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?