Hi everybody!! I am studing cohomology of sheaves on schemes. I have a question for you. Let $X$ be a noetherian scheme. Do you know if the cathegory QCoh(X) has enough injectives? I know that it has "enough flasque" and that Mod(X) has enough injective but what about QCoh(X)? Thank you
Yes. If $X$ is a Noetherian scheme, then the category of quasicoherent sheaves on $X$ has enough injectives. See Hartshorne's Algebraic Geometry, exercise III.3.6(a). 


Yes, it has enough injectives. As explained in Hartshorne's "Residues and Duality", the injective hull of a quasicoherent sheaf in Mod(X) is again quasicoherent. This is even more than you asked for. Less explicit, you can show that Qcoh(X) is a Grothendieck category (see Daniel Murfet's Notes) provided X is concentrated, and as such has enough injectives. 


First of all, let me quote the famous paper by Grothendieck "Sur quelques points d'algebre homologique" (Tohoku Math part 1 & part 2). According to Theorem 1.10.1 there, every abelian category satisfying (AB5) (equivalent to exactness of filtered direct limits) and with a generator has enough injectives, For every scheme $X$ it is well known that $Qco(X)$ is abelian and satisfies (AB5), see, for instance, EGA I, new edition, Corollaire (2.2.2)(iv) where it is proved that such a limit preserves quasicoherence. So, the only issue is the existence of a generator. For some time it was known that over a quasicompact quasiseparated scheme, $Qco(X)$ has a generator. For a nice geometric argument, consult the proof of Theorem (4) in Kleiman's "Relative duality for quasicoherent sheaves". Note that any noetherian scheme $X$ is quasicompact and quasiseparated, EGA I, (6.1.1) and (6.1.13). Surprisingly, using techniques from relative homological algebra, Enochs and Estrada proved in 2005 the existence of a generator for any scheme. See their paper "Relative homological algebra in the category of quasicoherent sheaves". (There was a previous unpublished proof by O. Gabber). Summing up, for any scheme $X$, the category $Qco(X)$ has enough injectives. 

