Why is an injective quasi-coherent sheaf's restriction to an open subset still an injective object? - MathOverflow most recent 30 from http://mathoverflow.net2013-05-21T14:06:25Zhttp://mathoverflow.net/feeds/question/6762http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/6762/why-is-an-injective-quasi-coherent-sheafs-restriction-to-an-open-subset-still-anWhy is an injective quasi-coherent sheaf's restriction to an open subset still an injective object?Taisong Jing2009-11-25T04:22:22Z2012-05-20T14:06:41Z
<p>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?</p>
http://mathoverflow.net/questions/6762/why-is-an-injective-quasi-coherent-sheafs-restriction-to-an-open-subset-still-an/6768#6768Answer by Jonathan Wise for Why is an injective quasi-coherent sheaf's restriction to an open subset still an injective object?Jonathan Wise2009-11-25T06:05:38Z2009-11-25T06:05:38Z<p>Restriction to an open subset has an exact left adjoint (extension by zero).</p>
http://mathoverflow.net/questions/6762/why-is-an-injective-quasi-coherent-sheafs-restriction-to-an-open-subset-still-an/6773#6773Answer by Adam Topaz for Why is an injective quasi-coherent sheaf's restriction to an open subset still an injective object?Adam Topaz2009-11-25T07:00:32Z2009-11-25T07:00:32Z<p>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)).</p>
<p>Maybe you need to assume that $X$ is locally Noetherian, or even Noetherian?</p>
http://mathoverflow.net/questions/6762/why-is-an-injective-quasi-coherent-sheafs-restriction-to-an-open-subset-still-an/6774#6774Answer by Taisong Jing for Why is an injective quasi-coherent sheaf's restriction to an open subset still an injective object?Taisong Jing2009-11-25T07:07:00Z2009-11-25T07:07:00Z<p>Hey, I am the asker for the question, I want to say two points:</p>
<ol>
<li><p>I forgot the Noetherian condition; X should be Noetherian;</p></li>
<li><p>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...</p></li>
</ol>
http://mathoverflow.net/questions/6762/why-is-an-injective-quasi-coherent-sheafs-restriction-to-an-open-subset-still-an/6778#6778Answer by Anatoly Preygel for Why is an injective quasi-coherent sheaf's restriction to an open subset still an injective object?Anatoly Preygel2009-11-25T09:57:58Z2009-11-25T09:57:58Z<p>The restriction-by-zero type arguments can actually be made to work, with some effort and an extra hypothesis. Suppose <code>$X$</code> is <em>locally Noetherian</em>, <code>$j: U \to X$</code> the inclusion of an open subscheme.</p>
<p>Let $Mod(X)$ and $QCoh(X)$ be the categories of $O_X$-modules, and quasi-coherent $O_X$-modules, respectively.</p>
<p>The "some effort" is the following Lemma</p>
<p><strong>Lemma</strong> 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.</p>
<p><strong>Pf</strong>: 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 <code>$I$</code> 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".</p>
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<p>Now, let's prove the result using the Lemma: If <code>$J$</code> 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 <code>$j^* J$</code> 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.</p>
<p>[Aside: On a <em>Noetherian</em> 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.]</p>