1
$\begingroup$

Given a commutative ring $A$ we denote by $A_{\mathrm{red}}$ the ring $A/\mathrm{Nil}(A)$ where $\mathrm{Nil}(A)$ is the nilradical of $A$. It is not difficult to see that this construction is functorial, that is, given a ring morphism $f:A\rightarrow B$ there is a natural induced morphism $f_\mathrm{red}:A_\mathrm{red}\rightarrow B_\mathrm{red}$.

Is it true that if $f:A\rightarrow B$ makes $B$ a flat $A$-algebra then $f_\mathrm{red}:A_\mathrm{red}\rightarrow B_\mathrm{red}$ makes $B_\mathrm{red}$ a flat $A_\mathrm{red}$-algebra?

$\endgroup$
3
  • 2
    $\begingroup$ This is not true in general and one example is due to Cowsik and Nori. See sciencedirect.com/science/article/pii/0021869376902374 $\endgroup$
    – Mohan
    Apr 14, 2020 at 21:57
  • $\begingroup$ This seems really interesting but I can see how can I obtain my example from it. It is related with the Hironaka's criterion for Cohen Macaulay? $\endgroup$
    – Walter Simon
    Apr 14, 2020 at 22:31
  • $\begingroup$ @Mohan Ok, know I understand. In the notation of the paper we have $A \hookrightarrow C$ is flat and $A$ is reduced but $A \hookrightarrow \varphi(C)=C_\mathrm{red}$ is not flat. Thanks!! $\endgroup$
    – Walter Simon
    Apr 14, 2020 at 23:15

0

Reset to default

You must log in to answer this question.