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

Question

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.

Answer 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.

Answer 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.

Answer 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.

Answer 4

Here's a suggestion : considering $u-1$, show that $u$ is algebraic over the caracteristic subfield (that is, the subfield generated by $1$). You then only have to rule out the case that the caracteristic subfield is $\Bbb Q$.

Unless $F=\Bbb Z/2\Bbb Z$ (and is thus finite), we must have $u-1\in F^*$ so that there is an integer $n\in \Bbb Z$ with $u-1=u^n$ (necessarily$n\neq 0,1$). From this it follows that $u$ annihiliates a polynomial with coefficients in the caracteristic subfield $\Bbbk$ (if $n<0$, just multiply the equation by a large power of $u$), that is, $u$ is algebraic. Thus $F=\Bbbk[u]$ is finite dimensional over $\Bbbk$. If $\Bbbk$ is finite we are done. But if it weren't, we'd have $\Bbbk=\Bbb Q$ and $\Bbb Q^*$ would be cyclic as a subgroup of the cyclic group $F^*$, which it isn't. Therefore there exists some prime number $p$ with $\Bbbk=\Bbb Z/p\Bbb Z$ and $F$ is finite.

One can directly show that the caracteristic $p$ of $F$ is nonzero. The case $p>2$ is easily dealt with, and only the case $p=2$ remains

Indeed, $\Bbb Q^*$ isn't cyclic, which prohibits it form being a subgroup of $F^*$, thus $p\neq 0$. If $p\geq 3$, then there exists $n$ with $u^n=2$. Since $2\in\Bbbk^*$, we have $2^{p-1}=1$, i.e. $u^{(p-1)n}=1$ and $F^*$ is finite. $F$ is finite because $F=F^*\cup\lbrace 0\rbrace$. So the argument above (considering $u-1$) is only necessary for the case $p=2$.