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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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    @DonAntonio Thanks. Theorem 1.2 of the paper treats finite extensions, but the infinite extension case follows trivially.2012-12-30

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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 $K=K_0$.