110.26 A locally closed subscheme which is not open in closed
This is a copy of Morphisms, 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.
The morphism Z \to X also gives an example of bad behaviour of scheme theoretic images of immersions. Namely, the arguments above show that the scheme theoretic image of the immersion Z \to X is X. On the other hand, we see
Z is not topologically dense in X, and
the scheme theoretic image of Z = Z \cap U \to U is just Z. This is not equal to U \cap X = U and hence formation of the scheme theoretic image in this case does not commute with restrictions to opens.
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