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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.

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    $\begingroup$ Why do you think that a direct sum of fields has nilpotent elements? I don't see any. $\endgroup$ May 27, 2023 at 16:10
  • $\begingroup$ @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...? $\endgroup$
    – NeoFanatic
    May 28, 2023 at 15:14

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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 $B(x_1,\dots x_n)$.

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  • $\begingroup$ Why is $A\otimes_FB\to K(A)\otimes_FK(B)$ injective? $\endgroup$
    – Kenta S
    May 27, 2023 at 16:24
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    $\begingroup$ @KentaS: $F$ is a field so $\otimes_F$ is exact in each variable (every $F$-module is flat). $\endgroup$ May 27, 2023 at 18:23
  • $\begingroup$ @Acrobatic Sorry for rather stupid question, but what is $K(A)$? $\endgroup$
    – NeoFanatic
    May 28, 2023 at 15:26
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    $\begingroup$ @NeoFanatic The field of fractions of $A$. $\endgroup$
    – Acrobatic
    May 28, 2023 at 15:34
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    $\begingroup$ @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. $\endgroup$
    – Acrobatic
    May 29, 2023 at 2:26

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