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Let $C$ be a projective curve (not necessarily reduced or irreducible). Let $\mathcal{F}, \mathcal{G}$ be $\mathcal{O}_C$-modules and $\phi:\mathcal{F} \to \mathcal{G}$ be a morphism of $\mathcal{O}_C$-modules. Suppose further that $\mathcal{F}$ is locally free. Is it true that the kernel of $\phi$ is also locally free?

Chen
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2 Answers2

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If you assume the curve smooth, then yes. This is because for a smooth curve, the local rings $\mathcal O_{X,x}$ are DVRs or fields, e.g. PIDs. In particular, a submodule of a free module over a PID is again free. Therefore, the exact sequence $$ 0 \to \ker \varphi \to \mathcal F \to \mathcal G $$ can be seen at the stalks as the sequence $$ 0 \to \ker \varphi_x \to \mathcal O_{X,x}^{\oplus I} \to \mathcal G_x. $$ where $I$ is some index set. Therefore $\ker \varphi_x$ is free for each $x \in X$, which means $\ker \varphi$ is locally free.

Of course not all curves are smooth, so with non-smooth curves you get counter-examples.

Another situation where the answer is "yes" is when $\varphi : \mathcal F \to \mathcal G$ is surjective and $\mathcal G$ is locally free. This is because looking at the exact sequence at the stalks, since $\mathcal G_x$ is free, it is projective, thus the exact sequence splits ; in particular, $\ker \varphi_x$ is projective. But a projective module over a local ring is free, so that $\ker \varphi_x$ is free for all $x \in X$.

Hope that helps,

Patrick Da Silva
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    The fact that the stalks $ker\varphi_x$ are $\mathcal{O}_{X, x}$-free at every $x$ does not necessarily imply that the sheaf $ker\varphi$ is locally free. See for instance Arapura's counterexample here: https://mathoverflow.net/questions/75150/sheaf-with-free-stalks/75161 – Flavius Aetius Dec 01 '19 at 05:07
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    @FlaviusAetius: That counter-example is certainly a disturbing one, but notice how irregular the example is. It is a theorem that when a $\mathcal O_X$-module $\mathcal M$ sits on a ringed space $X$, if $\mathcal M$ is locally of finite presentation, then it is locally free if and only if the stalks are free. I presume that your counter-example is pathological enough not to be locally of finite presentation, since neighborhoods of a point have stalks with uncountable rank. – Patrick Da Silva Dec 03 '19 at 17:59
  • I can give you a sketch of a proof. The reason why it works under this assumption is because you get the isomorphism $Hom(\mathcal M,\mathcal N)_x ~ Hom(\mathcal M_x, \mathcal N_x)$ in this case. This is clear in the free case, and by left-exactness of Hom, also in the lfp case. Therefore you can lift an isomorphism at the stalks to a local isomorphism. So if you know that the stalks are free, you get local isomorphisms with a free sheaf. You're working on a scheme, so coherent sheaves form an abelian category, which means the kernel is coherent, and in particular lfp by definition. – Patrick Da Silva Dec 03 '19 at 18:13
  • @FlaviusAetius: I think if we assume $\mathcal G$ quasi-coherent, then $\mathrm{ker} \varphi$ would be quasi-coherent and therefore locally of finite presentation, correct? Should I add this assumption together with the fact that the curve is smooth? – Patrick Da Silva Jul 15 '21 at 13:27
  • @PatrickDaSilva If you consider $\varphi\colon\mathcal{F}\to\underline{0}_C$ with $\mathcal{F}$ coherent not locally free, then $\ker(\varphi)$ has not to be locally free! – Armando j18eos Aug 25 '23 at 13:26
  • @Armandoj18eos: I assumed $\mathcal F$ locally free, so I don't know why you did that comment. Misread, maybe? – Patrick Da Silva Aug 28 '23 at 14:17
  • @PatrickDaSilva If you assume that $\mathcal{F}$ is locally free also in the last statement ("[...] Another situation where the answer is "yes" is when $\varphi\colon\mathcal{F}\to\mathcal{G}$ is surjective and $\mathcal{G}$ is locally free. [...]"): I'm agree! – Armando j18eos Aug 29 '23 at 18:54
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You ask if every submodule of a locally free module $F$ is again locally free. Of course this fails. First of all, this submodule doesn't have to be quasi-coherent. But even when it is quasi-coherent, and we therefore may pass to affine covers, it fails even for $F=\mathcal{O}$, i.e. not every quasi-coherent ideal is locally free. For this it would have to be zero or invertible.

Martin Brandenburg
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