## Flatly compactifiable morphisms

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Let $f:U \to S$ be a flat morphism. Let us say that $f$ is flatly compactifiable if there exists a proper morphism $\bar{f}:X \to S$ and a closed subscheme $Z \subset X$ such that

1) $U = X \setminus Z$ and $f = \bar{f}_{|U}$;

2) $\bar{f}$ is proper;

3) BOTH $X$ and $Z$ are flat over $S$.

My question is whether this notion already appeared in the literature and what is the correct name for it?

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I'm also interested in the same question with 3) replaced by: X is smooth over S and (Z,X) is strict relative normal crossings over S. – Dustin Clausen Oct 11 2010 at 17:09

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I suppose $f$ is finitely presented; heck, let's assume $S$ is noetherian. Now if $Z$ is to be flat then since it is also proper we see that its image in $S$ is both open and closed. So if we also assume $S$ is connected then $Z$ meets all fibers. That makes things sound problematic if $U$ is proper over a dense open in $S$ but not proper over $S$. Perhaps you have a more specific situation in mind? If so, please say more about that. – BCnrd Oct 11 2010 at 17:12
Brian, I think your comment is relevant to the problem "When is such a property likely to hold?" but this is not Sasha's question. – Laurent Moret-Bailly Oct 11 2010 at 18:09
Dear BCnrd, what do you mean by "things sound problematic"? Certainly, there are plenty of morphisms which do not admit such a compactification. For example $x:(A^2 \setminus (0,0)) \to A^1$ doesn't. My question was, what is the correct name for this notion and whether one can find something about it in the literature. – Sasha Oct 11 2010 at 18:19
Dear Sasha: Whoops, I misunderstood the question as to be asking for further hypotheses on $f$ which would suffice for the existence of such an $\overline{f}$, and so I was just mentioning a trivial example where the answer is negative and asking if you had specific classes of examples in mind for which you'd want an affirmative answer (so as to better guide appropriate kinds of hypotheses to seek). But this is not relevant to the question your were actually asking; sorry about that. – BCnrd Oct 11 2010 at 19:23

## 1 Answer

In general you can not expect that $f$ will be flat on $X-U$, but you have a locally finite stratification of $X$ for which $U$ is the dense stratum, and such that $f$ is flat over each strata. This notion is called "platification" (in French) and it was dealed with great details in the paper of Gruson and Raynaud "Techniques de platification d'un module".

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