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There seems to be a well-known result in commutative algebra that if $A$ is a commutative ring with unity and $M$ is a finitely generated $A$-module then any surjective endomorphism $f : M \to M$ is necessarily an automorphism. My reference for this fact is Eisenbud's GTM150 commutative algebra book Corollary 4.4 (or more directly, e.g., this).

Of course the above property fails for injective endomorphisms, as we have simple examples such as $A = k[x]$, $M = A$ with $f$ a multiplication map by $x$.

Now let $X$ be a projective variety over a field $k$ and $F$ a coherent sheaf on it. Say $f : F \to F$ is its endomorphism. If $f$ is surjective then the above commutative algebra fact says $f$ is an automorphism. My question is: if $f$ is injective, is $f$ an isomorphism?

My attempt was the following. Say we have a short exact sequence $0 \to F \to F \to C \to 0$. Fix any ample line bundle on $X$ and associate Hilbert polynomials to $F$ and $C$. By the short exact sequence, this forces the Hilbert polynomial associated to $C$ to be a constant $0$, forcing $C$ to be the zero sheaf. This also works when $f$ is surjective.

Is this proof correct? I'm a little worried since I couldn't find any reference for this.

Yoonjoo Kim
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I would like to answer the question by myself, to prevent any potential confusion by the statement (like I did).

My original attempt was a correct proof, and in fact it has the following equivalent commutative algebra formulation: let $S$ be a finitely generated graded algebra over $k$ (commutative with 1) and $M$ a finitely generated graded $S$-module. Then any injective (resp. surjective) graded $S$-module homomorphism $f : M \to M$ is automatically bijective.

The proof for this is simple, because each degree $M_i$ are finite dimensional $k$-vector spaces, so an injectivity (resp. surjectivity) of a $k$-linear map $f : M_i \to M_i$ implies it is bijective. It is interesting to compare this to the original non-graded commutative algebra version, which I believe is much more subtle.

I have no idea if the statement would still hold if $X$ is only proper over $k$.

Yoonjoo Kim
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