# Geometric fibers $\mathbb P^n$ + Vanishing of Brauer group implies projective bundle

Let $\pi: X\to Y$ be a projective flat morphism over a Noetherian integral scheme all of whose geometric fibers are isomorphic to $\mathbb P^n$ over the geometric point.

If there is a line bundle $\mathscr L$ that restricts to $\mathscr O(1)$ on the geometric fibers, then one can show that $\pi$ is a projective bundle (Exercise 28.1.L in Vakil's notes).

He then has a remark that you don't need the existence of such a line bundle if you assume that $Y$ is a smooth curve over an algebraically closed field by Tsen' Theorem. I assume the real meat of this remark is saying that when the generic fiber of $Y$ has trivial Brauer group, then $\pi$ is a projective bundle.

This makes some sense since the Brauer group classifies $\mathbb P^n$ torsors in some sense. Nevertheless, I don't quite see how to prove the remark. To be completely clear, let me state the theorem I am looking for:

Question:

Let $\pi: X\to Y$ be a projective flat morphism over a Noetherian integral scheme all of whose geometric fibers are isomorphic to $\mathbb P^n$ over the geometric point.

Assume furthermore that the generic point of $Y$ has trivial Brauer group. Then, is $\pi$ a projective bundle - ie, is there a vector bundle $\mathcal E$ over $Y$ such that $X \cong \operatorname{Proj}(\operatorname{Sym}(\mathcal E))$?

Perhaps this is not true - in that case, I would still be interested in a proof when $Y$ is a smooth curve over an algebraically closed field.

Definition. The (cohomological) Brauer group is $$\operatorname{Br}(X) := H^2_\mathrm{ét}(X,\mathbf{G}_m).$$

For simplicity, we work with varieties over an algebraically closed field $k$.

Definition. A proper, flat morphism $\pi\colon X \to Y$ between varieties over $k$ is a $\mathbf{P}^n$-fibration if for every closed point $y \in Y$, the fiber $X_{y}$ is isomorphic to $\mathbf{P}^n_{k}$.

We will show the following:

Theorem. If $\pi\colon X \to Y$ is a $\mathbf{P}^n$-fibration and $\operatorname{Br}(Y) = 0$, then there exists a rank $n+1$ locally free sheaf $\mathscr{E}$ on $Y$ such that $X \simeq \mathbf{P}(\mathscr{E})$ over $Y$.

I don't know how exactly the condition $k = \overline{k}$ could be weakened to a version of this result where only geometric fibers are known to be isomorphic to $\mathbf{P}^n$.

Note that if $Y$ is regular, then since $\operatorname{Br}(Y) \hookrightarrow \operatorname{Br}\bigl(K(Y)\bigr)$, where $K(Y)$ denotes the function field of $Y$ [Grothendieck, Cor. 1.10], the triviality of the Brauer group of the generic point is indeed sufficient for the conclusion to hold.

We first need some preliminary results.

Lemma. Let $(R,\mathfrak{m},k)$ be a complete local ring, let $X$ be proper and flat over $R$, and consider the cartesian diagram $$\require{AMScd} \begin{CD} X @<<< X_0\\ @VVV @VVV\\ \operatorname{Spec} R @<<< \operatorname{Spec} k \end{CD}$$ If $X_0 \simeq \mathbf{P}^n_k$, then $X \simeq \mathbf{P}^n_R$.

Proof of Lemma. This follows from general deformation theory [Sernesi, Cor. 1.2.15] since $H^1(\mathbf{P}^n,T_{\mathbf{P}^n}) = 0$. $\blacksquare$

I also have a direct proof of the Lemma in mind, if you would like to see it.

Proposition. If $\pi\colon X \to Y$ is a $\mathbf{P}^n$-fibration, then for every closed point $y \in Y$, there is an étale morphism $\varphi\colon V \to Y$ such that $y \in \operatorname{im}(\varphi)$, and an isomorphism $X \times_Y V \simeq V \times \mathbf{P}^n$ over $V$.

Proof of Proposition. Consider the commutative diagram of cartesian squares $$\begin{CD} X_y @>>> X_y^h @>>> X\\ @VVV @VVV @VVV\\ \operatorname{Spec} \widehat{\mathcal{O}}_{Y,y} @>>> \operatorname{Spec}\mathcal{O}^h_{Y,y} @>>> Y \end{CD}$$ Here, $\mathcal{O}^h_{Y,y}$ is the henselization $$\mathcal{O}^h_{Y,y} = \varinjlim_{(B,\mathfrak{q})} B_\mathfrak{q},$$ where the colimit is taken over all étale $\mathcal{O}_{Y,y}$-algebras $B$, and prime ideals $\mathfrak{q} \in \operatorname{Spec}B$ lying over the maximal ideal $y \in \operatorname{Spec} \mathcal{O}_{Y,y}$, such that $$\mathcal{O}_{Y,y}/\mathfrak{m}_y \longrightarrow B/\mathfrak{q}$$ is an isomorphism [Stacks, Tag 0BSK]. The Lemma implies $X_y \simeq \mathbf{P}^n \times \operatorname{Spec} \widehat{\mathcal{O}}_{Y,y}$. By the Artin Approximation theorem [Artin], we therefore see that $X_y^h \simeq \mathbf{P}^n \times \operatorname{Spec} \mathcal{O}^h_{Y,y}$. By the definition of henselization, there is an étale $\mathcal{O}_{Y,y}$-algebra $B$ as in the colimit above, such that $$X \times_Y \operatorname{Spec} B \overset{\sim}{\longrightarrow} \mathbf{P}^n \times \operatorname{Spec} B.\tag*{\blacksquare}$$

Proof of Theorem. We have an exact sequence $$0 \longrightarrow \mathbf{G}_m \longrightarrow \mathrm{GL}_{n+1} \longrightarrow \mathrm{PGL}_{n+1} \longrightarrow 0$$ of abelian sheaves in the étale topology on $Y$. The long exact sequence on cohomology looks like $$H^1_\mathrm{ét}(Y,\mathrm{GL}_{n+1}) \overset{\alpha}{\longrightarrow} H^1_\mathrm{ét}(Y,\mathrm{PGL}_{n+1}) \longrightarrow H^2_\mathrm{ét}(Y,\mathbf{G}_m) =: \operatorname{Br}(Y)$$ Since $\operatorname{Br}(Y) = 0$, the map $\alpha$ is surjective. Note that

• $H^1_\mathrm{ét}(Y,\mathrm{GL}_{n+1}) \cong H^1_\mathrm{Zar}(Y,\mathrm{GL}_{n+1})$ by a version of Hilbert's theorem 90 [Serre, Thm. 2], and is therefore in bijection with isomorphism classes of vector bundles on $Y$;
• $H^1_\mathrm{ét}(Y,\mathrm{PGL}_{n+1})$ is in bijection with étale locally trivial $\mathbf{P}^n$-fibrations;
• The map $\alpha$ is given by $\mathscr{E} \mapsto \mathbf{P}(\mathscr{E})$.

Now suppose we are given a $\mathbf{P}^n$-fibration $\pi\colon X \to Y$. The Proposition implies $\pi$ is represented by an element $\eta \in H^1_\mathrm{ét}(Y,\mathrm{PGL}_{n+1})$. But since $\alpha$ is surjective, we see that $\eta = \mathbf{P}(\mathscr{E})$ for some vector bundle $\mathscr{E}$ of rank $n+1$ on $Y$. $\blacksquare$

• I had worked out almost all of this except the lemma about spreading out from a fiber to the local ring. That is a great result! Let me try to prove it myself and I get back to you if I can't. Thanks! – Asvin Apr 8 '17 at 22:18
• @Asvin Great! I would be interested to know what the correct statement using geometric fibers when $k\ne\overline{k}$, if you worked that out. – Takumi Murayama Apr 8 '17 at 22:43
• A few remarks: you need Y to be regular to show injectivity of the Brauer groups right? About the geometric fibers, I guess everything is ok if we require that the brauer group of each residue field is trivial (and the geometric fibers are P^n). – Asvin Apr 10 '17 at 0:40
• In the proof of the proposition, don't we only use that the fiber over the separable closure of the residue field is $P^n$ since that is the residue field of the strict henselization? So I think we only require the geometric fibers (or maybe the separably closed fibers) to be $P^n$ . – Asvin Apr 10 '17 at 0:55
• @Asvin You are right about $Y$ needing to be regular (it got lost when I moved material around); I'll correct that! As for geometric fibers, saying that the Brauer group of each residue field is trivial and that the geometric fibers are $\mathbf{P}^n$ implies that all fibers are isomorphic to $\mathbf{P}^n$, I think. So we might as well assume all the fibers are isomorphic to $\mathbf{P}^n$. As for your remark about the Proposition, I really need to use the completion (the Lemma needs a complete local ring). Maybe you have another proof? – Takumi Murayama Apr 10 '17 at 1:26