Calculating canonical divisor in product of projective spaces. Let $X$ be an intersection of two divisors of bidegree $(a,b)$ and $(c,d)$ in $\mathbb{P^2}\times \mathbb{P^2}$. Then how can I find the canonical divisor $K_X$? 
I'm asking because I have no experience in working with bidegrees.
 A: As usual when you have an intersection like this (which I assume has the right dimension and so on), you can use the adjunction formula.
Adjunction says that we have an exact sequence of bundles on $X$
$$ 0 \rightarrow N_X^\vee \longrightarrow \Omega_{\mathbf P^2 \times \mathbf P^2 \mid X} \rightarrow \Omega_X \longrightarrow 0$$
where $N_X$ is the normal bundle of $X$.
Now taking top exterior powers and rearranging we get
$$K_X = K_{\mathbf P^2 \times \mathbf P^2 \mid X} \otimes \bigwedge^2 N_X.$$ 
We know that $K_{\mathbf P^2 \times \mathbf P^2 \mid X} = O(-3,-3)_{\mid X}$ so it remains to find $\bigwedge^2 N_X$. If we write $X=D_1 \cap D_2$ for the intersection of your two divisors, then 
\begin{align*}
N_X &= (N_{D_1} \oplus N_{D_2})_{\mid X} \\
&= (O(a,b) \oplus O(c,d) )_{\mid X}.
\end{align*} 
So $\bigwedge^2 N_X = O(a+c,b+d)_{\mid X}$. 
Putting everything together we get 
$$K_X = O(a+c-3,b+d-3)_{\mid X}.$$
Edit: Alex asked why my formula for $\bigwedge^2 N_X$ is true, so let me give some more detail on that. 
There is a general formula for the exterior powers of a direct sum, written out nicely for example in the answers here. In particular, it says that if $L_1$ and $L_2$ are line bundles, then 
$$\bigwedge^2 (L_1 \oplus L_2) = L_1 \otimes L_2.$$
Applied here that gives us
\begin{align*}\bigwedge^2 N_x &= O(a,b) \otimes O(c,d) \\
 &= \left(\pi_1^* O(a) \otimes \pi_2^* O(b) \right) \otimes \left( \pi_1^*O(c) \otimes \pi_2^* O(d) \right) \\
\end{align*}
where $\pi_1$, $\pi_2$ are the two projections $\mathbf P^2 \times \mathbf P^2 \rightarrow \mathbf P^2$. Rearranging the factors in the tensor product then gives what we are after. 
