Let $X$ be a Brauer-Severi variety over a field $k$ of characteristic $0$. In other words, suppose that $X_{\overline{k}} \cong \mathbb{P}_{\overline{k}}^n$.

I came across a statement that the map $X \times_k X \longrightarrow X$ sending an element of $X \times_k X$ to its first factor having a section (the diagonal map) implies that $X \times_k X \cong \mathbb{P}_X^n$ (the fiber product with $X$). Why is this the case? I tried to show that $X \times_k X$ has the universal property of $\mathbb{P}_X^n$, but I'm not sure why this is clear in the first place. Is there a simple way of thinking about this that I'm missing?

Some further questions:

  1. More generally, what are other situations where maps/projections from a product of Brauer-Severi varieties having a section implies that the product is isomorphic or birational to a "similar" fiber product with projective space?

  2. Are there suitable varieties $A$ over $k$ such that everything above works when we replace $\mathbb{P}_k^n$ with $A$?

  3. Vague generalization: When do maps with sections induce isomorphisms with some (fiber) product?


Given the answer below, the third question may be rephrased as follows:

When does the existence of a section imply a map is Zariski locally trivial? (e.g. Brauer-Severi varieties) How does this compare to the topological situation? (e.g. principal bundles with a global section)

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    $\begingroup$ Can you provide a reference for where you read this fact? Maybe there are some extra assumptions. Either that, or my answer is wrong :) $\endgroup$ Sep 19 '19 at 19:41
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    $\begingroup$ Thank you for the answer and comment! :) Yes, the example refers to Brauer-Severi schemes. I found this in Example 6.4.3 on p. 145 of the book "Motivic Integration" by Chambert-Loir, Nicaise, and Sebag. It is apparently an example of Kollar, but I can't seem to find the exact reference. This is not in one of the references by Kollar in the book, but I can find similar steps in the proof of Lemma 17 on p. 8 of "Conics in the Grothendieck ring" by Kollar (projection inducing a birational map). $\endgroup$
    – modnar
    Sep 19 '19 at 20:02
  • $\begingroup$ I should add: The example I have given in my answer also in fact occurs in Example 6 in the cited paper of Kollár. There he states that $C \times C$ is birational to $C \times \mathbb{P}^1$, but it is implicit in what he writes that they are not isomorphic. This all agrees with what is written in my answer. $\endgroup$ Sep 22 '19 at 20:19
  • $\begingroup$ Kollár only says that they have the same class in the Grothendieck ring of varieties. This is most likely what the other reference says, but I don't have it hand, and probably explains your confusion and why you thought they were isomorphic. $\endgroup$ Sep 22 '19 at 20:24
  • $\begingroup$ Yes, Kollar only says that they are the same class in the Grothendieck ring of varieties and only gives a birational map as you mentioned. This was the reason for mentioning birational maps in the first follow-up question. I'll mention the first couple of sentences from the example that confused me in the next comment. $\endgroup$
    – modnar
    Sep 22 '19 at 21:12

This is false. Take $X$ to be a smooth plane conic without a rational point. Consider the surface $$S = X \times X.$$ Over the algebraic closure this becomes isomorphic to $\mathbb{P}^1 \times \mathbb{P}^1$. Recall that $\mathrm{Pic}(\mathbb{P}^1 \times \mathbb{P}^1) = \mathbb{Z}^2$. Then one easily sees that $$\mathrm{Pic}(S) = \langle (2,0), (1,1), (0,2) \rangle.$$ The divisors $(2,0)$ and $(0,2)$ come from the trivial divisor and a closed point of degree $2$ on one factor. The class $(1,1)$ comes from the diagonal embedding. Note that there is no divisor of type $(1,2)$.

Now consider the surface $$S' = \mathbb{P}^1_k \times X.$$ Here $$\mathrm{Pic}(S') = \langle (1,0), (0,2) \rangle.$$ This has a divisor of type $(1,2)$, thus $S \not \cong S'$.

What happened?

You seem to be looking for the notion of Brauer-Severi schemes. The theory is similar to that of Brauer-Severi varieties, but over a base scheme rather than a field.

Let $Y$ be a scheme. A Brauer-Severi scheme over $Y$ is a proper morphism $P \to Y$ which is etale locally ismorphic to the trivial projective bundle $\mathbb{P}^n_Y \to Y$ for some $n$.

Then the following holds: $P \to Y$ has a section if and only if it is Zariski locally isomorphic to the trivial projective bundle. This means that $P$ is the projectivisation of some vector bundle on $Y$.

In the above counter-example, one takes $P = X \times X$ and $Y = X$. This has a section hence is the projectivisation of some vector bundle. But it is not the projectivisation of the trivial vector bundle.


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