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In my book, I encountered the following problem:

Prove that any division ring of characteristic $0$ contains a division subring which is isomorphic to $\mathbb{Q}.$

I looked this up online and found a similar property, except that instead of division (sub)ring it was about a (sub)field. I tried to recreate the proof I found here, but I didn't succeed because I lacked the commutative property a field has, and a division ring doesn't.

Now I'm wondering if this problem is true at all, or the authors simply meant "field" instead of "division ring".

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    $\begingroup$ The center of the division ring is a field of characteristic zero hence contains a subfield isomorphic to the field of rationals. $\endgroup$
    – Fabio Lucchini
    Commented Sep 8, 2018 at 14:01

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I'm going to assume rings have a unit. Then your division ring $D$ has a subring $R$ which is isomorphic to $\mathbb{Z}$. Why? Define $f: \mathbb{Z} \to D$ as $f(n) = n \cdot 1$ which is injective. So the image of $f$ is our desired subring. But now we have a division ring, which means that there must be in fact a subring isomorphic to $\mathbb{Q}$.

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