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?
2 Answers
In Allen's notation, take:
$R = k[t]$, $X = \operatorname{Spec} R$
$S = k[x,y]/(y^2  x^2(x1) )$, $Y= \operatorname{Spec} S$.
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 nonreduced. $X \to Y$ is a surjective map of varieties. $X \times_Y Z = \operatorname {Spec} R/RI = \operatorname{Spec} k[t]/(t^2+1)$, which is reduced.
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.

$\begingroup$ What about $C[x] \leftarrow C[x^2, x^3]: \phi$, the ideal $I = (x^2)C[x^2, x^3]$, and $s = x^3$? $\endgroup$ Jul 22, 2013 at 1:07

$\begingroup$ 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. $\endgroup$ Jul 22, 2013 at 2:21

$\begingroup$ The issue is that $\phi(s)$ can be zero in $R/R\phi(I)$, i.e. it could be that, even if $s$ is not in $I$, $\phi(s)$ is in $R\phi(I)$. $\endgroup$– GiulioMay 4, 2020 at 13:24