Base change of nilradical is nilradical

Question

Let $B \to A, B \to B'$ be injective, finite ring homomorphisms (finite means that $A$ and $B'$ are finite $B$-modules). Suppose that $A$ and $B$ are integral domains. Denote by $N$ the nilradical of $B'$. Is $N \otimes_B A$ the nilradical of $B' \otimes_B A$?

Answer 1

No: let $B=K$ be a field of characteristic $p>0$ and assume $y:=\sqrt[p]{x}\not\in K$ for some $x\in K$. Let $A:=K(y)=:B^\prime$. Then $N=0$ but

$ B^\prime\otimes_B A=K(y)\otimes_K K(y)\cong K(y)[Y]/(Y-y)^p, $

where $K(y)[Y]$ is the polynomial ring in one variable over $K(y)$.