Why is $D$ of characteristic $p \gt 0$ in this proof?

Question

From Section $7.2$ of Herstein's "Topics in Algebra" ($2^{\text{nd}}$ edition):

THEOREM $7.2.2$ (JACOBSON) $\;$ Let $D$ be a division ring such that for every $a \in D$ there exists a positive integer $n(a) \gt 1$, depending on $a$, such that $a^{n(a)} = a$. Then $D$ is a commutative field.

$\;$ Proof. $\;$ If $a \neq 0$ is in $D$ then $a^{n} = a$ and $(2a)^{m} = 2a$ for some integers $n, m \gt 1$. Let $s = (n-1)(m-1) + 1$; $s \gt 1$ and a simple calculation shows that $a^{s} = a$ and $(2a)^{s} = 2a$. But $(2a)^{s} = 2^{s}a^{s} = 2^{s}a$, whence $2^{s}a = 2a$ from which we get $(2^{s} - 2)a = 0$. Thus $D$ has characteristic $p > 0$.

Why does it follow from $(2^{s} - 2)a = 0$ that $D$ has characteristic $p > 0$?

Answer 1

In short, from the $a=1_D$ case we know $(2^s - 2)1_D=0$ for some $s>1$. Since $2^s - 2 >0$, we can conclude that $D$ has a positive characteristic, which should be a prime number because $D$ is an integral domain.

Recall that, by definition, the characteristic of $D$ is $m \geq 0 $ if and only if $\ker \phi = (m)$ where $\phi : \mathbb{Z} \rightarrow D$ is the ring homomorphism given by $ n \mapsto \underbrace{1_D + 1_D + \cdots + 1_D}_{n \text{ times}}$. We often write $\underbrace{1_D + 1_D + \cdots + 1_D}_{n \text{ times}} = n1_D$, i.e. $D$ is considered to be a $\mathbb{Z}$-algebra. In this way every ring has a natural $\mathbb{Z}$-algebra structure.

Let's review the equality $(2^s - 2)1_D=0$. Here $2^s - 2 \in \mathbb{Z}$ and $1_D \in D$. Thus $2^s - 2$ will never be zero for $s>1$.

Answer 2

As $D$ is a division ring and $a \neq 0$, $a$ has a multiplicative inverse. From $(2^{s} - 2)a = 0$, we get $(2^{s} - 2)1= 0$ and as $s \gt 1$, that the characteristic of $A$ divides $2^s-2$.