Lemma 1

Let D be a field and Z its center. Denote the characteristic exponent of Z by $p$. Suppose that for every element $a$ of D, there exists an integer $m\geqslant 0$ such that $a^{p^m}$ belongs to Z. Then the field D is commutative.

If $p= 1$, then D = Z. We therefore assume $p >1$.

Let us give a proof by contradiction and suppose that D is not commutative. Let $a$ be an element of D- Z and $q$ a power of $p$ such that $a^q$ belongs to Z. Denote the identity mapping on D by I and the inner automorphism $x\mapsto axa^{-1}$ from D to D associated with $a$ by $\sigma$; we have $\sigma^q= I$ because $a^q$ belongs to Z. We have $\sigma -I\not= 0$ because $a$ does not belong to Z, and we have $(\sigma -I)^q=\sigma^q-I = 0$ because Z has characteristic $p$. Let $f$ be the greatest natural number such that $(\sigma -I)^f\not= 0$; we have $f\geqslant 1$. Choose an element $c$ of D such that $(\sigma -I)^f(c)\not= 0$, and set

By construction, we have $y\not= 0$ and $\sigma (y) =y$; if we set $z=y^{-1}x$, then it follows that

and therefore $\sigma (z^{p^j}) = 1 +z^{p^j}$ for every natural number $j$. Choose an integer $m\geqslant 0$ such that $z^{p^m}$ belongs to the center Z of D; we have

This contradiction implies Lemma 1.