Here are some of the basic properties and computations of \operatorname{Ext} groups. • \operatorname{Ext}_R^0(A,B)\cong\operatorname{Hom}_R(A,B) for any R-modules A and B. • \operatorname{Ext}_R^i(A,B)=0 for all i>0 if the R-module A is
projective (for example,
free) or if B is
injective. • The converses also hold: • If Ext(
A,
B) = 0 for all
B, then
A is projective (and hence Ext(
A,
B) = 0 for all
i > 0). • If Ext(
A,
B) = 0 for all
A, then
B is injective (and hence Ext(
A,
B) = 0 for all
i > 0). • \operatorname{Ext}^i_{\Z}(A,B) = 0 for all i\geq 2 and all abelian groups A and B. • Generalizing the previous example, \operatorname{Ext}^i_R(A,B)=0 for all i\geq 2 if R is a
principal ideal domain. • If R is a commutative ring and u in R is not a
zero divisor, then :\operatorname{Ext}_R^i(R/(u),B)\cong\begin{cases} B[u] & i=0\\ B/uB & i=1\\ 0 &\text{otherwise,}\end{cases} :for any R-module B. Here B[u] denotes the u-torsion subgroup of B, \{x\in B:ux=0\}. Taking R to be the ring \Z of integers, this calculation can be used to compute \operatorname{Ext}^1_{\Z}(A,B) for any
finitely generated abelian group A. • Generalizing the previous example, one can compute \operatorname{Ext} groups when the first module is the quotient of a commutative ring by any
regular sequence, using the
Koszul complex. For example, if R is the
polynomial ring k[x_1,\ldots,x_n] over a field k, then \operatorname{Ext}_R^*(k,k) is the
exterior algebra S over k on n generators in \operatorname{Ext}^1. Moreover, \operatorname{Ext}_S^*(k,k) is the polynomial ring R; this is an example of
Koszul duality. • By the general properties of derived functors, there are two basic
exact sequences for \operatorname{Ext}. First, a
short exact sequence 0\rightarrow K\rightarrow L\rightarrow M\rightarrow 0 of R-modules induces a long exact sequence of the form ::0 \to \mathrm{Hom}_R(A,K) \to \mathrm{Hom}_R(A,L) \to \mathrm{Hom}_R(A,M) \to \mathrm{Ext}^1_R(A,K) \to \mathrm{Ext}^1_R(A,L) \to \cdots, :for any R-module A. Also, a short exact sequence 0\rightarrow K\rightarrow L\rightarrow M\rightarrow 0 induces a long exact sequence of the form ::0 \to \mathrm{Hom}_R(M,B) \to \mathrm{Hom}_R(L,B) \to \mathrm{Hom}_R(K,B) \to \mathrm{Ext}^1_R(M,B) \to \mathrm{Ext}^1_R(L,B) \to \cdots, :for any R-module B. • Ext takes
direct sums (possibly infinite) in the first variable and
products in the second variable to products. That is: ::\begin{align} \operatorname{Ext}^i_R \left(\bigoplus_\alpha M_\alpha,N \right) &\cong\prod_\alpha \operatorname{Ext}^i_R (M_\alpha,N) \\ \operatorname{Ext}^i_R \left(M,\prod_\alpha N_\alpha \right ) &\cong\prod_\alpha \operatorname{Ext}^i_R (M,N_\alpha) \end{align} • Let
A be a
finitely generated module over a commutative
Noetherian ring R. Then Ext commutes with
localization, in the sense that for every
multiplicatively closed set S in
R, every
R-module
B, and every integer
i, ::S^{-1} \operatorname{Ext}_R^i(A, B) \cong \operatorname{Ext}_{S^{-1} R}^i \left (S^{-1} A, S^{-1} B \right ). ==Ext and extensions==
Equivalence of extensions The \operatorname{Ext} groups derive their name from their relation 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\to B\to E\to A\to 0. Two extensions :0\to B\to E\to A\to 0 :0\to B\to E' \to A\to 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\to B\to A\oplus B\to A\to 0. There is a one-to-one correspondence between
equivalence classes of extensions of A by B and elements of \operatorname{Ext}_R^1(A,B). This can be made precise as follows. {{Proof| Fix a short exact sequence :0 \to M \to P \to A \to 0 where P is projective. Applying \operatorname{Hom}(-, B) yields the long exact sequence :\operatorname{Hom}(P, B) \to \operatorname{Hom}(M, B) \xrightarrow{\delta} \operatorname{Ext}(A, B) \to 0. Given x \in \operatorname{Ext}(A, B), choose \beta \in \operatorname{Hom}(M, B) such that \delta(\beta) = x. Consider the pushout of j: M \to P along \beta, given by the
cokernel of the map :M \to P \oplus B, \quad m \mapsto (j(m), -\beta(m)). Define X as this pushout object. This yields the commutative diagram: : Here, X \to A is induced by the map P \to A. The bottom row is an extension of A by B, denoted \xi, and the connecting map \delta ensures that \delta(\xi) = x, proving surjectivity. To show well-definedness on equivalence classes, suppose \beta' is another lift of x. Then there exists f \in \operatorname{Hom}(P, B) such that \beta' = \beta + f \circ j. If X' is the pushout of j and \beta', then an isomorphism X \cong X' is induced, making the extensions equivalent. Conversely, given an extension :0 \to B \to X \to A \to 0, the
lifting property of P gives a map \tau: P \to X fitting into the diagram : Here X is the pushout of j and \gamma. This shows that the map is injective. Thus, the set of equivalence classes of extensions of A by B is naturally isomorphic to \operatorname{Ext}(A, B). }} The trivial extension corresponds to the zero element of \operatorname{Ext}_R^1(A,B).
The Baer sum of extensions The
Baer sum is an explicit description of the abelian group structure on \operatorname{Ext}_R^1(A,B), viewed as the set of equivalence classes of extensions of A by B. Namely, given two extensions 0\to B\xrightarrow{f} E \xrightarrow{g} A\to 0 and 0\to B\xrightarrow{f'} E'\xrightarrow{g'} A\to 0, first form the
pullback over A, \Gamma = \left\{ (e, e') \in E \oplus E' \; | \; g(e) = g'(e')\right\}. Then form the
quotient module Y = \Gamma / \{(f(b), -f'(b)) \;|\;b \in B\}. The Baer sum of E and E' is the extension 0\to B\to Y\to A\to 0, where the first map is b \mapsto [(f(b), 0)] = [(0, f'(b))] and the second is (e, e') \mapsto g(e) = g'(e').
Up to equivalence of extensions, the Baer sum is commutative and has the trivial extension as identity element. The negative of an extension 0\rightarrow B\rightarrow E\rightarrow A\rightarrow 0 is the extension involving the same module E, but with the homomorphism B\rightarrow E replaced by its negative. ==Construction of Ext in abelian categories==