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Hartshorne, "Algebraic Geometry," Exercise II.7.1, reads:

Let $(X, \mathcal{O}_X)$ be a locally ringed space, and let $f : \mathscr{L} \to \mathscr{M}$ be a surjective map of invertible sheaves on $X$. Show that $f$ is an isomorphism.

To prove this, I note that a morphism of sheaves is injective (resp. surjective) iff it is injective (resp. surjective) on stalks. Thus $f_P$ is surjective for each $P \in X$. Since $\mathscr{L}$ and $\mathscr{M}$ are invertible, $\mathscr{L}_P \cong \mathscr{M}_P \cong \mathcal{O}_{X, P}$ as $\mathcal{O}_{X, P}$-modules. From commutative algebra, a surjective endomorphism of finitely-generated modules over any ring is in fact an automorphism, so $f$ is an isomorphism on stalks and thus an isomorphism of sheaves.

This argument sounds fine to me, but now I'm worried because I didn't use many of the conditions in the problem -- I could weaken it to an arbitrary (rather than locally-) ringed space, to any locally free sheaves of the same rank or even to something a bit weaker (namely that $\mathscr{L}$ and $\mathscr{M}$ are locally isomorphic), etc.

Have I missed something here? Or does the result hold in much wider generality without any modification?

Daniel McLaury
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3 Answers3

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This is elementary undergraduate algebra and has nothing to do with local rings nor Nakayama:

One reduces to show that given a commutative ring $A$, any surjective $A$-linear map $f:A\to A$ is in fact bijective.
Since $f(x)=f(x\cdot 1)=x\cdot f( 1)$, the map $f$ has the form $f(x)=u\cdot x$ (with $u:=f(1)$).
Since $f$ is surjective, there exists $v\in A$ with $1=f(v)=u\cdot v$ .
This implies that $u$ is invertible and $f$ is thus bijective with inverse $f^{-1}(y)= v\cdot y$ .

Georges Elencwajg
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    Hmm... If this is Hartshorne's intended argument, then I don't see why he stipulated that $(X, \mathcal{O}_X)$ be a locally ringed space, since it seems to work just fine for an arbitrary ringed space. If he expected the reader to apply Nakayama, as Zhen Lin hypothesizes, then that could explain the presence of the hypothesis. – Daniel McLaury Oct 16 '13 at 10:02
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    But, yes, this is surely the quickest and easiest-to-understand way of answering the question as stated! – Daniel McLaury Oct 16 '13 at 10:04
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    Dear Daniel, one possibility is that arbitrary ringed spaces are not really studied in the book and that Hartshorne didn't bother to state his exercise in the greatest possible generality. – Georges Elencwajg Oct 16 '13 at 10:13
  • @DanielMcLaury May be useful: Look at Georges' answer here. –  Oct 16 '13 at 12:10
  • @user38268, not only am I aware of that result, I'm using it in my argument above! – Daniel McLaury Oct 16 '13 at 12:42
  • Another formulation: if $f:A\twoheadrightarrow A$, let $I$ be the kernel then $A\simeq A/I$ but $I (A/I)=0$ so that $I=0$ in $A$ and so $f$ is an isomorphism. – Gabriel Soranzo Mar 27 '21 at 10:34
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Daniel McLaury, you are right. The same proof works in the following generality: If $X$ is a ringed space, $F,G$ are $\mathcal{O}_X$-modules such that for every $x \in X$ there is some isomorphism $F_x \cong G_x$ and this $\mathcal{O}_{X,x}$-module is finitely generated, then every epimorphism $F \to G$ is an isomorphism. The reason is that $F_x \to G_x$ is an isomorphism for every $x \in X$, by Nakayama. Of course this argument simplifies when $F,G$ are invertible, see Georges' answer.

Martin Brandenburg
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You have used that $f_P$ is an endomorphism. At the very least, you need to assume that $\mathcal{L}$ and $\mathcal{M}$ are locally isomorphic.

Andrew Dudzik
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