Quartic in $\mathbb{P}^2_k$ are not hyperelliptic

Question

Let fix an algebrically closed field $k$. It is easy to show that a curve of genus $3$ over $k$ is hyperelliptic or a quartic in $\mathbb{P}^2_k$.

I have some difficulties to prove that there not exists quartic in $\mathbb{P}^2_k$ that are hyperellictic.

If $Q$ is a quartic hyperellipric curve I suppose that can be useful work with the canonical map induced by the canonical divisor but I feel like someting it's not clear to me.

Can anyone help me with this claim?

Thank you

Answer 1

Here are some standard facts about hyperelliptic curves. Let $C$ be a hyperelliptic curve of genus $g\geq 2$ and let $f:C\to\mathbb{P}^1$ be the quotient map by the hyperelliptic involution. Then $\deg f=2$ and $K_C=f^*(\mathcal{O}_{\mathbb{P}^1}(g-1))$. In particular, $H^0(K_C)=H^0(\mathcal{O}_{\mathbb{P}^1}(g-1))$, via $f^*$. So, the morphism given by the canonical map factors through the projective line and thus $K_C$ is not very ample, in particular. For a quartic $C$ in the plane, $K_C=\mathcal{O}_C(1)$, which is very ample and thus $C$ can not be hyperelliptic.

Answer 2

Just to add to Mohan's nice answer:

If you read the discussion of hyperelliptic/canonical curves in Hartshorne's book, you will see that this is truly a dichotomy: a curve is either hyperelliptic or canonical, but not both. Thus, you just have to verify that any smooth plane quartic is canonically embedded, and this is what Mohan explains in his answer.