TITLE: Multiplicative Group of a Field
QUESTION [7 upvotes]: 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?

REPLY [2 votes]: The following more general statement is also true.
There is no commutative ring $R$ whose unit group $R^*$ is the Prüfer $p$-group.
For a proof, see  https://arxiv.org/pdf/1505.03508.pdf
In this paper, Keir Lockride and I  gave a complete classification of all indecomposable abelian groups which occur as the group of  units of a commutative ring. Note that a Prüfer $p$-group is indecomposable and it does not appear in our list.