TITLE: Singular curves of genus 1
QUESTION [7 upvotes]: Let $C$ be an irreducible curve of arithmetic genus $1$ over a field $k$ and with a double $k$-point $p\in C$.
Is $C$ rational over $k$?
If $C$ is a plane cubic the answer is positive since we can parametrize $C$ with the lines through $p$.
In general I would embed $C$ in some projective space $\mathbb{P}^n$ and I would project $C$ until I get a plane cubic with a node. I think this should be ok if $k$ is infinite but if $k$ is finite I do not know how to make sure that $C\subset \mathbb{P}^n$ can be birationally projected onto a plane nodal cubic.
Thank you.

REPLY [10 votes]: There's no problem over finite fields, but there is a problem over fields that have a nontrivial Brauer class. If you take a genus $0$ curve that's not rational (say a plane quadric), it will always have points over a degree $2$ extension (intersect with a line), and then you can glue two of them together to get a nodal curve of arithmetic genus $1$.
This curve will not be a nodal cubic, but you can embed it as a degree $4$ curve in $\mathbb P^3$.
For example, let $a$ be a quadratic nonresidue mod $p$, consider the curve in $\mathbb P^3$ with coordinates $x,y,z,w$ given by the equations $x^2 -a y^2 - p z^2$ and $xy = zw$ over $\mathbb Q_p$. The only rational point of this curve is $(x:y:z:w)=(0:0:0:1)$, which is a node, so the curve has arithmetic genus $1$, has a node, and isn't rational.