Isomorphism of $S^1$. 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.
 A: 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.
A: This is not an answer to the question. Here I am going to answer the following question. Does there exist a unitary-ring structure on $(\Bbb R/ \Bbb Z,+)$ following the comment section.
Suppose there is another operation $\bullet$ such that $(\Bbb R/\Bbb Z,+, \bullet)$ becomes a ring with $\bullet$-identity $e\in\Bbb R/\Bbb Z$ such that we have $k\in \Bbb N$ with $ 0=e+...+e(k$-times addition$)$, where $ 0$ is the additive identity of $(\Bbb R/ \Bbb Z,+)$.  Now, for any $x\in\Bbb R/\Bbb Z$ we have $$x+...+x=(e\bullet x)+...+(e\bullet x)=(e+...+e)\bullet x=0\bullet x=0,$$ which tells us ecah $x\in\Bbb R/\Bbb Z$ must have at most $k$-as additive order, a contradiction, since $(\Bbb R/\Bbb Z,+, \bullet)$ has elements of all orders. Above all addition represents $k$-times addition.

Above is a special case of the fact. If a group has elements of every order and
each element has finite order, then there is no unitary ring structure
on the group.


What is left is that does there exist a unitary-ring $(\Bbb R/\Bbb
 Z,+,*)$ w.r.t. some multiplication $*$ such that $*$-identity of
$(\Bbb R/\Bbb Z,+,*)$ has infinite additive order? In other words, $*$-identity is of the form $\alpha+\Bbb Z$ for some irrational $\alpha?$

Answer to second question: If $(\Bbb R/\Bbb Z,+,∗)$ were a ring with unit $u$, then since $\Bbb R/\Bbb Z$ is divisible there would exist an element $v$ such that $v+v=u$, but then $\overline{1/2}=u∗\overline{1/2}=(v+v)∗\overline{1/2}=v∗(\overline{1/2}+\overline{1/2})=v*\overline 1=v*\overline 0=\overline 0$, a contradiction.
