Ext functor: Information from Answers.com

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 See also
7 References

Definition and computation

Let $R$ be a ring and let $Mod R$ be the category of modules over R. Let $B$ be in $Mod R$ and set $T(B) = \operatorname{Hom}_{\mathrm{Mod}_R}(A,B)$ , for fixed $A$ in $Mod R$ . This is a left exact functor and thus has right derived functors $R n T$ . The Ext functor is defined by

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

This can be calculated by taking any injective resolution

$0 \rightarrow B \rightarrow I^0 \rightarrow I^1 \rightarrow \dots,$

and computing

$0 \rightarrow \operatorname{Hom}_{\mathrm{Mod}_R}(A,I^0) \rightarrow \operatorname{Hom}_{\mathrm{Mod}_R}(A,I^1) \rightarrow \dots.$

Then $(R n T)(B)$ is the homology of this complex. Note that $\operatorname{Hom}_{\mathrm{Mod}_R}(A,B)$ is excluded from the complex.

An alternative definition is given using the functor $G(A)=\operatorname{Hom}_{\mathrm{Mod}_R}(A,B)$ . For a fixed module B, this is a contravariant left exact functor, and thus we also have right derived functors $R n G$ , and can define

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

This can be calculated by choosing any projective resolution

$\dots \rightarrow P^1 \rightarrow P^0 \rightarrow A \rightarrow 0,$

and proceeding dually by computing

$0\rightarrow\operatorname{Hom}_{\mathrm{Mod}_R}(P^0,B)\rightarrow \operatorname{Hom}_{\mathrm{Mod}_R}(P^1,B) \rightarrow \dots.$

Then $(R n G)(A)$ is the homology of this complex. Again note that $\operatorname{Hom}_{\mathrm{Mod}_R}(A,B)$ is excluded.

These two constructions turn out to yield isomorphic results, and so both may be used to calculate the Ext functor.

Properties of Ext

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

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

The converse also holds: if $\operatorname{Ext}^1_{\mathrm{Mod}_R}(A,B)=0$ for all $A$ , then $\operatorname{Ext}^i_{\mathrm{Mod}_R}(A,B)=0$ for all $A$ , and $B$ is injective; if $\operatorname{Ext}^1_{\mathrm{Mod}_R}(A,B)=0$ for all $B$ , then $\operatorname{Ext}^i_{\mathrm{Mod}_R}(A,B)=0$ for all $B$ , and $A$ is projective.

$\operatorname{Ext}^n_{\mathrm{Mod}_R}(\bigoplus_\alpha A_\alpha,B)\cong\prod_\alpha\operatorname{Ext}^n_{\mathrm{Mod}_R}(A_\alpha,B)$

$\operatorname{Ext}^n_{\mathrm{Mod}_R}(A,\prod_\beta B_\beta)\cong\prod_\beta\operatorname{Ext}^n_{\mathrm{Mod}_R}(A,B_\beta)$

Ext and extensions

Ext functors derive their name from the relationship to extensions of modules. Given R-modules A and B, an extension of A by B is a short exact sequence of R-modules

$0\rightarrow B\rightarrow E\rightarrow A\rightarrow0.$

Two extensions

$0\rightarrow B\rightarrow E\rightarrow A\rightarrow0$

$0\rightarrow B\rightarrow E^\prime\rightarrow A\rightarrow0$

are said to be equivalent (as extensions of A by B) if there is a commutative diagram

An extension of A by B is called split if it is equivalent to the extension

$0\rightarrow B\rightarrow A\oplus B\rightarrow A\rightarrow0.$

There is a bijective correspondence between equivalence classes of extensions

$0\rightarrow B\rightarrow E\rightarrow A\rightarrow 0$

of $A$ by $B$ and elements of

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

Given two extensions

$0\rightarrow B\rightarrow E\rightarrow A\rightarrow 0$ and

$0\rightarrow B\rightarrow E^\prime\rightarrow A\rightarrow 0$

we can construct the Baer sum, by forming the pullback $Γ$ of E → A and E ' → 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 E and E'.

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.

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. Similarly, 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 $X'' n$ be the pushout of $X n$ and $X' n$ under $B$ . Then we define the Baer sum of the extensions to be

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

Ring structure and module structure on specific Exts

One more very useful way to view the Ext functor is this: when an element of $\operatorname{Ext}^n_{\mathrm{Mod}_R}(A,B)$ is considered as an equivalence class of maps $f: P_n\rightarrow B$ for a projective resolution $P *$ of $A$ ; so, then 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 $\operatorname{Ext}^*_{\mathrm{Mod}_R}(k,k)$ . The multiplication has quite a few equivalent interpretations, corresponding to different interpretations of the elements of $\operatorname{Ext}^*_{\mathrm{Mod}_R}(k,k)$ .

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 $\operatorname{Hom}_{\mathrm{Mod}_R}(P_*,P_*)$ , which is a differential graded algebra, with homology precisely $\operatorname{Ext}_{\mathrm{Mod}_R}(k,k)$ .

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 $\operatorname{Ext}^n_{\mathrm{Mod}_R}(A,B)$ is an exact sequence starting in $A$ and ending in $B$ . This is then spliced with an element in $\operatorname{Ext}^m_{\mathrm{Mod}_R}(B,C)$ , 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 $\operatorname{Ext}_{\mathrm{Mod}_R}^*(k,M)$ is a module over $\operatorname{Ext}^*_{\mathrm{Mod}_R}(k,k)$ , again for sufficiently nice situations.

Interesting examples

If $\mathbb ZG$ is the integral group ring for a group $G$ , then $\operatorname{Ext}^*_{\mathrm{Mod}_{\mathbb ZG}}(\mathbb Z,M)$ is the group cohomology $H * (G, M)$ with coefficients in $M$ .

For $\mathbb F_p$ the finite field on $p$ elements, we also have that $H^*(G,M)=\operatorname{Ext}^*_{\mathrm{Mod}_{\mathbb F_pG}}(\mathbb F_p,M)$ , and it turns out that the group cohomology doesn't depend on the base ring chosen.

If $A$ is a $k$ -algebra, then $\operatorname{Ext}^*_{\mathrm{Mod}_{A\otimes_k A^{op}}}(A,M)$ is the Hochschild cohomology $\operatorname{HH}^*(A,M)$ with coefficients in the module M.

If $R$ is chosen to be the universal enveloping algebra for a Lie algebra $\mathfrak g$ , then $\operatorname{Ext}^*_{\mathrm{Mod}_R}(R,M)$ is the Lie algebra cohomology $\operatorname{H}^*(\mathfrak g,M)$ with coefficients in the module M.

References

Gelfand, Sergei I.; Manin, Yuri Ivanovich (1999), Homological algebra, Berlin: Springer, ISBN 978-3-540-65378-3
Weibel, Charles A. (1994), An introduction to homological algebra, Cambridge Studies in Advanced Mathematics, 38, Cambridge University Press, MR 1269324, ISBN 978-0-521-55987-4, OCLC 36131259

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Ext functor

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