Derivations of central simple algebras

Posted: February 2, 2011 in Noncommutative Ring Theory Notes, Simple Rings
Tags: , , ,

We will assume again that $k$ is a field. We will denote by $Z(A)$ the center of a ring $A.$ The following result is one of many nice applications of the Skolem-Noether theorem (see the lemma in this post!). For the definition of derivations and inner derivations of a $k$-algebra see Definition 2 in this post.

Theorem. Every derivation of a finite dimensional central simple $k$-algebra $A$ is inner.

Proof. First note that, since $A$ is simple, $M_2(A)$ is simple. We also have $Z(M_2(A)) \cong Z(A) = k.$ Finally $\dim_k M_2(A) = 4 \dim_k A < \infty.$ Thus $M_2(A)$ is also a finite dimensional central simple $k$-algebra. Now, let $\delta$ be a derivation of $A.$ Define the map $f: A \longrightarrow M_2(A)$ by

$f(a) = \begin{pmatrix} a & \delta(a) \\ 0 & a \end{pmatrix},$

for all $a \in A.$ Obviously $f$ is $k$-linear and for all $a,a' \in A$ we have

$f(a)f(a') = \begin{pmatrix} aa' & \delta(a)a'+a \delta(a') \\ 0 & aa' \end{pmatrix} = \begin{pmatrix} aa' & \delta(aa') \\ 0 & aa' \end{pmatrix} = f(aa').$

So $f$ is a $k$-algebra homomorphism and hence, by the lemma in this post, there exists $v \in M_2(A)$ such that $v$ is invertible and $f(a)=vav^{-1},$ for all $a \in A.$ Let

$v = \begin{pmatrix} x & y \\ z & t \end{pmatrix}.$

So $f(a)v=va$ gives us

$\begin{cases} ax+\delta(a)z=xa \\ ay+\delta(a)t=ya \\ az=za \\ at=ta. \end{cases} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*)$

Since $(*)$ holds for all $a \in A,$ we will get from the last two equations that $z,t \in Z(A)=k.$ Since we cannot have $z=t=0,$ because then $v$ wouldn’t be invertible, one of $z$ or $t$ has to be invertible in $k$ because $k$ is a field. We will assume that $z$ is invertible because the argument is similar for $t.$ It now  follows, from the first equation in $(*)$ and the fact that $z^{-1} \in Z(A),$ that

$\delta(a)=(xz^{-1})a - a(xz^{-1}). \ \Box$