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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 $M$.

We prove $(i)\Rightarrow (iii)$, the main step in Ext-connected components determine blocks. Suppose a block of a finite-dimensional algebra $B=Re$ had more than one component of the graph of nonsplit extensions between its simple modules. Partition its simple types into one component and the union of the others. There are no cross-extensions, so part (a) decomposes the left regular $B$-module as $B=U_1\oplus U_2$. Both summands are nonzero: every simple $B$-module is a quotient of the regular module and therefore occurs among its composition factors.

The maximality in (a) makes these summands canonical. Every right multiplication is a left $B$-module endomorphism, so it preserves them. Thus they are two-sided ideals, and their projection $p:B\to U_1$ commutes with both left and right multiplication. Putting $f=p(e)$ givesSince both summands are nonzero, $f\ne 0,e$. This contradicts the primitivity of the central block idempotent $e$. There is therefore just one Ext component within each block, proving that simples in the same block satisfy $\boxed{S\sim T}$.