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Say $X \to Y$ is a surjective map of algebraic varieties, and $Z \subset Y$ is nonreduced. Then is the preimage $Z \times_Y X$ also nonreduced?

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2 Answers

up vote 8 down vote accepted

In Allen's notation, take:

$R = k[t]$, $X = \operatorname{Spec} R$

$S = k[x,y]/(y^2 - x^2(x-1) )$, $Y= \operatorname{Spec} S$.

with the map defined by:

$y = t(t^2+1)$

$x=(t^2+1)$

$I=(x)$, $Z = \operatorname {Spec} S/I$.

$S/I$ contains a nilpotent, $y$ so $Z$ is non-reduced. $X \to Y$ is a surjective map of varieties. $X \times_Y Z = \operatorname {Spec} R/RI = \operatorname{Spec} k[t]/(t^2+1)$, which is reduced.

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This is local, of course. So $R \leftarrow S : \phi$ is an injection of domains, and $I \leq S$ is a nonradical ideal; is $R \phi(I)$ a nonradical ideal of $R$?

Say $s$ descends to a nonzero nilpotent of $S/I$. Then $\phi(s)$ will likewise be nilpotent in $R / R\phi(I)$, and the injectivity says it will be nonzero.

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What about $C[x] \leftarrow C[x^2, x^3]: \phi$, the ideal $I = (x^2)C[x^2, x^3]$, and $s = x^3$? –  Vivek Shende Jul 22 '13 at 1:07
    
In fact, this is not true for every injection of domains! Let $R= k[x,y]$, $S = k[x,xy]$, $I= (x, x^2y^2)$. One needs at least to use the fact that the map is a surjection, not just an epimorphism. –  Will Sawin Jul 22 '13 at 2:21
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