3
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $\overline k$? $\endgroup$ – Asvin Jan 21 '18 at 21:20
  • $\begingroup$ @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 $X \to \mathbb{P}(H^0(X,O_X(1))^\vee)$. $\endgroup$ – Sasha Jan 21 '18 at 21:49
  • $\begingroup$ Thanks, I don't really know much about Hilbert schemes but it seems reasonable. $\endgroup$ – Asvin Jan 21 '18 at 23:38

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.