When does a fibration has the geometrically integral generic fiber?

Question

Backgrounds: Let $k$ be an algebraically closed field. A fibration $\pi:X \to B$ is a surjective morphism from a surface(= an integral smooth projective scheme of dimension $2$ over $k$) to a integral smooth projective curve(= projective scheme of dimension $1$ over $k$). Then $\pi$ is automatically a flat and projective morphism. For the generic point $\eta \in B$, $X_{\eta}:=\pi^{-1}(\eta)$ is called the generic fiber. $X_{\eta}$ is an integral regular projective scheme of dimension $1$ over the function field $K(B)$ of $B$. Its function field $K(X_{\eta})$ coincides with the function field $K(X)$ of $X$.

My question: Is the following proposition true?

The following are equivalent for a fibration $\pi:X \to B$.

1. The generic fiber $X_{\eta}$ is geometrically integral
2. $K(B)$ is algebraically closed in $K(X)$.
3. The natural homomorphism $\mathcal{O}_B \to f_*\mathcal{O}_X$ is an isomorphism.

$(1) \Leftrightarrow (2)$: It easily follows from the characterization of geometrically integral schemes (c.f. Proposition 5.51 of Gortz and Wedhorn ) and Lemma 7.2. of Badescu, which asserts that $K(X)/K(B)$ is automatically a separable extension under the condition of (2).

$(2) \Rightarrow (3)$: It holds since for any affine open $U \subset B$, $\mathcal{O}_B(U)$ is integrally closed in $K(B)$ and $\pi_*\mathcal{O}_X(U)$ is a finitely generated $\mathcal{O}_B(U)$-module.

The problem is $(3) \Rightarrow (1)$. By the Zariski's main theorem, $X_{\eta}$ is geometrically connected (c.f. Corollary 5.3.17. of Liu). Can we prove geometrically reducedness of $X_{\eta}$ or directly prove $K(B)$ is algebraically closed in $K(X)$?

I encounter with this question when I read Badescu's Algebraic surfaces to study ruled surfaces. The implication $(3) \Rightarrow (2)$ is used in the proof of Theorem 11.10.

Answer 1

Yes, the implication $(3) \Rightarrow (2)$ in your notation does hold, since $X$ is normal.

To make what is going on a bit clearer, I will show the following more general result, which is essentially due to Fujita when $X$ and $Y$ are varieties over an algebraically closed field [Fujita 1981, Theorem 1.20], taking into account the corrections in [Fujita 1982, Theorem 1.20']. The implication you are interested in is $(4) \Rightarrow (2) \Rightarrow (1)$.

Theorem. Let $f\colon X \to Y$ be a proper dominant morphism of integral locally noetherian schemes. Consider the following conditions:

1. $K(Y)$ is algebraically closed in $K(X)$ via the homomorphism induced by $f^\sharp\colon \mathcal{O}_Y \to f_*\mathcal{O}_X$ at the generic point of $X$.
2. $\operatorname{rank}_Y(f_*\mathcal{O}_X) = 1$.
3. Every general fiber $F$ of $f$ satisfies $h^0(F,\mathcal{O}_F) = 1$.
4. The morphism $f^\sharp\colon \mathcal{O}_Y \to f_*\mathcal{O}_X$ is an isomorphism.

Then, we have the following implications: $$\begin{array}{ccccc} (1) & \Rightarrow & (2) & \Leftarrow & (4)\\ & & \Updownarrow\\ & & (3) \end{array}$$

Moreover, if $X$ is normal, then $(2) \Rightarrow (1)$, and if instead $Y$ is normal, then $(2) \Rightarrow (4)$.

Proof. First, $(4) \Rightarrow (2)$ holds by localizing the isomorphism $f^\sharp\colon \mathcal{O}_Y \to f_*\mathcal{O}_X$ at the generic point of $Y$. Next, $(1) \Rightarrow (2)$ holds by [EGAIII$_1$, Corollaire 4.3.12], which shows that when $(1)$ holds, there exists an open subset $U \subseteq Y$ such that $$ f^\sharp\rvert_U\colon \mathcal{O}_Y\rvert_U \longrightarrow (f_*\mathcal{O}_X)\rvert_U $$ is an isomorphism. The implication $(2) \Rightarrow (3)$ holds by combining generic flatness [EGAIV$_2$, Théorème 6.9.1] and cohomology and base change (see [Illusie, Corollary 8.3.11]), and the implication $(3) \Rightarrow (2)$ holds since the generic fiber is a general fiber.

Now suppose that $X$ is normal. We show the contrapositive of $(2) \Rightarrow (1)$. Suppose that $K(Y)$ is not algebraically closed in $K(X)$, and consider the Stein factorization $$X \overset{f'}{\longrightarrow} Y' \overset{\nu}{\longrightarrow} Y$$ of $f$. Since $X$ is normal, $\nu$ is the normalization of $Y$ in $K(X)$. By the assumption that $K(Y)$ is not algebraically closed in $K(X)$, the morphism $\nu$ is not birational, and hence $\operatorname{rank}_Y(\nu_*\mathcal{O}_{Y'}) > 1$. On the other hand, the injection $f^\sharp\colon \mathcal{O}_Y \hookrightarrow f_*\mathcal{O}_X$ factors as $$\mathcal{O}_Y \overset{\nu^\sharp}{\longrightarrow} \nu_* \mathcal{O}_{Y'} \overset{f^{\prime\sharp}}{\longrightarrow} f_*\mathcal{O}_X$$ where both arrows are injective by the fact that $\nu$ and $f'$ are scheme-theoretically dominant, and hence $\operatorname{rank}_Y(f_*\mathcal{O}_X) > 1$ as well.

Finally, if instead $Y$ is normal, then $(2) \Rightarrow (4)$ holds since $\operatorname{rank}_Y(f_*\mathcal{O}_X) = 1$ implies that $$\operatorname{\mathbf{Spec}}_Y(f_*\mathcal{O}_X) \longrightarrow Y$$ is a finite birational morphism, and hence an isomorphism by the normality of $Y$. $\blacksquare$