Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$F$ is a field such that $F^*= F\setminus \{0\}$ is the cyclic multiplicative group. Prove that $F$ is a finite field.

I am struggling with this problem. We did not consider the material in class and I got stuck with this problem…

share|cite|improve this question
If the multiplicative group $F^*$ is cyclic, it has a generator $a\in F$. Play around with $a$ and see what you can find. Think of the prime subfield of $F$ (you know what that is?), and how it fits in with $a$. – Lubin Nov 19 '13 at 1:28

As another answer has mentioned, since the only cyclic groups are $\mathbb{Z}$ and $\mathbb{Z}_n$, the problem reduces to showing that there is no field $K$ with multiplicative group $\mathbb{Z}$. Every field has one of $\mathbb{Q},\mathbb{Z}_p$, $p$ prime, as a subfield. This means that one of $\mathbb{Q}^*,\mathbb{Z}_p^*$ must be a subgroup of $K^*\simeq\mathbb{Z}$.

The first case $\mathbb{Q}\subseteq K$ is impossible because every subgroup of $\mathbb{Z}$ is of the form $n\mathbb{Z}$, and in particular is cyclic, being generated by $n$. But $\mathbb{Q}^*$ is not cyclic, or even finitely generated! It's also impossible to get $\mathbb{Z}_p^*$ as a subgroup of $\mathbb{Z}$, simply because every element of $\mathbb{Z}_p^*$ has finite order, whereas no element of $\mathbb{Z}$ does other than $0$.

share|cite|improve this answer
This doesn't rule out $p=2$. – Derek Holt Nov 19 '13 at 9:07
Actually, I believe it does. I guess you're thinking of the multiplicative structure on the integers where $-1$ has order 2, but the infinite cyclic group is the integers under addition, so the proposed isomorphism $K^*\simeq\mathbb{Z}$ turns multiplication into addition. – Kevin Carlson Nov 19 '13 at 9:59

Well, if $F^*$ is cyclic, i.e. a unigenerated abelian group, we're in fantastic shape provided the group isn't $\mathbb{Z}$ (it's either that or $\mathbb{Z}/n\mathbb{Z}$, which is wonderfully finite).

So, what would happen if $\mathbb{F}^*$ were isomorphic to $\mathbb{Z}$ - this would mean that there's some element $x$ such that $x^i = x^j \Leftrightarrow i = j \in \mathbb{Z}$, and $F = 0 \cup x^i: i \in \mathbb{Z}$.

Could $F$ be characteristic 0? Well, we would have a map $\mathbb{Q} \rightarrow F$, and so we would have $2 = x^i, 3 = x^j$, extending to a map $\overline{\mathbb{Q} } \rightarrow \overline{F}$, where the bar denotes algebraic closure, we would have $x$ is some $\zeta_i 2^{1/i}$ and $\zeta_j 3^{1/j}$, where $\zeta_i$ is an $i^{th}$ root of unity; raising both sides to the $(ij)^{th}$ power, we have $2^j = 3^i$, a contradiction.

Could $F$ be characteristic $p$? Well, in all characteristics, we would know that $x + x^2 = x^i$, whence $F$ would be a unigenerated algebraic field extension of $\mathbb{F}_p$, whence finite.

share|cite|improve this answer

Let $a$ generate $F^*$. Assume $F$ is not finite. Then, $\langle a \rangle \cong \mathbb{Z}$. Since $\mathbb{Z}$ has no torsion, we have $1 = -1$. Thus, $\operatorname{char} F = 2$.

Since $a \neq 1$, $a + 1 = a - 1 \neq 0$ so $a + 1 = a^i$ for some $i \neq 1$. If necessary we can replace $a$ by $a^{-1}$, so we may assume that $i > 1$. Then,

$a = a^i + 1 = (a + 1)(1 + a + \ldots + a^{i - 1}) = a^i(1 + a + \ldots + a^{i - 1})$, so $1 + a + \ldots + a^{i - 2} = 0$. This gives $a^{i - 1} = 1$, a contradiction.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.