# 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!

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I think this 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. -- Also, unless your algebras are all commutative, you should write "A-bimodule" instead of "A-module". –  darij grinberg Aug 2 '11 at 10:48
(PS. Posting it as a comment, because I feel that having an openly accessible complete proof would be of more use.) –  darij grinberg Aug 2 '11 at 10:49

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.

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Ok! Thank you very much, darij grinberg! Firstly, sorry for my bad redaction. You're completely right, I should have written (A,A)-bimodule. Those are our structures. (I continue the comment in another post). –  muestrass Aug 6 '11 at 12:08
My next question is... I see clear the proof you give using that $ta=at, \forall a\in A$. The characterization that I know of a separable algebra A is in terms of the existence of an element $t\in A$ such that $\mu(t)=1$ (\mu:A^e=A\otimes A^{op}\rightarrow A) and $Jt=0$, where $J=Ker(\mu)$. I have read in many web pages that, in A separable algebra, $ta=at, \forall a\in A$, but nowadays I don't know why (yes, probably I'm blind!). So, I would like what happen with that condition. Thank you for your answers. –  muestrass Aug 6 '11 at 12:12
PD.: I want to vote up the three answers obtained until now (the yours and the Charles one, but I don't know how to do it!) –  muestrass Aug 6 '11 at 12:12
First of all, you should have written $t\in A\otimes A$ rather than $t\in A$. Second, your new question boils down to proving that $Jt=0$ (where $J=\mathrm{Ker}\mu$) is equivalent to $ta=at$ for all $a\in A$. To prove this, notice that $J$ is generated, as a left $A$-module, by the set $\left\lbrace a\otimes 1-1\otimes a\mid a\in A\right\rbrace$ (should I give a proof for that, or do you remember it now?). Hence, $Jt=0$ is equivalent to $\left(a\otimes 1-1\otimes a\right)t=0$. Now, it is important to know that $\left(a\otimes 1-1\otimes a\right)t$ does not mean the product of ... –  darij grinberg Aug 6 '11 at 19:18
... $a\otimes 1-1\otimes a$ with $t$ in $A\otimes A$, but the product of $a\otimes 1-1\otimes a$ with $t$ in $A\otimes A^{\mathrm{op}}$. (In fact, $J$ was considered as a submodule of $A\otimes A^{\mathrm{op}}$, not $A\otimes A$; otherwise your definition of "separable" would be wrong!) In terms of the $A$-$A$-bimodule structure on $A\otimes A$, this product is simply $at-ta$. –  darij grinberg Aug 6 '11 at 19:20

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

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Thank you for your answer, Charles, I'm just trying to find that book! –  muestrass Aug 6 '11 at 12:06