Ext

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

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

Define

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.

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

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

• $\mathrm{Ext}^1_R(M,N)$ 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.

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,

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:

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

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

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 $A=\mathbb Z/m$,

(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 $1$.)