# Reduced rings and tensor products

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.
• The tensor product is over some 'coefficients', some other ring that makes this two algebras over it, isn't it? In this case, what is that ring? Well, I guess you may want the answer for any coefficients ... Well, you said already below.
– OR.
Jul 30, 2013 at 5:51
• @RGB we may assume $A,B$ are $R$-algebra for some ring $R$. Jul 30, 2013 at 5:52
• Hint: If $I$ is a nil ideal of a ring $C$, then $C' = \left(C/I\right)'$. Try to find a nil ideal $I$ of $A \otimes_R B$ such that $\left( A\otimes_R B \right) / I \cong A' \otimes_R B'$. Jul 30, 2013 at 8:14
• @darij I think your assertion is wrong. Take $A=\mathbb{Z}$ and $I=2\mathbb{Z}$. Jul 30, 2013 at 8:43
• @User0829: But $I$ is not a nil ideal! Jul 30, 2013 at 8:50

This can be proved using the following universal property of $A'$: if $f \colon A \to S$ is any $R$-algebra homomorphism where $S$ is reduced, then $f$ factors uniquely as $f = g \circ p$ where $p \colon A \to A/\sqrt{0} = A'$ is the canonical homomorphism. Or more briefly: any $R$-algebra homomorphism from $A$ to a reduced $R$-algebra factors uniquely through $A'$. (I will let you work this out, but feel free to ask in the comments if you need some hints.)
The characterization of $A'$ above makes it unique up to unique isomorphism: If $T$ is any other algebra with a homomorphism $q \colon A \to T$ satisfying the same universal property (with $p$ replaced by $q$), then $T \cong A'$ (via a unique isomorphism that respects $p$ and $q$).
Thus, to prove that $(A \otimes_R B)' \cong (A' \otimes_R B')'$, it is enough to show that $(A' \otimes_R B')'$ equipped with the canonical homomorphism $q \colon A \otimes_R B \to A' \otimes_R B' \to (A' \otimes_R B')'$ satisfies the same universal property as $(A \otimes_R B)'$.
So let $S$ be a reduced ring with an algebra homomorphism $f \colon A \otimes_R B \to S$. The canonical homomorphisms $A,B \to A \otimes_R B$ induce $A,B \to A \otimes_R B \to S$. Since $S$ is reduced, these must factor uniquely through $A'$ and $B'$. This induces $A' \otimes_R B' \to S$. But again, as the latter algebra is reduced (and the former may not be), this factors uniquely through $g \colon (A' \otimes_R B')' \to S$. It's now a small chore to check that this provides a unique factorization $f = gq$.
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 $\sqrt K=\sqrt{(0)}$.