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I'm trying to solve this question:

In order to solve this question above, I found this function: $r/w\mapsto (r/s)/(w/s)$ such that $w/s\in T$, I almost proved this map is an isomorphism, I'm stuck just in the surjectivity part.

If we get an element $(r/s)/(w/s)$ of $T^{-1}(S^{-1}R)$, ok! However, an element of $T{^1}(S^{-1}R)$ can be for example $(r/s_1)/(w/s_2)$ with $s_1\neq s_2$

I need help in this part.

Thanks in advance.

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Hint: Note that in $S^{-1}R$ we have $(ws_1)/(s_1s_2) = w/s_2 \in T$, so $ws_1 \in S_*$. Now consider $rs_2/ws_1 \in S_*^{-1}R$.

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Consider an element $z$ of $T{^1}(S^{-1}R)$, we can write it in the form

$$ z=\frac{\frac{r}{s_1}}{\frac{w}{s_2}} $$

with $r,w\in R$, $s_1,s_2\in S$ and further $\frac{w}{s_2} \in T$.

Let $w'=s_1\frac{w}{s_2}$. Then $\frac{w'}{s_1}=\frac{w}{s_2} \in T$, so we deduce that $w' \in S_{*}$. Finally $z=\frac{r}{w'}$ with $r\in R,w'\in S_{*}$.

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  • $\begingroup$ How do you define this multiplication $w'=s_1\frac{w}{s_2}$? $\endgroup$ – user42912 May 28 '13 at 11:33
  • $\begingroup$ Second, are you saying that an element $w'$ of $S_*$ is equal than other element $s_1(w/s_2)$ of $T$? $\endgroup$ – user42912 May 28 '13 at 11:35
  • $\begingroup$ @user42912 $s_1$ and $\frac{w}{s_2}$ are both elements of the ring $S^{-1}R$, so multiplying them makes sense. $\endgroup$ – Ewan Delanoy May 28 '13 at 11:40
  • $\begingroup$ @user42912 The element in $T$ is $w/s_2$, not $s_1(w/s_2)$. $\endgroup$ – Ewan Delanoy May 28 '13 at 11:41

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