2
$\begingroup$

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?

Edit:

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)

$\endgroup$
  • 2
    $\begingroup$ Can you provide a reference for where you read this fact? Maybe there are some extra assumptions. Either that, or my answer is wrong :) $\endgroup$ – Daniel Loughran Sep 19 '19 at 19:41
  • 1
    $\begingroup$ 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. $\endgroup$ – Daniel Loughran 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. $\endgroup$ – Daniel Loughran 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
6
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.