# Genus of a plane curve cut out by a quintic

The following handout states that the plane curve $X=V_+(x^3yz+y^5+z^5)\subseteq\mathbb{P}_k^2$ is a genus $5$ curve, however, when I try and calculate the genus, I seem to find that it has genus $6$.

This is my work: Let $i:X\hookrightarrow\mathbb{P}_k^2$ be the closed immersion. Note that we have a short exact sequence $$0\to\mathscr{I}\to\mathscr{O}_{\mathbb{P}^2}\to i_*\mathscr{O}_X\to 0$$ where the ideal sheaf $\mathscr{I}$ is isomorphic to $\mathscr{O}_{\mathbb{P}^2}(-5)$ as an $\mathscr{O}_{\mathbb{P}^2}$-module, so that since $H^p(X,\mathscr{O}_X)\cong H^p(\mathbb{P}^2,i_*\mathscr{O}_X)$ by finiteness and by calculations of cohomology on projective space, we can take the long exact sequence to obtain that $H^1(X,\mathscr{O}_X)\cong H^2(\mathbb{P}^2,\mathscr{O}_{\mathbb{P}^2}(-5))$, and $\dim_k H^2(\mathbb{P}^2,\mathscr{O}_{\mathbb{P}^2}(-5))=6$, so that $g=h^1(\mathscr{O}_X)=6$.

Where did I make a mistake here?

## 1 Answer

The curve you describe is singular, so the arithmetic and geometric genera don't agree. You can see that the handout calculates the same 6 as you using the degree-genus formula, then subtracts 1 (this is $\frac{1}{2}r(r-1),$ where $r = 2$ is the multiplicity of the node), to get the geometric genus $\rho_g.$