Let $\mathcal{K}$ be a skew field such that there exists $p\geq 2$ prime with $$\underbrace{1+1+\ldots+1}_{p\text{ times}}=0$$ and for any $x\in\mathcal{K}$ there is a positive integer $n=n(x)$ so that $x^{p^n}\in Z(\mathcal{K})$. Prove that $\mathcal{K}$ is commutative.

Obviously we have to prove $\mathcal{K}=Z(\mathcal{K})$. One idea is that we can prove that $(x+1)^p=x^p+1$ or similar expressions with terms which commute, but I can't seem to find the right substitutions in these expressions.

  • $\begingroup$ What is your definition of a field? $\endgroup$ – Dbchatto67 Feb 11 at 10:33
  • $\begingroup$ The English-language standard includes commutativity in the axioms for a field. You're clearly not talking about that - to clarify, do you mean what is known as a "skew field"/"division ring", a ring in which all nonzero elements have multiplicative inverses? $\endgroup$ – jmerry Feb 11 at 10:33
  • $\begingroup$ Yes, that is correct. A skew field $\endgroup$ – Andrew V Feb 11 at 10:42
  • $\begingroup$ Already, the case when $K$ is finite is not quite easy (and known as Wedderburn's theorem). So, you should probably start there. $\endgroup$ – Mohan Feb 11 at 14:25
  • $\begingroup$ It should be skew-field in the title, not field. $\endgroup$ – Dietrich Burde Feb 11 at 15:21

For $a\in K$, define $\delta_a(x)= xa-ax$, put $\delta_a^1=\delta_a$ and further $\delta^k_a(x)=\delta_a(\delta^{k-1}_a(x))$.

Claim 1. $\displaystyle \delta^k_a(x)= \sum_{i=0}^k(-1)^i{k\choose i}a^ixa^{k-i}$.

Proof. Directly by induction on $k$. $\square$

Claim 2. (a) $\delta_a(x+y)=\delta_a(x)+\delta_a(y)$;

(b) $\delta_a(xy)= x\delta_a(y)+\delta_a(x)y$;

(c) for $x\neq 0$, $\delta_a(x)=0$ iff $\delta_a(x^{-1})=0$.

Proof. (a) Obvious. (b) $\delta_a(xy)= xya-axy= xya-xay+xay-axy= x\delta_a(y)+\delta_a(x)y$. (c) $\delta_a(x)=0$ iff $xa=ax$ iff $x^{-1}a=ax^{-1}$ iff $\delta_a(x^{-1})=0$. $\square$

Assume that $K\neq Z(K)$ and let $a\in K\smallsetminus Z(K)$. By the assumption $a^{p^n}\in Z(K)$ for some $n$, and let $n$ be the least such. Without loss, by changing $a$ with $a^{p^{n-1}}$, we may assume that $n=1$, i.e. $a^p\in Z(K)$. Since $a\notin Z(K)$, we can find $x$ such that $\delta_a(x)\neq 0$. By Claim 1, since the characteristic is $p$, $\delta_a^p(x)= xa^p-a^px$, so $\delta_a^p(x)=0$ because $a^p\in Z(K)$. Choose $k$, $1\leqslant k<p$, such that $\delta_a^k(x)\neq 0$ and $\delta_a^{k+1}(x)=0$. Let $y=\delta_a^{k-1}(x)\delta_a^k(x)^{-1}$. By Claim 2(b,c), $\delta_a(y)= \delta_a^{k-1}(x)\delta_a(\delta_a^k(x)^{-1})+\delta_a(\delta_a^{k-1}(x))\delta_a^k(x)^{-1}= \delta_a^{k-1}(x)\cdot 0+\delta_a^{k}(x)\delta_a^k(x)^{-1}=1$. Then $\delta_a(ya)= y\delta_a(a)+\delta_a(y)a=a$, i.e. $ya^2-aya=a$. Then $ya^2=a+aya$, so $a^{-1}(ya)a=1+ya$. By the assumption, for some $m$, $(ya)^{p^m}\in Z(K)$, so $a^{-1}(ya)^{p^m}a= (a^{-1}(ya)a)^{p^m}= (1+ya)^{p^m}= 1+(ya)^{p^m}$, and since $(ya)^{p^m}\in Z(K)$, we have $(ya)^{p^m}=1+(ya)^{p^m}$. A contradiction.


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