# Problem regarding proving a field to be finite. [duplicate]

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.

• 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 $\mathrm {char}\, F = 0$? – xbh Oct 28 '18 at 16:45
• Props for showing your work. But this is at least the third incarnation of this question. Fun that it is :-) – Jyrki Lahtonen Oct 28 '18 at 17:35

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 $$0$$.