Characterisation for separable extension of a field Can someone verify this for me.. or tell me what reference shows me this... is this true:

Let $k$ be a field. Then a field extension $K$ of $k$ is separable over $k$ iff for any field extension $L \supseteq k$ the Jacobson radical of the tensor product $K\otimes_k L$ is trivial.

I got this idea by looking at some definitions of separable algebras (which is not my field of research.. but somehow this definition got me intruiged). Anyone knows if this is true and why so? or maybe a reference or two about it?
 A: Let $k\subset K$ be a completely arbitrary extension of fields. This extension is said to be separable if equivalently
a) For all extensions $k\subset L$, the ring  $K \otimes _k L$ is reduced.  [A ring is reduced if $x^n=0 \Rightarrow x=0$] 
or
b) For some algebraically closed extension $k \subset \Omega$ , the ring $K\otimes _k \Omega$ is reduced.
If the extension $k\subset K$ is algebraic the above are equivalent to the more elementary definition
c) Every $x\in K$ is separable, i.e. the minimum polynomial of $x$ over $k$ is separable. [Which means that its roots are distinct in an algebraic closure $k^a$ of $k$].
This can all be found in Bourbaki's Algebra, chapter V.
A: You should take a look at Theorem 3.4 (p 85) of Farb and Dennis' book Noncommutative Algebra. The statement is:
Let $L/k$ be a finite extension of fields. Then $K\otimes _k L$ is semisimple for every field $K\supseteq k$ if and only if $L/k$ is a separable extension. 
That the tensor product is semisimple implies that its Jacobson radical vanishes. Conversely, any artinian ring with trivial Jacobson radical is semisimple. Therefore your equivalent formulation of separability is true.
