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?

Answer

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.

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

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