# Difference between revisions of "Math 750 -- Homological algebra -- Homeworks"

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− | The homework assignments refer to the following file, which also contains easier and harder exercises. | + | <!-- The homework assignments refer to the following file, which also contains easier and harder exercises. |

[[Media:Homological_algebra_problems.pdf|Math 750 problems and exercises]] | [[Media:Homological_algebra_problems.pdf|Math 750 problems and exercises]] | ||

;Homework 1: Section 1.1: 4, 5, 8, 10, 20, 21, 22, due Thursday, February 6th. | ;Homework 1: Section 1.1: 4, 5, 8, 10, 20, 21, 22, due Thursday, February 6th. | ||

− | ;Homework 2: Section 1.2: 2, 4, 6, 7, 8, 14, 15, due Thursday, February 20th. | + | ;Homework 2: Section 1.2: 2, 4, 6, 7, 8, 14, 15, due Thursday, February 20th. --> |

Homework 1, due Thursday, October 1: | Homework 1, due Thursday, October 1: |

## Revision as of 23:44, 12 February 2018

Homework 1, due Thursday, October 1:

a) Show by example that HoCh(A) is not always an abelian category.

b) Give an example of a bounded chain complex in Ch(Z-mod) which is acyclic but not split exact. (So both conditions of the Theorem we proved in class are needed.)

c) Prove that any acyclic complex of vector spaces over a field is split exact. (Do not assume that the complex is bounded!)

d) Prove that the full subcategory F of Ch^-(R-Mod) consisting of bounded above complexes of free R-modules satisfies the properties below:

-- the cone of a morphism between two objects in F is again in F;

-- if G : R-Mod -> A is an additive functor to an abelian category A, then G takes quasi-isomorphisms in F to quasi-isomorphisms in Ch(A);

-- if M is an R-module, a resolution of M is an object C of F together with a map C -> M[0] which is a quasi-isomorphism. Prove that maps M -> N of R-modules lift to maps of resolutions of the modules, uniquely up to homotopy.

e) Let f: M' -> M be an injective map of R-modules, and let M" denote the cokernel of f. Show that cone(f) is quasi-isomorphic to M"[0]. Are they always isomorphic in HoCh(R-mod)?

Homework 2, due Thursday, October 22:

a) Compute Ext^i_{Z/4Z}(Z/2Z, Z/2Z) and Tor_i^{Z/4Z}(Z/2Z, Z/2Z) for all i. (For Ext use projective resolutions in the first variable.)

b) Show that for any Z-modules M and N we have Ext^i_Z(M, N) = 0 for i >=2. We say that Z (the integers) has homological dimension 1.

c) Compute Ext^i_R(R, k) and Ext^i_R(k,k) where R = k[x1,...,xn] is the polynomial ring in n variables, k is a field, and k is regarded as an R module by the identification k = R/(x1, ..., xn). (Hint: you may want to look up the following commutative algebra topics on Wikipedia or in any standard textbook: regular sequence, Koszul resolution.) So the homological dimension of R is >=n. (It is in fact n.)

d) If F: A -> B is a right exact functor between abelian categories, and if A has enough projectives (so that LF is defined) then we shall say that an object X of A is F-acyclic if it satisfies R^i F(X) = 0 for i>0. For example any projective in A is F-acyclic.

-- show that if F is in fact exact, then any X in A is F-acyclic.

-- an F-acyclic resolution of an object Y of A is a complex

... -> X_n -> X_{n-1} -> ... -> X_0 -> 0

such that the complex is exact except at the last spot, where the homology is Y, and each X_i is F-acyclic.

Show that the homology H_i(F(X.)) is naturally isomorphic to L_i F(Y). (Thus derived functors can be computed using F-acylic resolutions, not only using projective resolutions.) In some situations this allows us to construct left derived functors even when there are not enough projectives, if we can identify enough F-acyclics. (Hint: break up the resolution into short exact sequences.)

e) (More difficult) Consider a short exact sequence of R-modules

0 -> M' -> M -> M" -> 0.

Let eta in Ext^1_R(M", M') be the image of the identity in Hom(M", M") under the map

Hom_R(M', M') -> Ext^1_R(M", M')

obtained by applying the Hom_R(--, M') functor to the above short exact sequence. Prove that eta=0 if and only if the short exact sequence we started with is split.

Homework 3, due Tuesday, Nov. 24:

a) Exercises 5.1.2 and 5.1.3 from Weibel's book.

b) Exercises A3.45, A3.49, A3.50 (parts a, b, c only) from Eisenbud's Commutative Algebra book.

c) Show that the Koszul complex associated to a regular sequence (f1,...,fn) in a *local* commutative ring R is self-dual after a degree shift, in the sense that K(f1,...,fn)^* is isomorphic to K(f1,...fn)[-n]. Here (...)^* denotes the dual complex, that is, the complex obtained by taking Hom_R(..., R).

Conclude that if the ideal I = (f1,...,fn) is cut out by a regular sequence of length n, we have Ext^n_R(R/I, R) = R/I, and all other Ext's are zero. (Hint: show that the complex K(f1) is self-dual in the above sense, and duality commutes with tensor products for complexes of free modules.)

Homework 4, due Tuesday Dec. 15:

Do the following exercises from Gelfand-Manin:

a) Exercise 1 on p. 163

b) Exercises 1 and 3 in the section beginning on p. 183

c) Exercises 1 and 5 in the section beginning on p. 214