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Theorem 7.4 (Fundamental theorem on symmetric functions). Let \( K \) be a field, and let \( \varphi \in K\left( {{t}_{1},\ldots ,{t}_{n}}\right) \) . Then \( \varphi \) is symmetric if and only if it is a rational function (with coefficients in \( K \) ) of the elementary symmetric functions \( {s}_{1},\ldots ,{s}_{n}... | Proof. Let \( F = K\left( {{t}_{1},\ldots ,{t}_{n}}\right) \), and let \( k = K\left( {{s}_{1},\ldots ,{s}_{n}}\right) \) be the subfield generated by the elementary symmetric functions over \( K \). Then \( F \) is a splitting field of the separable polynomial \( {P}_{n}\left( x\right) \) over \( k \). In particular \... | Yes |
Corollary 7.6. Let \( G \) be a finite group. Then there exists a Galois extension \( k \subseteq F \) such that \( {\operatorname{Aut}}_{k}\left( F\right) \cong G \) . | ## Proof. Exercise 7.4 | No |
Lemma 7.10. Every separable radical extension is contained in a Galois radical extension. | Proof. Let \( k \subseteq F \) be a separable radical extension. In particular \( k \subseteq F \) is finite and separable, so \( F = k\left( \alpha \right) \) for some \( \alpha \in F \) (Proposition 5.19). Let \( p\left( x\right) \) be the minimal polynomial of \( \alpha \) over \( k \) . The splitting field \( L \) ... | Yes |
Lemma 7.11. Let \( k \) be a field of characteristic 0, and let \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) be an irreducible polynomial. Then there exists a formula solving \( f\left( x\right) \) by radicals if and only if the splitting field of \( f\left( x\right) \) is contained in a Galois radical exten... | Proof. If the splitting field of \( f\left( x\right) \) is contained in a radical extension (Galois or not), then the roots may be written as combinations of field operations and radicals, as needed.\n\nFor the converse, assume \( f\left( x\right) \) is solvable by radicals. A formula for a root \( {t}_{1} \) can be tu... | Yes |
Lemma 7.13. Let \( k \subseteq F \) be a Galois extension, with \( \operatorname{char}k = 0 \) . Provided it has enough roots of \( 1, k \subseteq F \) is radical if and only if it is solvable. | Proof. Modulo the fundamental theorem of Galois theory, this is a straightforward generalization of Proposition 6.19\n\nIndeed, assume that \( k \subseteq F \) is radical:\n\n\[ k \subseteq k\left( {\delta }_{1}\right) \subseteq \cdots \subseteq k\left( {{\delta }_{1},\ldots ,{\delta }_{r}}\right) = F \]\n\nwith \( {\d... | Yes |
Corollary 7.16. Let \( k \) be a field of characteristic 0, and let \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) be an irreducible polynomial. Then \( f\left( x\right) \) is solvable by radicals if and only if its Galois group is solvable. | ## Proof. This is an immediate consequence of Lemma 7.11 and Proposition 7.14, | Yes |
Lemma 7.18. Let \( k \) be a field with \( \operatorname{char}k \neq 2 \). If \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) is a separable polynomial with roots \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) in its splitting field, and \( D \) is the discriminant of \( f\left( x\right) \), then the Galois group \(... | Every permutation of the roots fixes \( D \), so \( D \) must be fixed by the whole Galois group; therefore, \( D \in k \). Odd permutations move \( \Delta \) (if \( \operatorname{char}k \neq 2 \)), and even permutations fix it; therefore \( \Delta \) is fixed by the Galois group \( G \) (in other words, \( \Delta \in ... | Yes |
Example 7.19. Lemma 7.18 and a discriminant computation are all that is needed to compute the Galois group of an irreducible cubic polynomial \[ f\left( x\right) = {x}^{3} + a{x}^{2} + {bx} + c. \] | It would be futile to try to remember the discriminant \[ D = {a}^{2}{b}^{2} - 4{a}^{3}c - 4{b}^{3} + {18abc} - {27}{c}^{2}; \] but one may remember the trick of shifting \( x \) by \( a/3 \) (in characteristic \( \neq 3 \) ), with the effect of killing the coefficient of \( {x}^{2} \) : \[ f\left( {x - \frac{a}{3}}\ri... | Yes |
Let \( f\left( x\right) \in \mathbb{Q}\left\lbrack x\right\rbrack \) be an irreducible polynomial of degree \( p \), where \( p \) is prime. Assume that \( f\left( x\right) \) has \( p - 2 \) real roots and 2 nonreal, complex roots. Then the Galois group of \( f\left( x\right) \) is \( {S}_{p} \) . | Indeed, complex conjugation induces an automorphism of the splitting field and acts by interchanging the two nonreal roots, so the Galois group \( G \), as a subgroup of \( {S}_{p} \), contains a transposition. On the other hand, the degree of the splitting field (and hence \( \left| G\right| \) ) is divisible by \( p ... | Yes |
The operation \( {\operatorname{Aut}}_{k}\left( \_ \right) \) from Galois field extensions to groups is contravariantly functorial. | Indeed, if \( k \subseteq E \subseteq F \) is viewed as a morphism of two Galois extensions \( \left( {k \subseteq E\text{to}k \subseteq F}\right) \), we have a corresponding group homomorphism\n\n\[ \n{\operatorname{Aut}}_{k}\left( F\right) \rightarrow {\operatorname{Aut}}_{k}\left( E\right) \n\]\n\ndefined by restric... | Yes |
In [111]4.3] I have defined the spectrum of a commutative ring \( R \) , Spec \( R \), as the set of prime ideals of \( R \) . If \( R, S \) are commutative rings and \( \varphi : R \rightarrow S \) is a ring homomorphism, then the inverse image \( {\varphi }^{-1}\left( \mathfrak{p}\right) \) of a prime ideal \( \mathf... | \[ {\varphi }^{ * } : \operatorname{Spec}\left( S\right) \rightarrow \operatorname{Spec}\left( R\right) \] This assignment is clearly functorial, so we can view Spec as a contravariant functor from the category of commutative rings to Set. | Yes |
If \( \mathrm{C} \) is a category, and let \( X \) be an object of \( \mathrm{C} \). Then the assignments\n\n\[ \nA \mapsto {\operatorname{Hom}}_{\mathsf{C}}\left( {X, A}\right) \;,\;A \mapsto {\operatorname{Hom}}_{\mathsf{C}}\left( {A, X}\right) \n\]\n\ndefine, respectively, covariant and contravariant functors \( \ma... | For example, if\n\n\[ \nA\xrightarrow[]{\alpha }B\xrightarrow[]{\beta }C \n\]\n\nis a diagram in \( \mathrm{C} \), stare at\n\n\n\nEvery \( \alpha : A \rightarrow B \) determines a function\n\n\[ \n{\operatorname{Hom... | No |
Construct a category by taking the objects to be nonnegative integers and \( \operatorname{Hom}\left( {m, n}\right) \) to be the set of \( n \times m \) matrices with entries in a field \( k \), with composition defined by product of matrices (and suitable care concerning matrices with no rows or columns). | The resulting category is equivalent to the category of finite-dimensional \( k \) -vector spaces. Indeed, we obtain a functor from the former to the latter by sending \( n \) to the vector space \( {k}^{n} \) (endowed with the standard basis) and each matrix to the corresponding linear map. This functor is clearly ful... | Yes |
Recall (Definition VII 2.19) that the coordinate ring of an affine algebraic set over a field \( K \) is a reduced, commutative, finite-type \( K \) -algebra. We can define the category \( K \) -Aff of affine \( K \) -algebraic sets by prescribing that the objects be algebraic subsets of some affine \( K \) -space and ... | Thus, \( K \) -Aff is defined in such a way that the functor \( K \) -Aff \( {}^{op} \rightarrow K \) -Alg that maps an affine algebraic set \( S \) to its coordinate ring \( K\left\lbrack S\right\rbrack \) is an equivalence of the opposite category of \( K \) -Aff with the subcategory of reduced, commutative, finite-t... | Yes |
Example 1.11 (Equalizers and kernels). Let I again be a category with two objects \( \mathbf{1},\mathbf{2} \), but assume that morphisms look like this: \n\n--- | That is, add to the discrete category two ’parallel’ morphisms \( \alpha ,\beta \) from one of the objects to the other. A functor \( \mathcal{K} : \mathrm{I} \rightarrow \mathrm{C} \) amounts to the choice of two objects \( {A}_{1},{A}_{2} \) in \( \mathrm{C} \) and two parallel morphisms between them. Limits of such ... | No |
Claim 1.13. The limit \( \underset{i}{\overline{\lim }}{A}_{i} \) exists in \( R \) -Mod. | Proof. The product \( \mathop{\prod }\limits_{i}{A}_{i} \) consists of arbitrary sequences \( {\left( {a}_{i}\right) }_{i > 0} \) of elements \( {a}_{i} \in \) \( {A}_{i} \) . Say that a sequence \( {\left( {a}_{i}\right) }_{i > 0} \) is coherent if for all \( i > 0 \) we have \( {a}_{i} = {\varphi }_{i, i + 1}\left( {... | No |
Example 1.14. If \( \mathrm{C} = \) Set and all the \( {\psi }_{ij} \) are injective, we are talking about a 'nested sequence of sets':\n\n\[ \n{A}_{1} \subseteq {A}_{2} \subseteq {A}_{3} \subseteq {A}_{4} \subseteq \cdots \n\]\n\nthe direct limit of this sequence would be the ’infinite union’ \( \mathop{\bigcup }\limi... | More formally, \( \mathop{\bigcup }\limits_{i}{A}_{i} \) consists of equivalence classes of pairs \( \left( {i,{a}_{i}}\right) \), where \( {a}_{i} \in {A}_{i} \) and \( \left( {i,{a}_{i}}\right) \) is equivalent to \( \left( {j,{a}_{j}}\right) \) for \( i \leq j \) if \( {a}_{j} = {\psi }_{ij}\left( {a}_{i}\right) \) ... | Yes |
For all \( R \) -modules \( N, R{ \otimes }_{R}N \cong N \) . | Indeed, every \( R \) -bilinear \( R \times N \rightarrow P \) factors through \( N \) (as is immediately verified): where \( \otimes \left( {r, n}\right) = {rn} \) . By the uniqueness property of universal objects, necessarily \( N \cong R{ \otimes }_{R}N \) . | No |
Lemma 2.4. For all \( R \) -modules \( M, N, P \), there is an isomorphism of \( R \) -modules\n\n\[{\operatorname{Hom}}_{R}\left( {M,{\operatorname{Hom}}_{R}\left( {N, P}\right) }\right) \cong {\operatorname{Hom}}_{R}\left( {M{ \otimes }_{R}N, P}\right) . | Proof. As noted before the statement, every \( \alpha \in {\operatorname{Hom}}_{R}\left( {M,{\operatorname{Hom}}_{R}\left( {N, P}\right) }\right) \) determines an \( R \) -bilinear map \( \varphi : M \times N \rightarrow P \), by\n\n\[ \left( {m, n}\right) \mapsto \alpha \left( m\right) \left( n\right) .\n\nBy the univ... | No |
For every \( R \) -module \( N \), the functor \( {}_{ - }{ \otimes }_{R}N \) is left-adjoint to the functor \( {\operatorname{Hom}}_{R}\left( {N,}\right) \) . | Proof. The claim is that the isomorphism found in Lemma 2.4 is natural in the sense hinted at, but not fully explained, in [1.5] the interested reader should have no problems checking this naturality. | No |
Corollary 2.7. For any two sets \( A, B \) :\n\n\[ {R}^{\oplus A}{ \otimes }_{R}{R}^{\oplus B} \cong {R}^{\oplus A \times B}. \] | Indeed, 'distributing' the direct sum identifies the left-hand side with the direct sum \( {\left( {R}^{\oplus A}\right) }^{\oplus B} \), which is isomorphic to the right-hand side (Exercise III 6.5). For finitely generated free modules, this simply says that \( {R}^{\oplus m} \otimes {R}^{\oplus n} \cong {R}^{\oplus {... | No |
Corollary 2.8. For all \( R \) -modules \( N \) and all ideals \( I \) of \( R \) ,\n\n\[ \frac{R}{I}{ \otimes }_{R}N \cong \frac{N}{IN} \] | Indeed, \( {}_{ - }{ \otimes }_{R}N \) is right-exact; thus, the exact sequence\n\n\[ 0 \rightarrow I \rightarrow R \rightarrow \frac{R}{I} \rightarrow 0 \]\n\ninduces an exact sequence\n\n\[ I{ \otimes }_{R}N \rightarrow R{ \otimes }_{R}N \rightarrow \frac{R}{I}{ \otimes }_{R}N \rightarrow 0. \]\n\nThe image of \( I{ ... | Yes |
Corollary 2.9. For all ideals \( I, J \) of \( R \) ,\n\n\[ \frac{R}{I}{ \otimes }_{R}\frac{R}{J} \cong \frac{R}{I + J} \] | This follows immediately from Corollary 2.8 and the 'third isomorphism theorem', Proposition III 5.17. Indeed, \( {IR}/J = \left( {I + J}\right) /J \) . | Yes |
Example 2.10. \( \mathbb{Z}/m\mathbb{Z}{ \otimes }_{\mathbb{Z}}\mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}/\gcd \left( {m, n}\right) \mathbb{Z} \) | Indeed, \( \left( m\right) + \left( n\right) = \left( {\gcd \left( {m, n}\right) }\right) \) in \( \mathbb{Z} \) | No |
Multiplication by 2 gives an inclusion\n\n\[ \n{\mathbb{Z}}^{ \subset \cdot 2} \rightarrow \mathbb{Z} \n\]\n\nidentifying the first copy of \( \mathbb{Z} \) with the ideal (2) in the second copy. Tensoring by \( \mathbb{Z}/2\mathbb{Z} \) over \( \mathbb{Z} \) (and keeping in mind that \( R{ \otimes }_{R}N \cong N \) ),... | which sends both [0] and [1] to zero. This is the zero-morphism, and in particular it is not injective. | Yes |
Consider the affine algebraic set \( \mathcal{V}\left( {xy}\right) \) in the plane \( {\mathbb{A}}^{2} \) (over a fixed field \( k \) ) and the ’projection on the first coordinate’ \( \mathcal{V}\left( {xy}\right) \rightarrow {\mathbb{A}}^{1},\left( {x, y}\right) \mapsto x \) : | In terms of coordinate rings (cf. SVII 2.3), this map corresponds to the homomorphism of \( k \) -algebras:\n\n\[ k\left\lbrack x\right\rbrack \rightarrow \frac{k\left\lbrack {x, y}\right\rbrack }{\left( xy\right) } \]\n\ndefined by mapping \( x \) to the coset \( x + \left( {xy}\right) \) (this will be completely clea... | Yes |
Lemma 3.5. Suppose \( M \) is an \( R \) -module, \( N \) is an \( \left( {R, S}\right) \) -bimodule, and \( P \) is an S-module. Then there is a canonical isomorphism of abelian groups | Proof. Every element \( \alpha \in {\operatorname{Hom}}_{R}\left( {M,{\operatorname{Hom}}_{S}\left( {N, P}\right) }\right) \) determines a map\n\n\[ \varphi : M \times N \rightarrow P \]\n\nvia \( \varphi \left( {m,\_ }\right) \mathrel{\text{:=}} \alpha \left( m\right) ;\varphi \) is clearly \( \mathbb{Z} \) -bilinear.... | Yes |
Proposition 3.6. Let \( f : R \rightarrow S \) be a homomorphism of commutative rings. Then, with notation as above, \( {f}_{ * } \) is right-adjoint to \( {f}^{ * } \) and left-adjoint to \( {f}^{!} \) . In particular, \( {f}_{ * } \) is exact, \( {f}^{ * } \) is right-exact, and \( {f}^{!} \) is left-exact. | Proof. Let \( M \), resp., \( N \), be an \( R \) -module, resp., an \( S \) -module. Note that, trivially, \( {\operatorname{Hom}}_{S}\left( {S, N}\right) \) is canonically isomorphic to \( N \) (as an \( S \) -module) and to \( {f}_{ * }\left( N\right) \) (as an \( R \) -module). Thus \( {}^{15} \n\n\[ \n{\operatorna... | Yes |
Lemma 4.2. Let \( \varphi : {M}^{\ell } \rightarrow P \) be an \( R \) -multilinear function.\n\nIf \( \varphi \) is alternating, then for all \( \sigma \in {S}_{\ell } \), and all \( {m}_{1},\ldots ,{m}_{\ell } \), \n\n\[ \varphi \left( {{m}_{\sigma \left( 1\right) },\ldots ,{m}_{\sigma \left( \ell \right) }}\right) =... | Proof. For the first statement, it suffices to show that interchanging any two factors switches the sign of an alternating function (since transpositions generate the symmetric group). Since the other factors have no effect on this operation, this reduces the question to the case \( \ell = 2 \) . Therefore, we only hav... | Yes |
Lemma 4.3. Let \( R \) be a commutative ring, and let \( M \) be a free \( R \) -module of rank \( r \) . Then \( {\Lambda }_{R}^{\ell }\left( M\right) \) is a free \( R \) -module of rank \( \left( \begin{array}{l} r \\ \ell \end{array}\right) \) . | Proof. There are \( \left( \begin{array}{l} r \\ \ell \end{array}\right) \) sequences of indices \( {i}_{1},\ldots ,{i}_{\ell } \) satisfying \( 1 \leq {i}_{1} < \cdots < {i}_{\ell } \leq r \) , so we just need to show that the generators \( {e}_{{i}_{1}} \land \cdots \land {e}_{{i}_{\ell }} \) are linearly independent... | Yes |
For \( V = {k}^{4},{\Lambda }_{k}^{2}\left( V\right) \) has dimension \( \left( \begin{array}{l} 4 \\ 2 \end{array}\right) = 6 \) . On ’pure wedges’ \( {a}_{1} \land {a}_{2} \), the isomorphism \( {\mathbb{A}}_{k}^{2}\left( V\right) \rightarrow {k}^{6} \) works as follows. View the vectors \( {a}_{1},{a}_{2} \) as the ... | \[ A = \left( \begin{array}{ll} {a}_{1}^{1} & {a}_{2}^{1} \\ {a}_{1}^{2} & {a}_{2}^{2} \\ {a}_{1}^{3} & {a}_{2}^{3} \\ {a}_{1}^{4} & {a}_{2}^{4} \end{array}\right) \mapsto \left( \begin{array}{l} {a}_{1}^{1}{a}_{2}^{2} - {a}_{1}^{2}{a}_{2}^{1} \\ {a}_{1}^{1}{a}_{2}^{3} - {a}_{1}^{3}{a}_{2}^{1} \\ {a}_{1}^{1}{a}_{2}^{4}... | Yes |
The polynomial ring \( R\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) over any ring \( R \) carries a natural grading, given by the (ordinary) degree of polynomials. | We may write\n\n\[ R\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack = R \oplus \left\langle {{x}_{1},\ldots ,{x}_{n}}\right\rangle \oplus \left\langle {{x}_{1}^{2},{x}_{1}{x}_{2},\ldots ,{x}_{n}^{2}}\right\rangle \oplus \cdots .\n\] | Yes |
Lemma 4.9. Let \( S = {\bigoplus }_{i}{S}_{i} \) be a graded ring and let \( I \subseteq S \) be an ideal of \( S \) . Then the following are equivalent:\n\n(i) I is homogeneous;\n\n(ii) if \( s \in S \) and \( s = \mathop{\sum }\limits_{i}{s}_{i} \) is the decomposition of \( s \) into homogeneous elements \( {s}_{i} ... | Proof. (i) \( \Leftrightarrow \) (ii) is the very definition of homogeneous ideal; (ii) \( \Leftrightarrow \) (iii) is left to the reader (Exercise 4.10).\n\n(ii) \( \Rightarrow \) (iv): Assuming (ii) holds, define a grading on \( S/I \) by letting the piece of degree \( i \) consist of (0 and) the cosets of the elemen... | No |
The ideal \( I = \left( {y - {x}^{2}}\right) \) is not homogeneous in the ring \( k\left\lbrack {x, y}\right\rbrack \) , if this is given the grading by the usual degree. | indeed, \( y - {x}^{2} \in I \) while \( y \notin I \) , contradicting condition (ii) of Lemma 4.9. | No |
Lemma 4.12. Let \( {I}_{\mathbb{S}},{I}_{\mathbb{A}} \subseteq {\mathbb{T}}_{R}^{ * }\left( M\right) \) be the ideals respectively generated by all elements of the form \( \left( {m \otimes n - n \otimes m}\right) \) as \( m, n \in M \) and by elements of the form \( m \otimes m \) as \( m \in M \) . Then\n\n\[ \n{\mat... | Proof. We have observed in [4.3] that the kernel of a graded homomorphism is the direct sum of the kernels of the induced homomorphisms in each degree; the statement of the lemma then follows easily from the explicit descriptions of the kernels of the canonical projections from the tensor powers to the symmetric and ex... | No |
Proposition 4.16. Let \( R \) be a commutative ring, and let \( M \) be an \( R \) -module. Then for every \( R \) -algebra \( A \) and every \( R \) -module homomorphism \( \lambda : M \rightarrow A \) such that \( \lambda {\left( m\right) }^{2} = 0\forall m \in M \), there exists a unique homomorphism of \( R \) -alg... | Details may now safely be left to the reader (who may for example establish the first proposition from the universal property of tensor powers and then deduce the second and third from Lemma 4.12). | No |
The free case is particularly easy to understand. For instance,\n\n\[ \n{\mathbb{S}}_{R}^{ * }\left( {R}^{\oplus r}\right) \cong R\left\lbrack {{x}_{1},\ldots ,{x}_{r}}\right\rbrack \n\] | Indeed, the polynomial ring satisfies the appropriate universal property with respect to mapping to commutative rings (cf. [111,2.2] and [111,6.4]). Likewise, \( {\mathbb{T}}_{R}^{ * }\left( {R}^{\oplus r}\right) \) should be thought of as a ’noncommutative’ polynomial ring, in which the \( r \) inde-terminates do not ... | No |
Proposition 5.2. For every \( R \) -module \( N \), the functor \( {\operatorname{Hom}}_{R}\left( {\_, N}\right) \) is right-adjoint to itself. | Proof. Let \( L, M, N \) denote \( R \) -modules. Recall (cf. the considerations preceding Lemma 2.4) that \( R \) -bilinear maps\n\n\[ \varphi : L \times M \rightarrow N \]\n\nmay be identified with \( R \) -linear maps\n\n\[ L \rightarrow {\operatorname{Hom}}_{R}\left( {M, N}\right) \]\n\nBy the same token, they may ... | Yes |
Proposition 5.5. Let \( M \) be any \( R \) -module, and let \( F \) be a free \( R \) -module of finite rank. Then\n\n\[ \n{\operatorname{Hom}}_{R}\left( {M, F}\right) \cong {M}^{ \vee }{ \otimes }_{R}F.\n\] | Proof. By hypothesis \( F \cong {R}^{\oplus n} \cong {R}^{n} \) ; hence\n\n\( {\operatorname{Hom}}_{R}\left( {M, F}\right) \cong {\operatorname{Hom}}_{R}\left( {M,{R}^{n}}\right) \cong {\operatorname{Hom}}_{R}{\left( M, R\right) }^{n} \cong {\operatorname{Hom}}_{R}\left( {M, R}\right) { \otimes }_{R}{R}^{n} \cong {M}^{... | Yes |
Corollary 5.7. The dual of a free module is isomorphic to a product of copies of \( R \) :\n\n\[ \n{\left( {R}^{\oplus S}\right) }^{ \vee } \cong {R}^{S} \n\] \n\nIn particular, \( {\left( {R}^{n}\right) }^{ \vee } \cong {R}^{n} \) : if \( F \) is a free \( R \) -module of finite rank, then \( {F}^{ \vee } \cong F \) . | Proof. This follows from Lemma 5.6 and the fact that \( {R}^{ \vee } = {\operatorname{Hom}}_{R}\left( {R, R}\right) \) is isomorphic to \( R \) . | No |
To see that these isomorphisms do depend on the choice of the basis, consider the standard basis \( \left( {{\mathbf{e}}_{1},{\mathbf{e}}_{2}}\right) \) of \( {R}^{2} \) and the corresponding dual basis \( \left( {{\check{\mathbf{e}}}_{1},{\check{\mathbf{e}}}_{2}}\right) \) of \( {\left( {R}^{2}\right) }^{ \vee } \), a... | By definition\n\n\[ \n{\check{\mathbf{e}}}_{1}^{\prime }\left( {\mathbf{e}}_{2}^{\prime }\right) = 0 \n\]\n\nwhile\n\n\[ \n{\check{\mathbf{e}}}_{1}\left( {\mathbf{e}}_{2}^{\prime }\right) = {\check{\mathbf{e}}}_{1}\left( {{\mathbf{e}}_{1} + {\mathbf{e}}_{2}}\right) = 1 + 0 = 1. \n\]\n\nTherefore, \( {\check{\mathbf{e}}... | Yes |
Lemma 5.12. The duality functor is left-exact: every exact sequence\n\n\\[ \nL \rightarrow M \rightarrow N \rightarrow 0 \n\\]\n\nof \\( R \\) -modules induces an exact sequence\n\n\\[ \n0 \rightarrow {N}^{ \\vee } \rightarrow {M}^{ \\vee } \rightarrow {L}^{ \\vee } \n\\] | Proof. This is an immediate consequence of the left-exactness of Hom. | No |
Proposition 5.13. Let\n\n\[ \n0 \rightarrow M\xrightarrow[]{\;\mu \;}N\xrightarrow[]{\;\nu \;}P \rightarrow 0 \n\] \n\nbe an exact sequence of \( R \) -modules, with \( P \) free. Then the induced sequence \n\n\[ \n0 \rightarrow {P}^{ \vee }\xrightarrow[]{{\nu }^{ \vee }}{N}^{ \vee }\xrightarrow[]{{\mu }^{ \vee }}{M}^{... | Proof. Lemma 5.12 takes care of all but the surjectivity of the map \( {N}^{ \vee } \rightarrow {M}^{ \vee } \) induced from \( M \rightarrow N \) :\n\nThe question is whether every \( R \) -linear \( f : M \rightarrow R \) can be extended to an \( R \) -linear map \( g : N \rightarrow R \) so that \( f = g \circ \mu \... | Yes |
Lemma 5.15. Let \( A \) be the matrix representing a linear map \( \alpha : {R}^{n} \rightarrow {R}^{m} \) with respect to the standard bases. Then the dual map \( {\alpha }^{ \vee } : {\left( {R}^{m}\right) }^{ \vee } \rightarrow {\left( {R}^{n}\right) }^{ \vee } \) is represented by the transpose of \( A \) with resp... | The (easy) verification of this fact is left to the reader (Exercise 5.10). | No |
Proposition 5.16. Let \( R \) be an integral domain, and let \( M \) be an \( R \)-module. Then \( {M}^{ \vee } \) is torsion-free. | Proof. There is a surjection \( {R}^{\oplus S} \rightarrow M \), thus, an exact sequence\n\n\[ \n{R}^{\oplus T} \rightarrow {R}^{\oplus S} \rightarrow M \rightarrow 0.\n\]\n\nDualizing, \( {M}^{ \vee } \) is realized as the kernel of the induced map \( {R}^{S} \rightarrow {R}^{T} \) ; hence \( {M}^{ \vee } \) may be id... | No |
Lemma 6.2. An R-module \( P \) is projective if and only if for all epimorphisms of \( R \) -modules \( \mu : M \rightarrow N \), every \( R \) -linear map \( p : P \rightarrow N \) lifts to an \( R \) -linear map \( \widehat{p} : P \rightarrow M \). | Proof. This is straightforward. Since \( {\operatorname{Hom}}_{R}\left( {\_, Q}\right) \) is left-exact for all \( Q, Q \) is injective if and only if whenever a sequence\n\n\[ 0 \rightarrow L \rightarrow M \]\n\nis exact, then so is the induced sequence\n\n\[ {\operatorname{Hom}}_{R}\left( {M, Q}\right) \rightarrow {\... | No |
For example, assume that \( P \) is projective; then I claim that every exact sequence\n\n\[ 0 \rightarrow L\xrightarrow[]{\lambda }M\xrightarrow[]{\mu }P \rightarrow 0 \]\n\nsplits, in the sense that there is a submodule \( {P}^{\prime } \) of \( M \) such that \( \mu \) restricts to an isomorphism \( {P}^{\prime } \r... | Indeed, since \( P \) is projective, then the identity \( P\overset{ \equiv }{ \rightarrow }P \) lifts to a homomorphism \( \rho : P \rightarrow M \), and the reader can then verify that \( {P}^{\prime } = \rho \left( P\right) \) fits the requirement. Loosely speaking, in this situation we can simply replace \( M \) by... | No |
Proposition 6.4. An R-module \( P \) is projective if and only if it is a direct summand of a free module, that is, if and only if there exists a free \( R \) -module \( F \), an \( R \) - module \( K \), and an isomorphism \( K \oplus P \cong F \) . | Proof. Any set \( S \) of generators of \( P \) determines a surjection of the free module \( F = {R}^{\oplus S} \) onto \( P \) and hence an exact sequence\n\n\[ 0 \rightarrow K \rightarrow F \rightarrow P \rightarrow 0. \]\n\nAs observed above, such a sequence necessarily splits if \( P \) is projective; thus \( F \c... | No |
Let \( {P}_{1},{P}_{2} \) be projective \( R \) -modules. Then \( {P}_{1} \oplus {P}_{2} \) and \( {P}_{1}{ \otimes }_{R}{P}_{2} \) are projective. Projective modules are flat. | These statements follow easily from Proposition 6.4, the fact that \( \otimes \) is distributive with respect to \( \oplus \), and the fact that free modules are flat. | No |
Theorem 6.6. An R-module \( Q \) is injective if and only if every \( R \) -linear map \( f \) : \( I \rightarrow Q \), with \( I \) an ideal of \( R \), extends to an \( R \) -linear map \( \widehat{f} : R \rightarrow Q \) . | Proof. The 'only if' part of the statement is immediate from the definition of injective. To verify the ’if’ part, assume \( Q \) satisfies the stated extension condition, let \( L \subseteq M \) be any inclusion of \( R \) -modules, and let \( q : L \rightarrow Q \) be a given \( R \) -linear map:\n\n![cc115a52-9d62-4... | Yes |
Corollary 6.7. Let \( R \) be a PID. Then an \( R \) -module \( Q \) is injective if and only if it is divisible. | ## Proof. Exercise 6.14. | No |
Example 6.8. Viewed as abelian groups (i.e., \( \mathbb{Z} \) -modules), \( \mathbb{Q} \) and \( \mathbb{Q}/\mathbb{Z} \) are injective. More generally, if \( D \) is any divisible abelian group and \( K \subseteq D \), then \( D/K \) is injective. | Indeed, it is trivially divisible! | No |
Lemma 6.9. Let \( f : S \rightarrow R \) be a homomorphism of commutative rings, and let \( Q \) be an injective \( S \) -module. Then \( {f}^{!}\left( Q\right) \) is an injective \( R \) -module. | Proof. By adjunction (Lemma 3.5),\n\n\[ \n{\operatorname{Hom}}_{R}\left( {\_ ,{f}^{!}\left( Q\right) }\right) \cong {\operatorname{Hom}}_{S}\left( {{f}_{ * }\left( \_ \right), Q}\right) \n\]\n\nas functors \( R \) -Mod \( \rightarrow \mathrm{{Ab}} \) . Since \( {f}_{ * } \) is exact (Proposition 3.6) and \( {\operatorn... | Yes |
Corollary 6.12. Let \( M \) be an \( R \) -module. Then \( M \) can be identified with a submodule of an injective \( R \) -module. | Proof. I claim that it suffices to show that \( \mathbb{Z} \) -Mod has enough injectives, Indeed, this will show that there exists a divisible abelian group \( D \) such that \( M \subseteq D(M \) is in particular an abelian group); since \( R \) -linear maps are in particular \( \mathbb{Z} \) -linear,\n\n\[ M \cong {\... | Yes |
Proposition 6.14. An \( R \) -module \( P \) is projective if and only if \( {\operatorname{Ext}}_{R}^{1}\left( {P,\_ }\right) = 0 \), if and only if \( {\operatorname{Ext}}_{R}^{i}\left( {P,\_ }\right) = 0 \) for all \( i > 0 \) . | Proof. The second assertion: we have seen that \( Q \) is injective if \( {\operatorname{Ext}}_{R}^{1}\left( {\_, Q}\right) = 0 \) , and this is trivially the case if \( {\operatorname{Ext}}_{R}^{i}\left( {\_, Q}\right) = 0 \) for all \( i > 0 \) . So we just have to prove that \( {\operatorname{Ext}}_{R}^{i}\left( {\_... | Yes |
Lemma 1.3. A morphism \( \varphi : A \rightarrow B \) in an additive category is a monomorphism if and only if for all \( \zeta : Z \rightarrow A \) , \[ \varphi \circ \zeta = 0 \Rightarrow \zeta = 0. \] It is an epimorphism if and only if for all \( \beta : B \rightarrow Z \) , \[ \beta \circ \varphi = 0 \Rightarrow \... | Proof. This is simply because two morphisms with the same source and target are equal if and only if their difference in the corresponding Hom-set (which is an abelian group by hypothesis) is 0. | No |
Lemma 1.4. In any additive category, kernels are monomorphisms and cokernels are epimorphisms. | Proof. Let \( \varphi : A \rightarrow B \) be a morphism in an additive category \( \mathrm{A} \), and let \( \operatorname{coker}\varphi \) : \( B \rightarrow C \) be its cokernel. Let \( \gamma : C \rightarrow Z \) be a morphism such that \( \gamma \circ \operatorname{coker}\varphi = 0 \) . The composition \( \left( ... | No |
Lemma 1.5. Let \( \varphi : A \rightarrow B \) be a morphism in an additive category. Then \( \varphi \) is a monomorphism if and only if \( 0 \rightarrow A \) is its kernel, and \( \varphi \) is an epimorphism if and only if \( B \rightarrow 0 \) is its cokernel. | ## Proof. Let's do kernels this time.\n\nFirst assume \( \varphi : A \rightarrow B \) is a monomorphism. If \( \zeta : Z \rightarrow A \) is any morphism such that the composition \( Z \rightarrow A \rightarrow B \) is 0, then \( \zeta \) is 0 by Lemma 1.3, and in particular \( \zeta \) factors (uniquely) through \( 0 ... | No |
In an abelian category \( \mathrm{A} \), every cokernel is the cokernel of its kernel. | Let \( \varphi : A \rightarrow B \) be the cokernel of some morphism \( Z \rightarrow A \) ; since \( \mathrm{A} \) is abelian, \( \varphi \) has a kernel \( \iota : K \rightarrow A \) . The composition \( Z \rightarrow A \rightarrow B \) is 0, so \( Z \rightarrow A \) factors through \( \iota \) by definition of kerne... | Yes |
Lemma 1.9. Let \( \varphi : A \rightarrow B \) be a morphism in an abelian category \( \mathrm{A} \), and assume that \( \varphi \) is both a monomorphism and an epimorphism. Then \( \varphi \) is an isomorphism. | Proof. By Lemma 1.5 the kernel of \( \varphi \) is \( 0 \rightarrow A \), since \( \varphi \) is a monomorphism. Similarly, \( B \rightarrow 0 \) is a cokernel of \( \varphi \) . Further, \( \varphi \) is the cokernel of \( 0 \rightarrow A \) and the kernel of \( B \rightarrow 0 \), by Lemma 1.8 .\n\nNow consider the i... | Yes |
For instance, fibered products (or 'pull-backs') exist in any abelian category, just as in \( R \) -Mod (cf. Exercise III 6.10). Consider a diagram\nin an abelian category. The fibered product of \( A \) and \( B \) over \( C \) is an object \( A{ \times }_{C}B \) with morphisms to \( A \) and \( B \), completing the c... | The fibered product may be constructed in this context, just as in the particular case of \( R \) -Mod, as the kernel of the difference of the two morphisms\nwhere \( {\rho }_{A},{\rho }_{B} \) are the morphisms making \( A \times B \) a product. The reader will prove that these ’fiber squares’ preserve kernels, in the... | No |
Lemma 1.14. Let \( \varphi : A \rightarrow B \) be a morphism in an abelian category, and let \( \iota : K \rightarrow B \) be the kernel of the cokernel of \( \varphi \) . Then\n\n- \( \iota \) is a monomorphism;\n\n- \( \varphi \) factors through \( \iota \) ; and\n\n- \( \iota \) is initial with these properties. | By Lemma 1.4, \( \iota \) is a monomorphism. It is clear that \( \varphi \) factors through \( \iota \) : the composition \( A \rightarrow B \rightarrow \operatorname{coker}\varphi \) is the zero-morphism, so there is a naturally induced \( A \rightarrow K \) by the universal property of kernels. The more interesting p... | Yes |
Lemma 1.16. Let \( \varphi : A \rightarrow B \) be a morphism in an abelian category, and let \( \operatorname{im}\varphi : K \rightarrow B,\operatorname{coim}\varphi : A \rightarrow C \) be its image and coimage, respectively. Then the induced morphisms \( A \rightarrow K \) and \( C \rightarrow B \) are, respectively... | Proof. As usual, I will prove half of the statement and leave the other half to the reader (Exercise 1.20)\n\nTo verify that \( \bar{\varphi } : A \rightarrow K \) is an epimorphism, consider its image \( {K}^{\prime } \rightarrow K \) :\n\n\[ \n{K}^{\prime } \succ \xrightarrow[]{\text{ im }\bar{\varphi } = \ker \opera... | No |
For a slightly more interesting example, consider a diagram and the associated sequence obtained by letting \( A \oplus B \) play both roles of product and coproduct. Then - the diagram is commutative if and only if this sequence is a complex; | Indeed, the first assertion is trivial; | No |
Lemma 2.3. Let\n\n\n\nbe a fibered diagram in an abelian category, and assume \( \\varphi \) is an epimorphism. Then \( {\\varphi }^{\\prime } \) is also an epimorphism. | Proof. First, observe that if \( \\varphi : A \\rightarrow C \) is an epimorphism, so is the map \( A \\oplus B \\rightarrow \) \( C \) considered in Example 2.2. Since epimorphisms are cokernels in an abelian category and cokernels are cokernels of their kernels (Lemma 1.8), we see that \( A \\oplus B \\rightarrow C \... | Yes |
Lemma 2.5. \( z \sim 0 \Leftrightarrow z = 0 \). Further, a morphism \( \varphi : A \rightarrow B \) in \( \mathrm{A} \) is 0 if and only if \( \widehat{\varphi }\left( z\right) = 0 \) for all \( z \in \widehat{A} \). | Proof. According to the definition given above, \( z : Z \rightarrow A \) is equivalent to 0 if and only if there is an epimorphism \( W \rightarrow Z \) making the following diagram commute:\n\n\n\nSince \( W \right... | Yes |
Lemma 2.6. Let \( \varphi : A \rightarrow B \) be a morphism in A. Then\n\n- \( \varphi \) is a monomorphism if and only if \( \widehat{\varphi } \) is injective;\n\n- \( \varphi \) is an epimorphism if and only if \( \widehat{\varphi } \) is surjective. | Proof. As usual, I will propose a division of labor: the reader will prove the first statement (Exercise 2.6), and I will prove the second.\n\nAssume \( \varphi \) is an epimorphism, and let \( z : Z \rightarrow B \) represent an arbitrary ’element’ of \( \widehat{B} \) . Consider the fiber product:\n\n![cc115a52-9d62-... | No |
Lemma 2.7. With notation as above, let \( \varphi : A \rightarrow B \) be a morphism in a small abelian category \( \mathrm{A} \), and let \( \widehat{\varphi } : \widehat{A} \rightarrow \widehat{B} \) be the corresponding function of pointed sets. Let \( \ker \varphi : K \rightarrow A \), resp., \( \operatorname{im}\v... | Proof. These statements are very close to the universal properties satisfied by kernel and image.\n\nThe reader will verify the statement about the kernel (Exercise 2.7). For the image, recall that we have a decomposition of \( \varphi \) ,\n\n\[ \varphi : A \rightarrow I\overset{\operatorname{im}\varphi }{ \mapsto }B ... | No |
Proposition 2.8. Let \( \mathrm{A} \) be a small abelian category. Then a sequence\n\n\[ A\overset{\varphi }{ \rightarrow }B\overset{\psi }{ \rightarrow }C \]\n\nin \( \mathrm{A} \) is exact if and only if the corresponding sequence\n\n\[ \widehat{A}\xrightarrow[]{\widehat{\varphi }}\widehat{B}\xrightarrow[]{\widehat{\... | Proof. This now follows immediately from Lemma 2.7 exactness in A means that im \( \varphi = \ker \psi \), and exactness in \( {\operatorname{Set}}^{ * } \) means that the image of \( \widehat{\varphi } \) equals \( {\widehat{\psi }}^{-1}\left( 0\right) \) . By Lemma 2.7, these conditions are equivalent. | Yes |
Theorem 2.9 (Freyd-Mitchell theorem). Let \( \mathrm{A} \) be a small abelian category. Then there is a fully faithful, exact functor \( \mathrm{A} \rightarrow R \) -Mod for a suitable ring \( R \) . | This functor is fully faithful: this means that one can in fact construct morphisms in an arbitrary (small) abelian category by working with elements. Indeed, this amounts to constructing the appropriate morphisms in the ambient category \( R \) -Mod, and fullness guarantees that these morphisms ’already’ exist in A. | No |
Lemma 3.3. \( \mathrm{C}\left( \mathrm{A}\right) \) is an abelian category. | I will leave to the reader the careful verification of this fact (Exercise 3.3). In broad terms, morphisms between two given complexes form an abelian group, essentially because if \( {\alpha }^{i} \) and \( {\beta }^{i} : {M}^{i} \rightarrow {N}^{i} \) are both collections of morphisms making the appropriate diagram c... | No |
For every integer \( i \), the assignment\n\n\[ \n{H}^{i} : {M}^{ \bullet } \mapsto {H}^{i}\left( {M}^{ \bullet }\right) \n\]\n\ndefines an additive covariant functor \( \mathrm{C}\left( \mathrm{A}\right) \rightarrow \mathrm{A} \) . | Of course, the statement means that each \( {H}^{i} \) induces in a natural (and functorial) way homomorphisms of abelian groups\n\n\[ \n{\operatorname{Hom}}_{\mathrm{C}\left( \mathrm{A}\right) }\left( {{M}^{ \bullet },{N}^{ \bullet }}\right) \rightarrow {\operatorname{Hom}}_{A}\left( {{H}^{i}\left( {M}^{ \bullet }\rig... | Yes |
Theorem 3.5 (Long exact cohomology sequence). The sequence determined as above by a short exact sequence of complexes is an exact sequence. | Proof. The proof is a diagram chase, which everyone should perform once by oneself in his or her lifetime. So it is mostly left as an exercise for the reader (Exercise 3.9). But I will stress the extent to which \( {H}^{i} \) is exact ’on the nose’: part of the claim in this theorem is that if\n\n\[ 0 \rightarrow {L}^{... | No |
Proposition 4.1. There is an exact triangle\n\n\n\nwhere the connecting morphism \( \delta \) is the morphism induced by \( {\alpha }^{ \bullet } \) in cohomology. | Proof. The existence of the triangle is a direct consequence of Theorem 3.5 all we have to check is that the connecting morphism indeed agrees with the morphism\n\ninduced by \( {\alpha }^{ \bullet } \) . Chasing the diagram\n\n be a morphism of cochain complexes. Then the induced morphism \( {H}^{ \bullet }\left( {L}^{ \bullet }\right) \rightarrow {H}^{ \bullet }\left( {M}^{ \bullet }\right) \) is an isomorphism if and only if the mapping cone \( {MC}... | ## (Cf. Exercise 3.14.) | No |
The datum of a resolution \( {M}^{ \bullet } \) of an object \( A \) of an abelian category A, as in Definition 3.2 and with \( {M}^{i} = 0 \) for \( i > 0 \), is the same as the datum of a quasi-isomorphism\n\n\[ \n{M}^{ \bullet }\xrightarrow[]{\text{ q-iso. }}\iota \left( A\right) \n\]\n\nwhere \( \iota \) places \( ... | Thus, quasi-isomorphisms may be viewed as generalizations of more simpleminded resolutions. Also note that the mapping cone of a resolution as in Example 4.4 is obtained (as the reader should check) by shifting the complex 'one step to the left’ and completing it with \( A \), obtaining the exact complex:\n\n\[ \n\cdot... | No |
Let \( {M}^{ \bullet } \) be an exact complex in \( \mathrm{C}\left( \mathrm{A}\right) \) . Then the complex \( \mathcal{F}\left( {M}^{ \bullet }\right) \), obtained by applying \( \mathcal{F} \) to the objects and morphisms of \( {M}^{ \bullet } \), is a zero-object in \( \mathrm{D} \) . | To verify the first claim, note that since \( {M}^{ \bullet } \) is exact, the zero-morphism: \( {M}^{ \bullet } \rightarrow \) \( {M}^{ \bullet } \) is a quasi-isomorphism; hence it is mapped to an invertible morphism by \( \mathcal{F} \) :\n\n\[ \mathcal{F}\left( {M}^{ \bullet }\right) \underset{{\operatorname{id}}_{... | No |
Proposition 4.10. If \( {\alpha }^{ \bullet },{\beta }^{ \bullet } : {L}^{ \bullet } \rightarrow {M}^{ \bullet } \) are homotopic morphisms of complexes, then \( {\alpha }^{ \bullet },{\beta }^{ \bullet } \) induce the same morphisms on cohomology: \( {H}^{ \bullet }\left( {L}^{ \bullet }\right) \rightarrow {H}^{ \bull... | Proof. Let \( \bar{\ell } \in {H}^{i}\left( {L}^{ \bullet }\right) \) . Then \( \bar{\ell } \) is represented by an element \( \ell \in \ker \left( {d}_{{L}^{ \bullet }}^{i}\right) \), and its images in \( {H}^{i}\left( {M}^{ \bullet }\right) \) under the morphisms induced by \( {\alpha }^{ \bullet },{\beta }^{ \bullet... | Yes |
Corollary 4.11. Homotopy equivalent complexes have isomorphic cohomology. | Proof. Indeed, morphisms \( {\alpha }^{ \bullet } : {L}^{ \bullet } \rightarrow {M}^{ \bullet },{\beta }^{ \bullet } : {M}^{ \bullet } \rightarrow {L}^{ \bullet } \) such that \( {\beta }^{ \bullet } \circ {\alpha }^{ \bullet } \) and \( {\alpha }^{ \bullet } \circ {\beta }^{ \bullet } \) are both homotopic to the iden... | Yes |
Lemma 4.13. With \( \mathcal{F} \) as above, if \( {\alpha }^{ \bullet } \sim {\beta }^{ \bullet } \) in \( \mathrm{C}\left( \mathrm{A}\right) \), then \( \mathrm{C}\left( \mathcal{F}\right) \left( {\alpha }^{ \bullet }\right) \sim \mathrm{C}\left( \mathcal{F}\right) \left( {\beta }^{ \bullet }\right) \) in \( \mathrm{... | Proof. The second assertion follows from the first. The first is an immediate consequence of the fact that \( \mathcal{F} \) is additive. Indeed, if \( h \) is a homotopy between \( {\alpha }^{ \bullet },{\beta }^{ \bullet } : {L}^{ \bullet } \rightarrow {M}^{ \bullet } \), then\n\n\[ \n{\beta }^{i} - {\alpha }^{i} = {... | Yes |
Theorem 4.14. Let \( \mathcal{F} : \mathrm{A} \rightarrow \mathrm{B} \) be an additive functor between two abelian categories. If \( {L}^{ \bullet },{M}^{ \bullet } \) are homotopy equivalent complexes in \( \mathrm{C}\left( \mathrm{A}\right) \), then the cohomology complexes\n\n\[ \n{H}^{ \bullet }\left( {\mathrm{C}\l... | The proof of this statement is essentially immediate after all our preparatory work, so it is left to the reader (Exercise 4.16). | No |
Lemma 5.3. Let \( \mathrm{A} \) be an abelian category. Then the homotopic category \( \mathrm{K}\left( \mathrm{A}\right) \) of complexes is an additive category. | ## Proof. Exercise 5.1. | No |
Proposition 5.4. Let \( \mathcal{F} : \mathrm{C}\left( \mathrm{A}\right) \rightarrow \mathrm{D} \) be an additive functor such that \( \mathcal{F}\left( {\rho }^{ \bullet }\right) \) is an isomorphism in \( \mathrm{D} \) for all quasi-isomorphisms \( {\rho }^{ \bullet } \) in \( \mathrm{C}\left( \mathrm{A}\right) \) . ... | ## Proof. Exercise 5.2. | No |
Lemma 5.11. Let \( {P}^{ \bullet } \) be a complex of projective objects of an abelian category A such that \( {P}^{i} = 0 \) for \( i > 0 \), and let \( {L}^{ \bullet } \) be a complex in \( \mathrm{C}\left( \mathrm{A}\right) \) such that \( {H}^{i}\left( {L}^{ \bullet }\right) = 0 \) for \( i < 0 \). Let \( {\alpha }... | Proof. We have to construct morphisms \( {h}^{i} : {P}^{i} \rightarrow {L}^{i - 1} \) such that (*) \[ {\alpha }^{i} = {d}_{{L}^{ \bullet }}^{i - 1} \circ {h}^{i} + {h}^{i + 1} \circ {d}_{{P}^{ \bullet }}^{i}. \] Of course \( {h}^{i} = 0 \) necessarily for \( i > 0 \). For \( i = 0 \), use the fact that the morphism in... | Yes |
Corollary 5.12. Let \( {P}^{ \bullet } \) be a bounded-above cochain complex of projectives of an abelian category \( \mathrm{A} \), and let \( {L}^{ \bullet } \) be an exact complex in \( \mathrm{C}\left( \mathrm{A}\right) \). Then every morphism of complexes \( {P}^{ \bullet } \rightarrow {L}^{ \bullet } \) is homoto... | This follows immediately from (a harmless shift of) Lemma 5.11 since every morphism to an exact complex has no choice but to induce the zero-morphism in cohomology. | No |
Corollary 5.13. Let \( {P}^{ \bullet } \) (resp., \( {Q}^{ \bullet } \) ) be a bounded-above exact complex of projec-tives (resp., a bounded-below exact complex of injectives). Then \( {P}^{ \bullet } \) (resp., \( {Q}^{ \bullet } \) ) is homotopy equivalent to the zero-complex. | ## Proof. Exercise 5.12. | No |
Lemma 5.14. Let \( \mathrm{A} \) be an abelian category, and let \( {\rho }^{ \bullet } : {L}^{ \bullet } \rightarrow {M}^{ \bullet } \) be a quasi-isomorphism in \( \mathrm{C}\left( \mathrm{A}\right) \) . Let \( {P}^{ \bullet } \) be a bounded-above complex of projectives, and let \( {\alpha }^{ \bullet } : {P}^{ \bul... | Proof. Let \( {h}^{i} : {P}^{i} \rightarrow {M}^{i - 1} \) define a homotopy between \( {\rho }^{ \bullet } \circ {\alpha }^{ \bullet } \) and 0, so that \( - {\rho }^{i} \circ {\alpha }^{i} = {d}_{{M}^{ \bullet }}^{i - 1} \circ {h}^{i} + {h}^{i + 1} \circ {d}_{{P}^{ \bullet }}^{i} \) . Consider the mapping cone \( {MC... | No |
To see that \( {\alpha }^{ \bullet } \) may not be zero on the nose even if \( {\rho }^{ \bullet } \circ {\alpha }^{ \bullet } = 0 \) , look back again at Example 4.6: | Here \( {\rho }^{ \bullet } \) is a quasi-isomorphism, and \( {\rho }^{ \bullet } \circ {\alpha }^{ \bullet } = 0 \) . According to Lemma 5.14, the (nonzero) morphism \( {\alpha }^{ \bullet } \) is homotopic to 0 . (Indeed, a homotopy is immediately visible. What is it?) | No |
Proposition 5.16. Let \( \mathrm{A} \) be an abelian category, and let \( {L}^{ \bullet } \) be a complex in \( \mathrm{C}\left( \mathrm{A}\right) \) . Let \( {P}^{ \bullet } \) in \( {\mathrm{C}}^{ - }\left( \mathrm{P}\right) \) be a bounded-above complex of projectives, and let \( {\alpha }^{ \bullet } : {L}^{ \bulle... | Proof. Since \( {\alpha }^{ \bullet } : {L}^{ \bullet } \rightarrow {P}^{ \bullet } \) is a quasi-isomorphism, the mapping cone \( {MC}{\left( \alpha \right) }^{ \bullet } \) of \( \alpha \) is an exact complex (Corollary 4.2). Let \( {\rho }^{ \bullet } \) be the morphism of complexes\n\n\[{\rho }^{ \bullet } = \left(... | Yes |
Lemma 6.3. Let \( A \) be an object of an abelian category A. Let \( {M}^{ \bullet } \) be a resolution of \( A \), and let \( {P}^{ \bullet } \) be any complex in \( {\mathrm{C}}^{ \leq 0}\left( \mathrm{P}\right) \) . Let\n\n\[ \varphi : {H}^{0}\left( {P}^{ \bullet }\right) \rightarrow {H}^{0}\left( {M}^{ \bullet }\ri... | Proof. We have to define \( {\alpha }^{i} : {P}^{i} \rightarrow {M}^{i} \) for all \( i \) . Since \( {\alpha }^{i} = 0 \) necessarily for \( i > 0 \) , we may as well replace \( {M}^{ \bullet } \) with its truncated version (cf. Exercise 3.1) and then extend both \( {P}^{ \bullet } \) and this complex as follows:\n\n!... | Yes |
Proposition 6.4. Any two projective (resp., injective) resolutions of an object \( A \) of an abelian category \( \mathrm{A} \) are homotopy equivalent. | ## (This is also a direct consequence of Lemma 6.3.) | No |
Proposition 6.5. Let \( {A}_{0},{A}_{1} \) be objects of an abelian category \( \mathrm{A} \), and let \( {P}_{i}^{ \bullet } \) be a projective resolution of \( {A}_{i}, i = 0,1 \) . Then every morphism \( \varphi : {A}_{0} \rightarrow {A}_{1} \) in \( \mathrm{A} \) is induced by a morphism \( {\alpha }^{ \bullet } : ... | Proof. By hypothesis, \( \varphi \) is a morphism \( {H}^{0}\left( {P}_{0}^{ \bullet }\right) \rightarrow {H}^{0}\left( {P}_{1}^{ \bullet }\right) \) . The complex \( {P}_{0}^{ \bullet } \) consists of projectives, and \( {P}_{1}^{ \bullet } \) is a resolution of \( {A}_{1} \) ; therefore a lift \( {\alpha }^{ \bullet ... | Yes |
Theorem 6.6. Assume the abelian category A has enough projectives, and let \( {L}^{ \bullet } \) be a complex in \( {\mathrm{C}}^{ - }\left( \mathrm{A}\right) \) . Then there exists a bounded-above complex of projectives \( {P}^{ \bullet } \) and a quasi-isomorphism \( {P}^{ \bullet } \rightarrow {L}^{ \bullet } \), an... | Proof. The proof of this result is admittedly rather technical, as it involves many of the tools that we have developed.\n\nConstruction of \( {P}^{ \bullet } \) . We may assume that \( {L}^{ \bullet } \) is in \( {\mathrm{C}}^{ \leq 0}\left( \mathrm{\;A}\right) \)\n\n\[ \cdots \rightarrow {L}^{-2}\overset{{d}_{{L}^{ \... | Yes |
Theorem 6.7. Let \( \\mathrm{A} \) be an abelian category with enough projectives. Then the functor \( \\mathrm{K}^{ - }\\left( \\mathrm{P}\\right) \\rightarrow \\mathrm{D}^{ - }\\left( \\mathrm{A}\\right) \) is an equivalence of categories. | Theorem 6.7 is proven in full detail in any more complete treatment of homological algebra. Since we have not actually constructed \( \\mathrm{D}^{ - }\\left( \\mathrm{A}\\right) \\), we cannot really prove this statement here; but we now know enough to appreciate why Theorem 6.7 should be true, in the sense that \( \\... | No |
If \( {\rho }^{ \bullet } \) is a quasi-isomorphism in \( {\mathrm{C}}^{ - }\left( \mathrm{A}\right) \), then \( \mathcal{P}\left( {\rho }^{ \bullet }\right) \) is an isomorphism in \( {\mathrm{K}}^{ - }\left( \mathrm{P}\right) \). | For the first point, let \( {\rho }^{ \bullet } : {L}^{ \bullet } \rightarrow {M}^{ \bullet } \) be any morphism in \( {\mathrm{C}}^{ - }\left( \mathrm{A}\right) \) . By Theorem 6.6. \( {\rho }^{ \bullet } \) lifts to a morphism of resolutions: we have a diagram\n\n is (additive and) exact. Then I claim that \( \widehat{\mathrm{A}} \) is sent to \( \widehat{\mathrm{B}} \) by \( \mathrm{L}\mathcal{F} \) . | Indeed, let \( {P}^{ \bullet } \) be a complex in \( \widehat{\mathrm{A}} : {H}^{i}\left( {P}^{ \bullet }\right) = 0 \) for \( i \neq 0 \) . The image \( \mathrm{L}\mathcal{F}\left( {P}^{ \bullet }\right) \) is obtained by choosing a projective resolution \( {P}_{\mathcal{F}\left( {P}^{ \bullet }\right) }^{ \bullet } \... | No |
Every \( R \) -module \( N \) determines a functor \( \_ { \otimes }_{R}N : M \mapsto M{ \otimes }_{R}N \) (see SVIII 2.2). The left-derived functor of \( {}_{ - }{ \otimes }_{R}N \) is denoted \( {}_{ - }{ \otimes }_{R}N \) and acts \( {\mathrm{D}}^{ - }\left( {R\text{-Mod}}\right) \rightarrow {\mathrm{D}}^{ - }\left(... | indeed, the construction of \( {\operatorname{Tor}}_{i}^{R}\left( {M, N}\right) \) given in SVIII 2.4 matches precisely the ’concrete’ interpretation of the \( i \) -th left-derived functor given above. The reader may note that in SVIII 2.4 we used a free resolution of \( M \) ; free modules are projective, so this was... | No |
Similarly, \( {\operatorname{Hom}}_{R} \) admits a right-derived functor \( {\operatorname{RHom}}_{R} \), and its manifestations as the right-derived functors of \( {\operatorname{Hom}}_{R}\left( {M,\_ }\right) \) are the Ext modules: \( {\operatorname{Ext}}_{R}^{i}\left( {M, N}\right) \) is (isomorphic to) the \( i \)... | This projective resolution should really be viewed as an injective resolution in the opposite category \( R - {\operatorname{Mod}}^{op} \), since the functor \( {\operatorname{Hom}}_{R}\left( {\_, N}\right) \) is contravariant. | Yes |
Lemma 7.8. Let\n\n(*) \n\n\\[ \n0 \rightarrow L \rightarrow M \rightarrow N \rightarrow 0 \n\\] \n\nbe an exact sequence in an abelian category A with enough projectives. Assume \\( {P}_{L}^{ \bullet },{P}_{N}^{ \bullet } \\) are projective resolutions of \\( L, N \\), respectively. Then there exists an exact sequence ... | Proof. The hypotheses give us the solid part of the diagram \n\n \n\nand our task is to fill in the blanks with projective objects and morphisms so that all rows are exact, and the middle column is a resolution of \\... | No |
Corollary 7.9. Let\n\n\\[ \n{M}^{ \bullet } : \\;\\cdots \\rightarrow {M}^{-3} \\rightarrow {M}^{-2} \\rightarrow {M}^{-1} \\rightarrow {M}^{0} \\rightarrow 0 \n\\]\n\nbe a complex in an abelian category A with enough projectives. Then there is a complex of complexes:\n\n\\[ \n{P}_{{M}^{ \bullet }}^{ \bullet } : \\;\\c... | Proof. Break up \\( {M}^{ \bullet } \\) into short exact sequences\n\n\\[ \n0 \\rightarrow {K}^{i} \\rightarrow {M}^{i} \\rightarrow {I}^{i + 1} \\rightarrow 0 \n\\]\n\ntogether with exact sequences\n\n\\[ \n0 \\rightarrow {I}^{i} \\rightarrow {K}^{i} \\rightarrow {H}^{i} \\rightarrow 0 \n\\]\n\nwhere \\( {K}^{i} \\) i... | Yes |
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