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.

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 hedefines"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. -- Also, unless your algebras are all commutative, you should write "A-bimodule" instead of "A-module". – darij grinberg Aug 2 '11 at 10:48