Arithmetically Cohen-Macaulay curve on a quadric 
If $Y$ is a curve of bidegree $(a,b)$ on a smooth quadric surface $Q\subset \mathbb{P}^3$, how do we see that it is arithmetically Cohen-Macaulay (ACM, for short) iff $|a-b|\leq 1$?

If (like me) you aren't familiar with what ACM means, two equivalent conditions in this case are:
1) $H^1(\mathcal{I}_Y(\ell))=0$ for all $\ell\in \mathbb{Z}$, where $\mathcal{I}_Y$ is the ideal sheaf for $Y$ in $\mathbb{P}^3$. 
2) $k[x_0,\ldots,x_3]=\bigoplus_{n\in\mathbb{Z}}{H^{0}(\mathscr{O}_{P^3}(n))}\rightarrow \bigoplus_{n\in\mathbb{Z}} {H^0(\mathscr{O}_Y(n))}$ is surjective. 
(This is exercise 8.1(a) in Hartshorne's Deformation Theory, and the equivalent conditions are from Proposition 8.6.)
 A: We will use your condition (2). Since $Q$ is a hypersurface, it is ACM (by Hartshorne's Algebraic Geometry, III, Exc. 5.5(a)), so it suffices to show 
$$\bigoplus_{n \in \mathbf{Z}} H^0(Q,\mathcal{O}_Q(n)) \longrightarrow \bigoplus_{n \in \mathbf{Z}} H^0(Y,\mathcal{O}_Y(n))$$
is surjective if and only if $\lvert a - b \rvert \le 1$.
Remark. If you don't like the usage of the Künneth formula, you can also prove your statement using a lot of long exact sequences. An outline is in Hartshorne's Algebraic Geometry, III, Exc. 5.6, especially part (b), (3).
Consider the exact sequence
$$
  0 \longrightarrow \mathcal{O}_Q(n-a,n-b) \longrightarrow \mathcal{O}_Q(n) \longrightarrow \mathcal{O}_Y(n) \longrightarrow 0
$$
The long exact sequence on cohomology is
$$\cdots \longrightarrow H^0(Q,\mathcal{O}_Q(n)) \longrightarrow H^0(Y,\mathcal{O}_Y(n)) \longrightarrow H^1(Q,\mathcal{O}_Q(n-a,n-b)) \longrightarrow H^1(Q,\mathcal{O}_Q(n)) \longrightarrow \cdots$$
Now if $\lvert a - b \rvert \le 1$, then since $Q \cong \mathbf{P}^1 \times \mathbf{P}^1$, we have, by the Künneth formula,
\begin{align*}
H^1(Q,\mathcal{O}_Q(n-a,n-b)) &=
\left( H^0(\mathbf{P}^1,\mathcal{O}_{\mathbf{P}^1}(n-a)) \otimes H^1(\mathbf{P}^1,\mathcal{O}_{\mathbf{P}^1}(n-b)) \right)\\
&\quad\quad\oplus \left( H^1(\mathbf{P}^1,\mathcal{O}_{\mathbf{P}^1}(n-a)) \otimes H^0(\mathbf{P}^1,\mathcal{O}_{\mathbf{P}^1}(n-b)) \right)
\end{align*}
is zero, since if $\lvert a - b \rvert = \lvert (n-b) - (n-a) \rvert \le 1$, then in each summand, one of $H^0$ or $H^1$ must vanish by the cohomology of $\mathbf{P}^1$.
On the other hand, if $\lvert a - b \rvert \ge 2$, then suppose without loss of generality that $a > b$. Letting $n = b$, the long exact sequence from above becomes
$$\cdots \longrightarrow H^0(Q,\mathcal{O}_Q(b)) \longrightarrow H^0(Y,\mathcal{O}_Y(b)) \longrightarrow H^1(Q,\mathcal{O}_Q(b-a,0)) \longrightarrow H^1(Q,\mathcal{O}_Q(b)) \longrightarrow \cdots$$
But, again by the Künneth formula,
$$H^1(Q,\mathcal{O}_Q(b)) = \left( H^0(\mathbf{P}^1,\mathcal{O}_{\mathbf{P}^1}(b)) \otimes H^1(\mathbf{P}^1,\mathcal{O}_{\mathbf{P}^1}(b))\right)^{\oplus 2} = 0$$
while
$$H^1(Q,\mathcal{O}_Q(b-a,0)) = H^1(\mathbf{P}^1,\mathcal{O}_{\mathbf{P}^1}(b-a)) \ne 0$$
and so $H^0(Q,\mathcal{O}_Q(b)) \to H^0(Y,\mathcal{O}_Y(b))$ is not surjective.
