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 2


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:

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


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

$S/I$ 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.


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. $\endgroup$
    – Will Sawin
    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 $R\phi(I)$. $\endgroup$
    – Giulio
    May 4, 2020 at 13:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.