Ext - Knowlpedia

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

For $n\ge 0$ , the group $\operatorname{Ext}^n_R(M,N)$ is the $n$ -th right derived functor of $\operatorname{Hom}_R(M,-)$ , evaluated at $N$ .

Computation by a projective resolution

Equivalently, by deriving in the first (contravariant) variable, it is computed from a projective resolution of $M$ :

\operatorname{Ext}^n_R(M,N) =H^n\!\bigl(\operatorname{Hom}_R(P_\bullet,N)\bigr).

Computation from a projective resolution

Choose a projective resolution $P_\bullet \to M$ :

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

with each $P_i$ projective. Applying $\operatorname{Hom}_R(-,N)$ gives the 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.

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

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

Equivalent definition (via an injective resolution)

Alternatively, 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, $\operatorname{Ext}^0_R(M,N)\cong\operatorname{Hom}_R(M,N)$ .

Conceptual meaning

$\operatorname{Ext}^1_R(M,N)$ classifies extensions $0 \to N \to E \to M \to 0$ up to the usual equivalence relation; see Ext^1 classifies extensions.
Higher $\operatorname{Ext}^n$ groups can be viewed as higher obstructions.
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, so a projective resolution can be taken to have 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: $\operatorname{Ext}^1_{\mathbb Z}(\mathbb Z/n,A)\cong A/nA$

Take the standard projective resolution of $\mathbb Z/n$ :

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

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

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 $\operatorname{Hom}(\mathbb Z,A)\cong A$ , 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, $\operatorname{Ext}^1_{\mathbb Z}(\mathbb Z/n,\mathbb Z)\cong\mathbb Z/n$ .

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

Taking $A=\mathbb Z/m$ in the preceding example gives

\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, $\operatorname{Ext}^i_{\mathbb Z}(\mathbb Z/n,-)=0$ for $i\ge2$ , because $\mathbb Z/n$ has projective dimension $1$ .

Example 1: Vector spaces over a field

Example 2: Ext⁡Z1(Z/n,A)≅A/nA\operatorname{Ext}^1_{\mathbb Z}(\mathbb Z/n,A)\cong A/nAExtZ1​(Z/n,A)≅A/nA

Example 3: Ext⁡Z1(Z/n,Z/m)\operatorname{Ext}^1_{\mathbb Z}(\mathbb Z/n,\mathbb Z/m)ExtZ1​(Z/n,Z/m)

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

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