# Does a flat compactification always exist?

Let $$\pi:X\to S$$ be a separated flat morphism of finite type of Noetherian schemes. Does $$\pi$$ necessarily factor as an open immersion followed by a proper flat morphism? The analogue of this question with the word "flat" replaced by "smooth" has a negative answer (consider an elliptic curve over $$\mathbb{Q}_p$$ that has bad reduction).

This already fails if $$S$$ is regular of dimension $$3$$ and $$\pi$$ is quasi-finite. Indeed, let $$X$$ be a normal affine variety over $$\mathbf C$$ of dimension $$3$$ with an isolated non-Cohen–Macaulay singularity (e.g. an affine cone over a smooth projective surface $$Y$$ with $$H^1(Y,\mathcal O_Y) \neq 0$$). By Noether normalisation, there exists a finite surjection $$\pi \colon X \to S = \mathbf A^3,$$ without loss of generality taking the isolated singularity $$x_0 \in X$$ to the origin $$0 \in \mathbf A^3$$. Let $$U = X \setminus x_0$$, which is smooth by assumption, so $$\pi|_U$$ is flat by miracle flatness. Now I claim that $$\pi|_U \colon U \to \mathbf A^3$$ does not have a flat compactification.
Indeed, suppose $$U \hookrightarrow X' \stackrel{\pi'}\to S$$ is a factorisation into an open immersion and a proper flat morphism. Because $$\pi'$$ is flat and generically finite, it is quasi-finite, hence finite since it is proper. Since $$\pi'$$ is finite flat and $$S$$ is regular, we conclude that $$X'$$ is Cohen–Macaulay.
Let $$\bar U \subseteq X'$$ be the scheme-theoretic closure of $$U$$, and let $$V = S \setminus 0$$. Since $$\pi^{-1}(V) \subseteq U$$, we conclude that $$\bar U \setminus U$$ is supported on $$\pi'^{-1}(0)$$, in particular has dimension $$0$$. By Hartshorne's connectedness theorem, this implies that there are no other components in $$X'$$ (otherwise two components would meet only in a $$0$$-dimensional set), i.e. $$\bar U = X'$$ set-theoretically. Since $$X'$$ is generically reduced and Cohen–Macaulay, it is reduced, so $$\bar U = X'$$ scheme-theoretically.
In particular, $$X' \setminus U$$ is $$0$$-dimensional, so $$X'$$ is regular in codimension $$1$$ since the same holds for $$U$$. Since $$X'$$ is Cohen–Macaulay, this forces $$X'$$ normal, so it equals the normalisation of $$S$$ in $$K(U)$$, which is $$X$$. But $$X$$ is not Cohen–Macaulay, contradicting flatness of $$\pi'$$. $$\square$$
• is it true is $S$ is regular of Krull dimension 2? – user158636 Jun 12 '20 at 8:03
• Hmm, there is a positive result in the quasi-finite case if $S$ is regular of dimension $2$ and $X$ is normal. Because then you can take the normalisation, which is Cohen–Macaulay (this uses dimension $2$), hence flat. I'm not sure what happens if $X$ is not normal, or if $\pi$ is not quasi-finite. A thing to try is the projectivisation of a vector bundle (e.g. of rank $2$) that does not extend, but it's hard to get a grip on the possible compactifications (which is why I restricted to the quasi-finite case). – R. van Dobben de Bruyn Jun 12 '20 at 19:35
• Of course if $S$ is regular of dimension $1$ the result is positive (for arbitrary $\pi$): in this case flat just means torsion free, which can be arranged by taking the scheme-theoretic closure of $X$ in an arbitrary compactification. – R. van Dobben de Bruyn Jun 12 '20 at 19:47