If you are given an abelian group $\ (G, +)$, is there some algorithm to find **all** possible semigroups $\ (G, ×)$, such that $\ (G, +, ×)$ is a ring?  

If not, can you at least decide, if the ring must have zero divisors?



Finding a compatible multiplication for a given group
**If not, can you at least decide, if the ring must have zero divisors?**


It would be interesting to know if there is a good answer to this question. Here I will just make three remarks. I will call an abelian group $G$ *good* if it can be equipped with a bi-additive multiplication which makes it a ring with no nontrivial zero divisors.


1. By [this answer](https://mathoverflow.net/q/213052), there is an abelian group $G$ that is elementarily equivalent to the group of integers $\mathbb Z$ such that the only bi-additive operation on $G$ is the zero function. This means that $\mathbb Z$ is good, $G$ is not, yet $\mathbb Z$ and $G$ satisfy the same first-order sentences.
2. If $G$ is good and has a nonidentity element $g$ of finite additive order, then $G$ must be an elementary abelian $p$-group for some prime $p$. Conversely, any elementary abelian $p$-group is good. [For the first assertion, if $m\in\mathbb Z$, $g, h\in G-\{0\}$, and $mg=0$, then $0=(mg)h=g(mh)$, so by the goodness of $G$ we get $mh=0$. This shows that once one nonidentity element of $G$ has finite order, then all nonidentity elements have the same finite order, which must be prime. For the second assertion, any elementary abelian $p$-group supports a field structure, hence is good.]
3. This remark addresses the first interesting cases where the structure of $G$ is not decided by Remark 2, which are the cases where $G$ is torsion-free of rank one.


**Claim.** If $G$ is a torsion-free abelian group of rank one, then $G$ is good iff $G$ is isomorphic to the additive group of a unital subring of $\mathbb Q$.


Besides being a torsion-free abelian group of rank one, the additive group of any unital subring of $\mathbb Q$ has the following additional property: Call an element $g\in G$ *$m$-divisible* for a given positive integer $m$ if the equation $mx=g$ is solvable in $G$. Call $G$ $m$-divisible if every $g\in G$ is $m$-divisible. The additional property is: there is an element $u\in G$ such that for every prime $p$ we have ($u$ is $p$-divisible implies $G$ is $p$-divisible).


**Pf.** 
The *if* part of the claim is clear: the additive subgroup of a unital subring of $\mathbb Q$ is a good, torsion-free, abelian group of rank one.


It is easy to see why the additive groups of unital subrings of $\mathbb Q$ satisfy the additional property. If $G$ is the additive group of a unital subring of $\mathbb Q$, choose $u=1\in G$. Then if $px=1$ is solved by $x=x\_p$, for any $g\in G$ we get that $py=g$ is solved by $y=x\_p\cdot g$. 


Here is a sketch for why the additional property characterizes the subgroups of unital subrings of $\mathbb Q$ among all torsion-free abelian groups of rank one. If $G$ is torsion-free of rank one, then there is a group embedding $\varphi:G\to\mathbb Q$. If $u\in G$ is an element such that for every prime $p$ we have ($u$ is $p$-divisible implies $G$ is $p$-divisible), then we can scale the embedding to $\frac{1}{u}\varphi:G\to Q$ so that the image of $u$ is $1$. Now the property that for every prime $p$ we have ($u$ is $p$-divisible implies $G$ is $p$-divisible) translates into the property that the set $\frac{1}{u}\varphi(G)$ consists of all rational numbers whose denominators are divisible only by those primes for which $u$ is $p$-divisible. This is a unital subring of $\mathbb Q$.


What remains to explain is why if $G$ is torsion-free of rank one, and $G$ contains no element $u$ such that ($u$ is $p$-divisible implies $G$ is $p$-divisible), then $G$ is not good.


Argument:


This is a proof by contradiction, so assume that 
$G$ is torsion-free of rank one, and $G$ contains no element $u$ such that ($u$ is $p$-divisible implies $G$ is $p$-divisible), YET $G$ is nevertheless good.
Applying an embedding if necessary, assume also that $G$ is an additive subgroup of $\mathbb Q$ that contains $1$. Suppose that $G$ is NOT $p$-divisible for exactly the primes $p\_1<p\_2<\cdots$. Using the facts that $1\in G$ and $G$ is not $p\_i$-divisible, conclude that there is a nonnegative integer $e\_i$ such that $\frac{1}{p\_i^{e\_i}}\in G$, but no reduced fraction $\frac{r}{s}\in G$ has denominator divisible by $p\_i^{e\_i+1}$.


I claim that the $e\_i$'s will be positive infinitely often. For if there is some $k$ such that $e\_i=0$ when $i>k$, then 
$u=\frac{1}{p\_1^{e\_1}\cdots p\_k^{e\_k}}\in G$ is an element that is not $p$-divisible for the same primes as $G$ ($p\_1<p\_2<\dots$), and we have assumed that there is no such element.


Choose any $i$ where $e\_i>0$. 
Let $\ast$ denote a bi-additive multiplication on $G$ that witnesses that $G$ is good. Write $1\ast 1$ as $\frac{r}{s}$ in reduced form, and write 
$\frac{1}{p\_i^{e\_i}}\ast \frac{1}{p\_i^{e\_i}}$ as $\frac{r\_i}{s\_i}$ in reduced form. Since $1 = p\_i^{e\_i}\cdot \frac{1}{p\_i^{e\_i}}$ and $\ast$ is bi-additive, 
$$\frac{r}{s}=1\ast 1 = p\_i^{2e\_i}\cdot \left(\frac{1}{p^{e\_i}}\ast \frac{1}{p^{e\_i}}\right) = p^{2e\_i}\frac{r\_i}{s\_i}.$$
Since no reduced fraction in $G$ has denominator divisible by $p\_i^{e\_i+1}$, it follows that the numerator $r$ of $\frac{r}{s} = \frac{p^{2e\_i}r\_i}{s\_i}$ is divisible by $p\_i^{e\_i}$. As $i$ ranges we see that the fixed integer $r$ is divisible by infinitely many primes, a contradiction. \\\


