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 $2$?

  • $\begingroup$ arxiv.org/abs/math/0611606 seems relevant $\endgroup$ Nov 23 at 21:12
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    $\begingroup$ 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). $\endgroup$ Nov 23 at 22:19

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 $\gcd(4,9)=1$.

  • $\begingroup$ Thanks, that's a nice observation! $\endgroup$
    – Asvin
    Nov 23 at 23:13

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.


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