Show that if $R$ is a division ring and $S$ is a subring of $R$ then $S$ is also a division ring.

Question

I'm trying to do problem 7.1.7 in Dummit and Foote.

Problem: Prove that the centre of a ring $R$ is a subring of $R$ that contains the identity. Prove that the center of a division ring is a field.

I've already shown that $Z(R)$ is a subring, but my issue here is that I would like to show that if $S \subseteq R$ is a subring of a division field then $S$ must itself be a division ring. This would then show that the centre of a division ring is a field since it will be a commutative division ring. However I don't seem to be able to finish it off even though it seems basic. I know that an additive subgroup of $R$ that is also closed under multiplication and contains the multiplicative identity, but I can't seem to leverage this to show that any element in $S$ has a multiplicative inverse. As I'm sure it's basic a hint would be appreciated, thanks in advance for your help.

Answer 1

It is not true that any subring of a division ring is itself a division ring. Consider the integers as a subring of the rationals.

Answer 2

The key to the problem you ask in the body (not the title) is that if $x \in Z(R)$ and $x^{-1} \in R$, then $x^{-1} \in Z(R)$ also. Notice for any $r \in R$ and $x \in Z(R)$, we have $xr = rx$ so $$x^{-1}(xr)x^{-1} = x^{-1}(rx)x^{-1},$$ and, by re-associating and simplifying, we get $rx^{-1}=x^{-1}r$ for all $r \in R$. Thus $x^{-1} \in Z(R)$.

Thus $Z(R)$ is a commutative subring of $R$ closed under taking inverses and is thus a commutative division ring, i.e. a field.