Let $K$ be a non-perfect field of characteristic $2$. Let $T \subseteq K$ be a discrete valuation ring. Assume there exist $a,b \in K^{\times}$ such that the projective conic $C$ defined by $$aX^2 + b Y^2 + Z^2=0$$ contains no $K$-rational point (in particular $a,b$ and $ab$ are not squares in $K$). Then $C$ is non-smooth over $K$ at every point. However, $C$ is regular (see Exercise 4.3.22 (d) of Qing Liu's book Algebraic Geometry and Arithmetic Curves).

Question: Does $C$ have a regular projective model over $T$ (i.e. a regular fibered projective surface over $T$ with generic fiber isomorphic to $C$) ?

Every smooth curve over $K$ does admit a regular projective model over $T$, as is shown (for example) in Corollary 8.3.51 of Liu's aforementioned book. This seems to be based on Lipman's resolution of singularities of $2$-dimensional excellent Noetherian schemes, applied to a normal model of the curve over $T$, or rather some base change to the completion of $T$ (excuse the sloppy description, this may not be a very accurate description of what is really going on). In any case, it seems to be crucially used that the generic fiber of the normal model is not only regular, but smooth over $K$.

Nevertheless, this is a general statement, and it is a priori possible that in concrete cases of regular projective curves (such as all integral regular conics over $K$, which I am interested in) there do exist regular projective models over $T$, regardless of whether the regular curve is smooth over $K$ or not.


1 Answer 1


In general this isn't possible. The material that follows is a bit technical, but I do not have the time to explain all the details here now.

Suppose that $a, b \in T$ and that $a, b$ become squares in the completion of $T$ (see example below). Then the projective scheme $S$ defined by $a X^2 + b Y^2 + Z^2 = 0$ in $\mathbf{P}^2_T$ is $2$-dimensional with generic fibre $C$ and has the following property: for every point $s \in S$ lying over the closed point of $\text{Spec}(T)$ the local ring $\mathcal{O}_{S, s}$ is a domain with non-reduced completion (this is where we use that $a, b$ become squares in the completion of $T$).

If $C$ had a regular projective model $S'$ over $T$, then we could eleminate indeterminacies in the rational map from $S'$ to $S$ by a finite sequence of blowing ups in closed points of $S'$. Note that the blowing up of $S'$ in a closed point produces another regular surface. Thus we may assume there is a morphism $S' \to S$ over $T$ which is proper and birational. In particular $S' \to S$ is finite over an open $U \subset S$ which contains all codimension $1$ points. Let $s \in U$ be the generic point of an irreducible component of the closed fibre of $S$. The scheme $S' \times_S \text{Spec}(\mathcal{O}_{S, s})$ is affine, say equal to the spectrum of $A$. Then the inclusion $\mathcal{O}_{S, s} \subset A$ is finite and $A$ is regular. It follows that the $\mathfrak m_s$-adic completion $A^\wedge$ of $A$ is a finite product of dvrs in particular reduced. Since completion is exact on finite modules, we get $\mathcal{O}_{S, s}^\wedge \subset A^\wedge$ which is a contradiction with the non-reducedness seen above.

A reference for some of the things discussed above is the chapter on resolution of surfaces in the Stacks project, especially the section on dominating by quadratic transformations and the section on implied properties.

Remark. It seems possible to me that for "most" choices of $a$ and $b$ the curve $C$ does have a regular model, but I didn't try to prove or disprove it.

Example. Let $k = \mathbf{F}_2(s_1, s_2, s_3, \ldots, t_1, t_2, t_3, \ldots)$ be the purely transcendental extension of $\mathbf{F}_2$ on elements $s_1, s_2, s_3, \ldots, , t_1, t_2, t_3, \ldots$. Let $T \subset k[[x]]$ be the set of power series $\sum a_i x^i$ such that $[k^2(a_0, a_1, a_2, \ldots) : k^2] < \infty$. Then $T$ is the standard example of a non-Nagata dvr in char $2$. Let $a = 1 + \sum_{i \geq 1} s_i^2 x^{2i}$ and $b = 1 + \sum_{i \geq 1} t_i^2 x^{2i}$. Note that $a, b, ab$ are not squares in $T$ but become squares in the completion of $T$. We omit the verification that $C(K)$ is empty.

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    $\begingroup$ Just to give a reference: if one wants to verify that the example described by @darx does not admit a regular model, one can apply [Liu, Algebraic Geometry and Arithmetic Curves, Chapter 8, Theorem 3.50 (i) equivalent to (iii)]. $\endgroup$ Aug 20, 2019 at 9:37
  • $\begingroup$ Thank you both @darx for your detailed answer with example and Olivier Benoist for relating the example to the reference in Liu's book (I had not made the connection) ! This helped a lot ! Indeed, the conic defined by a and b as in the example wont be regular after base change to the completion of $K$ with respect to $T$ ! $\endgroup$
    – DGrimm
    Aug 20, 2019 at 16:36

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