How to show that a algebraic projective set of $\mathbb{P}^n\times\mathbb{P}^m$ has the form $V(F)$ if and only if all its irreducible components have dimension $n+m-1$. Moreover, if $F$ is square free of bidegree $(a,b)$, then $V(F)$ has bidegree $(a,b)$.
Prove that the arithmetic genus of a curve of bidegree $(a,b)$ in $\mathbb{P}^1\times\mathbb{P}^1$ is $(a-1)(b-1)$.
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In order to calculate the arithmetic genus we consider the short exact sequence $0 \to \mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1}(-a,-b) \to \mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1} \to i_{*}\mathscr{O}_X \to 0$, then taking the long exact sequence in cohomology and using the kunneth formula we get $H^1(\mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1})=0$ and $H^2(\mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1})=0$. We have $H^1(O_X)=H^2(\mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1}(-a,-b))$ and by Kunneth and knowing the cohomology of the projective space we get $dim{H^2(\mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1}(-a,-b))}=(a-1)(b-1)$ as wanted.