MathOverflow Asked by user158636 on November 3, 2021
Let $S$ be a geometrically connected smooth projective surface over $mathbb{Q}_p$. Can it be put in a proper flat $mathbb{Z}_p$-scheme with a geometrically integral special fiber?
Let $X$ be an integral regular scheme which is proper over $mathbb{Z}_p$. Assume that the special fibre $X_{mathbb{F}_p}$ is irreducible and let $k$ be the algebraic closure of $mathbb{F}_p$ in the function field of $X_{mathbb{F}_p}$. Then $k$ is a birational invariant of the generic fibre $X_{mathbb{Q}}$. (The field $k$ is something like the field of definition of the geometric irreducible components of $X_{mathbb{F}_p}$, with $k = mathbb{F}_p$ if and only if $X_{mathbb{F}_p}$ is geometrically irreducible).
This is a special case of a more general result given in [1, Cor. 2.3] (which also makes precise what I mean by "birational invariant of the generic fibre", also allows the special fibre to be reducible, and applies to general dvrs).
So consider the $mathbb{Z}_p$-scheme
$$ x^2 - ay^2 = p z^2$$
where $a in mathbb{Z}_p^times$ is a non-square (model of a plane conic). This is easily check to be integral regular and proper. But the special fibre is
$$x^2 - ay^2 equiv 0 bmod p$$ which has $k = mathbb{F}_p(sqrt{a})$ in the above notation. Therefore, by the above result, there is no regular integral proper model of the generic fibre with geometrically integral special fibre.
For a counter-example involving surfaces, just take the fibre product of the above scheme with $mathbb{P}^1_{mathbb{Z}_p}$.
Your question allowed arbitrary (not necessarily regular) models. If you are given a non-regular model $Y$ for the above conic, then you can perform resolution of singularities to obtain a regular model $widetilde{Y}$ (since $dim Y leq 3$). If $Y$ had a geometrically integral special fibre, then the special fibre of $widetilde{Y}$ would have an irreducible component which is geometrically integral, which is not allowed by the more general [1, Cor. 2.3].
[1] Skorobogatov - Descent on toric fibrations
Answered by Daniel Loughran on November 3, 2021
That is not true. There are probably shorter answers than the following. Let $K$ be a field, and denote a separable closure by $K^{text{sep}}$. Let $n>1$ be an integer.
Definition. A Severi-Brauer variety over $K$ of relative dimension $n-1$ is a proper, smooth $K$-scheme whose base change to $K^{text{sep}}$ is isomorphic to projective space of dimension $n-1$ over $K^{text{sep}}$.
There is a natural bijection between the set of $K$-isomorphism classes of Severi-Brauer varieties over $K$ of relative dimension $n-1$ and the subset of $text{Br}(K)[n]$ of those ed. $n$-torsion elements in the Brauer group of $K$ whose order index divides $n$. In particular, the identity element in this subset corresponds to the isomorphism class of projective space, i.e., the isomorphism class of any Severi-Brauer variety over $K$ of relative dimension $n-1$ that has a $K$-rational point.
For a finite, separable field extension $L/K$ of degree $d$, there are restriction and corestriction group homomorphisms, $$ text{Br}(K)to text{Br}(L), text{Br}(L)to text{Br}(K), $$ whose composition is the "multiplication by $d$" map.
Finally, for every finite extension of $mathbb{Q}_p$, local class field theory gives a natural isomorphism, $$ text{inv}_K:text{Br}(K) to mathbb{Q}/mathbb{Z}, $$ ed. and the index of every element equals the order of that element.
Now let $K$ be a finite extension of $mathbb{Q}_p$, and let $X_K$ be a Severi-Brauer variety over $K$ of relative dimension $n-1$ whose image in $text{Br}(K)[n]$ is a generator for this cyclic group of order $n$. By the restriction-corestriction homomorphisms, for a finite field extension $L/K$, the base change of $X_K$ over $L$ has an $L$-rational point only if the degree $d$ of the field extension is divisible by $n$.
On the other hand, if $X_K$ has a proper, flat model over the ring of integers of $K$ whose special fiber has nonempty smooth locus an irreducible component that is geometrically integral, then by the Lang-Weil estimates together with Hensel's Lemma, for every sufficiently large integer $d$, there is an unramifield field extension $L/K$ of degree $d$ such that the base change has an $L$-rational point. Therefore, there is no proper, flat model of $X_K$ over the ring of integers of $K$.
In particular, for the integer $n=3$, there exists a Severi-Brauer scheme over $mathbb{Q}_p$ of relative dimension $n-1=2$ that has no proper, flat model over $mathbb{Z}_p$ (ed. . . . with an irreducible component that is geometrically integral!).
Answered by Jason Starr on November 3, 2021
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