Let $R$ be a commutative ring. Let $P_{i}$, $1\leq i\leq n$ be prime ideals none of which are contained in each other. Let $S=R-(\cup_{i=1}^{n} P_{i})$. Then $S$ is a multiplicatively closed set and $S^{-1}R$ exists. Define $\pi:R\rightarrow S^{-1}R$ by $x\mapsto x/1$. Let $S_{i}=R-P_{i}$. Clearly $S\subset S_{i}$. All I want to do is explicitly compute $\pi(S_{i})$. I believe it should be $S^{-1}R-P_{i}\cdot S^{-1}R$. I have the inclusion $i(S_{i})\subset S^{-1}R-P_{i}\cdot S^{-1}R$. The other inclusion is eluding me. This is what I have so far. I think I am approaching it the wrong way but don't have any other idea how to do it.
If $r/s\in S^{-1}R-P_{i}\cdot S^{-1}R$ I aim to find $x\in S_{i}$ such that $r/s\sim x/1$ in $S^{-1}R$. If it were true then I would need to find $x\in S_{i}$ and $t\in S$ such that $t(r-sx)=0$. I have no idea how to find such elements $t$ and $x$.
Any help or a new appraoch would be much appreciated. Thanks.
This is not true, even if $n = 1$: consider $R = \mathbb{Z}$, $P_1 = (2)$. Then $\pi(S_1) = \{ \frac{n}{1} : n \text{ odd}\}$, whereas $S_1^{-1}R - P_1 \cdot S_1^{-1}R = \{\frac{n}{s} : n, s \text{ odd}\}$, which contains e.g. $\frac{1}{3}$, but $\frac{1}{3}$ is not of the form $\frac{m}{1}$ for $m$ odd (otherwise $s(3m - 1) = 0$ for some odd $s$, impossible).
What is true is that $W_i := S^{-1}R - P_i \cdot S^{-1}R$ is the saturation of $\pi(S_i)$, i.e. $W_i = \{y \in S^{-1}R \mid \exists x \in S^{-1}R, xy \in \pi(S_i) \}$. This holds because $W_i^{-1}(S^{-1}R) \cong R_{P_i} = S_i^{-1}R \cong \pi(S_i)^{-1}(S^{-1}R)$ by transitivity of localization, and $W_i$ is the complement of a prime ideal in $S^{-1}R$, hence is a a saturated subset of $S^{-1}R$.