Let $X$ be a connected normal scheme equipped with a proper flat morphism $f\colon X \rightarrow \mathrm{Spec }(R)$ with $R$ a discrete valuation ring and such that the fibers of $f$ are curves (i.e., purely of dimension $1$). Is $f$ necessarily projective?

The answer is 'yes' if the normality assumption is strengthened to regularity of $X$: this was proved in Lichtenbaum's thesis.

  • $\begingroup$ So you want to try to find a relatively ample divisor. There are several ways one might go about this: one is to try to lift an ample Cartier divisor on the special fiber; another is to take the closure of a point on the general fiber and twist by components of the special fiber to make this ample. Have you tried either of these? $\endgroup$ Commented Apr 6, 2015 at 2:50
  • $\begingroup$ @DanielLitt: No, I haven't tried any of these, but I don't believe that the question is the matter of simple tricks because Lichtenbaum's theorem is a serious result (and would be a special case of a positive answer to this question). Also, your second suggestion does not quite make sense to me because individual components of the special fiber need not be Cartier divisors (since $X$ is merely normal). $\endgroup$
    – Lisa S.
    Commented Apr 6, 2015 at 3:42
  • $\begingroup$ My second suggestion seems likely to work if in addition $X$ is factorial or Q-factorial (e.g. if X has rational singularities). You might also try to reduce to Lichtenbaum's case by resolving singularities via blowup (possible bc you're in the surface case), then contracting the exceptional divisor by imitating the proof of Castelnuovo's contraction theorem. $\endgroup$ Commented Apr 6, 2015 at 7:21
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    $\begingroup$ I think the answer is also yes if $R$ is henselian, if this helps. $\endgroup$ Commented May 6, 2015 at 19:25
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    $\begingroup$ @Lisa: I suppose you may do that. In a more down-to-earth way, you can pick a closed point $x$ of the closed fiber $X_0$ which is in the regular locus $U$, and then a (Weil) divisor $D$ through $x$, finite over $R$. If $R$ is henselian, $D$ splits as $D'\coprod D"$ where $x$ is the only point of $D'$ in the closed fiber, so $D'\subset U$. Repeat this with one $x$ in each component of $X_0$ and take the sum of the $D'$'s: this is an ample Cartier divisor. $\endgroup$ Commented May 7, 2015 at 5:45

2 Answers 2


I believe that you can form a counterexample in the same manner "as usual". Begin with a projective morphism $\pi:\mathcal{C}_R\to \text{Spec}(R)$ whose closed fiber $\mathcal{C}_k$ over the residue field $k$ is a smooth, geometrically connected curve of genus $g>0$. For simplicity, assume that $\text{Pic}(\mathcal{C}_R)$ is $\mathbb{Z}$ and assume that $\mathcal{C}_k$ has a $k$-point that is linearly independent from the image of $\text{Pic}(\mathcal{C}_R)$ in $\text{Pic}(\mathcal{C}_k)$.

For instance, let $d\geq 3$ be an integer, let $\mathbb{P}^N = \mathbb{P}H^0(\mathbb{P}^2,\mathcal{O}(d))$ be the projective space of all degree $d$ plane curves, let $D_q\subset \mathbb{P}^N$ be the hyperplane parameterizing degree $d$ plane curves that contain a specified point $q\in \mathbb{P}^2$, let $\mathcal{C} \to \mathbb{P}^N$ be the universal curve, and let $R$ be the stalk $\mathcal{O}_{\mathbb{P}^N,\eta_H}$ at the generic point $\eta_H$ of $H_q$. Let $\mathcal{C}_R$ be the base chance of $\mathcal{C}$ to $\text{Spec}(R)$. For the fraction field $K$ of $R$, $\mathcal{C}_K$ has Picard group generated by the restriction of $\mathcal{O}_{\mathbb{P}^2}(1)$, by a straightforward incidence correspondence argument. Yet, by construction, $\mathcal{C}_k$ contains the $k$-point $q$. Moreover, the divisor class of $\underline{q}$ is linearly independent from the restriction of $c_1(\mathcal{O}_{\mathbb{P}^2}(1))$ in $\text{Pic}(\mathcal{C}_k)$.

Anyway, now let $$ \nu:\mathcal{C}'_R \to \mathcal{C}_R $$ be the blowing up at the closed point $q$ (with its reduced structure).
The strict transform $\widetilde{\mathcal{C}}_k$ has normal sheaf $\mathcal{O}_{\mathcal{C}_k}(-\underline{q})$. By Grauert / Artin, there exists a contraction $$ \mu:\mathcal{C}'_R \to \mathcal{C}''_R $$ of $\widetilde{\mathcal{C}}_k$ in $\mathcal{C}'_R$. I claim that the projection morphism to $\text{Spec}(R)$ factors through $\mu$, $$ \pi'':\mathcal{C}''_R \to \text{Spec}(R). $$ If $\mathcal{C}''_R$ were projective, the pullback to $\mathcal{C}'_R$ of an ample invertible sheaf would restrict on $\mathcal{C}_K$ to an integer multiple of the generator of $\text{Pic}(\mathcal{C}_K)$. Yet on $\widetilde{\mathcal{C}}_k$ it restricts to the structure sheaf. Since the exceptional divisor on $\mathcal{C}'_R$ restricts on $\widetilde{\mathcal{C}}_k$ to $\mathcal{O}_{\mathcal{C}_k}(-\underline{q})$, ultimately this implies a linear relation between $\text{Pic}(\mathcal{C}_R)$ and the divisor class of $\underline{q}$. This contradiction implies that $\mathcal{C}''_R$ is not projective.

Edit. As Laurent Moret-Bailly points out, the surface $\mathcal{C}''$ produced above is usually an algebraic space, not a scheme.

  • $\begingroup$ If I remember correctly, Artin's contraction results give rise to algebraic spaces, not schemes (except, for instance, over a finite field). So, what kind of ``curve'' are we looking for? $\endgroup$ Commented Apr 6, 2015 at 15:14
  • $\begingroup$ @LaurentMoret-Bailly: Of course you are correct. This argument produces an algebraic space, not necessarily a proper scheme. $\endgroup$ Commented Apr 6, 2015 at 23:07

This is perhaps not the answer the poster is looking for, but Proposition 3.3 of Families of rationally simply connected varieties over surfaces and torsors for semisimple groups, by de Jong--He--Starr, proves something related. Indeed, an immediate consequence of this result is that if $\pi \colon C \to S$ is a proper, flat, and finitely presented morphism of algebraic spaces of relative dimension 1, then there exists an etale cover $S' \to S$ such that the pullback $\pi' \colon C\times_S S' \to S'$ is projective.

In particular, in the notation of the posed question: if $R$ is strictly henselian (e.g., complete with algebraically closed residue field), then $f$ is projective.

Note that it is not even necessary to assume that $X$ is normal for this this result to hold.

  • $\begingroup$ That result is indeed quite related to the OP's question. Unfortunately, de Jong, He and I cannot claim authorship of that result :) It is an old result that, 'etale locally on the base, every flat family of proper curves is (relatively) projective. $\endgroup$ Commented Apr 10, 2015 at 2:14

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