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In localization of a ring $R$ or a module $M$ over $R$ at a multiplicative subset $S$ of $R$, we define an equivalence relation on $R\times S$ or $M\times S$ and define addition and multiplication on equivalence classes. I think, if you don't define an equivalence relation, and define addition and multiplication in the same way, you would still get a ring (respectively an $R$-module). Although, I don't expect this structure to have the nice properties of localization, I was wondering if such a structure has been studied.

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So, you are trying to define addition on $R\times S$ by $(r,s)+(r',s') = (rs'+r's,ss')$. The identity element of this operation is $(0,1)$ (assuming $1\in S$). But if $S$ contains an element $s$ that is not a unit, then $(s,s)$ does not have an additive inverse: $(s,s)+(r,t) = (st+sr,st) = (0,1)$ if and only if $st=1$; and adding $(r,t)+(s,s)$ would require $ts=1$, so $s$ must be a unit. You only get an additive group if $S$ is a subgroup of the unit group of $R$. – Arturo Magidin Mar 14 '11 at 16:02
zero and one wouldnt be unique, nor would additive/multiplicative inverses... – yoyo Mar 14 '11 at 16:19
@Arturo: That's a good point. Do we get anything good under the restriction that $S$ is a subgroup of the unit group? – Yan Etor Mar 14 '11 at 16:21
For commutative rings, you get a homomorphism from this ring to $R$ via $(r,s)\mapsto rs^{-1}$, which is onto with kernel $\{0\}\times S$. Doesn't seem to me like it's terribly interesting then. – Arturo Magidin Mar 14 '11 at 16:29

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