Example 29.3.4. Here is an example of an immersion which is not a composition of an open immersion followed by a closed immersion. Let $k$ be a field. Let $X = \mathop{\mathrm{Spec}}(k[x_1, x_2, x_3, \ldots ])$. Let $U = \bigcup _{n = 1}^{\infty } D(x_ n)$. Then $U \to X$ is an open immersion. Consider the ideals

Note that $I_ n k[x_1, x_2, x_3, \ldots ][1/x_ nx_ m] = (1)$ for any $m \not= n$. Hence the quasi-coherent ideals $\widetilde I_ n$ on $D(x_ n)$ agree on $D(x_ nx_ m)$, namely $\widetilde I_ n|_{D(x_ nx_ m)} = \mathcal{O}_{D(x_ n x_ m)}$ if $n \not= m$. Hence these ideals glue to a quasi-coherent sheaf of ideals $\mathcal{I} \subset \mathcal{O}_ U$. Let $Z \subset U$ be the closed subscheme corresponding to $\mathcal{I}$. Thus $Z \to X$ is an immersion.

We claim that we cannot factor $Z \to X$ as $Z \to \overline{Z} \to X$, where $\overline{Z} \to X$ is closed and $Z \to \overline{Z}$ is open. Namely, $\overline{Z}$ would have to be defined by an ideal $I \subset k[x_1, x_2, x_3, \ldots ]$ such that $I_ n = I k[x_1, x_2, x_3, \ldots ][1/x_ n]$. But the only element $f \in k[x_1, x_2, x_3, \ldots ]$ which ends up in all $I_ n$ is $0$! Hence $I$ does not exist.

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Comment #4574 by Andy on

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