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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 ($D=Z(D)$).

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closed as off-topic by Xam, Stefan4024, quid Nov 25 '17 at 21:53

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    $\begingroup$ This conclusion would imply that any finite dimensional division algebra over $\mathbb{Q}$ is commutative, which is absurd. $\endgroup$ – Phil. Z Nov 25 '17 at 9:07
  • $\begingroup$ 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. $\endgroup$ – 1ENİGMA1 Nov 25 '17 at 9:44
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    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Nov 25 '17 at 11:56
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A counterexample is $\mathbb Q(i,j,k)$ inside the quaternions. (Use that every pure quaternion has a real square.)

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