sigma \) is \( p \) -singular, and \( \tau \) is \( p \) -regular, and \( \sigma ,\tau \) commute. It will suffice to prove that \[ f\left( x\right) \equiv f\left( \tau \right) \;\left( {\;\operatorname{mod}\;p}\right) \] Let \( H \) be the cyclic subgroup generated by \( ... | Yes |
Lemma 10.5. Let \( R = \mathbf{Z}\left\lbrack \zeta \right\rbrack \) be as before. If \( a \in \mathbf{Z} \) and \( a \in {pR} \) then \( a \in p\mathbf{Z} \) . | Proof. This is immediate from the fact that \( R \) has a basis over \( \mathbf{Z} \) such that 1 is a basis element. | No |
Lemma 10.6. Let \( \tau \) be p-regular in \( G \), and let \( T \) be the cyclic subgroup generated by \( \tau \) . Let \( C \) be the subgroup of \( G \) consisting of all elements commuting with \( \tau \) . Let \( P \) be a p-Sylow subgroup of \( C \) . Then there exists an element \( \psi \in {X}_{R}\left( {T \tim... | Proof. We note that the subgroup of \( G \) generated by \( T \) and \( P \) is a direct product \( T \times P \) . Let \( {\psi }_{1},\ldots ,{\psi }_{r} \) be the simple characters of the cyclic group \( T \), and assume that these are extended to \( T \times P \) by composition with the projection:\n\n\[ T \times P ... | Yes |
Lemma 10.7. Assume that the family of subgroups \( \left\{ {H}_{\alpha }\right\} \) covers \( G \) (i.e. every element of \( G \) lies in some \( {H}_{\alpha } \) ). Iff is a class function on \( G \) taking its values in \( \mathbf{Z} \), and such that all the values are divisible by \( n = \left( {G : 1}\right) \), t... | Proof. Let \( \gamma \) be a conjugacy class, and let \( p \) be prime to \( n \) . Every element of \( G \) is \( p \) -regular, and all \( p \) -subgroups of \( G \) are trivial. Furthermore, \( p \) -conjugacy is the same as conjugacy. Applying Lemma 10.6, we find that there exists in \( {V}_{R}\left( G\right) \) a ... | Yes |
Theorem 10.8. (Artin). Every character of \( G \) is a linear combination with rational coefficients of induced characters from cyclic subgroups. | Proof. In Lemma 10.7, let \( \left\{ {H}_{\alpha }\right\} \) be the family of cyclic subgroups of \( G \) . The constant function \( n.{1}_{G} \) belongs to \( {V}_{R}\left( G\right) \) . By Lemma 10.3, this function belongs to \( {V}_{\mathbf{Z}}\left( G\right) \), and hence \( n{X}_{\mathbf{Z}}\left( G\right) \subse... | No |
Lemma 10.9. Let \( p \) be a prime number, and assume that every p-elementary subgroup of \( G \) is contained in some \( {H}_{\alpha } \) . Then there exists a function \( f \in {V}_{R}\left( G\right) \) whose values are in \( \mathbf{Z} \), and \( \equiv 1\left( {\;\operatorname{mod}\;{p}^{r}}\right) \) . | Proof. We apply Lemma 10.6 again. For each \( p \) -class \( \gamma \), we can find a function \( {f}_{\gamma } \) in \( {V}_{R}\left( G\right) \), whose values are 0 on elements outside \( \gamma \), and \( ≢ 0{\;\operatorname{mod}\;p} \) for elements of \( \gamma \) . Let \( f = \sum {f}_{\gamma } \), the sum being t... | Yes |
Lemma 10.10. Let \( p \) be a prime number and assume that every p-elementary subgroup of \( G \) is contained in some \( {H}_{\alpha } \) . Let \( n = {n}_{0}{p}^{r} \) where \( {n}_{0} \) is prime to \( p \) . Then the constant function \( {n}_{0} \cdot {1}_{G} \) belongs to \( {V}_{\mathbf{z}}\left( G\right) \) . | Proof. By Lemma 10.3, it suffices to prove that \( {n}_{0} \cdot {1}_{G} \) belongs to \( {V}_{R}\left( G\right) \) . Let \( f \) be as in Lemma 10.9. Then\n\n\[ \n{n}_{0} \cdot {1}_{G} = {n}_{0}\left( {{1}_{G} - f}\right) + {n}_{0}f.\n\]\n\nSince \( {n}_{0}\left( {{1}_{G} - f}\right) \) has values divisible by \( {n}_... | Yes |
Theorem 10.11. (Brauer). Assume that for every prime number \( p \), every p-elementary subgroup of \( G \) is contained in some \( {H}_{\alpha } \) . Then \( X\left( G\right) = {V}_{\mathbf{Z}}\left( G\right) \) . Every character of \( G \) is a linear combination, with integer coefficients, of characters induced from... | Proof. Immediate from Lemma 10.10, since we can find functions \( {n}_{0} \cdot {1}_{G} \) in \( {V}_{\mathbf{z}}\left( G\right) \) with \( {n}_{0} \) relatively prime to any given prime number. | No |
Corollary 10.12. A class function \( f \) on \( G \) belongs to \( X\left( G\right) \) if and only if its restriction to \( {H}_{\alpha } \) belongs to \( X\left( {H}_{\alpha }\right) \) for each \( \alpha \) . | Proof. Assume that the restriction of \( f \) to \( {H}_{\alpha } \) is a character on \( {H}_{\alpha } \) for each \( \alpha \) . By the theorem, we can write\n\n\[ \n{1}_{G} = \mathop{\sum }\limits_{\alpha }{c}_{\alpha }{\operatorname{ind}}_{{H}_{\alpha }}^{G}\left( {\psi }_{\alpha }\right)\n\]\n\nwhere \( {c}_{\alph... | Yes |
Theorem 10.13. (Brauer). Every character of \( G \) is a linear combination with integer coefficients of characters induced by 1-dimensional characters of subgroups. | Proof. By Theorem 10.11, and the transitivity of induction, it suffices to prove that every character of a \( p \) -elementary group has the property stated in the theorem. But we have proved this in the preceding section, Corollary 9.5. | Yes |
Proposition 11.1. Let the notation be as above. Then the characters of the representations of \( G \) on \( E \) and on \( {E}^{\prime } \) are equal. | Proof. Let \( \left\{ {{v}_{1},\ldots ,{v}_{m}}\right\} \) be a basis of \( E \) over \( k \) . Then\n\n\[\n\left\{ {1 \otimes {v}_{1},\ldots ,1 \otimes {v}_{m}}\right\}\n\]\n\nis a basis of \( {E}^{\prime } \) over \( {k}^{\prime } \) . Thus the matrices representing an element \( \sigma \) of \( G \) with respect to ... | Yes |
Proposition 11.2. Let \( E, F \) be simple representation spaces for the finite group \( G \) over \( k \) . Let \( {k}^{\prime } \) be an extension of \( k \) . Assume that \( E, F \) are not \( G \) - isomorphic. Then no \( {k}^{\prime } \) -simple component of \( {E}_{{k}^{\prime }} \) appears in the direct sum deco... | Proof. Consider the direct product decomposition\n\n\[ k\left\lbrack G\right\rbrack = \mathop{\prod }\limits_{{\mu = 1}}^{{s\left( k\right) }}{R}_{\mu }\left( k\right) \]\n\nover \( k \), into a direct product of simple rings. Without loss of generality, we may assume that \( E, F \) are simle left ideals of \( k\left\... | Yes |
Corollary 11.3. The simple characters \( {\chi }_{1},\ldots ,{\chi }_{s\left( k\right) } \) of \( G \) over \( k \) are linearly independent over any extension \( {k}^{\prime } \) of \( k \) . | Proof. This follows at once from the proposition, together with the linear independence of the \( {k}^{\prime } \) -simple characters over \( {k}^{\prime } \) . | No |
Theorem 11.4. (Brauer). Let \( G \) be a finite group of exponent \( m \) . Every representation of \( G \) over the complex numbers (or an algebraically closed field of characteristic 0) is definable over the field \( \mathbf{Q}\left( {\zeta }_{m}\right) \) where \( {\zeta }_{m} \) is a primitive \( m \) -th root of u... | Proof. Let \( \chi \) be the character of a representation of \( G \) over \( \mathbf{C} \), i.e. an effective character. By Theorem 10.13, we can write\n\n\[ \chi = \mathop{\sum }\limits_{j}{c}_{j}{\operatorname{ind}}_{{S}_{j}}^{G}\left( {\psi }_{j}\right) ,\;{c}_{j} \in \mathbf{Z}, \]\n\nthe sum being taken over a fi... | Yes |
Lemma 12.1. The subgroup \( {C}_{K} \) is a maximal commutative semisimple subgroup. | Proof. If \( \alpha \in G{L}_{2}\left( F\right) \) commutes with all elements of \( {C}_{K} \) then \( \alpha \) must lie in \( F\left\lbrack {C}_{K}\right\rbrack \), for otherwise \( \{ 1,\alpha \} \) would be linearly independent over \( F\left\lbrack {C}_{K}\right\rbrack \), whence \( {\operatorname{Mat}}_{2}\left( ... | Yes |
Lemma 12.2. Every maximal commutative semisimple subgroup of \( G{L}_{2}\left( F\right) \) is a Cartan subgroup, and conversely. | Proof. It is clear that the split Cartan subgroup is maximal commutative semisimple. Suppose that \( H \) is a maximal commutative semisimple subgroup of \( G{L}_{2}\left( F\right) \) . If \( H \) is diagonalizable over \( F \), then \( H \) is contained in a conjugate of the split Cartan. On the other hand, suppose \(... | Yes |
Proposition 12.3. Let \( H \) be a Cartan subgroup of \( G{L}_{2}\left( F\right) \) (split or not). Then\n\n\( H \) is of index 2 in its normalizer \( N\left( H\right) \) . | Proof. We may view \( G{L}_{2}\left( F\right) \) as operating on the 2-dimensional vector space \( {V}^{\mathrm{a}} = {F}^{\mathrm{a}} \oplus {F}^{\mathrm{a}} \), over the algebraic closure \( {F}^{\mathrm{a}} \) . Whether \( H \) is split or not, the eigencharacters are distinct (because of the separability assumption... | Yes |
Theorem 12.6. The irreducible characters of \( G = G{L}_{2}\left( F\right) \) are as follows. | Proof. We have exhibited characters of four types. In each case it is immediate from our construction that we get the stated number of distinct characters of the given type. The dimensions as stated are immediately computed from the dimensions of induced characters as the index of the subgroup from which we induce, and... | No |
Let \( E \) be free of dimension \( n \) over \( R \). If \( r > n \) then \( \mathop{\bigwedge }\limits^{r}\left( E\right) = 0 \). Let \( \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) be a basis of \( E \) over \( R \). If \( 1 \leqq r \leqq n \), then \( {\bigwedge }^{r}\left( E\right) \) is free over \( R \), and the... | Proof. We shall first prove our assertion when \( r = n \). Every element of \( E \) can be written in the form \( \sum {a}_{i}{v}_{i} \), and hence using the formula \( x \land y = - y \land x \) we conclude that \( {v}_{1} \land \cdots \land {v}_{n} \) generates \( \mathop{\bigwedge }\limits^{n}\left( E\right) \). On... | Yes |
Proposition 1.2. Let\n\n\\[ \n0 \rightarrow {E}^{\prime } \rightarrow E \rightarrow {E}^{\prime \prime } \rightarrow 0 \n\\]\n\nbe an exact sequence of free R-modules of finite ranks \\( r, n \\), and \\( s \\) respectively. Then there is a natural isomorphism\n\n\\[ \n\varphi : {\\bigwedge }^{r}{E}^{\prime } \otimes {... | Proof. The proof proceeds in the usual two steps. First one shows the existence of a homomorphism \\( \\varphi \\) having the desired effect. The value on the right of the above formula is independent of the choice of \\( {u}_{1},\\ldots ,{u}_{s} \\) lifting \\( {w}_{1},\\ldots ,{w}_{s} \\) by using the alternating pro... | No |
Proposition 1.3. There is a natural isomorphism\n\n\[ \mathop{\bigwedge }\limits^{i}{E}^{\prime } \otimes \mathop{\bigwedge }\limits^{{n - i}}{E}^{\prime \prime } \rightarrow \mathop{\bigwedge }\limits_{i}^{n}E/\mathop{\bigwedge }\limits_{{i + 1}}^{n}E. \] | Proof. Let \( {x}_{1}^{\prime \prime },\ldots ,{x}_{n - i}^{\prime \prime } \) be elements of \( {E}^{\prime \prime } \), and lift them to elements \( {y}_{1},\ldots ,{y}_{n - i} \) of \( E \) . We consider the map\n\n\[ \left( {{x}_{1}^{\prime },\ldots ,{x}_{i}^{\prime },{x}_{1}^{\prime \prime },\ldots ,{x}_{n - i}^{\... | Yes |
Proposition 1.4. Let \( E = {E}^{\prime } \oplus {E}^{\prime \prime } \) be a direct sum of finite free modules. Then for every positive integer \( n \), we have a module isomorphism\n\n\[ \n\mathop{\bigwedge }\limits^{n}E \approx \mathop{\bigoplus }\limits_{{p + q = n}}\mathop{\bigwedge }\limits^{p}{E}^{\prime } \otim... | Proof. Each natural injection of \( {E}^{\prime } \) and \( {E}^{\prime \prime } \) into \( E \) induces a natural map on the alternating algebras, and so gives the homomorphism\n\n\[ \n\land {E}^{\prime } \otimes \land {E}^{\prime \prime } \rightarrow \land E\n\]\n\nwhich is graded, i.e. for \( p = 0,\ldots, n \) we h... | No |
Proposition 2.1. Let \( A \) be a \( p \times q \) matrix and \( B \) a \( q \times s \) matrix. Then\n\n\[{\left( AB\right) }^{\left( r\right) } = {A}^{\left( r\right) }{B}^{\left( r\right) }\;\text{ for }\;r \geqq 0.\] | If one uses the alternating products as mentioned above, the proof simply says that the matrix of the composite of linear maps with respect to fixed bases is the product of the matrices. If one does not use the alternating products, then one can prove the proposition by a direct computation which will be left to the re... | No |
Proposition 2.2. Let \( {R}^{q} \rightarrow E \rightarrow 0 \) be a representation of \( E \) as a quotient of \( {R}^{q} \), and let \( {x}_{1},\ldots ,{x}_{q} \) be the images of the unit vectors in \( {R}^{q} \) . Then \( {I}_{r}\left( x\right) \) is the ideal generated by all values\n\n\[ \lambda \left( {{w}_{1},\l... | Proof. This is immediate from the definition of the determinant ideal. | No |
Lemma 2.3. The Fitting ideal \( {F}_{k}\left( x\right) \) does not depend on the choice of generators \( \left( x\right) \) . | Proof. Let \( {y}_{1},\ldots ,{y}_{s} \) be elements of \( E \) . We shall prove that\n\n\[ \n{I}_{r}\left( x\right) = {I}_{r + s}\left( {x, y}\right) \n\] \n\nThe relations of \( \left( {x, y}\right) \) constitute a matrix of the form\n\n\[ \nW = \left( \begin{matrix} {a}_{11} & \cdots & {a}_{1q} & 0 & \cdots & 0 & \\... | Yes |
Proposition 2.5. Suppose that \( E \) can be generated by \( q \) elements. Then\n\n\[{\left( {\operatorname{ann}}_{R}\left( E\right) \right) }^{q} \subset F\left( E\right) \subset {\operatorname{ann}}_{R}\left( E\right) .\]\n\nIn particular, if \( E \) can be generated by one element, then\n\n\[F\left( E\right) = {\op... | Proof. Let \( {x}_{1},\ldots ,{x}_{q} \) be generators of \( E \) . Let \( {a}_{1},\ldots ,{a}_{q} \) be elements of \( R \) annihilating \( E \) . Then the diagonal matrix whose diagonal components are \( {a}_{1},\ldots ,{a}_{q} \) is a matrix of relations, so the definition of the Fitting ideal shows that the determi... | Yes |
Corollary 2.6. Let \( E = R/\mathfrak{a} \) for some ideal \( \mathfrak{a} \) . Then \( F\left( E\right) = \mathfrak{a} \) . | Proof. The module \( R/a \) can be generated by one element so the corollary is an immediate consequence of the proposition. | No |
Let\n\n\\[ \n0 \rightarrow {E}^{\prime } \rightarrow E \rightarrow {E}^{\prime \prime } \rightarrow 0 \n\\] \n\nbe an exact sequence of finite \( R \) -modules. For integers \( m, n \geqq 0 \) we have\n\n\\[ \n{F}_{m}\left( {E}^{\prime }\right) {F}_{n}\left( {E}^{\prime \prime }\right) \subset {F}_{m + n}\left( E\right... | Proof. We may assume \( {E}^{\prime } \) is a submodule of \( E \) . We pick generators \( {x}_{1},\ldots ,{x}_{p} \) of \( {E}^{\prime } \) and elements \( {y}_{1},\ldots ,{y}_{q} \) in \( E \) such that their images \( {y}_{1}^{\prime \prime },\ldots ,{y}_{q}^{\prime \prime } \) in \( {E}^{\prime \prime } \) generate... | Yes |
Proposition 2.8. Let \( {E}^{\prime },{E}^{\prime \prime } \) be finite \( R \) -modules. For any integer \( n \geqq 0 \) we have\n\n\[ \n{F}_{n}\left( {{E}^{\prime } \oplus {E}^{\prime \prime }}\right) = \mathop{\sum }\limits_{{r + s = n}}{F}_{r}\left( {E}^{\prime }\right) {F}_{s}\left( {E}^{\prime \prime }\right) .\n... | Proof. Let \( {x}_{1},\ldots ,{x}_{p} \) generate \( {E}^{i} \) and \( {y}_{1},\ldots ,{y}_{q} \) generate \( {E}^{\prime \prime } \) . Then \( \left( {x, y}\right) \) generate \( {E}^{\prime } \oplus {E}^{\prime \prime } \) . By Proposition 2.6 we know the inclusion\n\n\[ \n\sum {F}_{r}\left( {E}^{\prime }\right) {F}_... | Yes |
Corollary 2.9. Let\n\n\[ E = {\bigoplus }_{i = 1}^{s}R/{a}_{i} \]\n\nwhere \( {\mathfrak{a}}_{i} \) is an ideal. Then \( F\left( E\right) = {\mathfrak{a}}_{1}\cdots {\mathfrak{a}}_{s} \). | Proof. This is really a corollary of Proposition 2.8 and Corollary 2.6. | No |
Theorem 3.1. The pair \( \left( {J/{J}^{2}, d}\right) \) is universal for derivations of \( A \) . This means: Given a derivation \( D : A \rightarrow M \) there exists a unique \( A \) -linear map \( f : J/{J}^{2} \rightarrow M \) making the following diagram commutative. | Proof. There is a unique \( R \) -bilinear map\n\n\[ \nf : A \otimes A \rightarrow M\;\text{ given by }\;x \otimes y \mapsto {xDy}, \n\]\n\nwhich is \( A \) -linear by our definition of the \( A \) -module structure on \( A \otimes A \) . Then by definition, the diagram is commutative on elements of \( A \), when we ta... | Yes |
Theorem 3.2. There exists a unique sequence of \( R \) -homomorphisms\n\n\[ \n{d}_{i} : {\Omega }_{A/R}^{i} \rightarrow {\Omega }_{A/R}^{i + 1}\n\]\n\nsuch that for \( \omega \in {\Omega }^{i} \) and \( \eta \in {\Omega }^{j} \) we have\n\n\[ \nd\left( {\omega \land \eta }\right) = {d\omega } \land \eta + {\left( -1\ri... | The proof will be left as an exercise. | No |
Theorem 3.3. Let \( k \) be a field of characteristic 0, and let \( A = k\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) be the polynomial ring in \( n \) variables. Then the De Rham complex\n\n\[ 0 \rightarrow k \rightarrow A \rightarrow {\Omega }_{A/k}^{1} \rightarrow \cdot \cdot \cdot \rightarrow {\Omega }_{A... | Again the proof will be left as an exercise. Hint: Use induction and integrate formally. | No |
Theorem 4.1. Let \( g \) be a symmetric bilinear form on a finite dimensional vector space \( E \) over \( k \) . Then the Clifford algebra \( \left( {C\left( g\right) ,\rho }\right) \) exists. The map \( \rho \) in injective, and \( C\left( g\right) \) has dimension \( {2}^{n} \) over \( k \), if \( n = \dim E \) . | Proof. Let \( T\left( E\right) \) be the tensor algebra as in Chapter XVI,§7. In that algebra, we let \( {I}_{g} \) be the two-sided ideal generated by all elements\n\n\[ x \otimes x - g\left( {x, x}\right) \cdot 1\text{ for }x \in E. \]\n\nWe define \( {C}_{g}\left( E\right) = T\left( E\right) /{I}_{g} \) . Observe th... | No |
Theorem 2.1. Let\n\n\[ \n0 \rightarrow {E}^{\prime } \overset{f}{ \rightarrow } E \overset{g}{ \rightarrow } {E}^{\prime \prime } \rightarrow 0 \]\n\nbe an exact sequence of complexes with \( f, g \) of degree 0 . Then the sequence\n\nis exact. | This theorem is merely a special application of the snake lemma. | No |
Theorem 3.1. Let \( F \) be a complex, which is of even length if it is closed. Assume that \( \varphi \left( {F}^{i}\right) \) is defined for all \( i,\varphi \left( {F}^{i}\right) = 0 \) for almost all \( i \), and \( {H}^{i}\left( F\right) = 0 \) for almost all i. Then \( {\chi }_{\varphi }\left( F\right) \) is defi... | Proof. Let \( {Z}^{i} \) and \( {B}^{i} \) be the groups of \( i \) -cycles and \( i \) -boundaries in \( {F}^{i} \) respectively. We have an exact sequence\n\n\[ \n0 \rightarrow {Z}^{i} \rightarrow {F}^{i} \rightarrow {B}^{i + 1} \rightarrow 0. \n\]\n\nHence \( {\chi }_{\varphi }\left( F\right) \) is defined, and\n\n\... | Yes |
Theorem 3.3. Let\n\n\[ \n0 \rightarrow {E}^{\prime } \rightarrow E \rightarrow {E}^{\prime \prime } \rightarrow 0 \]\n\nbe an exact sequence of complexes, with morphisms of degree 0 . If the complexes are closed, assume that their length is even. Let \( \varphi \) be an Euler-Poincaré mapping on the category of modules... | Proof. We have an exact homology sequence\n\n\[ \n\rightarrow {H}^{\prime \prime \left( {i - 1}\right) } \rightarrow {H}^{\prime i} \rightarrow {H}^{i} \rightarrow {H}^{\prime \prime i} \rightarrow {H}^{\prime \left( {i + 1}\right) } \rightarrow \]\n\nThis homology sequence is nothing but a complex whose homology is tr... | Yes |
Theorem 3.4. Assume that \( \alpha \) satisfies the Euler-Poincaré condition. Then there is a map\n\n\[ \n\gamma : Q \rightarrow \mathbf{K}\left( Q\right) \n\]\n\nof \( \mathbf{Q} \) into an abelian group \( \mathbf{K}\left( \mathbf{Q}\right) \) having the universal property with respect to Euler-Poincaré maps defined ... | To construct this, let \( {F}_{\mathrm{{ab}}}\left( Q\right) \) be the free abelian group generated by the set of such \( \left\lbrack E\right\rbrack \) . Let \( B \) be the subgroup generated by all elements of type\n\n\[ \n\left\lbrack E\right\rbrack - \left\lbrack {E}^{\prime }\right\rbrack - \left\lbrack {E}^{\prim... | Yes |
Theorem 3.5. Let \( M \in \mathcal{Q}\left( \mathcal{C}\right) \) and suppose we have two resolutions\n\n\[ \n{L}_{M} \rightarrow M \rightarrow 0\;\text{ and }\;{L}_{M}^{\prime } \rightarrow M \rightarrow 0,\n\]\n\nby finite complexes \( {L}_{M} \) and \( {L}_{M}^{\prime } \) in \( \mathcal{C} \) . Then\n\n\[ \n\sum {\... | Proof. Take first the special case when there is an epimorphism \( {L}_{M}^{\prime } \rightarrow {L}_{M} \) , with kernel \( E \) illustrated on the following commutative and exact diagram.\n\n\n\nThe kernel is a com... | Yes |
Lemma 3.6. Given two epimorphisms \( u : M \rightarrow N \) and \( v : {M}^{\prime } \rightarrow N \) in \( \alpha \) , there exist epimorphisms \( F \rightarrow M \) and \( F \rightarrow {M}^{\prime } \) with \( F \) in \( \mathcal{C} \) making the following diagram commutative. | Proof. Let \( E = M{ \times }_{N}{M}^{\prime } \), that is \( E \) is the kernel of the morphism\n\n\[ M \times {M}^{\prime } \rightarrow N \]\n\n given by \( \left( {x, y}\right) \mapsto {ux} - {vy} \) . (Elements are not really used here, and we could write formally \( u - v \) instead.) There is some \( F \) in \( \... | Yes |
Theorem 3.7. The natural map\n\n\\[ \n\\epsilon : \\mathbf{K}\\left( \\mathcal{C}\\right) \\rightarrow \\mathbf{K}\\left( {\\mathcal{Q}\\left( \\mathcal{C}\\right) }\\right) \n\\]\n\nis an isomorphism. | Proof. The map is surjective because given a resolution\n\n\\[ \n0 \\rightarrow {F}_{n} \\rightarrow \\cdots \\rightarrow {F}_{0} \\rightarrow M \\rightarrow 0 \n\\]\n\nwith \\( {F}_{i} \\in \\mathcal{C} \\) for all \\( i \\), the element\n\n\\[ \n\\sum {\\left( -1\\right) }^{i}{\\gamma }_{\\mathrm{e}}\\left( {F}_{i}\\... | Yes |
Lemma 3.8. Given an exact sequence in \( Q\left( \mathcal{C}\right) \)\n\n\[ 0 \rightarrow {M}^{\prime } \rightarrow M \rightarrow {M}^{\prime \prime } \rightarrow 0 \]\n\nthere exists a commutative and exact diagram\n\n in \( \mathcal{C} \) to fit an exact and commutative diagram\n\n\n\nWe first select an epimorphism \( {L}^{\prime \prime } \rightarrow... | Yes |
Theorem 3.9. Let \( Q \) be the category of modules over a ring. Let \( P \) be the family of projective modules. Given an exact sequence of modules\n\n\[ 0 \rightarrow {E}^{\prime } \rightarrow E \rightarrow {E}^{\prime \prime } \rightarrow 0 \]\n\nif any two of \( {E}^{\prime }, E,{E}^{\prime \prime } \) admit finite... | Proofs in a more subtle case will be given in Chapter XXI, Theorem 2.7. | No |
Proposition 3.10. Let \( M \in \mathcal{Q}\left( \mathcal{C}\right) \) and let \( N \in \mathcal{Q} \) . Let\n\n\[ 0 \rightarrow {L}_{n} \rightarrow \cdots \rightarrow {L}_{0} \rightarrow M \rightarrow 0 \]\n\nbe a finite resolution of \( M \) by objects in \( \mathcal{C} \) . Then\n\n\[ {\operatorname{cl}}_{\mathrm{e}... | Proof. The formulas are immediate consequences of the definitions, and of Theorem 3.1. | No |
Theorem 3.12. Let \( E \) be a finite free module in \( \alpha \), of rank \( r \) . Then for all integers \( n \geqq 1 \) we have\n\n\[ \mathop{\sum }\limits_{{i = 0}}^{r}{\left( -1\right) }^{i}{\lambda }^{i}\left( E\right) {\sigma }^{n - i}\left( E\right) = 0 \]\n\nwhere by definition \( {\sigma }^{j}\left( E\right) ... | Proof. The first formula depends on the analogue for the symmetric product and the alternating product of the formula given in Proposition 1.1 of Chapter XIX. It could be proved directly now, but the reader will find a proof as a special case of the theory of Koszul complexes in Chapter XXI, Corollary 4.14. The power s... | No |
Theorem 4.1. Every module is a submodule of an injective module. | The proof will be given by dualizing the situation, with some lemmas. We first look at the situation in the category of abelian groups. If \( M \) is an abelian group, let its dual group be \( {M}^{ \land } = \operatorname{Hom}\left( {M,\mathbf{Q}/\mathbf{Z}}\right) \) . If \( F \) is a free abelian group, it is reason... | Yes |
Lemma 4.2. If \( T \) is divisible, then \( T \) is injective in the category of abelian groups. | Proof. Let \( {M}^{\prime } \subset M \) be a subgroup of an abelian group, and let \( f : {M}^{\prime } \rightarrow T \) be a homomorphism. Let \( x \in M \) . We want first to extend \( f \) to the module \( \left( {{M}^{\prime }, x}\right) \) generated by \( {M}^{\prime } \) and \( x \) . If \( x \) is free over \( ... | Yes |
Lemma 4.3. If \( T \) is a divisible abelian group, then \( {\operatorname{Hom}}_{\mathbf{Z}}\left( {A, T}\right) \) is injective in the category of \( A \) -modules. | Proof. It suffices to prove that if \( 0 \rightarrow X \rightarrow Y \) is exact in the category of \( A \) -modules, then the dual sequence obtained by taking \( A \) -homomorphisms into \( {\operatorname{Hom}}_{\mathbf{Z}}\left( {A, T}\right) \) is exact, that is the top map in the following diagram is surjective.\n\... | Yes |
Iff, \( g \) are homotopic, then \( f, g \) induce the same homomorphism on the homology \( H\left( E\right) \), that is\n\n\[ H\left( {f}_{n}\right) = H\left( {g}_{n}\right) : {H}^{n}\left( E\right) \rightarrow {H}^{n}\left( {E}^{\prime }\right) . \] | Proof. The lemma is immediate, because \( {f}_{n} - {g}_{n} \) vanishes on the cycles, which are the kernel of \( {d}^{n} \), and the homotopy condition shows that the image of \( {f}_{n} - {g}_{n} \) is contained in the boundaries, that is, in the image of \( {d}^{\prime \left( {n - 1}\right) } \) . | Yes |
Lemma 5.2. Consider two complexes:\n\n\n\nSuppose that the top row is exact, and that each \( {I}^{n}\left( {n \geqq 0}\right) \) is injective. Let \( \varphi : M \rightarrow {M}^{\prime } \) be a given homomorphism.... | Proof. By definition of an injective, the homomorphism \( M \rightarrow {I}^{0} \) via \( {M}^{\prime } \) extends to a homomorphism\n\n\[ {f}_{0} : {E}^{0} \rightarrow {I}^{0} \]\n\nwhich makes the first square commute:\n\n be a group and let \( R = \mathbf{Z}\left\lbrack G\right\rbrack \) be the group ring. Let \( Q \) be the category of \( G \) -modules, i.e. \( \mathbf{Q} = \operatorname{Mod}\left( R\right) \), also denoted by \( \operatorname{Mod}\left( G\right) \). For a \( G \) -module \( A \), let \( {A}^{G} \) be the s... | Its left derived functors give rise to the cohomology of groups. Some results from this special cohomology will be carried out in the exercises, as further examples of the general theory. | No |
Theorem 6.1. Let \( Q \) be an abelian category with enough injectives, and let \( F : \mathcal{Q} \rightarrow \mathcal{B} \) be a covariant additive left exact functor to another abelian category \( \mathfrak{B} \) . Then:\n\n(i) For each \( n \geqq 0,{R}^{n}F \) as defined above is an additive functor from \( \mathca... | We now describe how to construct the \( \delta \) -homomorphisms. Given a short exact sequence, we can find an injective resolution of \( {M}^{\prime }, M,{M}^{\prime \prime } \) separately, but they don't necessarily fit in an exact sequence of complexes. So we must achieve this to apply the considerations of \( §1 \)... | No |
Theorem 6.2. Let\n\n\[ \n0 \rightarrow M \rightarrow {X}^{0} \rightarrow {X}^{1} \rightarrow {X}^{2} \rightarrow \cdots \n\]\n\nbe a resolution of \( M \) by \( F \) -acyclics. Let\n\n\[ \n0 \rightarrow M \rightarrow {I}^{0} \rightarrow {I}^{1} \rightarrow {I}^{2} \rightarrow \cdots \n\]\n\nbe an injective resolution. ... | Proof. The existence of the morphism of complexes extending the identity on \( M \) is merely Lemma 5.2. The usual proof of the theorem via spectral sequences can be formulated independently in the following manner, shown to me by David Benson. We need a lemma. | No |
Lemma 6.3. Let \( {Y}^{i}\left( {i \geqq 0}\right) \) be \( F \) -acyclic, and suppose the sequence\n\n\[ 0 \rightarrow {Y}^{0} \rightarrow {Y}^{1} \rightarrow {Y}^{2} \rightarrow \cdots \]\n\nis exact. Then\n\n\[ 0 \rightarrow F\left( {Y}^{0}\right) \rightarrow F\left( {Y}^{1}\right) \rightarrow F\left( {Y}^{2}\right)... | Proof. Since \( F \) is left exact, we have an exact sequence\n\n\[ 0 \rightarrow F\left( {Y}^{0}\right) \rightarrow F\left( {Y}^{1}\right) \rightarrow F\left( {Y}^{2}\right) \]\n\nWe want to show exactness at the next joint. We draw the cokernels:\n\n has enough injectives. Then for any left exact functor \( F : \mathcal{Q} \rightarrow \mathcal{B} \), the derived functors \( {R}^{n}F \) with \( n \geqq 0 \) form a universal \( \delta \) -functor with \( F \approx {R}^{0}F \), which is erasable by injectives. Conversely, if \( G = {\left\... | If \( F \) is a left exact functor, then the \( {\left\{ {R}^{n}F\right\} }_{n \geqq 0} \) form a \( \delta \) -functor by Theorem 6.1. Furthermore, for any object \( A \), let \( u : A \rightarrow I \) be a monomorphism of \( A \) into an injective. Then \( {R}^{n}F\left( I\right) = 0 \) for \( n > 0 \) by Theorem 6.1... | Yes |
Proposition 7.3. Let\n\n\\[ \n0 \rightarrow Q \rightarrow {E}_{n - 1} \rightarrow \cdots \rightarrow {E}_{0} \rightarrow M \rightarrow 0 \n\\]\n\nbe an exact sequence, such that \\( {E}_{i} \in \\mathcal{E} \\) . Then we have an isomorphism\n\n\\[ \n{F}^{p}\\left( Q\\right) \\approx {F}^{p + n}\\left( M\\right) \\;\\te... | Proof. Let \\( Q = {Q}_{n} \\) . Also without loss of generality, take \\( p = 1 \\) . We may insert kernels and cokernels at each step as follows:\n\n\n\nThen shifting dimension with respect to each short exact sequ... | Yes |
Lemma 8.1. Let \( T \) be a bifunctor satisfying HOM 1, HOM 2. Let \( A \in \mathcal{Q} \) , and let \( {M}_{A} \rightarrow A \rightarrow 0 \), that is \[ \rightarrow {M}_{1} \rightarrow {M}_{0} \rightarrow A \rightarrow 0 \] be a \( T \) -exact resolution of \( A \) . Let \( {F}^{n}\left( B\right) = {H}^{n}\left( {T\l... | Proof. Given an exact sequence \[ 0 \rightarrow {B}^{\prime } \rightarrow B \rightarrow {B}^{\prime \prime } \rightarrow 0 \] we get an exact sequence of complexes \[ 0 \rightarrow T\left( {M,{B}^{\prime }}\right) \rightarrow T\left( {M, B}\right) \rightarrow T\left( {M,{B}^{\prime \prime }}\right) \rightarrow 0, \] wh... | Yes |
Proposition 8.2. Let \( T \) be a bifunctor satisfying HOM 1, HOM 2, HOM 3. Assume that \( \mathfrak{B} \) has enough injectives. Let \( A \in \mathcal{Q} \) . Let\n\n\[ \n{M}_{A} \rightarrow A \rightarrow 0 \]\n\nbe a T-exact resolution of \( A \) . Then the two \( \delta \) -functors\n\n\[ \nB \mapsto {R}^{n}T\left( ... | Proof. This comes merely from the universality of a \( \delta \) -functor erasable by injectives. | No |
Lemma 8.3. Let \( T \) satisfy HOM 1, HOM 2, and HOM 3. Assume that \( \mathfrak{B} \) has enough injectives. Let\n\n\[ 0 \rightarrow {A}^{\prime } \rightarrow A \rightarrow {A}^{\prime \prime } \rightarrow 0 \]\n\nbe a short exact sequence. Then for fixed \( B \), we have a long exact sequence\n\n\[ 0 \rightarrow T\le... | Proof. Let \( 0 \rightarrow B \rightarrow {I}_{B} \) be an injective resolution of \( B \) . From the exactness of the functor \( A \mapsto T\left( {A, J}\right) \), for \( J \) injective we get a short exact sequence of complexes\n\n\[ 0 \rightarrow T\left( {{A}^{\prime \prime },{I}_{B}}\right) \rightarrow T\left( {A,... | No |
Proposition 8.4. Let \( T \) satisfy HOM 1, HOM 2, HOM 3. Assume that \( \sigma \) has enough injectives. Fix \( B \in \mathcal{B} \) . Then the association\n\n\[ A \mapsto {R}^{n}T\left( {A, B}\right) \]\n\nis a contravariant \( \delta \) -functor on \( \mathcal{Q} \) which vanishes on \( T \) -exacts, for \( n \geqq ... | Proof. By Lemma 8.3 we know that the association is a \( \delta \) -functor, and it vanishes on \( T \) -exacts by Lemma 8.1. The last statement is then merely an application of the universality of erasable \( \delta \) -functors. | No |
Let \( Q = \$ \) be the category of modules over a ring. For fixed \( B \), let \( {\operatorname{ext}}^{n}\left( {A, B}\right) \) be the left derived functor of \( A \mapsto \operatorname{Hom}\left( {A, B}\right) \), obtained by means of projective resolutions of \( A \) . Then \[ {\operatorname{ext}}^{n}\left( {A, B}... | Proof. Immediate from Proposition 8.4. | No |
Proposition 8.6. Let \( T \) be a bifunctor satisfying HOM 1, HOM 2, HOM 3. Assume that \( \mathfrak{B} \) has enough injectives. Then the following conditions are equivalent:\n\nTE 1. \( A \) is \( T \) -exact.\n\nTE 2. For every \( B \) and every integer \( n \geqq 1 \), we have \( {R}^{n}T\left( {A, B}\right) = 0 \)... | Proof. Let\n\n\[ 0 \rightarrow B \rightarrow {I}^{0} \rightarrow {I}^{1} \rightarrow \]\n\nbe an injective resolution of \( B \) . By definition, \( {R}^{n}T\left( {A, B}\right) \) is the \( n \) -th homology of the sequence\n\n\[ 0 \rightarrow T\left( {A,{I}^{0}}\right) \rightarrow T\left( {A,{I}^{1}}\right) \rightarr... | Yes |
Proposition 8.7. Let \( T \) satisfy HOM 1, HOM 2, HOM 3. Assume that \( \mathbb{R} \) has enough injectives. Suppose that an object \( A \) admits a resolution\n\n\[ 0 \rightarrow {E}_{d} \rightarrow {E}_{d - 1} \rightarrow \cdots \rightarrow {E}_{0} \rightarrow A \rightarrow 0 \]\n\nwhere \( {E}_{0},\ldots ,{E}_{d} \... | Proof. By dimension shifting we conclude that \( Q \) has \( T \) -dimension 0, whence \( Q \) is \( T \) -exact by Proposition 8.6. | Yes |
Corollary 8.8. If there is a bifunctorial isomorphism \( T\left( {A, B}\right) \approx T\left( {B, A}\right) \) , and if \( B \) is \( T \) -exact, then for all \( A,{L}_{n}T\left( {A, B}\right) = 0 \) for \( n \geqq 1 \) . In short, T-exact implies acyclic. | Proof. Let \( {M}_{A} = {P}_{A} \) be a projective resolution in Proposition \( {8.2}^{\prime } \) . By hypotheses, \( X \mapsto T\left( {X, B}\right) \) is exact so \( {H}_{n}\left( {T\left( {P, B}\right) }\right) = 0 \) for \( n \geqq 1 \) ; so the corollary is a consequence of the proposition. | No |
Proposition 9.2. The homomorphism\n\n\[ \n{d}_{1} : {E}_{1}^{p} \rightarrow {E}_{1}^{p + 1} \n\]\n\nis the coboundary operator arising from the exact sequence\n\n\[ \n0 \rightarrow {F}^{p + 1}/{F}^{p + 2} \rightarrow {F}^{p}/{F}^{p + 2} \rightarrow {F}^{p}/{F}^{p + 1} \rightarrow 0 \n\]\n\nviewing each term as a comple... | Proof. Indeed, the coboundary\n\n\[ \n\delta : {E}_{1}^{p} = H\left( {{F}^{p}/{F}^{p + 1}}\right) \rightarrow H\left( {{F}^{p + 1}/{F}^{p + 2}}\right) = {E}_{1}^{p + 1} \n\]\n\nis defined on a representative cycle \( z \) by \( {dz} \), which is the same way that we defined \( {d}_{1} \) . | Yes |
Proposition 9.3. Let \( {FK} \) be a filtered complex. Then there exists a spectral sequence \( \left\{ {E}_{r}\right\} \) with:\n\n\[ \n{E}_{0}^{p, q} = {F}^{p}{K}^{p + q}/{F}^{p + 1}{K}^{p + q} \n\]\n\n\[ \n{E}_{1}^{p, q} = {H}^{p + q}\left( {{\operatorname{Gr}}^{p}K}\right) \n\]\n\n\[ \n{E}_{\infty }^{p, q} = {\oper... | The statement of Proposition 9.3 is merely a special case of Proposition 9.1, taking into account the extra graduation. | No |
Lemma 9.4. Let\n\n\[ \n0 \rightarrow {M}^{\prime } \rightarrow M \rightarrow {M}^{\prime \prime } \rightarrow 0 \]\n\nbe a short exact sequence. Let\n\n\[ \n0 \rightarrow {M}^{\prime } \rightarrow {I}_{{M}^{\prime }}\;\text{ and }\;0 \rightarrow {M}^{\prime \prime } \rightarrow {I}_{{M}^{\prime \prime }} \]\n\nbe injec... | Proof. The proof is the same as at the beginning of the proof of Theorem 6.1. | No |
Lemma 9.5. Given a complex \( C \) there exists a fully injective resolution of \( C \) . | Proof. We insert the kernels and cokernels in \( C \), giving rise to the short exact sequences with boundaries \( {B}^{p} \) and cycles \( {Z}^{p} \) :\n\n\[ \n0 \rightarrow {B}^{p} \rightarrow {Z}^{p} \rightarrow {H}^{p} \rightarrow 0 \n\]\n\n\[ \n0 \rightarrow {Z}^{p - 1} \rightarrow {C}^{p - 1} \rightarrow {B}^{p} ... | Yes |
Theorem 9.6. (Grothendieck spectral sequence). Let\n\n\\[ \nT : \\mathcal{Q} \\rightarrow \\mathcal{B}\\;\\text{ and }\\;G : \\mathcal{B} \\rightarrow \\mathcal{C} \n\\]\n\nbe covariant left exact functors such that if \\( I \\) is injective in \\( \\mathbf{Q} \\), then \\( T\\left( I\\right) \\) is G-acyclic. Then for... | Proof. Let \\( A \\) be an object of \\( \\alpha \\), and let \\( 0 \\rightarrow A \\rightarrow {C}_{A} \\) be an injective resolution. We apply \\( T \\) to get a complex\n\n\\[ \n{TC} : 0 \\rightarrow T{C}^{0} \\rightarrow T{C}^{1} \\rightarrow T{C}^{2} \\rightarrow \n\\]\n\nBy Lemma 9.5 there exists a fully injectiv... | Yes |
Lemma 9.7. The map \( {\chi }_{T} \) extends to a homomorphism\n\n\[ \mathbf{K}\left( {\mathfrak{F}}_{a}\right) \rightarrow \mathbf{K}\left( {\mathfrak{F}}_{a}\right) \] | Proof. Let\n\n\[ 0 \rightarrow {A}^{\prime } \rightarrow A \rightarrow {A}^{\prime \prime } \rightarrow 0 \]\n\nbe an exact sequence in \( \mathfrak{F} \) . Then we have the cohomology sequence\n\n\[ \rightarrow {R}^{i}T\left( {A}^{\prime }\right) \rightarrow {R}^{i}T\left( A\right) \rightarrow {R}^{i}T\left( {A}^{\pri... | Yes |
Theorem 9.8. Assume that \( T : \mathcal{Q} \rightarrow \mathcal{B} \) and \( G : \mathcal{B} \rightarrow \mathcal{C} \) satisfy the conditions CHAR 1 and CHAR 2. Also assume that \( T \) maps injectives to \( G \) -acyclics. Then\n\n\[ \n{\chi }_{G} \circ {\chi }_{T} = {\chi }_{GT} \n\] | Proof. By Theorem 9.6, the Grothendieck spectral sequence of the composite functor implies the existence of a filtration\n\n\[ \n\cdots \subset {F}^{p}{R}^{n}\left( {GT}\right) \left( A\right) \subset {F}^{p + 1}{R}^{n}\left( {GT}\right) \left( A\right) \subset \cdots \n\]\n\nof \( {R}^{n}\left( {GT}\right) \left( A\ri... | Yes |
Proposition 1.1. Let \( C \) be a complex such that \( {H}^{p}\left( C\right) \neq 0 \) only for \( 0 \leqq p \leqq n \) . Let \( \mathfrak{F} \) be a sufficient family of projectives. There exists a complex\n\n\[ 0 \rightarrow {K}^{0} \rightarrow {K}^{1} \rightarrow \cdots \rightarrow {K}^{n} \rightarrow 0 \]\n\nsuch ... | Proof. We define \( {f}_{m} \) by descending induction on \( m \) :\n\n\n\nWe suppose that we have defined a morphism of complexes with \( p \geqq m + 1 \) such that \( {H}^{p}\left( f\right) \) is an isomorphism for... | Yes |
Proposition 1.2. Let \( f : K \rightarrow C \) be a morphism of complexes, such that \( {K}^{p} \) , \( {H}^{p}\left( C\right) \) are \( \neq 0 \) only for \( p = 1,\ldots, n \) . Let \( \mathfrak{F} \) be a complete family, and assume that \( {K}^{p},{C}^{p} \) are in \( \mathfrak{F} \) for all \( p \), except possibl... | Before giving the proof, we define a new complex called the mapping cylinder of an arbitrary morphism of complexes \( f \) by letting\n\n\[ \n{M}^{p} = {K}^{p} \oplus {C}^{p - 1} \n\]\n\nand defining \( {\delta }_{M} : {M}^{p} \rightarrow {M}^{p + 1} \) by\n\n\[ \n{\delta }_{M}\left( {x, y}\right) = \left( {{\delta x},... | Yes |
Proposition 1.3. Let \( \mathfrak{F} \) be a complete family which is exact for \( T \) . Let \( f : K \rightarrow C \) be a morphism of complexes, such that \( {K}^{p} \) and \( {C}^{p} \) are in \( \mathfrak{F} \) for all \( p \), and \( {K}^{p},{H}^{p}\left( C\right) \) are zero for all but a finite number of \( p \... | Proof. Construct the mapping cylinder \( M \) for \( f \) . As in the proof of Proposition 1.2, we get \( H\left( M\right) = 0 \) so \( M \) is exact. We then start inductively from the right with zeros. We let \( {Z}^{p} \) be the cycles in \( {M}^{p} \) and use the short exact sequences\n\n\[ 0 \rightarrow {Z}^{p} \r... | Yes |
Theorem 2.1. Let \( M \) be a projective module. Then \( M \) is stably free if and only if \( M \) admits a finite free resolution. | Proof. If \( M \) is stably free then it is trivial that \( M \) has a finite free resolution. Conversely assume the existence of the resolution with the above notation. We prove that \( M \) is stably free by induction on \( n \) . The assertion is obvious if \( n = 0 \) . Assume \( n \geqq 1 \) . Insert the kernels a... | Yes |
Proposition 2.2. Let \( M \) be an A-module. Then \( M \) has a finite free resolution of length \( n \geqq 1 \) if and only if \( M \) has a stably free resolution of length \( n \) . | Proof. One direction is trivial, so we suppose given a stably free resolution with the above notation. Let \( 0 \leqq i < n \) be some integer, and let \( {F}_{i},{F}_{i + 1} \) be finite free such that \( {E}_{i} \oplus {F}_{i} \) and \( {E}_{i + 1} \oplus {F}_{i + 1} \) are free. Let \( F = {F}_{i} \oplus {F}_{i + 1}... | Yes |
Lemma 2.4. Let\n\n\[ \n0 \rightarrow K \rightarrow P \rightarrow M \rightarrow 0 \]\n\n\[ \n0 \rightarrow {K}^{\prime } \rightarrow {P}^{\prime } \rightarrow M \rightarrow 0 \]\n\nbe exact sequences where \( P,{P}^{\prime } \) are projective. Then there is an isomorphism\n\n\[ \nK \oplus {P}^{\prime } \approx {K}^{\pri... | Proof. Since \( P \) is projective, there exists a homomorphism \( P \rightarrow {P}^{\prime } \) making the right square in the following diagram commute.\n\n\n\n\n\nThen one can find a homomorphism \( K \rightarrow... | No |
Theorem 2.5. Let \( M \) be a module which admits a stably free resolution of length \( n \)\n\n\[ 0 \rightarrow {E}_{n} \rightarrow \cdots \rightarrow {E}_{0} \rightarrow M \rightarrow 0. \]\n\nLet\n\n\[ {F}_{m} \rightarrow \cdots \rightarrow {F}_{0} \rightarrow M \rightarrow 0 \]\n\nbe an exact sequence with \( {F}_{... | Proof. Insert the kernels and cokernels in each sequence, say\n\n\[ {K}_{m} = \operatorname{Ker}\left( {{E}_{m} \rightarrow {E}_{m - 1}}\right) \;\text{ if }\;m \neq 0 \]\n\n\[ {K}_{0} = \operatorname{Ker}\left( {{E}_{0} \rightarrow M}\right) \]\n\nand define \( {K}_{m}^{\prime } \) similarly. By Lemma 2.3, \( {K}_{m} ... | Yes |
Theorem 2.7. Let\n\n\\[ \n0 \rightarrow {M}^{\prime } \rightarrow M \rightarrow {M}^{\prime \prime } \rightarrow 0 \n\\]\n\nbe an exact sequence. If any two of these modules have a finite free resolution,\nthen so does the third. | Proof. Assume \( {M}^{\prime } \) and \( M \) have finite free resolutions. Since \( M \) is finite, it follows that \( {M}^{\prime \prime } \) is also finite. By essentially the same construction as Chapter XX, Lemma 3.8, we can construct an exact and commutative diagram where \( {E}^{\prime }, E,{E}^{\prime \prime } ... | No |
Theorem 2.9. (Serre). If \( k \) is a field and \( {x}_{1},\ldots ,{x}_{r} \) independent variables, then every finite projective module over \( k\left\lbrack {{x}_{1},\ldots ,{x}_{r}}\right\rbrack \) is stably free, or equivalently admits a finite free resolution. | Proof. By induction and Theorem 2.8 we conclude that every finite module over \( k\left\lbrack {{x}_{1},\ldots ,{x}_{r}}\right\rbrack \) is of finite stably free dimension. (We are using Theorem 2.1.) This concludes the proof. | No |
Theorem 3.1. (Horrocks). Let \( \left( {\mathfrak{o},\mathfrak{m}}\right) \) be a local ring and let \( A = \mathfrak{o}\left\lbrack x\right\rbrack \) be the polynomial ring in one variable over \( \mathfrak{o} \) . Let \( f \) be a unimodular vector in \( {A}^{\left( n\right) } \) such that some component has leading ... | Proof. (Suslin). If \( n = 1 \) or 2 then the theorem is obvious even without assuming that \( \mathfrak{o} \) is local. So we assume \( n \geqq 3 \) and do an induction of the smallest degree \( d \) of a component of \( f \) with leading coefficient 1 . First we note that by the Euclidean algorithm and row operations... | Yes |
Corollary 3.2. Let \( \mathfrak{o} \) be a local ring. Let \( f \) be a unimodular vector in \( \mathfrak{o}{\left\lbrack x\right\rbrack }^{\left( n\right) } \) such that some component has leading coefficient 1. Then \( f \sim f\left( 0\right) \) over \( \mathfrak{o}\left\lbrack x\right\rbrack \) . | Proof. Note that \( f\left( 0\right) \in {\mathfrak{o}}^{\left( n\right) } \) has one component which is a unit. It suffices to prove that over any commutative ring \( R \) any element \( c \in {R}^{\left( n\right) } \) such that some component is a unit is equivalent over \( R \) to \( {e}^{1} \), and this is obvious. | Yes |
Lemma 3.3. Let \( R \) be an entire ring, and let \( S \) be a multiplicative subset. Let \( x, y \) be independent variables. If \( f\left( x\right) \sim f\left( 0\right) \) over \( {S}^{-1}R\left\lbrack x\right\rbrack \), then there exists \( c \in S \) such that \( f\left( {x + {cy}}\right) \sim f\left( x\right) \) ... | Proof. Let \( M \in G{L}_{n}\left( {{S}^{-1}R\left\lbrack x\right\rbrack }\right) \) be such that \( f\left( x\right) = M\left( x\right) f\left( 0\right) \) . Then \( M{\left( x\right) }^{-1}f\left( x\right) = f\left( 0\right) \) is constant, and thus invariant under translation \( x \mapsto x + y \) . Let\n\n\[ G\left... | Yes |
Theorem 3.4. Let \( R \) be an entire ring, and let \( f \) be a unimodular vector in \( R{\left\lbrack x\right\rbrack }^{\left( n\right) } \), such that one component has leading coefficient 1 . Then \( f\left( x\right) \sim f\left( 0\right) \) over \( R\left\lbrack x\right\rbrack \) . | Proof. Let \( J \) be the set of elements \( c \in R \) such that \( f\left( {x + {cy}}\right) \) is equivalent to \( f\left( x\right) \) over \( R\left\lbrack {x, y}\right\rbrack \) . Then \( J \) is an ideal, for if \( c \in J \) and \( a \in R \) then replacing \( y \) by ay in the definition of equivalence shows th... | Yes |
Theorem 3.5. (Quillen-Suslin). Let \( k \) be a field and let \( f \) be a unimodular vector in \( k{\left\lbrack {x}_{1},\ldots ,{x}_{r}\right\rbrack }^{\left( n\right) } \). Then \( f \) has the unimodular extension property. | Proof. By induction on \( r \). If \( r = 1 \) then \( k\left\lbrack {x}_{1}\right\rbrack \) is a principal ring and the theorem is left to the reader. Assume the theorem for \( r - 1 \) variables with \( r \geqq 2 \), and put\n\n\[ R = k\left\lbrack {{x}_{1},\ldots ,{x}_{r - 1}}\right\rbrack .\n\]\n\nWe view \( f \) a... | No |
Theorem 3.6. Let \( A \) be a commutative ring which has the unimodular column extension property. Then every stably free module over \( A \) is free. | Proof. Let \( E \) be stably free. We use induction on the rank of the free modules \( F \) such that \( E \oplus F \) is free. By induction, it suffices to prove that if \( E \oplus A \) is free then \( E \) is free. Let \( E \oplus A = {A}^{\left( n\right) } \) and let\n\n\[ p : {A}^{\left( n\right) } \rightarrow A \... | Yes |
Proposition 4.1. Let \( I = \left( {{x}_{1},\ldots ,{x}_{r}}\right) \) be generated by a regular sequence in \( A \) . Then \( I/{I}^{2} \) is free of dimension \( r \) over \( A/I \) . | Proof. Let \( {\bar{x}}_{i} \) be the class of \( {x}_{i}{\;\operatorname{mod}\;{I}^{2}} \) . It suffices to prove that \( {\bar{x}}_{1},\ldots ,{\bar{x}}_{r} \) are linearly independent. We do this by induction on \( r \) . For \( r = 1 \), if \( \bar{a}\bar{x} = 0 \) , then \( {ax} = b{x}^{2} \) for some \( b \in A \... | Yes |
Lemma 4.2. Notation as above, the homomorphisms \( {f}_{p} \) define a morphism of Koszul complexes:\n\n\n\nand define an isomorphism if \( D \) is a unit in \( A \), for instance if \( \left( y\right) \) is a permut... | Proof. By definition\n\n\[ f\left( {{e}_{{i}_{1}}^{\prime } \land \cdots \land {e}_{{i}_{p}}^{\prime }}\right) = \left( {\mathop{\sum }\limits_{{j = 1}}^{r}{c}_{{i}_{1}j}{e}_{j}}\right) \land \cdots \land \left( {\mathop{\sum }\limits_{{j = 1}}^{r}{c}_{{i}_{p}j}{e}_{j}}\right) . \]\n\nThen\n\n\[ {fd}\left( {{e}_{{i}_{1... | Yes |
Proposition 4.3. There is a natural isomorphism\n\n\\[ \nK\\left( {{x}_{1},\\ldots ,{x}_{r}}\\right) \\approx K\\left( {x}_{1}\\right) \\otimes \\cdots \\otimes K\\left( {x}_{r}\\right) .\n\\] | Proof. The proof will be left as an exercise. | No |
Lemma 4.4. Let \( y \in A \) and let \( C \) be a complex as above. Then \( m\left( y\right) \) annihilates \( {H}_{p}\left( {C \otimes K\left( y\right) }\right) \) for all \( p \geqq 0 \) . | Proof. If \( \left( {v, w}\right) \) is a cycle, i.e. \( d\left( {v, w}\right) = 0 \), then from (7) we get at once that \( \left( {{yv},{yw}}\right) = d\left( {0,{\left( -1\right) }^{p}v}\right) \), which proves the lemma. | No |
Let \( {x}_{1},\ldots ,{x}_{r} \) be a regular sequence for \( M \) . Then \( {H}_{p}K\left( {x;M}\right) = 0 \) for \( p > 0 \) . (Of course, \( {H}_{0}K\left( {x;M}\right) = M/{IM} \) .) In other words, the augmented Koszul complex is exact. | Proof. We prove (a) by induction on \( r \) . If \( r = 1 \) then \( {H}_{1}\left( {x;M}\right) = 0 \) directly from the definition. Suppose \( r > 1 \) . We use the exact sequence of Theorem 4.5(a). If \( p > 1 \) then \( {H}_{p}\left( {x;M}\right) \) is between two homology groups which are 0, so \( {H}_{p}\left( {x;... | Yes |
Theorem 4.7. Let \( {x}_{1},\ldots ,{x}_{r} \) be a regular sequence in \( A \) . Then there is an isomorphism\n\n\[ \n{\varphi }_{x, M} : {H}^{r}\left( {\operatorname{Hom}\left( {K\left( x\right), M}\right) }\right) \rightarrow M/{IM} \n\] \n\nto be described below. | Proof. The module \( {K}_{r}\left( x\right) \) is 1-dimensional, with basis \( {e}_{1} \land \cdots \land {e}_{r} \) . Depending on this basis, we have an isomorphism\n\n\[ \n\operatorname{Hom}\left( {{K}_{r}\left( x\right), M}\right) \approx M \n\] \n\nwhereby a homomorphism is determined by its value at the basis ele... | Yes |
Lemma 4.8. If \( I = \left( x\right) = \left( y\right) \) where \( \left( x\right) ,\left( y\right) \) are two regular sequences, then we have a commutative diagram\n\n\n\nwhere all the maps are isomorphisms of \( A/... | The fact that we are dealing with \( A/I \) -modules is immediate since multiplication by an element of \( A \) commutes with all homomorphisms in sight, and \( I \) annihilates \( A/I \) .\n\nBy Proposition 4.1, we know that \( I/{I}^{2} \) is a free module of rank \( r \) over \( A/I \) . Hence\n\n\[ \mathop{\bigwedg... | Yes |
Theorem 4.10. Let \( {x}_{1},\ldots ,{x}_{r} \) be an \( M \) -regular sequence in \( A \) . Let \( I = \left( x\right) \) . Then \[ {\operatorname{Ext}}^{i}\left( {A/I, M}\right) = 0\;\text{ for }\;i < r. \] | Proof. For the proof, we assume that the reader is acquainted with the exact homology sequence. Assume by induction that \( {\operatorname{Ext}}^{i}\left( {A/I, M}\right) = 0 \) for \( i < r - 1 \) . Then we have the exact sequence \[ 0 = {\operatorname{Ext}}^{i - 1}\left( {A/I, M/{x}_{1}M}\right) \rightarrow {\operato... | Yes |
Theorem 4.11. Let \( I = \left( {{x}_{1},\ldots ,{x}_{r}}\right) \) be an ideal of \( A \) generated by a regular sequence of length \( r \) .\n\n(i) There is a natural isomorphism\n\n\[ \n{\operatorname{Tor}}_{i}^{A}\left( {A/I, A/I}\right) \approx \mathop{\bigwedge }\limits_{{A/I}}^{i}\left( {I/{I}^{2}}\right) ,\;\te... | These isomorphisms will follow from the next considerations.\n\nFirst we use again that the residue classes \( {\bar{x}}_{1},\ldots ,{\bar{x}}_{r}{\;\operatorname{mod}\;{I}^{2}} \) form a basis of \( I/{I}^{2} \) over \( A/I \) . Therefore we have a unique isomorphism of complexes\n\n\[ \n{\varphi }_{x} : K\left( x\rig... | No |
Lemma 4.12. Let \( I = \left( x\right) \supset {I}^{\prime } = \left( y\right) \) be two ideals generated by regular sequences of length \( r \) . Let \( f : K\left( y\right) \rightarrow K\left( x\right) \) be the morphism of Koszul complexes defined in Lemma 4.2. Then the following diagram is commutative: | Proof. We have\n\n\[ \n{\varphi }_{x} \circ \left( {f \otimes \operatorname{can}}\right) \left( {{e}_{{i}_{1}}^{\prime } \land \cdots \land {e}_{{i}_{p}}^{\prime } \otimes 1}\right) \n\] \n\n\[ \n= \mathop{\sum }\limits_{{j = 2}}^{r}{c}_{{i}_{1}j}{\bar{x}}_{j} \land \cdots \land \mathop{\sum }\limits_{{j = 1}}^{r}{c}_{... | Yes |
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