I am looking for a counter example to the fact that a faithfully flat morphism is an effective descent morphism for the category of quasicoherent sheaves when one forgets the quasicompact hypothesis. If possible, I'd like a counterexample on the descent of morphism aspects. In other words, I am looking for a morphism of scheme $X' \rightarrow X$, faithfully flat but not quasicompact, and two quasicoherent sheaves $F$ and $G$ on $X$, such that, calling $X''=X' \times_X X'$, $F'$, $F''$, $G'$, $G''$ thepullback of $F$,$G$ to $X'$, $X''$, the sequence of modules $$0 \rightarrow Hom_X(F,G) \rightarrow Hom_{X'}(F',G') \rightarrow Hom_{X''}(F'',G'')$$ is not exact (here the last morpheme is as it should the difference between the morphism induced by the two projections $X'' > X'$. I wrote it this way because I am not sure how to do a double arrow in mathoverflow's tex:)

$\begingroup$ You can use \rightrightarrows. $\endgroup$ – RP_ Apr 12 '13 at 13:38

2$\begingroup$ Martin: the $X'_i$ are not faithfully flat over $X$, because not surjective. And their disjoint union, of course, is not affine if there are infinitely many $X'_i$ (or even $10^10^10$ $X'_i$ if you're an ultrafinitist  end of joke). Seriously, the reason I think the hypothesis is needed is that Grothendick has it in SGA 1. OTH, I am not sure it is needed already for descent of morphisms of schemes, and not only for the subtler descent of sheaves. $\endgroup$ – Joël Apr 12 '13 at 13:49

3$\begingroup$ Argh, I spoiled my joke by forgetting some delimiters. It was bad anyway... $\endgroup$ – Joël Apr 12 '13 at 13:50
I think the following should be a counterexample.
Let $X=\text{Spec}(\mathbb{C}[T])$ be the affine line and $X'=\bigsqcup_{x\in\mathbb{C}}\text{Spec}(\mathbb{C}[T]_{(Tx)})$ be its faithfully flat cover built from all local rings at closed points. Let $F=\mathcal{O}_{X}$ and $G=\bigoplus_{x\in\mathbb{C}}\mathcal{O}_X/(Tx)\simeq\bigoplus_{x\in\mathbb{C}}\mathbb{C}_x$ be two quasicoherent sheaves on $X$.
Then, there is a natural quotient morphism $u':F'\to G'$ given by $\mathbb{C}[T]_{(Tx)}\to \mathbb{C}[T]/(Tx)$ on $\text{Spec}(\mathbb{C}[T]_{(Tx)})$. This morphism obviously satisfies the descent condition, as the only nontrivial conditions to check are on connected components of $X''$ that are isomorphic to the generic point of $X$, and $G''$ is zero on these components. However, this morphism does not come from a morphism $u:F\to G$ because $u(1)$ should have a nonzero component in every factor of the direct sum, which is impossible.