In this question, all varieties are supposed to be over an algebraically closed field $k$.
Hypothesis: X is a smooth projective surface and $f:X\longrightarrow \mathbb P^1$ is a morphism with we following properties (maybe some conditions are redundant but for completeness I write the complete list):
- $f$ is flat, proper and has a section.
- There is an open dense subset $U \subseteq\mathbb P^1 $ such that the fiber $X_u$ is a smooth projective curve (i.e. integral, separated scheme of finite type) for every $u\in U$.
- All fibers are irreducible (and hence connected).
- The singular fibers can have only one node as singularities (multiple nodes are not allowed)
Conclusions:
I'd like to show (if true) that all fibers are reduced. Pactically it remains to show that the singular fibers are reduced.
Best Answer
If $y \in \mathbb P^1$ is a (closed) point and $V$ is an affine n.h. of $y$, then we may find a function $a \in \mathcal O(V)$ which vanishes precisely at $y$. If we let $U = f^{-1}(V)$, then $U$ is an open set containing the fibre over $y$, and the fibre over $y$ is cut out by $f^* a \in \mathcal O(V)$. Thus this fibre is a local complete intersection, and in particular Cohen--Macaulay, and in particular $S_1$.
Now let $\sigma$ be the section of $f$. Since $f\circ \sigma = \text{id}_{\mathbb P^1}$, we see that $f$ induces a surjection from $T_{\sigma(y)}X$ to $T_{y}\mathbb P^1$, i.e. (in differential topology language) $f$ is a submersion at $\sigma(y)$, or in algebraic geometry language, $f$ is smooth in a n.h. of $\sigma(y)$. In particular, the fibre over $y$ is then smooth in a n.h. of $\sigma(y)$, and in particular, is reduced in a n.h. of $\sigma(y)$.
Thus this fibre, being irreducible (by assumption) is generically reduced.
A general theorem says that (for Noetherian rings, or equivalently, locally Noetherian schemes) being $R_0$ (i.e. reduced at all generic points) and $S_1$ is equivalent to being reduced. This applies here to let us conclude that the fibre over $y$ is reduced.