Let $S$ be a subset in a commutative ring $R$, such that:
Define a relation $\sim$ on the Cartesian product $R\times S$ through $(r_{1},s_{1})\sim(r_{2},s_{2})$ if $\exists s\in S$ such that:
$$s(r_{1}s_{2}-r_{2}s_{1})=0$$
How can $R\times S/\sim$ be made into a ring in a natural way?
Can you help me with this question. Thank you for helping!
On
It's basically the same construction one have to do to obtain rational numbers from integers.
The operation you are looking for are those defined by the following equations:
$$[(r_1,s_1)]+[(r_2,s_2)] = [(r_1s_2+r_2s_1,s_1s_2)]$$ $$[(r_1,s_1)][(r_2,s_2)] = [(r_1r_2,s_1s_2)]$$
of course you have to prove that these are well defined, i.e. the definition doesn't depend on the choice of representants: if $(r'_1,s'_1) \sim (r_1,s_1)$ and $(r'_2,s'_2) \sim (r_2,s_2)$ the we want that $$(r_1s_2+r_2s_1,s_1s_2) \sim (r'_1s'_2+r'_2s'_1,s'_1s'_2)$$ and $$(r_1r_2,s_1s_2) \sim (r'_1r'_2,s'_1s'_2)$$
If you choose to indicate by $\frac{r}{s}$ the equivalence class $[(r,s)]$ then these operation have the form $$\frac{r_1}{s_1}+\frac{r_2}{s_2} = \frac{r_1s_2+r_2s_1}{s_1s_2}$$ and $$\frac{r_1}{s_1}\frac{r_2}{s_2} = \frac{r_1r_2}{s_1s_2}$$ respectively (are very likely as the operation you put on $\mathbb Z \times \mathbb Z/\sim$ to obtain $\mathbb Q$).
You can easily prove that $\mathbb Q$ is exactly obtained as this kind of ring by letting $R=\mathbb Z$ and $S=\mathbb Z \setminus \{0\}$.
This is the so called localization of $R$ with respect to the multiplicative subset $S$.
$R \times S / \sim $ inherits a ring structure in the following way: $$\frac{r_{1}}{s_{1}} + \frac{r_{2}}{s_{2}} = \frac{r_{1}s_{2} + r_{2}s_{1}}{s_{1}s_{2}}$$ $$\frac{r_{1}}{s_{1}} \cdot \frac{r_{2}}{s_{2}} = \frac{r_{1}r_{2}}{s_{1}s_{2}}$$ where $\frac{r_{1}}{s_{1}}$ is the equivalence class $(r_{1}, s_{1})$. Of course one has to prove that these operations are well defined, i.e. they don't depend on the equivalence classes' representatives