Let $A \subseteq B$ be commutative noetherian rings.
I have found the following claim: **"Separability implies smoothness"** with the following explanation:
"The natural thing is to prove that a separable algebra is a regular homomorphism (and so smooth if it is of finite type). In general ...".

Remark: I am only interested in finitely generated algebras, so did not quote the rest of the explanation.

However, it seems to me, at least with my definitions, that this claim is false:
If $A \subseteq B$ is separable, then (assuming the claim is true) $A \subseteq B$ is smooth, then by a result of Grothendieck, $A \subseteq B$ is flat, but this seems too much (especially in view of 
[Examples on pages 95-97][1] and [Corollary 9][2]; I do not see why separability should imply flatness of $B$ over $A$. Notice that actually, this is what the explanation suggests: it says that separability implies $A \to B$ is a regular homomorphism, and by definition, a regular homomorphism is flat).

I will really appreciate if one can either post a sketch of proof/tell me where I can find a proof, or a counterexample. 

Actually, I have already posted this question [here][3], but got no comments thus far.


  [1]: http://link.springer.com/chapter/10.1007%2F978-94-015-8555-2_5
  [2]: http://ac.els-cdn.com/0021869380902331/1-s2.0-0021869380902331-main.pdf?_tid=73b8ca36-2846-11e5-8b8e-00000aab0f6b&acdnat=1436672055_edd3c1b73a42fb02541d5db701287e09
  [3]: http://math.stackexchange.com/questions/1358096/separability-implies-smoothness