Does there exist an infinite non-commutative division ring with finite center?
1 Answer
Yes.
The following construction is attributed to Hilbert in "A first course in noncommutative rings" by T.Y. Lam., p. 217.
Let $k$ be a field with an automorphism $\sigma$. Write $D = k((x,\sigma))$ for the noncommutative ring of formal Laurent series $\sum_{i = n}^\infty a_ix^i$ with twisted multiplication rule $xa = \sigma(a) x$ for $a\in k$. Then $D$ is a division ring. If $k_0$ is the fixed field of $\sigma$, then either
$Z(D) = k_0$ or $Z(D) = k_0((x^s))$
depending on whether $\sigma$ has infinite order or finite order $s$, respectively.
Thus all we need to do is choose $\sigma$ and $k$ such that $k_0$ is a finite field and $\sigma$ has infinite order. This can be done by choosing say the Frobenius automorphism of $\overline{\mathbb{F}}_p$.