Let $R$ be a ring and let $A$ be a $R$-algebra. $A$ is separable over $R$ if and only if the localisation $A_P$ is separable over $R_P$ for all $P\in Spec(R)$.
An exercise about separable algebras
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abstract-algebra
noncommutative-algebra
1 Answers
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I assume that the ring $R$ is commutative (else I don't think "separable algebra" makes sense.)
a) It is always true that $A$ separable over $R$ implies $A_P$ separable over $R_P$ for all $P\in Spec(R)$. More generally separability is preserved under base-change: if $A$ is separable over $R$ then, for any commutative $R$-algebra $S$, the $S$-algebra $S\otimes_R A$ is $S$-separable.
b) For the converse you need some supplementary hypothesis.
If for all $P\in Spec(R)$ the algebra $A_P$ is separable over $R_P$, then $A$ separable over $R$ under any of the following two assumptions:
i) $A$ is a finitely generated $R$-module.
ii) $A$ is commutative and a finitely generated $R$-algebra.
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0Perfect, tha$n$k you very much! – 2011-06-26