Let $X$ be a quasi-projective variety, $Y$ a projective variety, and $f:X \rightarrow Y$ an open immersion. If $\mathcal{F}$ is a locally free coherent sheaf, what can be said about $f_\ast \mathcal{F}$? Is it coherent? Is it torsion free? Is it reflexive?
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$\begingroup$ $F=\mathcal{O}_X$ is a counterexample for the first and the third question. $\endgroup$– Martin BrandenburgCommented Sep 14, 2010 at 11:42
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$\begingroup$ You also might find Section 1, of ``Generalized Divisors on Gorenstein Schemes'' a useful read. In particular Prop 1.11 and Thm 1.12. $\endgroup$– Karl SchwedeCommented Sep 14, 2010 at 15:34
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3 Answers
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About your new question:
Let $Y$ be a projective variety and let $X\subset Y$ be an open subset with complement the closed subset $S:=Y\setminus X$. Call $f:X\hookrightarrow Y$ the inclusion.
Let $\mathcal F$ be an algebraic coherent sheaf without torsion on $X$.
Theorem (Serre-Grothendieck) Suppose that $Y$ is normal and that $S$ has codimension $\geq 2$. Then the sheaf $f_\ast \mathcal F$ is coherent.
Serre, Prolongement de faisceaux analytiques cohérents, Ann.Inst.Fourier 16 (1966), 363-374
answered Sep 14, 2010 at 13:47
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$\begingroup$ This is close to Torsten's answer, but Serre supposes that the sheaf is without torsion, rather than reflexive.The article also considers the analytic case. $\endgroup$ Commented Sep 14, 2010 at 14:04
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$\begingroup$ In the context of locally free F this statement (replacing projective variety with integral scheme) follows immediately from algebraic Hartogs' lemma $\endgroup$– TomoCommented Apr 28, 2018 at 3:51
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Dear Yemon,
a)The sheaf $f_\ast \mathcal{F}$ is not coherent in general since its stalk will not be finitely generated over the local ring of a point of $Y\setminus X$. For example take $P$ a point of $\mathbb P^1=Y$ and $X= \mathbb P^1 \setminus P=\mathbb A^1$. Then for $\mathcal F =\mathcal O_X$, you get $(f_\ast \mathcal{F})_P= Rat(Y)$
b) The direct image $f_\ast \mathcal{F}$ will be torsion free because an inductive limit of torsion free modules over a domain is torsion free ( I assume that variety means in particular integral scheme.)
c) I'm not sure reflexive is a reasonable concept for a non-coherent sheaf.
answered Sep 14, 2010 at 11:50
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$\begingroup$ Thank you for your answer. My question was motivated by the fact that I would like to construct a reflexive coherent sheave $\mathcal{G}$ on $Y$ such that $\mathcal{G}|_X = \mathcal{F}$. Is it possible? What if I suppose that $Y$ is normal and codim($Y\setminus X) \geq 2$? $\endgroup$ Commented Sep 14, 2010 at 12:50
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$\begingroup$ This is different, you may always extend any coherent sheaf on $X$ to some coherent sheaf on $Y$ and then take its double dual. Under your supplementary conditions such an extension is equal to the direct image (which in particular is coherent and reflexive). $\endgroup$ Commented Sep 14, 2010 at 12:56
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$\begingroup$ Indeed, for the part on finding a coherent sheaf on $Y$ that restricts to $F$ on $X$, take a look at Hartshorne, chapter II, exercise 5.15 where this construction is done step by step. $\endgroup$ Commented Sep 14, 2010 at 15:16
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By the way, assuming by varieties, you mean irreducible varieties, then for the second question, the answer is yes.
For the torsion free-ness, suppose that $r \in H^0(U, O_X)$ kills some non-zero element $z \in H^0(U, f_* \mathcal{F}) = H^0(U \cap X, \mathcal{F})$. By restriction, $r$ is a non-zero element of $H^0(X \cap U, \mathcal{O}_Z)$. We still have $rz = 0$ even in this setting, and so by restricting to an affine cover of $X$, it still happens. This will contradict the torsion-freeness (and thus in particular the locally-freeness) of $\mathcal{F}$.
answered Sep 14, 2010 at 15:30