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)


I'd like to show (if true) that all fibers are reduced. Pactically it remains to show that the singular fibers are reduced.

  • $\begingroup$ 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

1 Answer 1


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.

  • $\begingroup$ 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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