Here is the question I wanted to answer letter(c) of it:
Let $R$ be a ring (with 1), and let $H$ be the cyclic subgroup of $(R,+)$ generated by $1.$ The characteristic of $R$ is $n$ if $H\cong \mathbb{Z}_{n},$ and is $0$ if $H \cong \mathbb{Z}.$\ (a) Show that, if $a \in R$ has additive order $m,$ then $m$ divides the characteristic of $R.$\ (b) Prove or disprove: If the characteristic of $R$ is $0,$ then $(R,+)$ is torsion free.\ (c) Is there a ring $R$ such that $(R,+) \cong S^{1}$ ?
Here is a solution to letter $(c)$:
The answer is No. And here is the justification:
$S^{1}$ is defined as $\{ z \in \mathbb{C}^{*}\ :\ |z|=1 \}$ and if $a,b \in S^1,$ then $|a| = |b| = 1$ but $|a + b| \leq |a| + |b| = 2$ and so $a + b$ is not necessarily in $S^1.$\\
On the other hand, $(R, +)$ form an additive group and so it is closed under addition.
Hence, $$(R,+) \ncong S^1.$$
Here is a hint to the correct solution:
Assume $na = 0$ and $nb = 1$ for $a,b \in R, n \in \mathbb{Z}^{+}.$ Then $$a = a. (nb) = (na). b = 0 .b = 0 $$
Also, I found a solution here Isomorphism of $S^1$.
My questions are:
1- In the link I mentioned, the author is using the idea that $S^1$ is divisible, I do not know why that is correct, could anyone explain that for me please?
2-Also, I do not understand the general idea of the solution, could anyone explain it for me please?
3- My understanding is that we are using the idea that $R$ is a ring with unit so it has multiplicative and additive identity, but why $a$ can not be zero in the hint?