TITLE: Reduced rings and tensor products
QUESTION [8 upvotes]: I assume all rings are commutative with identity. Denote $A':=A/ \sqrt{0}$ for convenience.
Question is simple: 

For a given ring $R$ and $R$-algebras $A,B$, does this isomorphism $(A\otimes_R B)'\cong (A'\otimes_R B')'$ hold?

With specific (easy) examples, it seems to be true. How can I show that in general setting?

This question arises from the equality $(X_{\text{red}}\times_S Y_{\text{red}})_{\text{red}}\cong (X\times_S Y)_{\text{red}}$ where $X,Y$ are $S-$schemes. In the proof, the author says that equality comes from the induced map $X_{\text{red}}\times_S Y_{\text{red}}\to X\times_S Y$ by the canonical maps $X_{\text{red}}\to X$ and $Y_{\text{red}}\to Y$. But I cannot understand how can we get the isomorphism just from the induced map.

REPLY [2 votes]: Maybe @darijgrinberg's suggestions deserve to be an answer as long as they provide a very short one. 
One can write $A/I\otimes_RB/J$ as $(A\otimes_RB)/K$, where $K$ is the ideal of $A\otimes_RB$ generated by $a\otimes b$ with $a\in I$ or $b\in J$. In the particular case when $I=N(A)$ and $J=N(B)$ we have $A'\otimes_RB'=(A\otimes_RB)/K$ with $K$ generated by $a\otimes b$ with $a\in A$ or $b\in B$ nilpotent. Then we have $$(A'\otimes_RB')'=\left((A\otimes_RB)/K\right)'=(A\otimes_RB)',$$ where the last equality holds since $K$ is a nil ideal, that is, an ideal with $\sqrt K=\sqrt{(0)}$.