Field structure of a given set

Let $R$ be a ring. (Not necessarily commutative).

For every pair $(a,b)$ where $a , b \in R$ and $b \neq 0$, we define $S_{ab}$ as the set of pairs $(m,n) \in R \times R$ such that $an=bm$.

Let $F$ be the set of all distinct sets $S_{ab}$.

Further, operations $+$ and $*$ are defined on elements of $F$ as:

$$S_{ab}+S_{mn} = S_{xy}\;\; , \;\;\; x= an + mb, \;\;\; y=bn$$

$$S_{ab}*S_{mn} = S_{xy}\;\; ,\;\;\;\; x = am,\;\;\;\; y = bn.$$ Which of the following conditions is necessary and sufficient to make $F$ a field?

a) $R$ is a commutative ring but not an integral domain

b) $R$ is an integral domain but not a commutative ring

c) $R$ is a commutative integral domain

d) None of the above

This question was posed by a friend of mine. I guess it's from some entrance exam.

I believe it would be a field if $R$ is a commutative integral domain with $S_{0a}$ as the additive identity and $S_{aa}$ as the unity. The commutativity is needed for the unity to be well-defined and the integral domain structure for the additive identity. But I am still not very sure. The sets and elements mixture is confusing me.

• Yes, you are right – Vishesh Sep 11 '16 at 17:11
• @Vishesh Then you should add the above info to your question. – DonAntonio Sep 11 '16 at 17:22
• @DonAntonio. Done – Vishesh Sep 11 '16 at 17:24

Note that the way the question is posed, a) and b) are immediately wrong: If we obtain a field when $R$ is commutative, but not integral - how should the construction suddenly fail when $R$ is "accidentally" integral? Similar for b).
Assume $F$ is a field. If $ab=0$ with $a\ne 0\ne b$, then $S_{0,a}*S_{0,b}$ is not even defined. Hence $R$ is necessarily an integral domain. Let $u,v\in R$. If $u=0$ or $v=0$ then trivially $uv=vu$. Hence we may assume $u\ne 0\ne v$. We have $$S_{u,u}+S_{-v,v}=S_{uv-vu,uv}$$ and $$S_{v,v}+S_{-v,v}=S_{0,v^2}.$$ Note that for $x\ne 0$, $S_{x,x}=\{\,(y,y)\mid y\in R\,\}$, hene $S_{u,u}=S_{v,v}$ and so $S_{0,v^2}=S_{uv-vu,v^2}$ Clearly $S_{0,v^2}=\{\,(x,0)\mid x\in R\,\}$. Hence $(uv-vu)\cdot x = v^2\cdot 0=0$ for all $x\in R$. As $R$ is integral, we conclude $uv-vu=0$. Hence $R$ is necessarily commutative.
On the other hand, if $R$ is a commutative integral domain, one readily verifies the "usual suspects": $S_{0,a}=0$, $S_{a,a}=1$, $S_{-a,b}=-S_{a,b}$, $S_{b,a}=S_{a,b}^{-1}$. And of course we have associativity and distributivity exactly as we know it from fractions. Thus commutative integral domain is sufficient.