Let $A$ be a simple $k$-algebra, $e\in A$ be a primitive and $f\in A$ a not necessarily primitive idempotent. A text I am currently trying to understand seems to implicitly assume that $\dim eA$ divides $\dim fA$, which does not seem obvious to me.
Clearly, we may assume w.l.o.g. that $e$ is a summand of $f$ (since we may consider $1-f$ instead of $f$ otherwise). But how do we continue from here?
Edit: I should have mentioned that my rings are assumed to be unital and associative, and, if necessary, Artinian.
I will assume you mean $A$ is associative and unital, and artinian, if that last point is not something you want it could be more. Hence, for every $e\in A$, $eA$ is a (right) $A$-module because it is right ideal. Since $A$ is simple, its modules are semisimple, i.e. products of simples. Now $e$ is primitive so $eA$ is simple. Associative algebras have one simple module up to isomorphism. So $fA$ is a product of modules isomorphic to $eA$. Hence, $\dim fA=m\cdot \dim eA$.