# All subrings of ring are division rings, then is a field.

My professor put this problem in a list of $$200$$ problems for the examen:

A ring $$R$$ is called periodic if for each $$x\in R$$ exist $$n\geq 2$$ such that $$x^n=x$$. Starting from the Jacobson Theorem (All periodic rings are commutative) conclude that:

• Any finite division ring is a field.
• If $$R$$ is a ring such that any subring is a division ring, then $$R$$ is a field.

The first one is the Wedderburn theorem, and I found proofs in books and in internet, but the second one I don't know how to proceed with it. I would appreciate if anyone can give me a hint or a reference to read the proof.

• I'm no expert, but for any $x\in R$, can't you look at a subring containing $x$, it's a division ring so $x^{-1}$ is in the subring, and is therefore in $R$? I'm unsure if there's a similar notion to "principal ideal generated by $x$" for subrings, but if there is it should be that easy.
– Mark
Jan 17, 2017 at 4:58
• @Mark: It is trivial that $R$ is a division ring (it is a subring of itself). The tricky part is to show it is commutative. Jan 17, 2017 at 4:59

Hint: First show $R$ must have positive characteristic. Then show that every element of $R$ must be algebraic over the prime field. Now conclude that $R$ is periodic.
Since $R$ is a subring of itself, it is a division ring. If $R$ had characteristic $0$, $\mathbb{Z}$ would be a subring, but $\mathbb{Z}$ is not a division ring. Thus $R$ has characteristic $p$ for some $p>0$. If $x\in R$ were transcendental over $\mathbb{F}_p$, then $\mathbb{F}_p[x]\subset R$ would be a polynomial ring, which is not a division ring. Thus every element of $R$ is algebraic over $\mathbb{F}_p$. In particular, the subring generated by any single element of $R$ is finite, so any nonzero element has finite order. It follows that that $R$ is a periodic ring, and hence commutative by Jacobson's theorem.
• Sorry, I wrote it wrong. What I meant was that if it has zero characteristic, then if I consider an element of the ring, the generated ring by it will be infinite, so that can't happen because of the finiteness of $R$. Jan 17, 2017 at 5:35
• The ring $R$ is not assumed to be finite anywhere. The prime field of a given characteristic is the minimal field of that characteristic, i.e. either $\mathbb{Q}$ or $\mathbb{F}_p$. Jan 17, 2017 at 6:12