How to get $eA$ faithful from $Ae$ faithful?

Question

Let $A$ be a finite dimensional algebra over a field K. Suppose $e$ is an idempotent in $A$ such that $D(Ae)\cong eA$, where $D=Hom_K(-,K)$ is the duality. How to get $eA$ faithful from $Ae$ faithful?

Answer 1

Nothing mysterious here: it all stems from the definition of the $A$ action on $D(Ae)$:

For all $f\in D(Ae)$ and $a\in A$, the map $f\cdot a$ is given by $(f\cdot a)(x):=f(ax)$ for all $x\in Ae$.

Therefore if there exists nonzero $r\in A$ such that $rAe=\{0\}$, you would also have that $(f\cdot r)(x)=0$ for any choice of $f\in D(Ae)$. This shows that $r$ annihilates $D(Ae)$ and (since $eA$ is isomorphic) $eA$.

The other direction is nearly as simple. You would just like to conclude that if $r\in A$ is nonzero and $(f\cdot r)(x)=0$ for every $f$, then in fact $rAe=\{0\}$. Suppose $0\neq rae\in rAe$. Extend $\{rae\}$ to a basis of $Ae$ and let $f$ be the linear functional mapping $rae$ to $1$ and all other basis elements to $0$. Now $f\cdot r$ is also a linear functional, just as $f$ is, but it isn't zero since $(f\cdot r)(ae)=rae\neq 0$. This is a contradiction and the upshot is that apparently $rAe=\{0\}$ in the first place.