## 109.29 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

$f_*\mathcal{O}_ X(D(a)) = \prod \nolimits _{n \in \mathbf{N}} A_ a$

Hence, in order for $f_*\mathcal{O}_ X$ to be quasi-coherent we would need

$\prod \nolimits _{n \in \mathbf{N}} A_ a = \left(\prod \nolimits _{n \in \mathbf{N}} A\right)_ a$

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

$A = k[t, z, x_1, x_2, x_3, \ldots ]/(tx_1z, t^2x_2^2z, t^3x_3^3z, \ldots )$

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

$0 \to f_*\mathcal{O}_ X \to \mathcal{O}_ Y \oplus \mathcal{O}_ Y \xrightarrow {(1, -1)} j_*\mathcal{O}_ V$

where $j : V \to Y$ is the inclusion morphism. Since

$A \longrightarrow \prod A_{x_ n}$

is injective (details omitted) we see that $\Gamma (Y, f_*\mathcal{O}_ X) = A$. On the other hand, the kernel of the map

$A_ t \longrightarrow \prod A_{tx_ n}$

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.

Lemma 109.29.1. 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.

Proof. See discussion above. $\square$

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