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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!

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I don't think the sides make any difference here. The path that should make you most comfortable: Decompose $A^{opp}$ as $I \oplus B$. Alternatively just write your homomorphisms on the correct side and then there is no "opp" business. Lam's First Course in Noncomm Rings has a nice treatment of Artin-Wedderburn as does Jacobon's Basic Algebra. – Jack Schmidt May 14 '14 at 14:36
Thanks for the comment @JackSchmidt. I'll try and have a look at those books. Apologies if this is a stupid question, and if I'm missing something obvious, but what do you mean by "I don't think the sides make any difference here"? What's bothering me is that the decomposition into $I \oplus B$ is based on how elements of $A$ act on $A$ by multiplication on the right, but then when we consider $I P_i$ we're saying something about how they act on the left. Is there an intrinsic way to define $I$ and $B$? (I'm not very familiar with the general theory, only what is done in the notes.) – Jez May 14 '14 at 18:52
$I$ is the sum of all two-sided nilpotent ideals, the (Wedderburn or nil or Jacobson) “radical” of $A$; it does not have a “side”. $B$ is a complement, not uniquely defined even on one side. For most things you won't need $B$ itself, just $A/I \cong B$. The $P_i$ are called the PIMs or projective indecomposable modules or principal indecomposable modules. – Jack Schmidt May 14 '14 at 19:18
$p_iP_i$ are called the blocks, or block ideals and are uniquely defined. Those are the things you probably already understand. Breaking up the blocks is somewhat arbitrary, the specific submodules $P_i$ are not well defined, only their isomorphism type. – Jack Schmidt May 14 '14 at 19:21
@JackSchmidt Thank you! If only I'd been told that $I$ was the Jacobson radical then things would have been much clearer. It isn't at all obvious to me (although I've been through a proof and understand it) that the various different left and right definitions of the Jacobson radical coincide, so no wonder I was having trouble relating the left action to the right. Thanks again. – Jez May 14 '14 at 19:27

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