THAN YOU NEED to KNOW ABOUT EXT GROUPS If a and G Are

MORE THAN YOU NEED TO KNOW ABOUT EXT GROUPS

If A and G are abelian groups, then Ext(A, G) is an abelian group. Like Hom(A, G), it is a covariant functor of G and a contravariant functor of A.

If we generalize our point of view a little so that now A and G are R-modules for some ring R, then n 0 we get not two functors but a whole sequence, ExtR(A, G), with ExtR(A, G) = HomR(A, G). In 1 n the special case R = Z, module means abelian group and ExtR is called Ext and ExtR is trivial for n > 1. I will explain the deﬁnition and some key properties in the case of general R.

To be deﬁnite, let’s suppose that R is an associative ring with 1, and that module means left module. If A and G are two modules, then the set HomR(A, G) of homomorphisms (or R-linear maps) A → G has an abelian group structure. (If R is commutative then HomR(A, G) can itself be viewed as an R-module, but that’s not the main point.)

We ﬁx G and note that A 7→ HomR(A, G) is a contravariant functor of A, a functor from R-modules ∗ to abelian groups. As long as G is ﬁxed, we sometimes denote HomR(A, G) by A .

The functor also takes sums of morphisms to sums of morphisms, and (therefore) takes the zero map to the zero map and (therefore) takes trivial object to trivial object. Thus, given an exact sequence A → B → C we get a sequence A∗ ← B∗ ← C∗ of groups and homomorphisms in which the composition is again zero. We may ask whether the new sequence is exact. The answer is no in general, but it is yes if C is trivial. In other words, the dualization functor takes surjections to injections. It is also yes if B → C is surjective; in other words an exact sequence A → B → C → 0 yields an exact sequence A∗ ← B∗ ← C∗ ← 0. It is also yes if A → B is a split injection. Thus a split exact sequence

0 → A → B → C → 0 yields a (split) exact sequence 0 ← A∗ ← B∗ ← C∗ ← 0.

Free resolutions: If A is an R-module, then it is always possible to make an exact sequence of modules 0 ← A ← F0 ← F1 ← F2 ← ... in which each module Fn is free. In order to do so, we ﬁrst make a free module F0 having a basis big enough to map onto a generating set of A, then make a free module F1 having a basis big enough to map onto a generating set of the kernel of F0 → A, then make a free module F2 having a basis big enough to map onto a generating set of the kernel of F1 → F0, and so on.

1 Of course, in the case when R = Z it is always possible to choose the resolution in such a way that Fn is trivial for all n > 1, because after choosing F0 and a surjection F0 → A we are free to choose F1 to be the kernel. (This works because in this case submodules of free modules are always free.) Say that such a free resolution has length one.

In the case when R is a ﬁeld, all modules are free and we can do even better and make a free resolution of length zero: F0 = A and Fn = 0 for all n > 0.

Here is an example where a free resolution must have length greater than one: Let R be the polynomial ring Z[T ]. Let A be the groupZ/p for some prime p, made into an R-module by setting T x = 0 for all x ∈ A. Here is a free resolution of length two:

0 ← Z/p ← Ra ← Rb ⊕ Rc ← Rd ← 0

Ra means a free R-module with basis element consiting of one element a. The maps are

a 7→ 1, b 7→ pa, c 7→ T a, d 7→ T b − pc.

We will see below that no shorter free resolution is possible.

(There are also cases when a free resolution of inﬁnite length is required.)

Note that when you have a free resolution then

... ← 0 ← F0 ← F1 ← F2 ← ... ∼ is a chain complex F• of free modules and that “its homology is A” in the sense that H0(F•) = A while Hn(F•) = 0 for all other n

∗ Applying the functor Hom(−,G) to this chain complex of modules, we get a cochain complex F• of abelian groups ∗ ∗ ∗ ... → 0 → F0 → F1 → F2 → ... The main thing we want to do is show that the cohomology of this complex is independent, up to isomorphism, of the choice of free resolution.

The key is the following lemma:

0 0 0 LEMMA If F• is a free resolution of A and F• is a free resolution of A and h : A → A is a map, ˜ 0 then (existence) there exists a chain map h : F• → F• inducing h on H0, in other words, such that 0 0 ˜ the map from H0(F•) = A to H0(F•) = A induced by H is h. And (uniqueness) any two such chain maps are related by some chain homotopy.

(In the lemma, all maps, chain maps, and chain homotopies are R-linear.)

The proof is in Hatcher.

A ﬁrst consequence of the lemma is that the cohomology of Hom(F•,G) is independent, up to 0 isomorphism, of the chosen free resolution of A: If F• and F• are two free resolutions of A, apply

2 0 0 the lemma with A = A and h equal to the identity map to get a chain map F• → F• inducing 0 the identity A → A, then apply it again to get a map F• → F• doing the same, then (applying the uniqueness part of the lemma twice) conclude that these are chain homotopy inverses. Now the functor HomR(−,G) applied to the two chain maps yields cochain maps in both directions ∗ 0∗ between F• and F• , and also applied to the chain homotopies it yields cochain homotopies to show that the cochain maps are homotopy inverses. This leads to a canonical isomorphism between ∗ 0∗ n the cohomology of F• and that of F• , making ExtR(A, G) well-deﬁned up to isomorphism.

Logically speaking, it is better to make it well-deﬁned than just well-deﬁned up to isomorphism, and practically speaking it is good to make it functorial. We can proceed as follows:

A A To every module A, associate a particular free resolution F as follows. Let F0 be the free module A having the set A as basis and consider the obvious surjective module map : F0 → A. Let A F1 be the free module having the set kernel() as basis and consider the obvious module map A A A ∂ : F1 → F0 . Let F2 be the free module having the set kernel(∂) as basis . . . .

˜ A B One sees easily that a module map h : A → B yields a map h : F• → F• of free resolutions, in such n A n a way that compositions and identities are preserved, so that A 7→ H (Hom, F• ,G) = Ext (A, G) becomes a functor. Take this as the deﬁnition. For computing an example (or for other reasons) one may use a free resolution other than this canonical one, by the discussion above. By a similar discussion again relying the lemma, one can also ﬁgure out the map of Ext groups induced by a given map of modules, even if the Ext groups have been calculated using other free resolutions.

Of course, the freedom to use other resolutions is also what tells us that Extn(A, G) is trivial for Z n > 1.

The fact that the Z[T ]-module Z/p in the example above has no free resolution of length less than 2 1 follows from the fact that Ext (Z/p, Z/p) (which you can compute from the given resolution) is nontrivial.

Here is another point of view: Consider the categories and functors

{R − modules} ← {resolutions} → {abgroups}

Here on the left the objects are R-modules and the maps are module homomorphisms, in the middle the objects are free resolutions (i.e. chain complexes F• of R-modules with Fn trivial for n < 0, Fn free for all n, and Hn(F•) trivial for all n 6= 0) and maps are chain homotopy classes of R-linear chain maps, and on the right the objects are abelian groups and the maps are homomorphisms. n The left functor is H0 and the right functor is H Hom(−,G).

The lemma shows that the left arrow is an equivalence of categories, so that it has an inverse up n to natural isomorphism. ExtR can be deﬁned as the right arrow composed with such an inverse. The construction A 7→ F A is an explicit choice of right inverse (and inverse up to isomorphism).

Here are some related ideas:

In general an R-module P is called projective if HomR(P, −) preserves exactness, equivalently if HomR(P, −) takes surjections to surjections. Every free module is projective. Every projective

3 module is a direct summand of a free module. (Proof: Choose a surjection f : F → P from some free module to P . Because P is projective, there is an element s ∈ HomR(P,F ) that maps to 1 ∈ HomR(P,P ) when you compose with f. Therefore f ◦ s = 1 and f is a split surjection.)

Ext groups can be computed using projective resolutions instead of free resolutions.

n If P is projective then ExtR(P,G) = 0 for every n > 0 and every module G. Conversely, if 1 ExtR(P,G) = 0 for every module G then P is projective.

n ∼ n−1 If 0 ← A ← P ← K ← 0 is exact and P is projective then ExtR(A, G) = ExtR (K,G) for every G and every n > 1. (Use a projective resolution of K to make a projective resolution of A.)

A module I is called injective if the contravariant functor HomR(−,I) preserves exactness, equivalently if HomR(−,I) takes injections to surjections. That is, I is injective if whenever A is a submodule of B then every module map from A to I can be extended to a module map from B to I. A Zorn’s lemma argument shows that for this to be true it is suﬃcient if it is true in the case when B is R itself (so that A is left ideal). In the case R = Z this condition on I is equivalent to saying that for every integer m > 0, for every element x ∈ I, there is an element y ∈ I such that my = x. So for Z-modules injective is the same as divisible.

In general, every module G admits an injective map G → I to some injective module I, and therefore for every G there is an injective resolution

0 → G → I0 → I1 → I2 → ...

n th This leads to a diﬀerent deﬁnition of ExtR(A, G) as the n cohomology of the cochain complex HomR(A, I•).