Ext functor

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In mathematics, the Ext functors of homological algebra are derived functors of Hom functors. They were first used in algebraic topology, but are common in many areas of mathematics.

1 Definition and computation
2 Properties of Ext
3 Ext and extensions
- 3.1 Ext in abelian categories
4 Ring structure and module structure on specific Exts
5 Interesting examples
6 Reference

[edit] Definition and computation

More precisely, write $\mathcal C = \mathfrak{Mod}(R)$ for the category of modules over $R$ , a ring. Let $B$ be in $\mathcal C$ and set $T(B) = \operatorname{Hom}_{\mathcal C}(A,B)$ , for fixed $A$ in $\mathcal C$ . (This is a left exact functor and thus has right derived functors $R n T$ ). To this end, define

$\operatorname{Ext}_R^n(A,B)=(R^nT)(B),$

i.e., take an injective resolution

$J(B)\leftarrow B\leftarrow 0,$

compute

$\operatorname{Hom}_{\mathcal C}(A,J(B))\leftarrow\operatorname{Hom}_{\mathcal C}(A,B)\leftarrow0,$

and take the cohomology of this complex.

Similarily, we can view the functor $G(A)=\operatorname{Hom}_{\mathcal C}(A,B)$ for a fixed module B as a contravariant left exact functor, and thus we also have right derived functors $R n G$ by instead of the injective resolution used above choosing a projective resolution $P (B)$ , and proceeding dually by calculating from

$P(A)\rightarrow A\rightarrow 0,$

compute

$\operatorname{Hom}_{\mathcal C}(P(A),B)\leftarrow\operatorname{Hom}_{\mathcal C}(A,B)\leftarrow0,$

and then take the cohomology.

These two constructions turn out to yield isomorphic results, and so both may be used for calculation of Ext.

[edit] Properties of Ext

The Ext functor exhibits some convenient properties, that are useful in computations.

$\operatorname{Ext}^i_{\mathcal C}(A,B)=0$ for $i > 0$ if either $B$ is injective or $A$ is projective.

The inverse also holds: if $\operatorname{Ext}^1_{\mathcal C}(A,B)=0$ for all $A$ , then $\operatorname{Ext}^i_{\mathcal C}(A,B)=0$ for all $0$ and $B$ is injective, and if $\operatorname{Ext}^1_{\mathcal C}(A,B)=0$ for all $B$ , then $\operatorname{Ext}^i_{\mathcal C}(A,B)=0$ for all $0$ and $A$ is projective.

$\operatorname{Ext}^n_{\mathcal C}(\bigoplus_\alpha A_\alpha,B)\cong\prod_\alpha\operatorname{Ext}^n_{\mathcal C}(A_\alpha,B)$

$\operatorname{Ext}^n_{\mathcal C}(A,\prod_\beta B_\beta)\cong\prod_\beta\operatorname{Ext}^n_{\mathcal C}(A,B_\beta)$

[edit] Ext and extensions

Ext functors take their name from their relationship to extensions. Given $R$ -modules $A$ and $B$ , there is a bijective correspondence between equivalence classes of extensions

$0\rightarrow B\rightarrow C\rightarrow A\rightarrow 0$

of $A$ by $B$ and elements of

$\operatorname{Ext}_R^1(A,B).$

Given two extensions

$0\rightarrow B\rightarrow C\rightarrow A\rightarrow 0$ and

$0\rightarrow B\rightarrow C'\rightarrow A\rightarrow 0$

we can construct the Baer sum, by forming the pullback $Γ$ of $C\rightarrow A$ and $C'\rightarrow A$ . We form the quotient $Y = Γ / Δ$ , with $\Delta=\{(-b,b):b\in B\}$ . The extension

$0\rightarrow B\rightarrow Y\rightarrow A\rightarrow 0$

thus formed is called the Baer sum of the extensions $C$ and $C'$ .

The Baer sum ends up being an abelian group operation on the set of equivalence classes, with the extension

$0\rightarrow B\rightarrow A\oplus B\rightarrow A\rightarrow 0$

acting as the identity.

[edit] Ext in abelian categories

This identification enables us to define $\operatorname{Ext}^1_{\mathcal C}(A,B)$ even for abelian categories $\mathcal C$ without reference to projectives and injectives. We simply take $\operatorname{Ext}^1_{\mathcal C}(A,B)$ to be the set of equivalence classes of extensions of $A$ by $B$ , forming an abelian group under the Baer sum. Similarily, we can define higher Ext groups $\operatorname{Ext}^n_{\mathcal C}(A,B)$ as equivalence classes of n-extensions

$0\rightarrow B\rightarrow X_n\rightarrow\cdots\rightarrow X_1\rightarrow A\rightarrow0$

under the equivalence relation generated by the relation that identifies two extensions

$0\rightarrow B\rightarrow X_n\rightarrow\cdots\rightarrow X_1\rightarrow A\rightarrow0$ and

$0\rightarrow B\rightarrow X'_n\rightarrow\cdots\rightarrow X'_1\rightarrow A\rightarrow0$

if there are maps $X_m\rightarrow X'_m$ for all $m$ in $1,2,.., n$ so that every resulting square commutes.

The Baer sum of the two n-extensions above is formed by letting $X'' 1$ be the pullback of $X 1$ and $X' 1$ over $A$ , and $Y n$ be the quotient of the pushout of $X n$ and $X' n$ under $B$ by the skew diagonal, as above. Then we define the Baer sum of the extensions to be

$0\rightarrow B\rightarrow Y_n\rightarrow X_{n-1}\oplus X'_{n-1}\rightarrow\cdots\rightarrow X_2\oplus X'_2\rightarrow X''_1\rightarrow A\rightarrow0.$

[edit] Ring structure and module structure on specific Exts

There is one more very useful way to view the Ext functor. If we start with the viewpoint that an element of $\operatorname{Ext}^n_{\mathcal C}(A,B)$ is an equivalence class of maps $f: P_n\rightarrow B$ for a projective resolution $P *$ of $A$ , we can pick a long exact sequence $Q *$ ending with $B$ and lift the map $f$ using the projectivity of the modules $P m$ to a chain map $f_*: P_*\rightarrow Q_*$ of degree -n. It turns out that homotopy classes of such chain maps correspond precisely to the equivalence classes in the definition of Ext above.

Under sufficiently nice circumstances, such as when the ring $R$ is a group ring, or a k-algebra, for a field k or even a noetherian ring k, we can impose a ring structure on $\operatorname{Ext}^*_{\mathcal C}(k,k)$ . The multiplication has quite a few equivalent interpretations, corresponding to different interpretations of the elements of $\operatorname{Ext}^*_{\mathcal C}(k,k)$ .

One interpretation is in terms of these homotopy classes of chain maps. Then the product of two elements is precisely the composition of the corresponding representatives. We can choose a single resolution of $k$ , and do all the calculations inside $\operatorname{Hom}_{\mathcal C}(P_*,P_*)$ , which is a differential graded algebra, with homology precisely $\operatorname{Ext}_{\mathcal C}(k,k)$ .

Another interpretation, not in fact relying on the existence of projective or injective modules is that of Yoneda splices. Then we take the viewpoint above that an element of $\operatorname{Ext}^n_{\mathcal C}(A,B)$ is an exact sequence starting in $A$ and ending in $B$ . This is then spliced with an element in $\operatorname{Ext}^m_{\mathcal C}(B,C)$ , by replacing

$\rightarrow X_1\rightarrow B\rightarrow 0$ and $0\rightarrow B\rightarrow Y_n\rightarrow$

with

$\rightarrow X_1\rightarrow Y_n\rightarrow$

where the middle arrow is the composition of the functions $X_1\rightarrow B$ and $B\rightarrow Y_n$ .

These viewpoints turn out to be equivalent whenever both make sense.

Using similar interpretations, we find that $\operatorname{Ext}_{\mathcal C}^*(k,M)$ is a module over $\operatorname{Ext}^*_{\mathcal C}(k,k)$ , again for sufficiently nice situations.

[edit] Interesting examples

If $R$ is chosen to be the integral group ring $\mathbb ZG$ for a group $G$ , then $\operatorname{Ext}^*_{\mathcal C}(\mathbb Z,M)$ is the group cohomology $H * (G, M)$ .

If $R$ is chosen to be the modular group ring $(\mathbb Z/p\mathbb Z)G$ for a group $G$ , then $\operatorname{Ext}^*_{\mathcal C}(\mathbb Z/p\mathbb Z,M)$ is also $H * (G, M)$ . In fact, it turns out that the group cohomology does not depend on which ring is chosen for base ring for the group ring.

If $R$ is chosen to be $A\otimes_k A^{op}$ for a $k$ -algebra $A$ , then $\operatorname{Ext}^*_{\mathcal C}(A,M)$ is the Hochschild cohomology $\operatorname{HH}^*(A,M)$ .

If $R$ is chosen to be the universal enveloping algebra for a Lie algebra $\mathfrak g$ , then $\operatorname{Ext}^*_{\mathcal C}(A,M)$ is the Lie algebra cohomology $\operatorname{H}^*(\mathfrak g,M)$ .