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

Question

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.

Answer 1

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.

A full proof is hidden below:

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.