Example 31.9.8. Let $R = \prod _{n \in \mathbf{N}} \mathbf{F}_2$. Let $I \subset R$ be the ideal of elements $a = (a_ n)_{n \in \mathbf{N}}$ almost all of whose components are zero. Let $\mathfrak m$ be a maximal ideal containing $I$. Then $M = R/\mathfrak m$ is a finite flat $R$-module, because $R$ is absolutely flat (More on Algebra, Lemma 15.104.6). Set $S = \mathop{\mathrm{Spec}}(R)$ and $\mathcal{F} = \widetilde{M}$. The closed subschemes of Lemma 31.9.6 are $S = Z_{-1}$, $Z_0 = \mathop{\mathrm{Spec}}(R/\mathfrak m)$, and $Z_ i = \emptyset $ for $i > 0$. But $\text{id} : S \to S$ does not factor through $(S \setminus Z_0) \amalg Z_0$ because $\mathfrak m$ is a nonisolated point of $S$. Thus Lemma 31.9.7 does not hold for finite type modules.
Comments (0)