Jacobson's formula ($p$ -powers and Lie brackets)

The treatment here essentially follows [1].

Let $p$ be a prime number. Let $A$ be a (not necessarily commutative, but unital, associative) algebra over $\mathbb{F}_p$ . We may also regard $A$ as a Lie algebra over $\mathbb{F}_p$ , the bracket product being the ordinary commutator. We would like to obtain a formula for

$$(a+b)^p$$

for $a,b\in A$ with the help of Lie brackets. To do that, we first introduce an transcendent element (=variable) $T$ which commutes with any element of $A$ . Then we expand $(T a+b)^p$ in terms of $T$ . Namely,

$$(T a+b)^p=T^p a^p+ b^p +\sum_{j=1}^{p-1} T^j s_j(a,b)$$ (1)

where $s_j(a,b)$ is a non-commutative polynomial in $a,b$ . Our task then is to find a nice formula for $s_j$ .

A first thing to do is to differentiate the equation (1) by $T$ .

$$\sum_{k=0}^{p-1} (T a + b)^k a (T a +b)^{p-1-k} =\sum_{j=1}^{p-1} j T^{j-1} s_j(a,b).$$ (2)

To compute the left hand side, we use a nice trick. For any element $x\in A$ , we denote by $\lambda(x)$ (respectively, $\rho(x)$ ) an operator defined by the left multiplication (respectively, the right multiplication) of $x$ . That is,

$$\lambda(x): A \ni f \mapsto x f \in A,$$

$$\rho(x): A \ni f \mapsto f x \in A.$$

It should be noted that for all $x,y\in A$ , $\lambda(x)$ and $\rho(y)$ always commute. Indeed, we have

$$\lambda(x)\rho(y)z=x(zy)=(xz)y=\rho(y)\lambda(x).$$

So from an ordinary result on commutative algebra, we have

$$(\lambda(a)-\rho(a))^p=\lambda(a)^p -\rho(a)^p=\lambda(a^p)-\rho(a^p).$$

We define $\operatorname{ad}(a)$ to be

$$\operatorname{ad}(a)=\lambda(a)-\rho(a).$$

Then the equation above may be written as

$$(\operatorname{ad}(a))^p=\operatorname{ad}(a^p).$$

Another interesting formula is the following.

$$(\lambda(a)-\rho(a))^{p-1}=\sum_{j=0}^{p-1} \lambda(a)^j \rho(a)^{p-1-j}$$

(To verify that it holds, we notice that for any commutative variable $T,U$ , an identity

$$(T-U)^p= T^p-U^p=(T-U)\sum_{j=0}^{p-1} T^j U^{p-1-j}$$

holds.) The above formula may then be rewritten as

$$(\operatorname{ad}(a))^{p-1}(b)=\sum_{j=0}^{p-1} a^j b a^{p-1-j}.$$

By suitable substitutions, we thus have

$$(\operatorname{ad}(T a+b ))^{p-1}(a)=\sum_{j=0}^{p-1} (T a+ b)^j a (T a+b)^{p-1-j}.$$ (3)

comparing the equations (2) and (3), we see that each $s_j(a,b)$ belongs to the Lie sub algebra of $A$ generated by $a,b$ .

To sum up, we have obtained the following proposition.

PROPOSITION 6.2 (Jacobson's formula)   Let $p$ be a prime number. Let $A$ be an algebra over $\mathbb{F}_p$ (which is not necessarily commutative, but unital, associative as we always assume.) Then for any elements $a,b\in A$ , we have

$$(a+b)^p=a^p+b^p+\sum_{j=1}^{p-1} s_j(a,b)$$

where $s_j$ is a universal polynomial in $a,b$ given in the following manner.

$$s_j(a,b)=(1/j)\operatorname{coeff}((\operatorname{ad}(T a+b))^{p-1}a,T^{j-1}).$$

(Here, $\operatorname{coeff}(\bullet, T^j)$ denotes the coefficient of $T^j$ in $\bullet$ .) In particular, $s_j(a,b)$ belongs to the Lie algebra generated by $a,b$ .

An important corollary is the following.

COROLLARY 6.3   Let $p$ be a prime number. Let $A,B$ be algebras over a ring $R$ of characteristic $p$ . Let $L$ be a Lie subalgebra of $A$ (that means, $R$ -submodule which is closed under commutators.) Let $\phi: A\to B$ be a $R$ -linear map and assume that

$$[\phi(x),\phi(y)]=\phi([x,y])$$

holds for any $x,y\in L$ . Let us define a map $\psi:L\to B$ by

$$\psi(x)=\phi(x)^p-\phi(x^p).$$

Then the map $\psi$ is $p$ -linear. That means, $\psi$ satisfies

$$\psi(f x+g y)=f^p \psi(x)+g^p \psi(y) \qquad (\forall x,y \in L, \forall f,g \in R).$$

PROOF.. The only thing which needs to be verified is the additivity of $\psi$ .

$$\psi(x+y)$$

$$= (\phi(x)+\phi(y))^p-\phi((x+y)^p)$$

$$= \left(\phi(x)^p+\phi(y)^p+\sum_j s_j(\phi(x),\phi(y))\right)$$

$$-\left (\phi(x^p)+\phi(y)^p+\sum_j \phi(s_j(x,y))\right)$$

$$= \psi(x)+\psi(y)$$

$$+ \left (\sum_j (s_j(\phi(x),\phi(y))-\phi(s_j(x,y))\right)$$

$$= \psi(x)+\psi(y)$$

(Note that each $s_j$ commutes with $\phi$ because it is built only with commutators.) $\qedsymbol$