Ext functor

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.

Definition and computation

Let $R$ be a ring and let $mathrm\{Mod\}\_R$ be the category of modules over "R". Let $B$ be in $mathrm\{Mod\}\_R$ and set $T(B)\; =\; operatorname\{Hom\}\_\{mathrm\{Mod\}\_R\}(A,B)$, for fixed $A$ in $mathrm\{Mod\}\_R$. (This is a left exact functor and thus has right derived functors $R^nT$). To this end, define

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

i.e., take an injective resolution

:$J(B)leftarrow\; Bleftarrow\; 0,$

compute

:$operatorname\{Hom\}\_\{mathrm\{Mod\}\_R\}(A,J(B))leftarrowoperatorname\{Hom\}\_\{mathrm\{Mod\}\_R\}(A,B)leftarrow0,$

and take the cohomology of this complex.

Similarly, we can view the functor $G(A)=operatorname\{Hom\}\_\{mathrm\{Mod\}\_R\}(A,B)$ for a fixed module B as a contravariant left exact functor, and thus we also have right derived functors $R^nG$, but instead of the injective resolution used above, choose a projective resolution $P(A)$, and proceed dually by calculating from:$P(A)\; ightarrow\; A\; ightarrow\; 0,$

compute

:$operatorname\{Hom\}\_\{mathrm\{Mod\}\_R\}(P(A),B)leftarrowoperatorname\{Hom\}\_\{mathrm\{Mod\}\_R\}(A,B)leftarrow0,$

and then take the cohomology.

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

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.

*

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Ext and extensions

Ext functors derive their name from the relationship to extensions. Given $R$-modules $A$ and $B$, there is a bijective correspondence between equivalence classes of extensions :$0\; ightarrow\; B\; ightarrow\; C\; ightarrow\; A\; ightarrow\; 0$of $A$ by $B$ and elements of:$operatorname\{Ext\}\_R^1(A,B).$

Given two extensions:$0\; ightarrow\; B\; ightarrow\; C\; ightarrow\; A\; ightarrow\; 0$ and:$0\; ightarrow\; B\; ightarrow\; C\text{'}\; ightarrow\; A\; ightarrow\; 0$we can construct the Baer sum, by forming the pullback $Gamma$ of $C\; ightarrow\; A$ and $C\text{'}\; ightarrow\; A$. We form the quotient $Y=Gamma/Delta$, with $Delta=\{(-b,b):bin\; B\}$. The extension:$0\; ightarrow\; B\; ightarrow\; Y\; ightarrow\; A\; ightarrow\; 0$ thus formed is called the Baer sum of the extensions $C$ and $C\text{'}$.

The Baer sum ends up being an abelian group operation on the set of equivalence classes, with the extension:$0\; ightarrow\; B\; ightarrow\; Aoplus\; B\; ightarrow\; A\; ightarrow\; 0$ 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"

:$0\; ightarrow\; B\; ightarrow\; X\_n\; ightarrowcdots\; ightarrow\; X\_1\; ightarrow\; A\; ightarrow0$

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

:$0\; ightarrow\; B\; ightarrow\; X\_n\; ightarrowcdots\; ightarrow\; X\_1\; ightarrow\; A\; ightarrow0$ and

:$0\; ightarrow\; B\; ightarrow\; X\text{'}\_n\; ightarrowcdots\; ightarrow\; X\text{'}\_1\; ightarrow\; A\; ightarrow0$

if there are maps $X\_m\; ightarrow\; X\text{'}\_m$ 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 $X\text{'}\_1$ over $A$, and $X"\_n$ be the pushout of $X\_n$ and $X\text{'}\_n$ under $B$. Then we define the Baer sum of the extensions to be:$0\; ightarrow\; B\; ightarrow\; X"\_n\; ightarrow\; X\_\{n-1\}oplus\; X\text{'}\_\{n-1\}\; ightarrowcdots\; ightarrow\; X\_2oplus\; X\text{'}\_2\; ightarrow\; X"\_1\; ightarrow\; A\; ightarrow0.$

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 $f:\; P\_n\; ightarrow\; B$ 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 $f\_*:\; P\_*\; ightarrow\; Q\_*$ 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 :$ightarrow\; X\_1\; ightarrow\; B\; ightarrow\; 0$ and $0\; ightarrow\; B\; ightarrow\; Y\_n\; ightarrow$with:$ightarrow\; X\_1\; ightarrow\; Y\_n\; ightarrow$where the middle arrow is the composition of the functions $X\_1\; ightarrow\; B$ and $B\; ightarrow\; Y\_n$.

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 $operatorname\{Ext\}^*\_\{mathrm\{Mod\}\_\{Aotimes\_k\; A^\{op\}(A,M)$ 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

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* | year=1994