TITLE: Does there exist a unital ring whose underlying abelian group is $\mathbb{Q}^*$?
QUESTION [5 upvotes]: Let $\mathbb{Q}^*$ be the group of units of the rational numbers.  
Does there exist a unital ring whose underlying additive group is $\mathbb{Q}^*$?
I don't really have a gut feeling yea or nea.  
Hope I'm not overlooking something obvious.
Thanks!

REPLY [9 votes]: By unique prime factorization of integers, the multiplicative group $\newcommand{\Q}{\mathbb{Q}}\Q^\times$ is isomorphic to the additive group of $\newcommand{\Z}{\mathbb{Z}}(\Z/2\Z) \times \bigoplus_{p \text{ prime}} \Z$. The isomorphism comes from the fact that every rational number can be written uniquely as $\pm \prod_{p \text{ prime}} p^{a_p}$ for some integers $a_p$, only finite many of which are nonzero; and conversely, every such expression corresponds to a rational number. But now we run into a difficulty: $(\Z/2\Z) \times \bigoplus_{p \text{ prime}} \Z$ with the obvious multiplication doesn't have a multiplicative identity, so while it's an associative $\Z$-algebra, it's not a ring.
But, there is a non-obvious multiplication that works! The additive group of the polynomial ring $\Z[x]$ is isomorphic to a countable direct sum of copies of $\Z$, and so (upon choosing a bijection between the natural numbers and the set of prime numbers), the additive group of $(\Z/2\Z) \times \Z[x]$ is isomorphic to the additive group $(\Z/2\Z) \times \bigoplus_{p \text{ prime}} \Z$.
Composing these isomorphisms, we see that the group of units $\Q^\times$ is isomorphic to the additive group of the ring $(\Z/2\Z) \times \Z[x]$.