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Let $\mathcal{A}$ be an abelian category that is also $k$-linear, where $k$ is some algebraically closed field. Let $X$ be a simple object in $\mathcal{A}$. What can we say about $\mathrm{Aut}(X)$? I mean it is a subgroup of the ring $\mathrm{Hom}(X,X)$, but perhaps we can say more because of simplicity . . . ?

Edit: Will $\mathrm{Hom}(X,X)$ ever be finite-dimensional over $k$?

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    $\begingroup$ The endomorphisms of $X$ form a division ring with $k$ in its centre (Schur's Lemma), and the automorphisms are just the multiplicative group of non-zero elements therein. $\endgroup$
    – Dave Benson
    Commented Apr 8 at 21:20
  • $\begingroup$ Why are the non-zero elements always invertible? $\endgroup$
    – Bobby-John Wilson
    Commented Apr 8 at 21:22
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    $\begingroup$ Otherwise the kernel and image are proper subobjects. $\endgroup$
    – LSpice
    Commented Apr 9 at 3:45
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    $\begingroup$ In your edit, that's the other part of Schur's Lemma. If $k$ is algebraically closed then division rings that are finite dimensional over $k$ are equal to $k$. $\endgroup$
    – Dave Benson
    Commented Apr 9 at 7:42
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    $\begingroup$ Thanks for the comments! So the key is Schur's lemma for abelian categories. I found some nice notes here: sites.math.rutgers.edu/~jjs435/garts/… $\endgroup$
    – Bobby-John Wilson
    Commented Apr 9 at 11:19

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