# Let $X\subset\mathbb{P}^1\times\mathbb{P}^1$ be cut out by $x^2=xy^6+y^3+1$. Then what is $h^0(\omega_X)$?

Fix a field $k$ of characteristic $0$, and let $X\subset\mathbb{P}_k^1\times_k\mathbb{P}_k^1$ be defined by the type $(2,6)$-equation $x_0^2 y_1^6 - x_0 x_1 y_0^6 + x_1^2 y_0^3 y_1^3 + x_1^2 y_1^6\in k[x_0,x_1,y_0,y_1]$, so it can be shown that $X$ is a nonsingular curve in $X$, and basic computations of homology using the invertible sheaves $\mathscr{O}_{\mathbb{P}^1\times\mathbb{P}^1}(d,e)$ give us that $h^1(\mathscr{O}_X)=5$, so that $X$ has arithmetic genus $5$.

I wish to prove by computation that in this case, the geometric genus $h^0(\omega_X)$ is also $5$, where $\omega_X=\Omega_{X/k}^1$. However, I've hit a snag when I try to gain any explicit information about the sheaf $\Omega_{X/k}$ – quite simply, I don't know ay quick and easy way to calculate that sheaf. My first thought was to try and use the closed immersion $i:X\hookrightarrow\mathbb{P}^1\times\mathbb{P}^1$ and the exact sequence $$i^*\Omega_{\mathbb{P}_k^1\times\mathbb{P}_k^1/k}\to\Omega_{X/k}\to 0$$ since it is not so difficult to calculate that $$\Omega_{\mathbb{P}_k^1\times\mathbb{P}_k^1/k}\cong\mathscr{O}_{\mathbb{P}^1\times\mathbb{P}^1}(-2,0)\oplus\mathscr{O}_{\mathbb{P}^1\times\mathbb{P}^1}(0,-2)$$ but this only tells me that $\Omega_{X/k}$ can be thought of as a quotient sheaf, it doesn't give me any hard information about the sheaf itself. What is the best way to calculate $h^0(\omega_X)$ here?

Adjunction formula says $$\omega_X = \omega_{\mathbb{P}^1 \times \mathbb{P}^1}([X])\vert_X.$$ So, $\omega_X = (\mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1}(-2,-2) \otimes_{\mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1}} \mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1}(2,6))\vert_X = \mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1}(0,4)\vert_X$, You can compute its cohomology by the Koszul complex $$0 \to \mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1}(-2,-6) \to \mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1} \to \mathscr{O}_X \to 0.$$ When tensored by $\mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1}(0,4)$, it gives $$0 \to \mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1}(-2,-2) \to \mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1}(0,4) \to \mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1}(0,4)\vert_X \to 0.$$ The cohomology exact sequence gives $$H^0(X,\omega_X) = H^0(X,\mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1}(0,4)\vert_X) = H^0(\mathbb{P}^1\times \mathbb{P}^1,\mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1}(0,4)) = H^0(\mathbb{P}^1,\mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1}(4)).$$ This is a 5-dimensional space.