The Stacks project

Example 28.3.4. 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

\[ I_ n = (x_1^ n, x_2^ n, \ldots , x_{n - 1}^ n, x_ n - 1, x_{n + 1}, x_{n + 2}, \ldots ) \subset k[x_1, x_2, x_3, \ldots ][1/x_ n]. \]

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.


Comments (3)

Comment #4574 by Andy on

Does anything change if ? In other words, does the matter?

Comment #4575 by Andy on

I mean, if , but is removed? or

Comment #4576 by on

Yes both of those work too. What is especially weird about the closed subscheme in the example the way we have it now, is that is some "positive distance" away from the origin because it is a zero dimensional scheme with only one point, namely (and the thickness around this point is finite too). So this way the example is a little bit like the standard way of explaining why the unit sphere in infinite dimensions isn't a compact space.


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