Ext

The right derived functors of Hom; measures extension classes and failure of exactness of Hom.

Ext

Let $R$ be a ring and let $M,N$ be (left) $R$-modules .

Definition (via a projective resolution)

Choose a projective resolution $P_\bullet \to M$ , i.e.

\cdots \to P_2 \xrightarrow{d_2} P_1 \xrightarrow{d_1} P_0 \to M \to 0

with each $P_i$ projective. Apply the functor $\mathrm{Hom}_R(-,N)$ to obtain a cochain complex

0 \to \mathrm{Hom}_R(P_0,N) \xrightarrow{d_1^*} \mathrm{Hom}_R(P_1,N) \xrightarrow{d_2^*} \mathrm{Hom}_R(P_2,N) \to \cdots.

Define

\mathrm{Ext}^n_R(M,N) \;:=\; H^n\!\big(\mathrm{Hom}_R(P_\bullet,N)\big),

where $H^n(-)$ denotes cohomology .

This construction is independent of the chosen resolution (up to canonical isomorphism), and is functorial: $\mathrm{Ext}^n_R(-,N)$ is contravariant in the first variable and $\mathrm{Ext}^n_R(M,-)$ is covariant in the second.

Equivalent definition (via an injective resolution)

If $R\text{-Mod}$ has enough injectives, choose an injective resolution $N \to I^\bullet$ and set

\mathrm{Ext}^n_R(M,N) \;:=\; H^n\!\big(\mathrm{Hom}_R(M,I^\bullet)\big).

In either approach, $\mathrm{Ext}^0_R(M,N)\cong \mathrm{Hom}_R(M,N)$ .

Conceptual meaning

$\mathrm{Ext}^1_R(M,N)$ classifies extensions $0 \to N \to E \to M \to 0$ up to the usual equivalence relation; see Ext^1 classifies extensions .
Higher $\mathrm{Ext}^n$ can be viewed as higher obstructions and are the values of the right derived functors of $\mathrm{Hom}$ is left exact .
A short exact sequence in either variable yields a long exact sequence in Ext , which is a special case of the long exact sequence for derived functors .

Examples

Example 1: Vector spaces over a field

Let $k$ be a field and $V,W$ be $k$ -vector spaces. Every $k$ -module is free (hence projective and injective), so any projective resolution can be taken to be length $0$ . Therefore,

\mathrm{Ext}^n_k(V,W)=0 \quad (n>0), \qquad \mathrm{Ext}^0_k(V,W)=\mathrm{Hom}_k(V,W).

Example 2: $\mathrm{Ext}^1_{\mathbb Z}(\mathbb Z/n, A)\cong A/nA$

Take the standard projective resolution of $\mathbb Z/n$ as a $\mathbb Z$ -module:

0 \to \mathbb Z \xrightarrow{\times n} \mathbb Z \to \mathbb Z/n \to 0.

Apply $\mathrm{Hom}_{\mathbb Z}(-,A)$ :

0 \to \mathrm{Hom}(\mathbb Z/n,A) \to \mathrm{Hom}(\mathbb Z,A)\xrightarrow{\times n} \mathrm{Hom}(\mathbb Z,A)\to \mathrm{Ext}^1(\mathbb Z/n,A)\to 0.

Using $\mathrm{Hom}(\mathbb Z,A)\cong A$ , the last map shows

\mathrm{Ext}^1_{\mathbb Z}(\mathbb Z/n,A)\;\cong\; \mathrm{coker}(A \xrightarrow{\times n} A)\;\cong\; A/nA.

In particular, $\mathrm{Ext}^1_{\mathbb Z}(\mathbb Z/n,\mathbb Z)\cong \mathbb Z/n$ .

Example 3: $\mathrm{Ext}^1_{\mathbb Z}(\mathbb Z/n,\mathbb Z/m)\cong \mathbb Z/\gcd(n,m)$

From Example 2 with $A=\mathbb Z/m$ ,

\mathrm{Ext}^1_{\mathbb Z}(\mathbb Z/n,\mathbb Z/m)\cong (\mathbb Z/m)/n(\mathbb Z/m)\cong \mathbb Z/\gcd(n,m).

(For these cyclic $\mathbb Z$ -modules, $\mathrm{Ext}^i_{\mathbb Z}(\mathbb Z/n,-)=0$ for $i\ge 2$ because $\mathbb Z/n$ has projective dimension $1$ .)