# Does there exist a relative compactification with flat boundary?

Let $S$ be the spectrum of a DVR, $X/S$ a flat finite type affine scheme. Fix a closed immersion $X\subseteq \mathbb{A}^n_S$ and consider the closure $\bar X$ of $X$ in $\mathbb{P}^n_S$. Then $\bar{X}$ is flat over $S$, as $X$ is dense in it. However, the boundary $Y = \bar X - X$ does not have to be flat over $S$ (see the examples below).

Does every point of the closed fiber of $X$ have an open neighborhood $X'$ which can be compactified to a projective $\bar X'/S$ with $Y' := \bar X' - X'$ flat over $S$?

Example. Take $X$ to be the generic point of $S$, $\bar X = S$ and $Y$ is the closed point of $S$. But as the closed fiber $X$ is empty, the condition is trivially satisfied.

Example. Let $X \subseteq{A}^2_S$ be given by the equation $x = \pi y^2$ where $\pi$ is a uniformizer, then $\bar X \subseteq \mathbb{P}^n$ is given by $xz = \pi y^2$ and $Y$ is given by $z = 0 = \pi y^2$, a union of the $z=0$ line in the closed fiber and the section $(1:0:0)$, so not flat. Of course, $X$ is isomorphic to $\mathbb{A}^1_S$, so we can find a compactification for which $Y$ is flat, namely $\bar X = \mathbb{P}^1_S$.

• There is an example in the book "Néron models", §6.7, Lemma 6 (remove the non-contractibe irreducible component in the closed fiber and let $X$ be an affine open subscheme with non empty closed fiber). The DVR is not henselian. On the other hand, if S is henselian, and $X$ is normal of relative dimension $1$, then the answer to your question is yes by contracting superfluous irreducible components (i.e. those of positive dimension) in the boundary, see op. cit. – Cantlog Mar 2 '14 at 17:26
• Thank you for the useful reference! Do you believe this should be true for $S$ henselian and $X$ of higher dimension? – Piotr Achinger Mar 4 '14 at 2:18
• It would suffice to contract vertical divisors (in the boundary), but I think this is impossible in general in higher dimension. – Cantlog Mar 4 '14 at 20:45

I believe the answer is no. All of the following will be relative to an algebraically closed field $k$. You may assume $k$ has characteristic $0$, but I think that is unnecessary (use the work of Stefan Schröer).

Let $g',g''>2$ be positive, distinct integers. Let $g$ be $g'+g''$. Let $\overline{M}_g$ denote the Deligne-Mumford moduli stack of stable curves of (arithmetic) genus $g$. Let $\overline{M}_g^o\subset \overline{M}_g$ be the maximal open substack that is a scheme. In particular $\overline{M}_g^o$ intersects the boundary divisor $\Delta_{g',g''}$ since $g'\neq g''$; denote the intersection by $\Delta_{g',g''}^o$. Denote the universal curve over $\overline{M}_g^o$ by $$\pi:\mathcal{C}\to \overline{M}_g^o.$$ Over $\Delta_{g',g''}^o$, the restriction of the universal curve is a union of two proper closed subschemes, $\mathcal{C}'$ and $\mathcal{C}''$, intersecting along divisors, $Z'\subset \mathcal{C}'$, respectively $Z''\subset \mathcal{C}''$, such that $$(\pi':\mathcal{C}'\to \Delta_{g',g''}^o,Z'), \ \text{resp.} \ (\pi'':\mathcal{C}''\to \Delta_{g',g''},Z''),$$ is a family of stable, $1$-pointed curves of (arithmetic) genus $g'$, resp. $g''$.

Let $\delta_{g',g''}$ denote the generic point of $\Delta_{g',g''}^o$, let $R$ be $\mathcal{O}_{\overline{M}_g^o,\delta_{g',g''}}$, the local ring of $\overline{M}_g^o$ at this codimension $1$ point. Of course $R$ is a DVR. Let $S$ be $\text{Spec}(R)$ with its natural morphism to $\overline{M}_g^o$. Denote the closed point of $S$ by $0$, and denote the generic point by $\eta$. Let $$\pi_S:\mathcal{C}_S\to S$$ be the base change of $\pi$ over $S$. Denote the closed fiber by $C_0$, with its two irreducible components $C'_0$ and $C''_0$ as above. Let $X$ be $\mathcal{C}_S \setminus C''_0$.

By way of contradiction, assume that there exists a Zariski open subset $X'\subset X$ that intersects $C'_0$ and that has a projective compactification $\overline{X}'_S$ over $S$ whose boundary $Y'$ is flat over $S$. In particular, $Y'$ is finite over $S$. Denote by $$u:\widetilde{X}\to \overline{X}'$$ the normalization of $\widetilde{X}$. In particular, the generic fiber $\widetilde{X}_\eta$ is isomorphic to the generic fiber $\widetilde{C}_\eta$, i.e., $\widetilde{X}_\eta \cong \mathcal{C}_\eta$. Moreover, because the closed fiber $\widetilde{X}_0$ is some compactification of an open subset of the genus $g'>0$ curve $C'_0$, in particular $\widetilde{X}_0$ contains no genus $0$ curves. Moreover, $\widetilde{C}_S$ is regular. Thus, by Abhyankar's lemma, etc., the isomorphism of generic fibers extends to an $S$-morphism, $$v :\mathcal{C}_S \to \widetilde{X},$$ that is automatically projective. Moreover, this morphism contracts the irreducible component $C''_0$ of the closed fiber of $\mathcal{C}_S$. In particular, for every invertible sheaf $\mathcal{L}$ on $\widetilde{X}$, $v^*\mathcal{L}$ restricts to the trivial invertible sheaf on $C''_0$. In particular, since $\widetilde{X}$ is projective, there exists an invertible sheaf $\mathcal{L}$ that is ample, so that $v^*\mathcal{L}$ is big.

This is impossible. By Franchetta's conjecture / Harer's theorem, the Picard group of $\mathcal{C}_\eta$ is generated by the relative dualizing sheaf, and the Picard group of $\mathcal{C}_S$ is generated by the relative dualizing sheaf together with the invertible sheaf $\mathcal{O}_{\mathcal{C}_S}(\underline{C'}_0)$. The restrictions of these invertible sheaves to $C''_0$ are $\omega_{C''_0}(\underline{Z}'')$ and $\mathcal{O}_{C''_0}(\underline{Z}'')$. But these are linearly independent in $\text{Pic}(C''_0)$, i.e., the universal curve over the generic point of $\Delta_{g',g''}$. Since $v^*\mathcal{L}$ has trivial restriction to $C''_0$, $v^*\mathcal{L}$ is trivial. In particular, the restriction of $v^*\mathcal{L}$ to the generic fiber $\widetilde{C}_\eta$ is trivial. However, by construction, it is also big. This is a contradiction, proving that $\overline{X}'$ does not exist.

Observe the following points. First, $S$ is not Henselian. Second, the residue field of $S$ is not algebraically closed. Third, this argument only works if $\overline{X}'$ is a projective scheme; the argument fails if $\overline{X}'$ is only a proper algebraic space. So, if you really need something like this to hold, there are still directions to explore.

• Wow, thank you! The last part was quite subtle. – Piotr Achinger Mar 2 '14 at 5:14
• Yes, I'm still interested in the case $S$ (strictly) Henselian. – Piotr Achinger Mar 2 '14 at 5:31