5
$\begingroup$

Suppose I have a genus 1 curve $C$ over a field $k$. If $C$ has a point, then we can embed it into the projective plane by a Weierstrass equation. Now let us suppose that $C$ does not have a point (so that it is a non trivial torsor for it's Picard group).

Can I still embed $C$ into the projective plane? I guess not but there is apparently a theorem of Lang-Tate that we can always find an effective divisor of some degree over $k$ (what is a reference?) so we can embed it into some high dimensional projective space.

Can we always embed C into a Severi Brauer variety of dimension $2$?

Improve this question
$\endgroup$
2
  • $\begingroup$ arxiv.org/abs/math/0611606 seems relevant $\endgroup$
    – Chris Wuthrich
    Nov 23, 2021 at 21:12
  • 4
    $\begingroup$ If a curve of genus $1$ embeds into the projective plane then it does so as a cubic. Hence the answer to your question is no unless the curve has a divisor of degree $3$ (in which case the answer is yes, using sections of that divisor). $\endgroup$
    – Noam D. Elkies
    Nov 23, 2021 at 22:19

2 Answers 2

Reset to default
11
$\begingroup$

Suppose that $C \subset X$ is a smooth projective curve of genus $1$ embedded in a Brauer-Severi surface over a field $k$. We have $C^2 = 9$ since this holds after passing to the algebraic closure, where it is embedded as a curve of degree $3$ in the projective plane. So we deduce that $C$ admits a divisor of degree $9$.

It thus just suffices to write down a curve of genus $1$ without a divisor of degree $9$. The example of Piotr Achinger works here. The given curve has a divisor of degree $4$ and cannot have a divisor of degree $9$, since otherwise it would have a divisor of degree $1$ as $\gcd(4,9)=1$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks, that's a nice observation! $\endgroup$
    – Asvin
    Nov 23, 2021 at 23:13
10
$\begingroup$

The answer to the first question is no. Let $k=\mathbb{R}$ be the real numbers. Then every cubic in $\mathbf{P}^2_k$ has a rational point. Take the genus $1$ curve $C$ in $\mathbf{P}^3_k$ obtained by intersecting the two quadrics $x_0^2+x_1^2+x_3^2+x_4^2=0$ and some other randomly chosen quadric so that the intersection is smooth. Then $C$ does not have a rational point. I don't know if it embeds into Brauer-Severi.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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