Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be a quasi-projective variety, $Y$ a projective variety, and $f:X \rightarrow Y$ be an open immersion. If $\mathcal{F}$ is a locally free coherent sheave, what can be said about $f_\ast \mathcal{F}$? Is it coherent? Is it torsion free? Is it reflexive?

share|improve this question
    
$F=\mathcal{O}_X$ 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

4 Answers 4

up vote 1 down vote accepted

Dear Yemon, about your new question:

Let $X$ be a projective variety and $Y=X\setminus S$ an open subset (with inclusion denoted $f:Y\to X$). Let $\mathcal F$ be an algebraic coherent sheaf without torsion on $Y$.

Theorem (Serre-Grothendieck) Suppose that $X$ 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

share|improve this answer
    
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

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.

share|improve this answer

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($Y\setminus X) \geq 2$?

share|improve this answer
3  
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 $\mathcal{F}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.