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Let $k$ be a field. Let $A$ be a commutative algebra over $k$. We say $A$ is geometrically reduced over $k$ if $A\otimes_k k'$ is reduced for every extension $k'$ of $k$.

Let $K$ be an algebraic extension of $k$. It is well-known that $K$ is geometrically reduced over $k$ if $K$ is separable over $k$. Conversely suppose $K$ is geometrically reduced over $k$. $K$ is separable over $k$?

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If the extension $\,K/k\,$ is infinite then yes, as can be checked in this very nice paper: . In fact, it is just definition. – DonAntonio Dec 30 '12 at 13:10
@DonAntonio: Doesn't theorem 1.2 say just that for finite extensions? – tomasz Dec 30 '12 at 13:15
Well, but there's a condition on non-zero nilpotent elements there, @tomasz – DonAntonio Dec 30 '12 at 13:17
@DonAntonio: I don't understand. The condition means exactly the same as being geometrically reduces, no? – tomasz Dec 30 '12 at 13:18
Perhaps I'm missing something here, @tomasz, but theorem 1.2 requires that tensor product to have no non-zero nilpotent elements. – DonAntonio Dec 30 '12 at 13:19
up vote 4 down vote accepted

Let $K_0$ be the separable closure of $k$ in $K$. Let $L$ be any extension of $K_0$. Then $$(K\otimes_{K_0} L) \hookrightarrow (K\otimes_{K_0} L)\otimes_{k} K_0 \simeq K\otimes_k L.$$ So $K\otimes_{K_0} L$ is reduced.

Let $p\ge 0$ be the characteristic of $k$. We can suppose $p>0$ (otherwise any extension is separable). If $K\neq K_0$, there exists $\alpha\in K\setminus K_0$ such that some power $a=\alpha^{p}\in K_0$. Consider $L=K_0[\alpha]\subseteq K$. Then $L\simeq K_0[X]/(X^p-a)$ and $$ K_0[\alpha]\otimes_{K_0} L\simeq K_0[X]/(X^p-a)=K_0[X]/(X-\alpha)^p$$ is not reduced. This is a contradiction because on the other hand $K_0[\alpha]\otimes_{K_0} L\subseteq K\otimes_{K_0} L$ must be reduced. So $K=K_0$.

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