# Is the pushforward of a locally free sheaf by an open immersion coherent?

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?

• $F=\mathcal{O}_X$ is a counterexample for the first and the third question. – Martin Brandenburg Sep 14 '10 at 11:42
• You also might find Section 1, of Generalized Divisors on Gorenstein Schemes'' a useful read. In particular Prop 1.11 and Thm 1.12. – Karl Schwede Sep 14 '10 at 15:34

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

• 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. – Georges Elencwajg Sep 14 '10 at 14:04
• In the context of locally free F this statement (replacing projective variety with integral scheme) follows immediately from algebraic Hartogs' lemma – Tomo Apr 28 '18 at 3:51
• See also Tag 0BK3. – Minseon Shin Oct 23 '19 at 17:19

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.

• 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($Y\setminus X) \geq 2$? – Yemon Dai Sep 14 '10 at 12:50
• 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). – Torsten Ekedahl Sep 14 '10 at 12:56
• 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. – Karl Schwede Sep 14 '10 at 15:16

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 $\mathcal{F}$.