Injective map on a coherent sheaf on a projective scheme must be an automorphism

Question

Given a coherent sheaf $E$ on a projective scheme $X$ over a field and an endomorphism $f:E \rightarrow E$, show that if $f$ is injective then it is an isomorphism. Give a counterexample to this statement if $X$ is not projective.

I found the following on page $108$ in Friedman's Algebraic Surfaces and Holomorphic Vector Bundles:

Let $X$ be a scheme, proper over a field k, and let $\mathcal{F}$ be a coherent sheaf on $X$. Then an injective map $\varphi$ from $\mathcal{F}$ to itself is an isomorphism.

(Since $\operatorname{Hom}(\mathcal{F},\mathcal{F})$ is finite-dimensional, $\varphi$ satisfies a polynomial equation, and since $\varphi$ is injective we can assume that this polynomial has a nonzero constant term. Clearly, $\varphi$ satisfies such a polynomial equation on each fibre $\mathcal{F}/m_x\mathcal{F}$, which is a finite-dimensional vector space. Thus, on each fibre $\varphi$ is injective and therefore surjective, so it is surjective by Nakayama's lemma.)

I have the following questions about this:

1) What $\textbf{exactly}$ does he mean when he says that $\operatorname{Hom}(\mathcal{F},\mathcal{F}$) is finite-dimensional?

2) Which part of Friedman's argument requires properness? Since the second part of the original question seems to imply that coherence is not enough, I am assuming that some part of the proof must be subtly using properness. The only part I can see requiring this condition would be the finite dimensionality mentioned above, as every other step seems (to me) to work just fine after the first statement. But I have no proof for why the finite-dimensionality would require properness.

3) I have the following solution in mind for the counter-example asked in the second part of the question:

$X = $ Affine Line $\mathbb{A}^1_k$ and $\mathcal{F} = \mathcal{O}_X$. Then the endomorphism $f$ that squares every section is certainly not an isomorphism (as it is not surjective).

Does this work as a counterexample and are there any other (interesting) counter-examples?

Thank you very much!

Answer 1

1. $\operatorname{Hom}(\mathcal{F},\mathcal{F})$ is a vector space over $k$, the base field. The claim is that this is a finite-dimensional vector space.

2. Finite-dimensionality of $\operatorname{Hom}(\mathcal{F},\mathcal{F})$ is implied by being proper over a field. First we remember that $\operatorname{Hom}(\mathcal{F},\mathcal{G})$ is the global sections of the hom-sheaf $\mathcal{Hom(F,G)}$, which is coherent if $\mathcal{F}$ and $\mathcal{G}$ are. Since coherent sheaves on schemes which are proper over a field have finite-dimensional spaces of global sections (this is just the statement that the proper pushforwards of a coherent sheaf is coherent), we have the result.

Contrast this with the situation over the affine line: $\operatorname{Hom}(\mathcal{O}_{\Bbb A^1_k},\mathcal{O}_{\Bbb A^1_k}) = k[x]$ which is not finite-dimensional as a vector space over $k$.

3. Continuing with the example above from the case of the affine line, consider multiplication by any nonzero nonunit of $k[x]$ as an endomorphism of $\mathcal{O}_{\Bbb A^1_k}$. This is a counterexample.