# A closed subscheme of an open subscheme that is not an open subscheme of a closed subscheme?

A morphism $f: V \rightarrow X$ of schemes is a locally closed immersion if it can be factored into a closed immersion followed by an open immersion. It is not hard to show that if $f$ is an open immersion followed by a closed immersion, then it is a locally closed immersion, but the converse is at the very least not clear (to me). For a number of reasons, this choice as the definition of locally closed immersion (rather than the opposite) is the right one (e.g. it is then not hard to see that compositions of locally closed immersions are locally closed immersions).

Is there some $f: V \rightarrow X$ that can be factored into a closed immersion followed by an open immersion, that cannot be factored into an open immersion followed by a closed immersion?

Warning: it isn't too hard to show that there is no example with $V$ reduced or with $f$ quasicompact, so any example has to be a little strange-looking. Back-story: I've been confronted with this question when learning algebraic geometry with a class (conventionally known as "teaching"); it seems a natural question. And any counterexample would likely be a handy example to have for other reasons as well: a very limited stock of counterexamples tends to refine my intuition, and to warn me what can go wrong.

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I second FCs' suggestions: why have you not just asked Brian? – Emerton Jan 28 '10 at 21:29

Hi Ravi,

There is an example in Tag 01QW in Johan's stacks project.

Jarod

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Added link. Nice example! – David Speyer Jan 28 '10 at 22:53
This is just beautiful. – Ravi Vakil Feb 1 '10 at 20:46

Does this work?

Let $k$ be a field, and let $R$ be the subring of $\prod_{i=1}^{\infty} k[t]$ consisting of those sequences that stabilize. Let $I$ be the ideal comprised of those sequences that stabilize at $0$. Set $X=\mathrm{Spec}(R)$, and let $U$ be the complement of $V(I)$ in $X$. (The inclusion $U\to X$ is quasiseparated, but not quasicompact; here $j_*\mathcal{O}_U$ is not quasicoherent.)

If $e_i$ denotes the sequence that is $0$ except in the $i$th spot, where it is $1$, then the $e_i$'s generate $I$, and since $e_ie_j=\delta_{ij}$, we deduce that $U=\coprod_{i=1}^{\infty}X_{e_i}=\coprod_{i=1}^{\infty}\mathbf{A}_k^1$.

Now consider the ideal $(t^i)\subset k[t]$ as an ideal sheaf on each $X_{e_i}$; these glue to give a quasicoherent ideal sheaf $I$ on $U$, which in turn defines a closed subscheme $Z$ of $U$, which is open in $X$.

I claim that there is no closed subscheme $V$ of $X$ such that $Z$ is open in $V$. If there were, it would be $V=V(J)$ for an ideal $J\subset R$ such that $J\cdot R_{e_i}=(t^i)$. But this is impossible, since any element $f=(f_i)\in J$ would have to have $f_i\in (t^i)$ but would nevertheless have to stabilize. So it would have to stabilize at zero; hence $J\subset I$, so $V(I)\subset V(J)$, whence we ... EDIT: do not have a contradiction.

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It looks as though I took too long to finish my answer. The one from de Jong's book seems to be generally of the some kind, but possibly more understandable. – Clark Barwick Jan 28 '10 at 23:55
Sorry: the argument of my last paragraph demonstrates nothing. I think my example simply does not work. Maybe there's a different ideal sheaf on $U$ that does work? I don't see it at the moment. – Clark Barwick Jan 29 '10 at 3:19
Could you delete this answer? – Colin McLarty Mar 8 at 17:37