multiplicative group of infinite fields I am stuck with the proof of below expression;
"If F is an infinite field, then no infinite subgroup of F* (the multiplicative group of F) is cyclic."
anyone can help?
 A: Any infinite field $F$ of characteristic zero has infinite cyclic subgroups of $F^\times$. Any such field would contain a copy of $\Bbb Q$ so it suffices to give examples for that. Take $x^{\Bbb Z}$ for any $x\in\Bbb Q\setminus\{-1,0,1\}$.
In characteristic $p$ one can show a field's group of units $F^\times$ has an infinite cyclic subgroup if and only if it has an element $T$ transcendental over the prime subfield ${\Bbb F}_p$. If such an element exists then take $T^{\Bbb Z}$ to be the subgroup. If no such element exists then every element of $F$ is algebraic over ${\Bbb F}_p$ and hence a root of unity (by the theory of finite fields) hence generates a finite group.
Thus it is incorrect to say "no infinite subgroup of an infinite field's group of units is cyclic."
Suppose $R$ is a ring such that $R^\times$ is infinite cyclic. There can be no nontrivial roots of unity, so therefore $-1=1$ and ${\rm char}\,R=2$. If $R^\times=\langle x\rangle$ then $x^{-1}\in R$ and hence ${\Bbb F}_2[x,x^{-1}]\subseteq R$. Our above argument shows $x$ is transcendental over ${\Bbb F}_2$. One may show the only units of ${\Bbb F}_2[x,x^{-1}]$ are in $x^{\Bbb Z}$. Therefore $R={\Bbb F}_2[T,T^{-1}]$ is the minimal example of such a ring where $R^\times\cong\Bbb Z$ in the sense that it injects into all rings with this property.
