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. The name "Ext" comes from group theory, as the Ext functor is used in group cohomology to classify abelian group extensions.^[1]

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) = Hom_R(A,B), for fixed A in Mod_R. This is a left exact functor and thus has right derived functors RⁿT. The Ext functor is defined by

\operatorname {Ext} _{R}^{n}(A,B)=(R^{n}T)(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} _{R}(A,I^{0})\rightarrow \operatorname {Hom} _{R}(A,I^{1})\rightarrow \dots .

Then (RⁿT)(B) is the homology of this complex. Note that Hom_R(A,B) is excluded from the complex.

An alternative definition is given using the functor G(A)=Hom_R(A,B). For a fixed module B, this is a contravariant left exact functor, and thus we also have right derived functors RⁿG, and can define

\operatorname {Ext} _{R}^{n}(A,B)=(R^{n}G)(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} _{R}(P^{0},B)\rightarrow \operatorname {Hom} _{R}(P^{1},B)\rightarrow \dots .

Then (RⁿG)(A) is the homology of this complex. Again note that Hom_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.

Ext and extensions

Equivalence of 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\rightarrow 0.

Two extensions

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

0\rightarrow B\rightarrow E^{\prime }\rightarrow A\rightarrow 0

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

Note that the Five Lemma implies that the middle arrow is an isomorphism. An extension of A by B is called split if it is equivalent to the trivial extension

0\rightarrow B\rightarrow A\oplus B\rightarrow A\rightarrow 0.

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).

The Baer sum of extensions

Given two extensions

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

0\rightarrow B\rightarrow E^{\prime }\rightarrow A\rightarrow 0

we can construct the Baer sum, by forming the pullback over $A$ ,

$\Gamma =\left\{(e,e')\in E\oplus E'\;|\;g(e)=g'(e')\right\}.$

We form the quotient

$Y=\Gamma /\{(f(b),0)-(0,f'(b))\;|\;b\in B\}$ ,

that is, we mod out by the relation $(f(b)+e,e')\sim (e,f'(b)+e')$ . The extension

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

where the first arrow is $b\mapsto [(f(b),0)]=[(0,f'(b))]$ and the second $(e,e')\mapsto g(e)=g'(e')$ thus formed is called the Baer sum of the extensions E and E'.

Up to equivalence of extensions, the Baer sum is commutative and has the trivial extension as identity element. The extension 0 → B → E → A → 0 has for opposite the same extension with exactly one of the central arrows turned to their opposite eg the morphism g is replaced by -g.

The set of extensions up to equivalence is an abelian group that is a realization of the functor Ext1
R(A, B)

Construction of Ext in abelian categories

The above identification enables us to define Ext1
Ab(A, B) even for abelian categories Ab without reference to projectives and injectives (even if the category has no projectives or injectives). We simply take Ext1
Ab(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 Extn
Ab(A, B) as equivalence classes of n-extensions, which are exact sequences

0\rightarrow B\rightarrow X_{n}\rightarrow \cdots \rightarrow X_{1}\rightarrow A\rightarrow 0

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

\xi \colon 0\rightarrow B\rightarrow X_{n}\rightarrow \cdots \rightarrow X_{1}\rightarrow A\rightarrow 0

\xi '\colon 0\rightarrow B\rightarrow X'_{n}\rightarrow \cdots \rightarrow X'_{1}\rightarrow A\rightarrow 0

if there are maps X_m → X′_m for all m in {1, 2, ..., n} so that every resulting square commutes, i.e. if there is a chain map X: $\xi$ → $\xi$ '.

The Baer sum of the two n-extensions above is formed by letting $X_{1}^{\prime \prime }$ be the pullback of $X 1$ and $X_{1}^{\prime }$ over A, and $X_{n}^{\prime \prime }$ be the pushout of $X n$ and $X_{n}^{\prime }$ under B; see Weibel, §3.4 (but remark there are some errata). 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\rightarrow 0.

Further properties of Ext

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

Exti
R(A, B) = 0 for i > 0 if either B is injective or A projective.
A converse also holds: if Ext1
R(A, B) = 0 for all A, then Exti
R(A, B) = 0 for all A, and B is injective; if Ext1
R(A, B) = 0 for all B, then Exti
R(A, B) = 0 for all B, and A is projective.
$\operatorname{Ext}^n_R \Bigl(\bigoplus_\alpha A_\alpha,B \Bigr)\cong\prod_\alpha\operatorname{Ext}^n_R(A_\alpha,B)$
$\operatorname {Ext} _{R}^{n}{\Bigl (}A,\prod _{\beta }B_{\beta }{\Bigr )}\cong \prod _{\beta }\operatorname {Ext} _{R}^{n}(A,B_{\beta })$

Ring structure and module structure on specific Exts

One more very useful way to view the Ext functor is this: when an element of Extn
R(A, B) = 0 is considered as an equivalence class of maps f: P_n → 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_* → 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 over a field k, or an augmented k-algebra, we can impose a ring structure on Ext*
R(k, k). The multiplication has quite a few equivalent interpretations, corresponding to different interpretations of the elements of Ext*
R(k, k).

One interpretation is in terms of these homotopy classes of chain maps. Then the product of two elements is represented by the composition of the corresponding representatives. We can choose a single resolution of k, and do all the calculations inside Hom_R(P_*,P_*), which is a differential graded algebra, with cohomology precisely Ext_R(k,k).

The Ext groups can also be interpreted in terms of exact sequences; this has the advantage that it does not rely on the existence of projective or injective modules. Then we take the viewpoint above that an element of Extn
R(A, B) is a class, under a certain equivalence relation, of exact sequences of length n + 2 starting with B and ending with A. This can then be spliced with an element in Extm
R(C, A), by replacing ... → X₁ → A → 0 and 0 → A → Y_n → ... with:

\cdots \rightarrow X_{1}\rightarrow Y_{n}\rightarrow \cdots

where the middle arrow is the composition of the functions X₁ → A and A → Y_n. This product is called the Yoneda splice.

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

Using similar interpretations, we find that Ext*
R(k, M) is a module over Ext*
R(k, k), again for sufficiently nice situations.

Interesting examples

If Z[G] is the integral group ring for a group G, then Ext*
Z[G](Z, M) is the group cohomology H*(G,M) with coefficients in M.

For F_p the finite field on p elements, we also have that H*(G,M) = Ext*
F_p[G](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 Ext*
A ⊗_k A^op(A, M) is the Hochschild cohomology HH*(A,M) with coefficients in the A-bimodule M.

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

References

↑ "nLab:Ext". nLab. Retrieved 2015-07-23. It derives its name from the fact that the derived hom between abelian groups classifies abelian group extensions of A by K. (This is a special case of the general classification of principal ∞-bundles/∞-group extensions by general cohomology/group cohomology.)

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. ISBN 978-0-521-55987-4. OCLC 36131259. MR 1269324.

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