Let $D$ be a division ring. Show that if every $a \in D$ is algebraic over the prime subfield of $D$ then $D$ is commutative

Question

Let $D$ be a division ring. Show that if every $a \in D$ is algebraic over the prime subfield of $D$ then $D$ is commutative ($D=Z(D)$).

This conclusion would imply that any finite dimensional division algebra over $\mathbb{Q}$ is commutative, which is absurd. — Phil. Z, Nov 25 '17 at 09:07
Hint :Jacobson says that for every $a\in R$ there exists an $n\in \mathbb{N}$ such that $a^{n}=a$, then $R$ is commutative. — 1ENİGMA1, Nov 25 '17 at 09:44
Chiming in with others. This is extremely and absurdly wrong. For example all the elements of the non-commutative division rings listed in the answers in this thread are algebraic over the prime subfield. — Jyrki Lahtonen, Nov 25 '17 at 11:56

score 2 · Answer 1 · answered Nov 25 '17 at 11:47

2

A counterexample is $\mathbb Q(i,j,k)$ inside the quaternions. (Use that every pure quaternion has a real square.)

answered Nov 25 '17 at 11:47

Bart Michels

1 Answers1