Let $k$ be a field of characteristic $p > 0$, $G$ a finite group, and $(P, e)$ a Brauer pair of $FG$ (i.e., a pair of a $p$-subgroup of $G$ and a centrally primitive idempotent of $FC_G(P)$). Denote the Brauer homomorphism by $$ \operatorname{Br}_P \colon (FG)^P \twoheadrightarrow FC_G(P). $$ I'd like to know why a primitive idempotent $i \in (FG)^P$ with $\operatorname{Br}_P(i)e \neq 0$ remains primitive under the image of $\operatorname{Br}_P$. (cf. Proof of Theorem 2.10.16 in this textbook.)
It seems that the decomposition $$ (FG)^P = FC_G(P) \oplus \ker \operatorname{Br}_P $$ has something to do with, but I couldn't see how.
In fact, the following more general fact is true.
Let $A$ be a finite dimensional algebra, $I\lhd A$ an ideal, and $\pi:A\to A/I$ the quotient map. If $e\in A$ is a primitive idempotent, then either $\pi(e)=0$ or $\pi(e)\in A/I$ is a primitive idempotent.
I think the easiest way to see this, assuming basic facts about modules for finite dimensional algebras, is to use the fact that an idempotent $e$ is primitive if and only if the projective (right) $A$-module $eA$ is indecomposable, which is the case if and only if the largest semisimple quotient of $eA$ is simple.
So if $\pi(e)$ is non-zero but not primitive, then $\pi(e)(A/I)$ has a non-zero semisimple quotient that is not simple. But $\pi(e)(A/I)=\pi(eA)$ is a quotient (considered as an $A$-module) of $eA$, and so $eA$ has a non-zero semisimple quotient that is not simple, and so $e$ is not primitive.