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$X$ is a algebraic variety. $\mathcal{F}$ is a coherent sheaf. $U\subset X$ is an open dense subset. $\mathcal{F}|_U$ is locally free on $U$.

Can we conclude that $\mathcal{F}$ is locally free on $X$ ?

Shuhang
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    It's worth noting that for any coherent sheaf $\mathcal{F}$ on an integral scheme X of finite type over a field, there exists a dense open subset $U\subseteq X$ such that $\mathcal{F}|_U$ is locally free. All you need to show this is that being locally free is an open condition (see Hartshorne exercises in section II.5) and then notice $\mathcal{F}$ is free at the generic point. – freeRmodule Nov 29 '16 at 07:54

1 Answers1

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No.
Take $X=\mathbb A^2_k\:$ and for $\mathcal F\subset \mathcal O_X$ take the ideal sheaf of some rational point $P\in X$.
If $U$ is the open dense subset $U=X\setminus \{P\}\subset X$, then $\mathcal F\vert U=\mathcal O_U$ is free of rank one.
Nevertheless $\mathcal F$ is not locally free because its stalk $\mathcal F_P$ is not free as an $\mathcal O_P$-module.

Edit: proof of non-freeness of $\mathcal F_P$
Say $P=(0,0)$ Then $\mathcal O_P=k[X,Y]_{\langle X,Y\rangle }$ and the stalk $\mathcal F_P$ is the maximal ideal $\mathcal F_P=\langle X,Y\rangle \mathcal O_P\subset \mathcal O_P$.
Since $\mathcal O_P$ is of dimension $2$ its maximal ideal is of height $2$ and thus cannot be generated by a single element (Krull). Hence $\mathcal F_P$ is a non-principal ideal and is thus not free as an $\mathcal O_P$-module.

Georges Elencwajg
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  • Can you please say why the stalk is not free over the local ring ? – Rene Schipperus Nov 28 '16 at 23:50
  • Oh, I think I know, you can cross multiply the generators to get a relation. – Rene Schipperus Nov 29 '16 at 00:00
  • Dear @Rene, I have added a proof of non-freeness in an Edit. – Georges Elencwajg Nov 29 '16 at 00:12
  • Thanks, is it also valid to argue by coherence, that $(x,y)$ is not a free $k[x,y]$ module ? – Rene Schipperus Nov 29 '16 at 00:18
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    Dear @Rene, I'm not sure what you mean: the sheaf $\mathcal F$ is certainly coherent, and yet not locally free. Another route is to prove that $\mathcal F_P$ is not $\mathcal O_P$-flat. Another yet is to invoke that over $\mathbb A^2$ locally free sheaves are free etc. – Georges Elencwajg Nov 29 '16 at 00:49
  • Sorry, what I meant was to ask if locally free implied that $M$ would be a free $A$ module, at least for coherent sheaves over an affine scheme. – Rene Schipperus Nov 29 '16 at 00:57
  • Yes, your last option seems to indicate that. – Rene Schipperus Nov 29 '16 at 01:02
  • @ReneSchipperus: Locally free as a coherent sheaf implies that the associated module is f.g. and projective, not necessarily free. (A famous conjecture of Serre, proved by Quillen and Suslin, says that over affine space, i.e. the case when $A = k[x_1,\dots,x_n]$, projective implies free. But this is a special property of affine spaces, and is not true for more general affine varieties/schemes.) – tracing Nov 29 '16 at 02:34
  • @tracing Thank-you very much for that. – Rene Schipperus Nov 29 '16 at 02:43
  • Short note on the geometric picture here: The ideal sheaf of a closed subscheme is always free on the open complement (which is dense if the ambient scheme is irreducible). It is locally free if and only if the closed subscheme is a divisor, i.e. of codimension $1$. – MooS Nov 29 '16 at 08:53
  • Thank You all for the answer and discussions ! – Shuhang Nov 29 '16 at 22:25