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?



Smooth projective surface with geometrically integral reduction
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!).


