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$k$ 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!

scsnm
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  • Have you seen https://en.wikipedia.org/wiki/Schur%27s_lemma#Statement_and_Proof_of_the_Lemma (Formulation in the language of modules)? Maybe it helps – 1123581321 Aug 29 '20 at 16:53

2 Answers2

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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 $K=k$.

egreg
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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.

MAS
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    You've shown that the centre of $\textrm{End}_A\left(S\right)$ is a field. You haven't shown that it is $k$. The conclusion isn't even consistent: suppose $k$ is the complex numbers, then $A$ is also a $\mathbb{Q}$ algebra and your argument would make the centre both $\mathbb{C}$ and $\mathbb{Q}$. – Joshua Tilley Aug 29 '20 at 18:39
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    If you want a finite dimensional example, replace $\mathbb{C}$ with a finite extension of $\mathbb{Q}$. – Joshua Tilley Aug 29 '20 at 18:41