Lemma 110.30.1. Schemes, Lemma 26.24.1 is sharp in the sense that one can neither drop the assumption of quasi-compactness nor the assumption of quasi-separatedness.
110.30 Pushforward of quasi-coherent modules
In Schemes, Lemma 26.24.1 we proved that f_* transforms quasi-coherent modules into quasi-coherent modules when f is quasi-compact and quasi-separated. Here are some examples to show that these conditions are both necessary.
Suppose that Y = \mathop{\mathrm{Spec}}(A) is an affine scheme and that X = \coprod _{n \in \mathbf{N}} Y. We claim that f_*\mathcal{O}_ X is not quasi-coherent where f : X \to Y is the obvious morphism. Namely, for a \in A we have
Hence, in order for f_*\mathcal{O}_ X to be quasi-coherent we would need
for all a \in A. This isn't true in general, for example if A = \mathbf{Z} and a = 2, then (1, 1/2, 1/4, 1/8, \ldots ) is an element of the left hand side which is not in the right hand side. Note that f is a non-quasi-compact separated morphism.
Let k be a field. Set
Let Y = \mathop{\mathrm{Spec}}(A). Let V \subset Y be the open subscheme V = D(x_1) \cup D(x_2) \cup \ldots . Let X be two copies of Y glued along V. Let f : X \to Y be the obvious morphism. Then we have an exact sequence
where j : V \to Y is the inclusion morphism. Since
is injective (details omitted) we see that \Gamma (Y, f_*\mathcal{O}_ X) = A. On the other hand, the kernel of the map
is nonzero because it contains the element z. Hence \Gamma (D(t), f_*\mathcal{O}_ X) is strictly bigger than A_ t because it contains (z, 0). Thus we see that f_*\mathcal{O}_ X is not quasi-coherent. Note that f is quasi-compact but non-quasi-separated.
Proof. See discussion above. \square
Comments (0)