Kodaira dimension of a ruled surface

I am trying to solve 21.5.J in Vakil’s FOAG.

The exercise is about showing that $X:=C\times_{\mathbb C} \mathbb P_{\mathbb C}^1$ is neither Fano, Calabi-Yau, nor general type where $C$ is a smooth complex curve of genus $>1$.

Since $\Omega_X = \text{pr}_1^\ast\Omega_C\oplus\text{pr}_2^\ast\Omega_{\mathbb P^1}$, we have $K_X = \text{pr}_1^\ast K_C\otimes \text{pr}_2^\ast K_{\mathbb P^1}$ and I can show that $X$ is not Fano nor CY.

To show that $X$ is not general type, I need to show that the Kodaira dimension of $X$ is not $2$. The Kodaira dimension is defined as the least integer $m$ such that $h^0(X,K_X^j)/j^m$ is bounded for $j>0$. Some google search tells me $H^0(X, K_X^j) = 0$ for all $j>0$, so I tried to show this as follows.

Let $s$ be a section of $K_X^j$. Then, for any point $p\in C$, pulling back $s$ to $p\times \mathbb P_{\mathbb C}^1$, we have $s=0$ since we can see that $K_X=\text{pr}_1^\ast K_C\otimes \text{pr}_2^\ast K_{\mathbb P^1}$ pulls back to $\mathcal O_{\mathbb P^1}\otimes K_{\mathbb P^1}=\mathcal O(-2)$ by considering the compositions
p\times \mathbb P^1\to C\times\mathbb P^1 \to \mathbb P^1\\
p\times \mathbb P^1\to C\times\mathbb P^1\to C

Therefore, $s$ pulls back to zero for any choice of $p\in C$.

Considering affine open subsets $\text{Spec} A, \text{Spec}B$ of $C,\mathbb P^1$, respectively, the above translates to algebra as the following: given a $A\otimes B$ module $M$ and $s\in M$, we have $s\in \mathfrak m M$ for all maximal ideals $\mathfrak m$ of $A$.

(i) Can we conclude that $s=0$ from here easily in general? Or since $M$ corresponds to an invertible sheaf, should I only consider the form $M=A\otimes \mathbb C[x]$ (or possibly something like $M=A\otimes \mathbb C(x)$)?

(ii) Is there a simpler way to solve this problem?

(More General Question) (iii) Some google search also tells me that for two irreducible complex projective varieties $X,Y$, the Kodaira dimension of $X\times Y$ is the sum of those of $X$ and $Y$. How do I show this?


Let me grab out this old question and answer (ii). If I’m not overlooking anything, then the following is an easier approach.

It’s not too hard to check $\mathscr{K}_{C \times_{\mathbb{C}} \mathbb{P}_{\mathbb{C}}^1} \cong \mathscr{K}_{C} \boxtimes \mathscr{K}_{\mathbb{P}^1_{\mathbb{C}}}$. Then, a Künneth formula computation (18.2.8) yields $$h^0(C \times_{\mathbb{C}} \mathbb{P}_{\mathbb{C}}^1, \mathscr{K}_{C}^{\otimes j} \boxtimes \mathscr{K}_{\mathbb{P}^1_{\mathbb{C}}}^{\otimes j}) = h^0(C, \mathscr{K}_C^{\otimes j})h^0(\mathbb{P}_{\mathbb{C}}^1, \mathscr{K}_{\mathbb{P}_\mathbb{C}^1}^{\otimes j}) = 0$$
for $j > 0$ because $\mathscr{K}_{\mathbb{P}_\mathbb{C}^1} \cong \mathcal{O}_{\mathbb{P}_{\mathbb{C}}^1}(-2)$.

On a side note: Vakil has just posted two pages about the Künneth formula one week ago. See here.

Source : Link , Question Author : JWL , Answer Author : Qi Zhu

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