Let the $F$ is a field and $F^\#=F-\{0\}$
And $\alpha$ and $\beta$ are algebraic over $F$
It is definitely that
$F$ is a finite $\Rightarrow$ $F^\#$ is a cyclic group.
(IF $F$ is a just field, The above theorem is not true. Like the example $K=F(\alpha,\beta)$ since $K^\#$ is not cyclic.)
But the question is What if the case, the simple extension?
IF the $E=F(\alpha)$ is a simple extension of the F
Is it true '$E^\#$ is a cyclic group?
(I guess its counterexample is $Q(\sqrt 2))$, But Couldn't sure that my example is true.)
How do you think about that?
Any answer would help.
Thank you.
Theorem. Let $F$ be a field. The multiplicative group of $F$ is cyclic if and only if $F$ is finite.
Proof. Suppose $F$ is finite of order $n$. The multiplicative group of $F$ is a finite abelian group, and so can be written (uniquely) as $$F^*\cong C_{m_1}\oplus\cdots \oplus C_{m_k}$$ where $C_r$ is the cyclic group of order $r$, and we have $1\lt m_1$, $m_1|m_2|\cdots|m_k$ and $m_1m_2\cdots m_k = n-1$. We want to show that $k=1$.
In particular, every element of $F^*$ has multiplicative order dividing $m_k$, and hence the polynomial $x^{m_k}-1\in F[x]$ hs $n-1$ roots. Since $F$ is a field, the polynomial has at most $m_k$ roots, so $n-1\leq m_k\leq n-1$. Thus, $m_k=n-1$ which requires $k=1$, as claimed. That is, $F^*$ is cyclic.
Conversely, suppose that $F$ is infinite. If $F$ is of characteristic $p\gt 0$, then it contains a copy of the finite field with $p$ elements; this field contains an element of multiplicative order $p-1$ by Fermat’s Little Theorem, and so $F^*$ cannot be infinite cyclic. Thus, $F^*$ is not cyclic. If the characteristic of $F$ is $0$, then $F$ contains a copy of $\mathbb{Q}$. But then the multiplicative subgroup generated by $2$ and that generated by $3$ intersect trivially, which means that $F^*$ cannot be infinite cyclic (as any two nontrivial subgroups of the infinite cyclic group have nontrivial intersection). Thus, $F^*$ is not cyclic. $\Box$