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The multiplicative group $F^{\times}=F\setminus \{0\}$ of a field is abelian, and it may contain torsion elements, may contain torsion free elements, or both may occur, as can be seen from the examples of any finite field, $\mathbb{Q},\mathbb{R}$ and $\mathbb{C}$. The Prüfer $p$-group is a (proper) subgroup of $\mathbb{C}^{\times}$. The question I would like to ask is

Question: Is there an infinite field $F$ such that $F^{\times}$ is isomorphic to the Prüfer $p$-group?

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Nice question! The answer is no.

If $F$ has characteristic zero it contains a copy of $\mathbb{Q}$, so $F^{\times}$ has torsion-free elements; hence $F$ has positive characteristic $p$. If $F^{\times}$ is isomorphic to a Prüfer $\ell$-group for some prime $\ell$, then $F$ contains all $\ell$-power roots of unity, and $\mathbb{F}_p^{\times}$ must consist of $\ell$-power roots of unity, so in particular $p \neq \ell$.

Let $k$ be a positive integer. Since $F$ contains all $\ell^k$-th roots of unity, $F$ contains $\mathbb{F}_{p^n}$ where $n$ is the smallest positive integer such that $\ell^k | p^n - 1$. Then $F^{\times}$ has a subgroup of order $p^n - 1$, which is necessarily a power of $\ell$. By choosing $k$ large enough, this is impossible by Zsigmondy's theorem.

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Note that already the assumption that $F^{\times}$ is torsion implies that $F$ must be an algebraic extension of a finite field, and from here it is possible to use knowledge of the absolute Galois groups of finite fields to classify all of the possibilities. –  Qiaochu Yuan May 30 '13 at 4:25
    
Could you explain how $\mbox{F}^{\times}_p$ enters in the first paragraph? I don't know much field theory, so it's probably something obvious. –  Zach L. May 30 '13 at 4:49
    
@Zach: if $L$ is a subfield of $F$, then $L^{\times}$ is a subgroup of $F^{\times}$. To say that $F$ has characteristic $p$ is equivalent to saying that $F$ contains $\mathbb{F}_p$ as a subfield, so in particular $F^{\times}$ contains $\mathbb{F}_p^{\times}$ as a subgroup. –  Qiaochu Yuan May 30 '13 at 4:52
    
That wasn't my issue, so my first comment was poorly worded. What I wasn't putting together was that if $p=l$, then $x^l-1$ has only one solution. Now I'm wondering why all $\mbox{F}_{p^n}$ must lie in $F$ if $l^k | p^n - 1$. I assume it's for a similar reason. –  Zach L. May 30 '13 at 15:56
    
@Zach: this follows from standard facts about finite fields. If $\alpha$ is algebraic over $\mathbb{F}_p$, then the subfield of the algebraic closure generated by $\alpha$ is $\mathbb{F}_{p^n}$ where $n$ is the order of $\alpha$ under the action of the Frobenius map. For a primitive $\ell^k$-th root of unity, it follows that $n$ is the smallest positive integer such that $\ell^k | p^n - 1$. –  Qiaochu Yuan Jun 1 '13 at 7:33
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