1
$\begingroup$

The question is -

If multiplicative group $F^\ast$ of nonzero elements of a field $F$ is cyclic, then $F$ is finite.

I have proved this problem partially, which goes as follows-
$F^\ast = F\backslash\{0\}$, let $F^\ast =<\alpha>$ where $\alpha\in F^\ast$
Now, if we can prove $F^\ast$ if finite, then we are done. Let us assume on contrary $F^\ast$ is infinite.
Case 1: If $\operatorname{Char}(F)=p\ne0$. Then $F=F_p(\alpha)$ where $F_p\simeq \Bbb{Z}_p$.
Now, $1+\alpha\in F$, if $1+\alpha=0\implies \alpha^2=1\implies O(a)$ is finite, contracdiction.
So, $1+\alpha\ne 0\implies1+\alpha\in F^\ast\implies1+\alpha=\alpha^r$ for some $r\in\Bbb{Z}\implies \alpha$ is root of the polynomial $x^r-x-1\in F_p[x]\implies \alpha$ is algebraic over $F_p\implies [F_p(\alpha):F_p]$ is finite(say $m$)$\implies |F_p(\alpha)|=p^m\implies |F|=p^m\implies F^\ast$ is finite, contradiction.
Case 2: Now, if $\operatorname{Char}(F)=0$. Then as per hint provided on the book
Since $-1\in F^\ast$ and $\operatorname{Char}(F)=0$, we get a strict positive or strict negative integer $k$ such that $\alpha^k=-1\implies \alpha^{2k}=1$, which contradicts the fact that order of $\alpha$ is infinite.
But, I can't understand how using $\operatorname{Char}(F)=0$, we can get such strict positive or strict negative integer $k$(i.e. $k$ is nonzero integer) such that $\alpha^k=-1$?
Can anybody clear up my doubts? Thanks for assistance in advance.

$\endgroup$
  • 1
    $\begingroup$ Is it because $-1 \in \langle \alpha \rangle $, then there is some $k \in \Bbb Z$ that $-1 = \alpha^k$ and this $k \neq 0$? Then $\alpha^{2k} =1$ and $\mathrm{order} (\alpha) \leqslant |2k| <\infty$ but $\mathrm {char}\, F = 0$? $\endgroup$ – xbh Oct 28 '18 at 16:45
  • $\begingroup$ Props for showing your work. But this is at least the third incarnation of this question. Fun that it is :-) $\endgroup$ – Jyrki Lahtonen Oct 28 '18 at 17:35
1
$\begingroup$

The definition of $\langle \alpha \rangle$ is the set of all elements $\{ \alpha^k : k \in \mathbb Z\}$. Since $-1 \in F^* = \langle \alpha \rangle$, we necessarily have $-1 = \alpha^k$ for some $k\in \mathbb Z$. The only thing that remains is to ensure that $k\ne 0$: this is obvious since $\alpha^0 = 1$ and $-1 \ne 1$ in characteristic $0$.

| cite | improve this answer | |
$\endgroup$

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