X is a Noetherian scheme, F is an injective object in the category of quasi-coherent sheaves on X. U is an open subset of X. Why F's restriction on U is still an injective object in the category of quasi-coherent sheaves on U?
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1$\begingroup$ Regarding answers below involving the "extension by zero" functor from sheaves on U to sheaves on X, extending a <i>quasicoherent</i> sheaf by zero does not usually result in a quasicoherent sheaf on X, so it is not even a functor from QCoh(U) to QCoh(X). This subtlety is what makes the question interesting. (I will keep this comment posted even if the answers it refers to are removed, since this is a very common mistake.) $\endgroup$– Andrew CritchCommented Nov 25, 2009 at 6:42
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2$\begingroup$ I suggest deleting the answers below so the question displayed accurately as unanswered, and following Dave's suggestion that a record of mistakes be kept to learn from, I'll leave my comment above for posterity ;) $\endgroup$– Andrew CritchCommented Nov 25, 2009 at 6:50
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$\begingroup$ Agreed. Also, Tiasong's answer below should really be a comment (I'll delete this once it is moved). $\endgroup$– David Zureick-BrownCommented Nov 25, 2009 at 6:59
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4 Answers
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The restriction-by-zero type arguments can actually be made to work, with some effort and an extra hypothesis. Suppose $X$ is locally Noetherian, $j: U \to X$ the inclusion of an open subscheme.
Let $Mod(X)$ and $QCoh(X)$ be the categories of $O_X$-modules, and quasi-coherent $O_X$-modules, respectively.
The "some effort" is the following Lemma
Lemma If $X$ is locally Noetherian, then the injective objects in $QCoh(X)$ are precisely the injective objects of $Mod(X)$ which are quasi-coherent as sheaves of modules.
Pf: Any injective object of $Mod(X)$ which is quasi-coherent must certainly be injective in the smaller category $QCoh(X)$. For the converse, it suffices to show that any injective object $I$ of $QCoh(X)$ injects into some $I'$ which is a quasi-coherent injective object of $Mod(X)$, for then $I$ will be a retract of $I'$ and so injective in $Mod(X)$. This seems tricky, but is proved in Theorem 7.18 of Hartshorne's "Residues and duality".
Now, let's prove the result using the Lemma: If $J$ is an injective object in $QCoh(X)$, then the hard direction of the Lemma implies that it is injective in $Mod(X)$. The restriction-by-zero argument applies in this category, allowing us to conclude that $j^* J$ is injective in $Mod(U)$. It's clearly quasi-coherent, so applying the easy direction of the Lemma we see that it is injective in $QCoh(U)$ as desired.
[Aside: On a Noetherian scheme, any quasi-coherent sheaf is a union of its coherent subsheaves and one can "extend" coherent sheaves on U to coherent sheaves on X (see e.g., Hartshorne Ex. II.5.15). Using these facts, one should be able to give a more direct argument in the Noetherian case.]
answered Nov 25, 2009 at 9:57
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1$\begingroup$ Thanks for telling us the lemma! This solves the problem. As for the last paragraph, I have tried that way, but the difficulty is, although one can always extend the sheaf, it's not obvious that one can always extend the sheaf homomorphism $\endgroup$– TJCMCommented Nov 25, 2009 at 15:27
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This is false is general. In particular, if $X = SpecA$ is affine, this would imply that given an injective $A$-module $M$ and $f \in A$, one would have $M_f$ is injective over $A_f$; this is FALSE in general (see, for example, "Localization of Injective Modules" by Everett C. Dade (it's in Journal of Algebra ~ April 1981)).
Maybe you need to assume that $X$ is locally Noetherian, or even Noetherian?
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Restriction to an open subset has an exact left adjoint (extension by zero).
answered Nov 25, 2009 at 6:05
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$\begingroup$ This is not a functor from QCoh(U) to QCoh(X), so it's not any kind of adjoint to the restriction functor. $\endgroup$ Commented Nov 25, 2009 at 6:43
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Hey, I am the asker for the question, I want to say two points:
I forgot the Noetherian condition; X should be Noetherian;
I lose the cookie so I can't log in that account any more; I don't know how to add comment or reply others... Why I don't have that button...
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$\begingroup$ You should register for the site so you don't need to rely on cookies. $\endgroup$– user1884Commented Nov 25, 2009 at 7:29
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$\begingroup$ I recommend getting an OpenID at myopenid.com , as I've found it more reliable than logging in with google ID. $\endgroup$ Commented Nov 25, 2009 at 7:45
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$\begingroup$ I've merged your accounts (I found three of them), so you should now have a single registered user. In particular, you should be able to edit your question to include the Noetherian hypothesis, and you have enough reputation to comment on anything you like. $\endgroup$ Commented Nov 25, 2009 at 14:47