Proving that all fibers of a morphism $\varphi:X\to\Bbb{P}^1$ are connected

Question

Let $X$ be a rational elliptic surface over an algebraically closed field $k$ and $\pi:X\to\Bbb{P}^1$ its elliptic fibration, which I assume is relatively minimal.

If $D$ is a nef divisor such that $D^2=0$ and $C\cdot K_X=-2$, then I know how to prove that the linear system $|D|$ is a base point free pencil which induces a morphism $\varphi:X\to\Bbb{P}^1$.

I was also able to prove that all fibers of $\varphi$ are connected. I won't go into the details, but the proof essentially uses intersection numbers and goes like this: if $F$ is a fiber of $\varphi$ with connected components $F_1,...,F_n$, then $F_i\cdot K_X=-2$ for each $i$, therefore $-2=F\cdot K_X=\sum_{i=1}^nF_i\cdot K_X=-2n$, which gives $n=1$.

Even though the proof works, it's pretty convoluted and relies on many peculiarities of the problem (for example, that components of reducible fibers of $\pi$ are rational $(-2)$-curves).

I wonder if there's a simpler, more general way to prove this. I came across the following result from Hartshorne's book (Corollary III.11.3):

Let $f:X\to Y$ be a projective morphism of Noetherian schemes and assume that $f_*\mathcal{O}_X\simeq\mathcal{O}_Y$. Then $f^{-1}(y)$ is connected for all $y\in Y$.

I suspect $\varphi$ should be projective (although I can't prove it) and I have no idea if indeed $\varphi_*\mathcal{O}_X\simeq \mathcal{O}_{\Bbb{P}^1}$, nor if it is practical to prove it. Any suggestion is welcome. Thank you!

Answer 1

Yes, there are more general ways to prove this sort of thing. The main tool is Stein factorization:

Lemma (Stacks 03H2). Let $S$ be a scheme. Let $f:X\to S$ be a proper morphism. There exists a scheme $S'$ and a factorization of $X\stackrel{f}{\to} S$ as $X\stackrel{f'}{\to} S'\stackrel{\pi}{\to} S$ with the following properties:

1. the morphism $f'$ is proper with geometrically connected fibers,
2. the morphism $\pi:S'\to S$ is integral (even better, finite if $S$ is locally noetherian),
3. we have $f'_*\mathcal{O}_X=\mathcal{O}_{S'}$,
4. we have $S'=\underline{\operatorname{Spec}}_S(f_*\mathcal{O}_X)$, and
5. $S'$ is the normalization of $S$ in $X$.

This proof comes about through finiteness of cohomology for proper morphisms plus the theorem on formal functions, so it's well-positioned to make an appearance at the end of Hartshorne chapter III once you've seen a good bit of material around cohomology.

This lets us prove some nice results about what has to happen with the number of connected components. For instance, with Stein factorization in our tool belt, we may prove the following:

Lemma (Stacks Project 0BUI): Let $X\to S$ be a flat proper morphism of finite presentation Let $n_{X/S}$ be the function on $S$ counting the number of geometric connected components of fibres of $f$. Then $n_{X/S}$ is lower semi-continuous.

This result lets us deal with your problem. Using the definition of an elliptic fibration from your previous post, the map $\pi:S\to\Bbb P^1$ is proper (eg by 01W6 or Hartshorne exercise II.4.8) and flat (eg by Hartshorne proposition III.9.7) so our result above applies. Since a function $f$ is lower semi-continuous iff the sets $\{x\in X\mid f(x) \leq a\}$ are closed for any $a$, we can pick $a=1$ and use the fact that almost all fibers are curves of genus one to see that the only possible values for the number of geometrically connected components of fibers of $\pi$ are $0$ and $1$. But being a clopen map to an integral scheme, $\pi$ is surjective, so every fiber has one geometric connected component and is therefore connected.