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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.

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  • $\begingroup$ If your fibers are reduced, then they are singular, and the locus of singularities is dense (so definitely not just a node). But your last condition is that every singularity on the fibre is isolated. So all fibres are reduced. $\endgroup$ Commented Jan 18, 2015 at 15:07
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    $\begingroup$ Even if you leave out the last condition, the reducedness follows from the fact that the image of your section (assumed to exist) lies in the smooth locus of $f$. So every fibre has a smooth point. Since they are all assumed to be irreducible, they are generically reduced (meaning reduced on a dense open), and thus reduced. (Probably this is the same answer as tracing gave.) $\endgroup$ Commented Jan 18, 2015 at 15:09
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    $\begingroup$ @AriyanJavanpeykar: In your first comment, you mean "If your fibres are non-reduced". Also, you're right, my answer is just a long-winded version of your comment! (I guess I gave some detail as to why the section lies in the smooth locus, and why fibres are automatically $S_1$, so that generically reduced implies reduced.) $\endgroup$
    – tracing
    Commented Jan 18, 2015 at 21:08
  • $\begingroup$ Dear @tracing thank you for the correction. Also, your answer is certainly more clear than what I wrote. $\endgroup$ Commented Jan 19, 2015 at 7:34
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    $\begingroup$ @Dubious I miswrote some things in my last comment as well. If $X$ is irreducible and generically non-reduced, then the locus of smoothness is empty. Of course, you can have an irreducible variety with only one non-reduced local ring. Then the locus of smoothness is non-empty (if $X$ has more than one point). $\endgroup$ Commented Jan 28, 2015 at 13:05

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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.

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  • $\begingroup$ What do you mean by $S^1$ (or $S_1$)? $\endgroup$
    – Dubious
    Commented Jan 18, 2015 at 5:30
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    $\begingroup$ @Dubious tracing is using Rn and Sn as defined (for instance) in math.nagoya-u.ac.jp/~takahashi/ncr.pdf. See also Serre's criterion for normality. $\endgroup$ Commented Jan 18, 2015 at 15:04
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    $\begingroup$ @Dubious: As Ariyan Javanpeykar indicates, I'm using the notions $R_n$ and $S_n$; they are defined for example in Hartshorne somewhere in Chapter III (I think), where he discussed Serre's criterion for normality. Serre's criterion is that, for a Noetherian ring, normal is equivalent to $R_1 + S_2$. An easier thing to prove, but also very useful, is that (again for a Noetherian ring) is that reduced is equivalent to $R_0 + S_1$; and this is what I am using in my answer. Basically, a scheme like $\mathbb C[x,y]/(xy,y^2)$ can't arise as the fibre of a map from a smooth surface to a ... $\endgroup$
    – tracing
    Commented Jan 18, 2015 at 21:11
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    $\begingroup$ ... smooth curve; these fibres are automatically local complete intersections, hence Cohen--Macaulay, hence $S_n$ for all $n$, and so to check that a fibre is reduced, it suffices to check that it is generically reduced (which is precisely condition $R_0$). $\endgroup$
    – tracing
    Commented Jan 18, 2015 at 21:12

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