Blocks of a finite-dimensional algebra via primitive central idempotents

Let $F$ be a field and $A$ a finite-dimensional unital $F$-algebra with centre $Z(A):=\{z\in A:za=az\text{ for all }a\in A\}$ (Algebras over a commutative ring, central structure maps, and algebra homomorphisms). A central idempotent of $A$ is an idempotent $e\in Z(A)$; it is primitive when $e\mathrm{\ne }0$ and $e$ is not the sum of two nonzero orthogonal central idempotents.

Then $Z(A)$ is a finite-dimensional commutative $F$-algebra, so the identity $1\in Z(A)$ admits a decomposition $1=e_1+\cdots +e_k$ into pairwise orthogonal primitive central idempotents (The identity of a finite-dimensional algebra decomposes into primitive central idempotents), and this decomposition is unique up to order (Decompositions of the identity into primitive central idempotents are unique). A block of $A$ is the two-sided ideal $B=eA=Ae$ cut out by a primitive central idempotent $e$ (the two descriptions agree because $e$ is central).

The blocks are pairwise orthogonal two-sided ideals: $B_i{B}_j\subseteq e_{i}A{e}_{j}A=e_i{e}_{j}A=0$ for $i\mathrm{\ne }j$, and $e_i\in B_i$ is the identity of $B_i$ since $e_i(e_{i}a)=e_{i}a$ and $(a{e}_i)e_i=a{e}_i$. Moreover $A=B_1\oplus \cdots \oplus B_k$ as $F$-algebras with units $e_i$: every $a\in A$ is $a=({\sum }_i{e}_i)a={\sum }_i{e}_{i}a$ with $e_{i}a\in B_i$, and if ${\sum }_i{b}_i=0$ with $b_i\in B_i$ then multiplying by $e_j$ gives $b_j=e_j{b}_j={\sum }_i{e}_j{b}_i=0$, so the sum is direct. Each $B_i$ is an indecomposable two-sided ideal: if $B_i=J\oplus J^{\mathrm{\prime }}$ were a direct sum of nonzero two-sided ideals, the projections of $e_i$ onto $J$ and $J^{\mathrm{\prime }}$ would be nonzero orthogonal central idempotents of $A$ summing to $e_i$, as is seen by applying to $e_i$ the $A$-bimodule projections $B_i\to J$ and $B_i\to J^{\mathrm{\prime }}$, which satisfy $\pi (ax)=a\pi (x)$ and $\pi (xa)=\pi (x)a$; this contradicts primitivity of $e_i$.

For every (not necessarily finite-dimensional) $A$-module $M$ one has $M={⨁}_{i=1}^k{e}_{i}M$ as $A$-modules: ${\sum }_i{e}_i=1$ acts as the identity on $M$, so $M={\sum }_i{e}_{i}M$, and if ${\sum }_i{m}_i=0$ with $m_i=e_i{m}_i$ then applying $e_j$ gives $m_j=0$; each $e_{i}M$ is $A$-stable because $e_i$ is central. A simple $A$-module $S$ satisfies $e_{i}S\mathrm{\ne }0$ for exactly one $i$, and then $S=e_{i}S$ lies in the corresponding block: $S=({\sum }_i{e}_i)S={\sum }_i{e}_{i}S$ forces some $e_{i}S\mathrm{\ne }0$, and if $e_{i}S\mathrm{\ne }0\mathrm{\ne }{e}_{j}S$ with $i\mathrm{\ne }j$ then simplicity gives $e_{i}S=S=e_{j}S$, whence $S=e_{i}S=e_i(e_{j}S)=(e_i{e}_j)S=0$, a contradiction. Nothing here asserts that a block is semisimple or simple, and the construction is choice-free.