Ext¹ classifies extensions - Imported Knowlpedia Content

Let $R$ be a ring and $A,C$ left $R$ -modules.

An extension of $C$ by $A$ is a short exact sequence

0 \longrightarrow A \xrightarrow{i} E \xrightarrow{p} C \longrightarrow 0.

Two extensions

0\to A \xrightarrow{i} E \xrightarrow{p} C \to 0, \qquad 0\to A \xrightarrow{i'} E' \xrightarrow{p'} C \to 0

are equivalent if there exists an $R$ -module isomorphism $\phi:E\to E'$ such that $\phi\circ i=i'$ and $p' \circ \phi = p$ (i.e. a commutative diagram with identity on $A$ and $C$ ).

Let $\mathrm{Ext}(C,A)$ denote the set of equivalence classes of extensions of $C$ by $A$ .

Theorem (classification by Ext $^1$ ). There is a natural bijection

\mathrm{Ext}^1_R(C,A)\ \cong\ \mathrm{Ext}(C,A),

where $\mathrm{Ext}^1_R(C,A)$ is the degree-1 Ext group (the first right derived functor of Hom; see derived functor).

Remarks

Moreover:

The zero element of $\mathrm{Ext}^1_R(C,A)$ corresponds to the split extension $E\cong A\oplus C$ .
The abelian group structure on $\mathrm{Ext}^1_R(C,A)$ corresponds to the Baer sum of extensions (constructed via pullback/pushout in the category of modules).

Cross-links: projective resolutions, injective resolutions, Hom is left exact.

Examples

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

Use the projective resolution from existence of projective resolutions:

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

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

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

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

0\to A \to E \to \mathbb Z/n\mathbb Z \to 0.

The class $0\in A/nA$ corresponds to the split extension $E\cong A\oplus \mathbb Z/n\mathbb Z$ .

Example 2: $\mathrm{Ext}^1_{k[x]}(k,k)\cong k$ and the module $k[x]/(x^2)$

Let $R=k[x]$ and $k\cong R/(x)$ . Using the free resolution

0\to R \xrightarrow{\cdot x} R \to k \to 0,

applying $\mathrm{Hom}_R(-,k)$ yields a map $k \xrightarrow{0} k$ (since $x$ acts as $0$ on $k$ ), hence

\mathrm{Ext}^1_{k[x]}(k,k)\cong k.

A concrete non-split extension representing a nonzero class is

0 \longrightarrow (x)/(x^2) \longrightarrow k[x]/(x^2) \longrightarrow k[x]/(x) \longrightarrow 0,

where $(x)/(x^2)\cong k$ and $k[x]/(x)\cong k$ as $k[x]$ -modules. The split class corresponds to $k\oplus k$ (with $x$ acting by $0$ on each summand).

Example 3: Over a field, every extension splits

If $R=k$ is a field, all $k$ -modules are projective (and injective), so

\mathrm{Ext}^1_k(V,W)=0

for all vector spaces $V,W$ . Equivalently, every short exact sequence $0\to W\to E\to V\to 0$ of vector spaces splits (choose a linear section $V\to E$ ).

Example 1: ExtZ1(Z/nZ,A)≅A/nA\mathrm{Ext}^1_{\mathbb Z}(\mathbb Z/n\mathbb Z, A)\cong A/nAExtZ1​(Z/nZ,A)≅A/nA

Example 2: Extk[x]1(k,k)≅k\mathrm{Ext}^1_{k[x]}(k,k)\cong kExtk[x]1​(k,k)≅k and the module k[x]/(x2)k[x]/(x^2)k[x]/(x2)

Example 3: Over a field, every extension splits

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

Example 2: $\mathrm{Ext}^1_{k[x]}(k,k)\cong k$ and the module $k[x]/(x^2)$