Tensor product $A \otimes_{F} B$ of finitely generated algebras $A, B$ with no zero divisors over a field has no nilpotents

Question

I am trying to prove that if $A, B$ are two finitely generated algebras with no zero divisors over field $F$ of characteristic $0$, then $A \otimes_{F} B$ has no nilpotents. How to prove this? I only could understood why assumption that characteristic $0$ is important, because if we have $\text{char}~F = p$, then $\mathbb{F}_{p^n} \otimes \mathbb{F}_{p^n} = \mathbb{F}_{p^n} \oplus \ldots \oplus \mathbb{F}_{p^n}$ -- direct sum of $n$ copies of $\mathbb{F}_{p^n}$, where the tensor product $\otimes$ is taken over $\mathbb{F}_{p^n}$. This direct sum obviously has nilpotent elements, so assumption of zero characteristic is nessesary.

Answer 1

We can assume that $A,B$ are fields since we have embedding of rings $A\otimes_FB\to K(A)\otimes_FK(B)$. Suppose that $F\subset E=F(x_1,\dots,x_n)\subset E(\alpha)=A$, $x_i$ transcendental over $F$, $\alpha$ algebraic over $E$. Then $A\otimes_FB=E(\alpha)\otimes_EE\otimes_FB\subset E(\alpha) \otimes_EB(x_1,\dots x_n)$ and the last ring is reduced since $\alpha$ is separable over $B(x_1,\dots x_n)$.