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Is an algebraic space over a DVR, whose special fibre (and all its infinitesimal neighborhood) and generic fibre are schemes, actually a scheme?

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  • $\begingroup$ if this is true, could anyone please leave references? $\endgroup$
    – Heer
    Jul 4, 2013 at 13:25
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    $\begingroup$ Maybe it is implicit in your use of parentheses, but I'd like to just point out that the scheme property is inherited by the infinitesimal fibers without any hypotheses on the algebraic space. More generally, if $S$ is an algebraic space such that $S_{\rm{red}}$ is a scheme then $S$ is a scheme. (This is not so easy to prove when $S$ is not locally noetherian, but it is true.) $\endgroup$
    – user61789
    Jul 4, 2013 at 16:12

2 Answers 2

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The answer is negative. Let me give a counter-example by modifying the example of a singular complex algebraic surface that is not a scheme given by Knutson in [Algebraic Spaces, p.21-22].

Let me work over $\mathbb{C}$. Consider the pencil in $\mathbb{P}^2$ generated by two smooth cubic curves $C$ and $C'$ intersecting transversally in nine points $P_1,\dots, P_9$. Blowing up these nine points, we get a morphism $X\to\mathbb{P}^1$ whose fiber over $0$ is the elliptic curve $X_0=C$. Choose, as usual, an inflection point $O\in C$ as the origin of the group law on $C$.

Let $\hat{X}$ be the blow-up of $X$ in a tenth point $Q\in X_0=C$ (chosen generic, so that no multiple of $Q$ is in the subgroup of $C$ generated by $P_1,\dots, P_9$ : here, we use the uncountability of the base field). At this point, the strict transform of $X_0$ in $\hat{X}$ has negative self-intersection and may be contracted to a point $y$ in a surface $Y\to\mathbb{P}^1$. Let $T$ be the local ring of $\mathbb{P}^1$ at $0$. I claim that $Y_T\to T$ is the counter-example we are looking for.

First, the fibers of $Y_T\to T$ (and their infinitesimal neighbourhoods) are schemes because they are one-dimensional [Algebraic Spaces V 4.9]. However, if $Y_T$ were a scheme, it would be possible to find a curve $D$ in $Y$ intersecting the tenth exceptional divisor, but not containing $y$. Its strict transform in $\mathbb{P}^2$ would be a plane curve meeting $C$ only at $P_1,\dots, P_9, Q$. By the choice of $Q$, this plane curve would meet $C$ only at $P_1,\dots, P_9$. This contradicts the fact that $D$ intersects the tenth exceptional divisor.

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Olivier's example is perfect, but let me just point out that counterexamples are easier to construct if you allow nonseparated spaces.

For example, start with two DVRs $R\hookrightarrow R'$ (with spectra $X'\to X$) such that $R'$ is finite free of rank $2$ over $R$, and the fraction field extension $K\hookrightarrow K'$ is separable.

The constant group scheme $(\mathbb{Z}/2\mathbb{Z})_X$ acts naturally on $X'$. Let $G$ be the open subgroup scheme obtained by removing the nontrivial point over the closed point of $X$. Put $Y:=X'/G$. The morphism $Y\to X$ is an isomorphism on the generic points, but has the same closed fiber as $X'$.

If $R\hookrightarrow R'$ is split unramified, then $Y$ is the familiar ``$X$ with the closed point doubled'' wich is a scheme. Otherwise, the closed fiber of $Y\to X$ is a one-point scheme, which has no affine (or even separated) Zariski neighborhood in $Y$. In particular, $Y$ is not a scheme.

You can check that $Y$ is locally separated (in Artin's sense: the diagonal map is an immersion) iff $R\hookrightarrow R'$ is unramified.

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  • $\begingroup$ what is the formal completion of $Y$ along the special fibre? Is it $\mathrm{Spf} \hat{R'}$ $\endgroup$
    – Heer
    Nov 8, 2013 at 5:43
  • $\begingroup$ I think your example is a particular case of the construction in page 10 of Knutson's book. Let me use the same notation as in that book, now: if I take two closed subscheme $T_1$ and $T_2$ with the same underlying topological space of $X$, do we get the same algebraic space $X_1'=X_2'$ ? $\endgroup$
    – Heer
    Nov 8, 2013 at 5:49
  • $\begingroup$ I feel very confused about the relation between an algebraic space $X$ over a DVR and its formal completion $\hat{X}$ along the special fibre. Here I suppose that $X$ is not a scheme, and $\hat{X}$ is actually a formal scheme. I would like to have some intuitive understanding. Thank you very much. $\endgroup$
    – Heer
    Nov 8, 2013 at 5:51
  • $\begingroup$ The most confusing thing is the following: a non-scheme algebraic space over a dvr could have a formal completion along special fibre a formal scheme (if this is ture). @Laurent Moret-Bailly $\endgroup$
    – Heer
    Nov 13, 2013 at 15:18
  • $\begingroup$ @Heer: the projection $X'\to Y$ is etale, with trivial residue field extension(s) over the closed point of $X$. So yes, the formal completion is $\mathrm{Spf}(\widehat{R'})$. $\endgroup$ Nov 28, 2013 at 11:11

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