What is the center of $End_A(S)$? $k$ is a field and $A$ a finite-dimensional algebra over $k$, $S$ is a simple $A$-module.
I know that $End_A(S) \subseteq End_k(S)$ and $End_k(S)$ has a center isomorphic to $k$. What is the center of $End_A(S)$? It contains $k$, for sure.
This is a question I came up with. If it doesn't even make sense, please let me know! Thank you!
 A: The endomorphism ring of a simple module is a division ring, because a nonzero endomorphism is necessarily bijective.
Call $D$ this division ring, which is also a $k$-algebra and is finite dimensional over $k$, because $S$ is finite dimensional over $k$.
The center $K$ of a division ring is easily seen to be a field, hence we can state that $K$ is a finite extension of $k$.
Which one? Any finite extension of $k$ is possible. Indeed, if $K$ is a finite extension of $k$, you can take $A=S=K=D$.
If $k$ is algebraically closed, then $K=k$.
A: The centre of $\text{End}_A(S)$ is a field.. For,
the endomorphism ring of a simple module is a division ring.
Now we will show that the centre $Z=Z(\text{End}_A(S))$ is exactly the field $k$ because for all $0 \neq x \in Z(\text{End}_A(S))$,  we always have $x^{-1} \in \text{End}_A(S)$. But we have to show is that the inverse $x^{-1} \in Z(\text{End}_A(S)).$ And this holds as  for $x \in Z(\text{End}_A(S))$, we have \begin{align*} &gx=xg \  \forall g \in \text{End}_A(S), \\ \Rightarrow &x^{-1}gxx^{-1}=x^{-1}xgx^{-1} \\
 \Rightarrow &x^{-1}g=gx^{-1}, \end{align*} which says $x^{-1} \in Z(\text{End}_A(S))$. Hence for $x \in Z(\text{End}_A(S))$ implies $x^{-1} \in Z(\text{End}_A(S))$. Further, note that centre of a ring is itself a ring and infact a commutative ring.
Thus $Z(\text{End}_A(S))$ being commutative ring as well having property that it consists of inverses of all its non-zero elements $(xx^{-1}=1)$, implies that $Z(\text{End}_A(S))$ is a field.
Edit: Now we have to check whether $Z(\text{End}_A(S))  =k$ or not. For this, see the answer by @egreg.
