Why must a field whose a group of units is cyclic be finite?

Solution 1:

As $F=F^\times\cup\{0\}$, the field $F$ is finite if and only if $F^\times$ is finite. The real question here is not "can the field be finite if its group of units is cyclic," but "can the field be infinite?"

Suppose $F^\times=\langle g\rangle$ is infinite generated by $g$. The characteristic must be $2$, since otherwise the element $-1\ne1$ has order $2$, but there is no element of order $2$ in $\Bbb Z$. Clearly $F=\Bbb F_2(g)$, by noting the containment goes both ways. If $g$ is transcendental then $g$ and $g+1$ are multiplicatively independent, so $F^\times$ must be generated by more than one element, a contradiction. If $g$ is algebraic over $\Bbb F_2$, then $\Bbb F_2(g)$ is finite-dimensional over $\Bbb F_2$, hence $F$ is finite, another contradiction.

Solution 2:

Hints: Let $u$ be a generator for the multiplicative group $F^*$

• If $\mathrm{char}F\neq2$, then $-u\neq u$ is a unit, and hence $-u=u^t$ for some integer $t$. Show that this implies that the order of $u$ is finite (Seth has an even better idea here).
• If $\mathrm{char}F=2$, then you can show that the even powers of $u$ form a subfield $K$ (think: Frobenius automorphism). Unless $u$ is of a finite odd order this subfield is a proper one. If $K$ is infinite, then it easily follows ($F$ is a vector space of dimension $2$ over $K$ and in $K^2$ there are infinitely many distinct lines through the origin) that $2=[F^*:K^*]=\infty$ which is absurd.

Solution 3:

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.