Long exact sequence for Ext

Let $R$ be a ring. Recall that Ext is a right-derived functor of Hom (left exactness of Hom ), constructed using an injective resolution (or a projective resolution in the first variable). See also derived functor .

Theorem (long exact sequence in Ext, both variables)

(A) Short exact sequence in the first variable (contravariant)

Let

be a short exact sequence of left $R$-modules, and let $B$ be a left $R$-module. Then there are natural connecting maps

(see connecting homomorphism ) and a natural long exact sequence

This is a special case of the general long exact sequence for derived functors .

(B) Short exact sequence in the second variable (covariant)

Let

be a short exact sequence of left $R$-modules, and let $A$ be a left $R$-module. Then there is a natural long exact sequence

Examples

1. Computing $\operatorname{Ext}^1_{\mathbb Z}(\mathbb Z/n, A)\cong A/nA$.
   Start from

   $$0\to \mathbb Z \xrightarrow{\cdot n}\mathbb Z \to \mathbb Z/n \to 0.$$

   Apply $\operatorname{Hom}_{\mathbb Z}(-,A)$. Since $\operatorname{Hom}_{\mathbb Z}(\mathbb Z,A)\cong A$, the relevant piece of the long exact sequence is

   $$A \xrightarrow{\cdot n} A \to \operatorname{Ext}^1_{\mathbb Z}(\mathbb Z/n,A)\to 0,$$

   so

   $$\operatorname{Ext}^1_{\mathbb Z}(\mathbb Z/n,A)\cong \operatorname{coker}(\cdot n:A\to A)\cong A/nA.$$

   (Also $\operatorname{Ext}^i_{\mathbb Z}(\mathbb Z/n,-)=0$ for $i\ge 2$ because $\mathbb Z/n$ has a length-1 projective resolution.)

2. Computing $\operatorname{Ext}^1_{\mathbb Z}(\mathbb Z/n, \mathbb Z/m)\cong \mathbb Z/\gcd(n,m)$.
   Take $A=\mathbb Z/m$ in the previous example:

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

3. Dual numbers: $\operatorname{Ext}^1_{k[\varepsilon]/(\varepsilon^2)}(k,k)\cong k$.
   Let $R=k[\varepsilon]/(\varepsilon^2)$, $k=R/(\varepsilon)$, and use the projective resolution

   $$0\to R\xrightarrow{\cdot\varepsilon}R\to k\to 0.$$

   Applying $\operatorname{Hom}_R(-,k)$ yields the cochain complex

   $$0\to \operatorname{Hom}_R(R,k)\xrightarrow{(\cdot\varepsilon)^\ast}\operatorname{Hom}_R(R,k)\to 0,$$

   and $(\cdot\varepsilon)^\ast=0$ because $\varepsilon$ acts as $0$ on $k$. Hence

   $$\operatorname{Ext}^1_R(k,k)\cong \operatorname{coker}(0:k\to k)\cong k.$$