# Infinite subfields of division algebras with finite center

Assume we are given a non-commutative division algebra $D$ over a finite field $\mathbb F_q$, with the center of $D$ equal to $\mathbb F_q$. Clearly $D$ must be infinite dimensional over its center.

Does $D$ necessarily contain an element of infinite multiplicative order? I.e. an element $x$ such that $x^n\neq 1$ for all positive integers $n$ ?

Any hints on a proof or a counterexample would be appreciated.

If there are examples where all elements have finite order, I wonder about the following weaker question:

Does $D$ contain an infinite subfield? (Here subfield of course means that it must be commutative.)

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It is not possible that $D^{\times}$ is a torsion group unless $D$ is a field.

Let $D$ be a division ring of characteristic $p$. Denote the center of $D$ by $Z(D)$.

Lemma 1: Every finite subgroup of $D^{\times}$ is commutative.

Proof: The $\mathbb F_q$-algebra generated by the subgroup is a finite subring. One easily sees that it has to be a subfield. Indeed, multiplication by non-zero elements is injective and hence surjective since the ring is finite. This implies that every non-zero element has a multiplicative inverse. We conclude that the subfield (and hence the subgroup) is commutative by Wedderburns theorem. q.e.d.

Herstein’s Lemma: Suppose $a$ is a noncentral, torsion element of $D^\times$. Then, there exists $y \in D^{\times}$ such that $yay^{−1} = a^i \neq a$ for some $i\geq 2$.

You can find a proof in "A First Course in Noncommutative Rings" by T.Y. Lam.

Theorem: Let $D$ be a division ring such that for any $a\neq 0$, there exists $n\geq 1$ such that $a^n = 1.$ Then, $D$ is a field.

Proof: Assume that $D$ is not a field. Let $0 \neq a \not \in Z(D)$. By Herstein's Lemma, there exists $y \in D^{\times}$ such that $yay^{-1} \neq a$. Since $y$ is torsion and normalizes the subgroup generated by $a$, $G=\langle a,y \rangle$ is a finite subgroup of $D^{\times}$. By Lemma 1 the group $G$ is commutative. But $ya \neq ay$ and we get a contradiction. q.e.d.

It is even enough to assume that all additive commutators are of finite order. This is sometimes called Jacobson’s commutativity theorem. A proof can also be found in the book by T.Y. Lam.

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Excellent, thank you very much! –  Max Horn Dec 4 '10 at 20:38