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Suppose X, Y, Z are k-varieties and $f: X \to Z$ factors through $f': X \to Y$ and $g: Y \to Z$. Suppose all of f, f', g are surjective. Assume that for $z \in Z$, the fibre $f^{-1} (z)$ is reduced. Then, is the fibre $g^{-1} (z)$ always reduced?

If not, when will it be true?

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Counterexample: $Spec k \to Spec k[X]/(X^2) \to Spec k$. –  Martin Brandenburg Jan 23 '10 at 19:49
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Sorry for being imprecise. Here $k$-varieties mean, integral $k$-schemes of finite type over $k$, where the middle scheme isn't probably fitting into the case. Do you know any other example where X, Y, Z are all k-varieties? –  Jinhyun Park Jan 23 '10 at 19:53
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2 Answers

How about: $Y$ is the cubic curve $y^2=x^2+x^3$ in $\mathbb A^2$ minus the point $(-1,0)$, $g:Y\to Z$ is the projection to the $x$-axis, and $X$ is the normalization of $Y$, which would be $\mathbb P^1$ minus 2 points.

The fiber $g^{-1}(0)$ is one non-reduced point. The fiber $f^{-1}(0)$ is two reduced points.

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Nice example! We can notice that here $X\to Y$ is not flat (a normalization map is never flat unless it is an isomorphism). –  Qing Liu Jan 24 '10 at 16:33
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The fiber $X_z$ is equal to $X\times_Y Y_z$. So if $X\to Y$ is faithfully flat, then $X_z$ reduced implies that $Y_z$ is reduced.

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