# Embedding torsors of elliptic curves into projective space

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 $$2$$?

• arxiv.org/abs/math/0611606 seems relevant Nov 23 at 21:12
• 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). 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 $$\gcd(4,9)=1$$.
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.