4

Let $X/k$ be a variety over a field such that $X_{\overline k} \cong \mathbb P^n$ over $\overline k$ for some $n$. Suppose moreover that $X$ has a rational $k$-point $P$. Then, I know that $X \cong \mathbb P^n$ over $k$. The argument I know goes as follows:

$(X,P)$ is a twist of $(\mathbb P^n,(1,0,\dots,0))$ and the automorphism group of this object is an extension of $\mathrm{GL}_n(\overline k)$ by ${\overline k}^\times$ and therefore has trivial first cohomology and hence no non trivial twists.

In the case of $n=1$ however, there is a more "geometric" proof that explicitly constructs the required isomorphism. First, embed $X$ in $\mathbb P^2$ over $k$ using the dual of the canonical divisor and then project onto a line in $\mathbb P^2$. This map is in fact an isomorphism.

Is there a similar construction in the general case? Can we explicitly construct the required isomorphism? I would also be interested in other proofs of this fact.

Asvin
  • 8,489

1 Answers1

3

First, let us check that the dual Severi--Brauer variety $X^\vee$ is trivial. Indeed, $X$ has a $k$-rational point, hence $X^\vee$ has a $k$-rational hyperplane, hence the line bundle $O_{X^\vee}(1)$ is defined over $k$, hence $X^\vee$ is trivial. Now it remains to note that the dual of the trivial Severi--Brauer variety is itself trivial.

Sasha
  • 20,727
  • Ah, I did not know that one can define the dual of a Severi-Brauer variety but I think I see one way to construct it: Can we construct $X^\vee$ by base changing $X$ to $\overline k$, taking duals and then descending to $k$ by Galois descent? Also, I suppose the reason the line bundle defined by the hyperplane is ample and gives an isomorphism is again by testing over $\overline k$? – Asvin Jan 21 '18 at 21:20
  • @user404577: This is one possibility. Another is to directly consider the Hilbert scheme of hyperplanes in $X$. To see that the existence of $O_X(1)$ implies $X$ is trivial, just note that there is a natural isomorphism $X \to \mathbb{P}(H^0(X,O_X(1))^\vee)$. – Sasha Jan 21 '18 at 21:49
  • Thanks, I don't really know much about Hilbert schemes but it seems reasonable. – Asvin Jan 21 '18 at 23:38
  • @Sasha Would you mind answer this question about your answer? Thank! –  Nov 15 '19 at 04:07