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.)


1 Answer 1


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.

  • 1
    $\begingroup$ Maybe it is worthwhile to emphasize that $\mathcal O_Q(b)=\mathcal O_Q(b,b)$ in the notation of Takumi's nice answer above. Anyway, +1. $\endgroup$ Commented Apr 22, 2016 at 9:27

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