Show Fiber Product of Rational Elliptic Surfaces is Calabi-Yau

Question

In a handful of contexts people study Calabi-Yau threefolds formed by taking the fiber product of two rational elliptic surfaces. I can't find any detailed explanation of why such geometries are actually Calabi-Yau, so I think it's just a straightforward computation which I don't fully understand.

Let $\pi: S \to \mathbb{P}^{1}$ and $\pi' : S' \to \mathbb{P}^{1}$ be two rational elliptic surfaces, and define their fiber product

$$X = S \times_{\mathbb{P}^{1}} S'.$$

With some mild assumptions on $S$ and $S'$, $X$ should be a Calabi-Yau threefold, and I'm hoping someone can help me complete the proof of this. In other words, I want to see that $\omega_{X}=0$ or $K_{X}=0$ (however, note that in general, $X$ will certainly not be smooth).

I believe one should start by considering the obvious map induced by $\pi$ and $\pi'$

$$f: S \times S' \to \mathbb{P}^{1} \times \mathbb{P}^{1}.$$

We can then write $X$ as the pullback of the diagonal $\Delta \subset \mathbb{P}^{1} \times \mathbb{P}^{1}$,

$$X = f^{*} \Delta.$$

So we can realize $X$ as a hypersurface in $S \times S'$, so you should then be able to apply the adjunction formula:

$$\omega_{X} = \omega_{S \times S'}|_{X} \otimes \mathcal{N}_{X/S \times S'},$$

where $\mathcal{N}_{X/S \times S'}$ is the normal bundle of $X$ in $S \times S'$. However, I'm sort of stuck on how to proceed -- How can one explicitly handle both of the two factors in the above tensor product and show they somehow cancel to give 0?

Answer 1

The diagonal $\Delta $ is linearly equivalent to $\{p\}\times \mathbb{P}^1 +\mathbb{P}^1\times \{p\} $ for any $p$ in $\mathbb{P}^1$. Therefore $X$ is the zero locus in $S\times S'$ of a section of $L:=\pi^*\mathcal{O}(1) \boxtimes \pi'^*\mathcal{O}(1) $. On the other hand, standard theory of elliptic surfaces gives $\omega _S=\pi ^*\mathcal{O}(-1) $ and $\omega _{S'}=\pi' ^*\mathcal{O}(-1) $, therefore $\omega _{S\times S'}= L^{-1}$. Then the adjunction formula gives indeed $\omega_X\cong \mathcal{O}_{X}$.

Answer 2

$S\times_{\mathbb{P}^1}S'$ is a complete intersection in $\mathbb{P}^1\times \mathbb{P}^2\times \mathbb{P}^2$: it is given by two equations of degree $(1,3,0)$ and $(1,0,3)$. The canonical bundle is trivial by the adjunction formula.