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

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.

• Why do you think that a direct sum of fields has nilpotent elements? I don't see any. May 27, 2023 at 16:10
• @JyrkiLahtonen yes, i made a mistake and sum of fields doesn't have nilpotent (if i calculated this tensor product over $\mathbb{F}_p$ correctly), so caharcteristic isn't important...? May 28, 2023 at 15:14

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 $$B(x_1,\dots x_n)$$.
• Why is $A\otimes_FB\to K(A)\otimes_FK(B)$ injective? May 27, 2023 at 16:24
• @KentaS: $F$ is a field so $\otimes_F$ is exact in each variable (every $F$-module is flat). May 27, 2023 at 18:23
• @Acrobatic Sorry for rather stupid question, but what is $K(A)$? May 28, 2023 at 15:26
• @NeoFanatic The field of fractions of $A$. May 28, 2023 at 15:34
• @NeoFanatic If $\operatorname{char}k\gt 0$, $\alpha$ may not be separable over $B(x_1,\dots,x_n)$. For example if $A=B=\mathbb{F}_p(t), F=\mathbb{F}_p(t^p),$ then $A\otimes_FB=A[x]/(x-t)^p$ and $(x-t)\neq 0$ is nilpotent. May 29, 2023 at 2:26