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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 $P\in Spec(R)$.

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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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I'm not quite sure what you know: depending on that, the "exercise" is easy or quasi-impossible to solve! Anyway, the statement you gave is not quite correct in that it has no finiteness assumption for the hard implication b) . Now you at least have precise statements... – Georges Elencwajg Jun 26 '11 at 13:45
Ok, thanks.. can I say $A_P=A\otimes_R R_P$ for all $P\in Spec(R)$? or something like that? – user11428 Jun 26 '11 at 16:04
Yes, the notation $A_P=A\otimes_R R_P$ is perfectly correct. – Georges Elencwajg Jun 26 '11 at 16:24
Perfect, thank you very much! – user11428 Jun 26 '11 at 18:01

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