TITLE: Why must a field whose a group of units is cyclic be finite?
QUESTION [21 upvotes]: Let $F$ be a field and $F^\times$ be its group of units. If $F^\times$ is cyclic, then show that $F$ is finite.

I'm a bit stuck. I know that I can represent $F^\times = \langle u \rangle$ for some $u \in F^\times$ and that we must have that $|F^\times| = o(u)$, where $o(u)$ denotes the order of $u$ in $F^\times$.  I tried assuming $o(u) = \infty$, but I'm not sure exactly where to go from there.
I was wondering if I could get a hint.

REPLY [5 votes]: Such a field cannot have characteristic $0$, since then $\mathbb{Q}^{\times}$ would be a subgroup of the multiplicative group.
If the field is of characteristic $p$, then the embedding of $\mathbb{F}_p$ makes the field an $\mathbb{F}_p$-algebra, and so it must be a quotient of $\mathbb{F}_p[x]$ (the free $\mathbb{F}_p$-algebra on one generator). But every quotient of $\mathbb{F}_p[x]$ by a nonzero ideal is finite. Therefore, such a field does not exist.