I'm working through some lecture notes on the representation theory of a finite dimensional algebra $A$ (associative, unital, over an algebraically closed field $k$), and have got stuck on a particular part.
Earlier in the notes he covers Krull-Schmidt, and the fact that if $M$ is a finite dimensional (left) $A$-module then $\mathrm{End}_AM$ decomposes as a $k$-vector space direct sum $I \oplus B$, where $I$ is a two-sided nilpotent ideal and $B$ is a subalgebra isomorphic to a product of matrix algebras (whose dimensions come from the multiplicities of the indecomposable summands of $M$). He also discusses the relationship between decompositions of $A$ as a (left or right) $A$-module and decompositions of $1 \in A$ as a sum of orthogonal idempotents. I'm happy with all of this stuff.
But then he suggests applying these results to $A$ itself, thought of as a left $A$-module. He takes a decomposition $A \cong \bigoplus_{i=1}^r p_i P_i$, with the $P_i$ pairwise non-isomorphic indecomposables, and then says:
1. There exists a decomposition of $k$-vector spaces $A = I \oplus B$ where $I$ is a nilpotent two-sided ideal and $B$ is a subalgebra isomorphic to $\Pi_{i=1}^r \mathrm{Mat}_{p_i}(k)$
2. For each $i$ the $A$-module $S_i := P_i/IP_i$ is simple, and every simple $A$-module is isomorphic to a unique $S_i$.
My problem is that I can't get my head around which way all of the actions go, and the proof doesn't give any details; it just refers back to the earlier results. Surely if we apply them to our module decomposition of $A$ then we get a decomposition of $\mathrm{End}_A(A) \cong A^{opp}$. We can carry this to $A$ via the identity map on the underlying vector spaces (and still get a two-sided nilpotent ideal and a subalgebra isomorphic to a product of matrix algebras), but then why should $I$ do anything sensible when multiplying $P_i$ on the left? It was defined in terms of how it acted by multiplication on the right (as left $A$-module endomorphisms).
If, on the other hand, we treat $A$ as a right $A$-module, so its endomorphism algebra is $A$ itself, then the decomposition into the $P_i$s no longer fits. Do you have to get involved with the corresponding idempotent decomposition of $1$, to pass between left- and right-module decompositions?
I tried asking the author of the notes (they're handwritten and so not available online) but didn't get a particularly helpful response. This is really frustrating me and it would be great to shed some light on it!