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$F$ 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…

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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

3 Answers 3

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 $0$.

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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.

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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.

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