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For a finite-dimensional associative algebra, form a graph on its simple modules, joining $S,T$ when ${\mathrm{Ext}}^1(S,T)$ or ${\mathrm{Ext}}^1(T,S)$ is nonzero. Its connected components are exactly the simple modules belonging to each block of a finite-dimensional algebra. They are also the components generated by sharing a Jordan–Hölder factor occurrence in an indecomposable representation that is a projective module.

Indeed, Ext separation of finite-length modules would split the regular module of a block into canonical summands if that block had two graph components. Right multiplication preserves these summands, so its projection supplies a nontrivial central idempotent, contradicting the definition of a block. An indecomposable projective belongs to one block. Finally, a nonsplit extension $0\to U\to V\to W\to 0$ of simple modules is a quotient of the projective cover of $W$: a lift onto $W$ must contain $U$, since otherwise the extension splits. Thus its two endpoints share factors in an indecomposable projective.

For the preliminary definitions, a block is $Re$ for a primitive central idempotent $e$, and a module belongs to it when $eM=M$. For a semisimple algebra, the Artin–Wedderburn theorem identifies its blocks with the factors $M_{n_i}(D_i)$ in its product decomposition. Each factor is a block of a finite-dimensional algebra.

We prove Ext separation of finite-length modules. The extension hypothesis is ${\mathrm{Ext}}^1(S,T)={\mathrm{Ext}}^1(T,S)=0$ for simples in opposite sets. First, for a simple $S\in {\mathcal{C}}_1$ and a module $N$ of finite composition length with factors in ${\mathcal{C}}_2$, we have ${\mathrm{Ext}}^1(S,N)=0$. Induct on the length of $N$: for $0\to N^{\prime }\to N\to T\to 0$ with $T$ simple, the long exact sequence of the Ext functor containswhose outer groups are zero. The same argument works with the two sets interchanged.

Now induct on the length of $M$, with $M=0$ immediate. Take a simple quotient in $0\to N\to M\to S\to 0$. By induction, $N=N_1\oplus N_2$ with factors in the respective sets. Suppose $S\in {\mathcal{C}}_1$; the other case is symmetric. Quotienting by $N_1$ givesThis splits by the preceding Ext vanishing. Let $U_1$ be the inverse image in $M$ of the chosen complementary copy of $S$. Then $U_1\cap N_2=0$, $U_1+N_2=M$, and $0\to N_1\to U_1\to S\to 0$. Thus $U_1$ has only factors in ${\mathcal{C}}_1$, while $U_2=N_2$ has only factors in ${\mathcal{C}}_2$.

If two finite-length modules have factors in disjoint sets, any homomorphism between them is zero: a nonzero image would, by the Jordan–Hölder theorem, have a simple factor belonging to both sets. For any submodule $V$ of $M$ with factors in ${\mathcal{C}}_1$, projection onto $U_2$ is consequently zero, so $V\subset U_1$. The analogous argument applies to $U_2$. HenceIn particular both summands are preserved by every endomorphism of $M$.