In section 1.6 of Lectures on Algebraic Categorification by Mazorchuk, it is mentioned that if $A$ and $B$ are finite dimensional unital $k$-algebras, ($k$ a field) and $M$ is simple as an $A\oplus B$-module, then $M$ is in fact either simple as an $A$-module or a $B$-module. Why is this?
I'm a bit confused by a few points. First, I'm pretty sure $A\oplus B$ really is the algebra with underlying set $\{(a,b):a\in A,\ b\in B\}$, with pointwise operations, and is not notation for the coproduct $A\otimes_k B$, since that case is treated in the next paragraph.
But then identifying $A$ with the subset $(A,0)$ and $B$ with $(0,B)$, one has $AB=BA=0$. If $M$ is not a simple $A$-module, there is $m\in M$ nonzero such that $0\neq Am\neq M$. But since $M$ is a simple $A\oplus B$-module, $$ M=(A\oplus B)(Am)=Am+BAm=Am $$ since $M$ is a simple $A\oplus B$-module, a contradiction. Doing the same argument with $B$ seems to show $M$ is both simple as an $A$-module and as a $B$-module. Is my reasoning flawed? I find it strange that the word 'or' is used instead of 'and' in that case in the original claim, even if it's still technically correct.
In this set-up, each $M$-module has the form $M=M_1\oplus M_2$ where $M_1$ is an $A$-module and $M_2$ is a $B$-module. Then $M$ is simple iff either $M_1$ is a simple $A$-module and $M_2=0$ or the other way round.
The ring $A\oplus B$ has central idempotents $e_1=(1,0)$ and $e_2=(0,1)$ with $e_1+e_2=1$. Then $M=M_1\oplus M_2$ with $M_i=e_i M$. The action is given by $(a,b)(m_1,m_2)=(am_1,bm_2)$.
I think the flaw in your argument is that $M$ is in general not an $A$-module (or a $B$-module) or at least not a unital $A$-module, since the identity element of $A$ does not in general act trivially on $M$. In $A\oplus B$ the identity of $A$ is in effect the idempotent $e_1=(1,0)$.