Prove a specific skew field K is commutative

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.

• What is your definition of a field? – Dbchatto67 Feb 11 at 10:33
• 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? – jmerry Feb 11 at 10:33
• Yes, that is correct. A skew field – Andrew V Feb 11 at 10:42
• Already, the case when $K$ is finite is not quite easy (and known as Wedderburn's theorem). So, you should probably start there. – Mohan Feb 11 at 14:25
• It should be skew-field in the title, not field. – 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 $$k$$, $$1\leqslant k, such that $$\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.