Addition and multiplication in localization of a ring

Question

Let $R$ be a ring and $S\subset R$ a multiplicatively closed subset. Define the localization of $R$ at $S$ as: $R_S:=\frac{R[t_s\ :\ s\in S]}{<st_s-1\ :\ s\in S>}$ and let $r/s$ denote the image of $rt_s$ in $R_S$. I am stuck in showing that $a/s+b/r=(ar+bs)/(sr)$ and $(a/s)(b/r)=(ab)/(sr)$. i.e that $at_s+bt_r-art_{sr}-bst_{sr}$ is in $<st_s-1\ :\ s\in S>$

Answer 1

Since $\psi:R[t_s:s\in S]\to R_S$ is a ring homomorphism, $$a/s+b/r=\psi(at_s)+\psi(bt_r)=\psi(at_s+bt_r),$$ and since $\psi(rt_r)=1$ and $\psi(st_s)=1$, it follows by the multiplicativity of $\psi$ that $$\psi(at_s+bt_r)=\psi(at_srt_r+bt_rst_s)=\psi((ar+bs)t_rt_s)=(ar+bs)/rs,$$ where we make use of the observation that $\psi(t_st_r)=\psi(t_{sr})$. This observation follows from the fact that $S$ is multiplicative, and that both $t_st_r$ and $t_{sr}$ are inverses to $sr$ in $R_S$, and hence are equal (this is a general fact, i.e. if $ab=1$ and $ac=1$ in an abelian group (written multiplicatively), then $c=abc=b$.)

For the other statement, $$(a/s)(b/r)=\psi(at_s)\psi(bt_r)=\psi((ab)(t_st_r))=\psi((ab)t_{sr})=ab/sr,$$ where I again used the above observation.

You can rearrange the above to show that the appropriate elements are in the kernel if you prefer to see it that way.