Let $k$ be a field and let $A$ be a $k$-algebra. Denote by $End_A(A)$ the set of all $A$-homomorphisms of the regular $A$-module $A$ into itself. Fix $a ∈ A$, and define the $A$-module homomorphism $r_a : A → A$ by $r_a(x) = x*a$.
$\\$ Show that $End_A(A)$ = {$r_a$ | $a ∈ A$}
$\\$ Answer: $\\$ Obviously I can see that, for any $a$ we have $r_a$ is in $End_A(A)$, but how do I go about the other direction? Namely why is it that any endomorphism at all has this very simple form? I just don't see it...
If $f\in\operatorname{End}_A(A)$ then by definition $f(ax)=af(x)$ for all $a\in A$ (and $f$ is also additive). Now let $b=f(1)$, then for all $a\in A$ one has $f(a)=f(a1)=af(1)=ab=r_b(a)$, so $f=r_b$, identically.