TITLE: Isomorphism of $S^1$.
QUESTION [6 upvotes]: Define $S^1 = \{ z \in \mathbb{C^*} \: : \: |z| = 1 \}$. My question is, does there exist some ring $R$ such that $(R, + )$ is isomorphic to $S^1$? My intuition says no but how would I go about showing that.

REPLY [8 votes]: The answer is no: $S^1$ is divisible, meaning for any $x\in S^1$ and any $n\in \mathbb Z$, there is $y$ with $y^n = x$. If we write $S^1$ additively, this says $ny = x$.
In particular, if $e$ is the multiplicative unit of a ring structure on $S^1$, $e$ is divisible, which in particular implies that $n\cdot e$ is invertible for every $n$: let $y$ satisfy $n\cdot y =e$, then $e= n\cdot y = (n\cdot e)y$.
In particular, if $n\cdot x = 0$, then $(n\cdot e) x= 0$, so that $x=0$: that would imply $S^1$ is torsion-free, which it's not.