2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.