Show that every curve of genus 2 can be expressed as a fourth degree plane curve possessing a double point.
This curve is of course a hyperelliptic curve. In order to find a map into $\mathbb{P}^3$, I considered through complete linear system. For $\omega$ a holomorphic differential with $D=(\omega )= p + q$ (possibly equal), $\mathrm{dim}L(D+2p) := l(D+2p) = 2+1-2 = 3$ by Riemann-Roch. Therefore we have 3 function in $L(D+2p)$ and a map $\phi : p \mapsto [f_1(p),f_2(p),f_3(p)]$. However I am not sure whether it maps to a fourth degree curve, or it just maps to second degree curve with mapping degree 2.
Best Answer
This map is a double covering of a plane conic if and only if $p$ is one of the six Weierstrass points.
Indeed, if the map is a double covering of a conic then the preimage of a point on the conic is an element of the hyperelliptic linear system, hence the preimage of a line on the plane is two such. This gives a linear equivalence $$ D + 2p \sim 2D, $$ hence $2p \sim D$, which means that $p$ is a Weierstrass point. The converse is analogous.