Long exact sequence for Ext - Imported Knowlpedia Content

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)

Let

0 \longrightarrow A' \xrightarrow{u} A \xrightarrow{v} A'' \longrightarrow 0

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

\delta^n:\operatorname{Ext}_R^{n}(A',B)\longrightarrow \operatorname{Ext}_R^{n+1}(A'',B) \quad (n\ge 0),

(see connecting homomorphism) and a natural long exact sequence

0\to \operatorname{Hom}_R(A'',B)\to \operatorname{Hom}_R(A,B)\to \operatorname{Hom}_R(A',B) \to \operatorname{Ext}_R^{1}(A'',B)\to \operatorname{Ext}_R^{1}(A,B)\to \operatorname{Ext}_R^{1}(A',B) \to \operatorname{Ext}_R^{2}(A'',B)\to \cdots .

Let

0 \longrightarrow B' \xrightarrow{u} B \xrightarrow{v} B'' \longrightarrow 0

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

0\to \operatorname{Hom}_R(A,B')\to \operatorname{Hom}_R(A,B)\to \operatorname{Hom}_R(A,B'') \to \operatorname{Ext}_R^{1}(A,B')\to \operatorname{Ext}_R^{1}(A,B)\to \operatorname{Ext}_R^{1}(A,B'') \to \operatorname{Ext}_R^{2}(A,B')\to \cdots .

Examples

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.)

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

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