# Affine $k$-algebras and reduced rings

In my question, $k$ is a field, and all $k$-algebras are assumed to be finitely generated. A $k$-algebra $A$ is said to be affine if $A \otimes_k \overline{k}$ is reduced. I am trying to prove the following:

If $A$ is an affine $k$-algebra, and $B$ is a reduced ring containing $k$, then $A \otimes_k B$ is reduced.

Since $A$ is $k$-flat, we can regard $A \otimes_k B_0$ as a subring of $A \otimes_k B$ for any subring $B_0$ of $B$ containing $k$. So without loss of generality we can assume that $B$ is finitely generated.

We can write $A = k[X]/I$ and $B = k[Y]/J$ for some ideals $I$ and $J$ of the polynomial rings $k[X], k[Y]$. Clearly $A$ is reduced, since $A \simeq A \otimes_k k$ is embedded in $A \otimes_k \overline{k}$, which is reduced. So $I$ and $J$ are radical ideals. Writing $I$ as an intersection of its finitely many minimal primes, we can embed $A$ into a direct sum of fields (take the quotient fields of all the $k[X]/\mathfrak p$).

The same can be done with $B$. So the tensor product $A \otimes_k B$ is isomorphic to a subring of a direct sum of tensor products of fields over $k$. In order to show that direct sum is reduced, it seems like we need to know something about the quotient fields of the $k[X]/\mathfrak p$, where $\mathfrak p$ is a minimal prime containing $I$. Somehow the fact that $A \otimes_k \overline{k}$ is reduced is coming into play here.

The one thing you're missing is that if $A \otimes_k \bar k$ is reduced, then actually $A \otimes_k L$ is reduced for any field extension $k \subseteq L$.