An R-algebra A is R-separable if and only if all derivations are inner. Hello everybody. 
I'm readying about derivations. It is very very known fact that all derivations $\delta: A\rightarrow M$ (A R-algebra, M A-module) are inner when the algebra is R-separable. 
Someone knows some reference to see the proof, please?
Thank you!
 A: Unless your algebras are all commutative, you should write "A-bimodule" instead of "A-module". So the correct result is:
Proposition. Let $A$ be a unital $R$-algebra (where $R$ is a commutative ring). Then, $A$ is a separable $R$-algebra if and only if every derivation from $A$ to an $A$-$A$-bimodule is inner.
Proof of the Proposition: $\Longrightarrow$: Assume that $A$ is a separable $R$-algebra. Then, there exists a $t\in A\otimes A$ (where all tensor products are over $R$) satisfying $\mu\left(t\right) = 1$ (where $\mu : A \otimes A \to A$ is the multiplication morphism of the $R$-algebra $A$) and $at=ta$ for every $a\in A$ (where we are using the standard $A$-$A$-bimodule structure on $A\otimes A$). Now, let $M$ be an $A$-$A$-bimodule, and $d: A\to M$ a derivation. Since $t\in A\otimes A$ is a tensor, we can write it in the form $t=\sum\limits_{i=1}^n t_i\otimes s_i$ for some $n\in\mathbb N$ and some $t_1,t_2,...,t_n\in A$ and $s_1,s_2,...,s_n\in A$. Then, $\mu\left(t\right)=\sum\limits_{i=1}^n t_i s_i$, so that $\mu\left(t\right)=1$ becomes $\sum\limits_{i=1}^n t_i s_i = 1$. On the other hand, every $a\in A$ satisfies $at=ta$. Since $t=\sum\limits_{i=1}^n t_i\otimes s_i$, this rewrites as $\sum\limits_{i=1}^n at_i\otimes s_i = \sum\limits_{i=1}^n t_i\otimes s_ia$. Applying the map $d\otimes \mathrm{id}$ to this equation, we get $\sum\limits_{i=1}^n d\left(at_i\right)\otimes s_i = \sum\limits_{i=1}^n d\left(t_i\right)\otimes s_ia$. Applying the action map $A\otimes M\to M,\ a\otimes m\mapsto am$ to this equation, we get $\sum\limits_{i=1}^n d\left(at_i\right)s_i = \sum\limits_{i=1}^n d\left(t_i\right)s_ia$. Thus,
$0 = \sum\limits_{i=1}^n d\left(at_i\right)s_i - \sum\limits_{i=1}^n d\left(t_i\right)s_ia$
$= \sum\limits_{i=1}^n \left(\underbrace{d\left(at_i\right)}_{=d\left(a\right)t_i+ad\left(t_i\right)\text{ (since }d\text{ is a derivation)}}s_i - d\left(t_i\right)s_ia\right)$
$= \sum\limits_{i=1}^n \left(d\left(a\right)t_is_i + ad\left(t_i\right)s_i - d\left(t_i\right)s_ia\right)$
$= d\left(a\right) \underbrace{\sum\limits_{i=1}^n t_is_i}_{=1} + a\sum\limits_{i=1}^n d\left(t_i\right)s_i - \sum\limits_{i=1}^n d\left(t_i\right)s_i a$
$= d\left(a\right) + a\sum\limits_{i=1}^n d\left(t_i\right)s_i - \sum\limits_{i=1}^n d\left(t_i\right)s_i a$.
Hence, $d\left(a\right) = - a\sum\limits_{i=1}^n d\left(t_i\right)s_i + \sum\limits_{i=1}^n d\left(t_i\right)s_i a$. In other words, $d\left(a\right) = au-ua$ where $u = -\sum\limits_{i=1}^n d\left(t_i\right)s_i$. This shows that $d$ is an inner derivation. We have thus proven that every derivation from $A$ into an $A$-$A$-bimodule is inner. The $\Longrightarrow$ direction of the Proposition is now shown.
$\Longleftarrow$: Assume that every derivation from $A$ into an $A$-$A$-bimodule is inner. Let $\mu : A \otimes A \to A$ be the multiplication morphism of the $R$-algebra $A$. Consider the $A$-$A$-bimodule $A\otimes A$; then, $\mathrm{Ker}\mu$ is a sub-bimodule of $A\otimes A$ (since $\mu$ is an $A$-$A$-bimodule map, as can be easily seen). Consider the map $\delta:A\to \mathrm{Ker}\mu,\ a\mapsto a\otimes 1-1\otimes a$. This map $\delta$ is a derivation (as can be easily shown by computation), so it is inner (by the assumption that every derivation from $A$ into an $A$-$A$-bimodule is inner). This means that there exists some $u\in \mathrm{Ker}\mu$ such that $\delta\left(a\right)=au-ua$ for every $a\in A$. Consider this $u$. Let $t=1\otimes 1-u$. Then, $\mu\left(u\right)=0$ (since $u\in\mathrm{Ker}\mu$) and $\mu\left(1\otimes 1\right)=1$ yield $\mu\left(t\right)=1$. On the other hand, every $a\in A$ satisfies
$at-ta = a\left(1\otimes 1-u\right)-\left(1\otimes 1-u\right)a$  (since $t=1\otimes 1-u$)
$= \left(\underbrace{a\left(1\otimes 1\right)}_{=a\otimes 1}-\underbrace{\left(1\otimes 1\right)a}_{=1\otimes a}\right) - \underbrace{\left(au-ua\right)}_{=\delta\left(a\right)=a\otimes 1-1\otimes a}$
$= \left(a\otimes 1-1\otimes a\right) - \left(a\otimes 1-1\otimes a\right) = 0$,
so that $at=ta$.
Thus there exists a $t\in A\otimes A$ such that $\mu\left(t\right)=1$ and such that $at=ta$ for every $a\in A$. This means that the $R$-algebra $A$ is separable. This proves the $\Longleftarrow$ direction of the Proposition. The Proposition is thus shown.
Note that a fact slightly stronger than our above proposition (by virtue of holding for nonunital $R$-algebras as well) is Theorem 5 in: Gerhard Hochschild, On the Cohomology Theory for Associative Algebras, The Annals of Mathematics, Second Series, Vol. 47, No. 3 (Jul., 1946), pp. 568-579. Notice that his algebras are not necessarily unital, and that he defines "separable" by "the first Hochschild cohomology vanishes" (i. e. "all derivations are inner"). But Theorem 5 shows that his version of separability is equivalent to what is nowadays considered one of the definitions of separability.
A: You should take a look at Théorie de la descente et algèbres d'Azumaya, M.-A. Knus and M. Ojanguren, Théorème 1.4 p 73-74.
This contains all the criteria you need and the non-commutative/commutative case
