# If $B$ is a flat $A$-algebra then $B_{\mathrm{red}}$ is a flat $A_{\mathrm{red}}$-algebra

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?

• This is not true in general and one example is due to Cowsik and Nori. See sciencedirect.com/science/article/pii/0021869376902374 Apr 14, 2020 at 21:57
• 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? Apr 14, 2020 at 22:31
• @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!! Apr 14, 2020 at 23:15