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This is from Hungerford's Algebra, Page 148 Exercise 2.

Let $S$ be a multiplicative subset of a commutative ring $R$ with identity and let T be a multiplicative subset of the ring $S^{-1}R$. Let $S_*=\{r\in R \mid r/s\in T for\ some\ s\in S\}$.Then $S_*$ is a multiplicative subset of $R$ and there is a ring isomorphism $S_*^{-1}R\cong T^{-1}\left(S^{-1}R\right)$

I've figured out how to construct the isomorphsim and verify it, i.e. define $r/s_*\mapsto (r/s)/(s_*/s)$,where $s\in S$ is an element s.t. $s_*/s\in T$.But through all my work (showing that the map is well-defined; showing that it is a monomorphism as well as an epimorphism) I didn't use the condition that the identity exists in $R$. Could you tell me in which part of my work should the identity be used (or it's an unnecessary condition)? I'm appreciative for any help from you.

Here is my proof.

(a)Show that the map $f$ is well-defined.Assume that $r_1/s_{*1}=r_2/s_{*2}$.Thus there exists $t\in S$ such that $t(r_1s_{*2}-r_2s_{*1})=0$.We should show that $f\left(r_1/s_{*_1}\right)=f\left(r_2/s_{*_2}\right)$, i.e.$(r_1/s_1)/(s_{*1}/s_1)=(r_2/s_2)/(s_{*2}/s_2)$, where$s_i\in S$ such that $s_{*i}/s_i\in T$.This is equivalent to find an $s\in S$ such that $ss_1s_2(r_1s_{*2}-r_2s_{*1})=0$.Since $R$ is commutative, we can simply let $t=s$.

(b)Show that $f$ is a homomorphsim.Here I just typed out the process of verifying that $f$ preserves the structure of the additive group because it is similar for multiplication. We shall show that $f\left((r_1/s_{*1})+(r_2/s_{*2})\right)=(r_1/s_{1})/(s_{*1}/s_1)+(r_2/s_{2})/(s_{*2}/s_2)$,which is followed by$f\left((r_1/s_{*1})+(r_2/s_{*2})\right)=f\left((r_1s_{*2}+r_2s_{*1})/(s_{*1}s_{*2})\right)= \\\left((r_1s_{*2}+r_2s_{*1})/(s_{1}s_{2})\right)/(s_{*1}s_{*2}/s_1s_2)=(r_1/s_1)/(s_{*1}/s_1)+(r_2/s_2)/(s_{*2}/s_2)=f(r_1/s_{*1})+f(r_2/s_{*2})$

The process of showing $f$ is injective is almost the same as that of showing $f$ is well-defined.

(c)Show that $f$ is surjective.For any $(r/s)/(r_0/s_0)=(rs_0/ss_0)/(r_0s/ss_0)\in T^{-1}(S^{-1}R)$, we can pick $rs_0/r_0s$ (note that $r_0s/ss_0=r_0/s_0\in T $, and $ss_0\in S$, so $r_0s\in S_*$) , so $f$ maps it to what we desired.

Arturo Magidin
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hhh
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    Without the existence of the identity, nothing in your question would even make sense. Localisation is about making ring elements invertible. Being invertible is defined in terms of the identity. What would $S^{-1}R$ even mean without it? – Tzimmo Sep 11 '24 at 13:41
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    To be fair to OP, Hungerford develops most of the elementary localization results without assuming $R$ has identity. If memory serves [it's been 25 years] in doing so the ring $S^{-1}R$ will still have an identity (of the form $s/s$) so invertibility questions make sense in the localization. – Randall Sep 11 '24 at 13:50
  • @Randall Yes, that's exactly how he develops those results. – hhh Sep 11 '24 at 14:04
  • To really get an answer to your question, you would need to provide your actual argument. Can't really check if it's right or not without it. – Randall Sep 11 '24 at 14:16
  • @Randall OK.I edited my post just now. – hhh Sep 11 '24 at 15:25
  • If I recall correctly, most of these results do not actually require the ring to have an identity to work. An identity simplifies a lot of the computations, but is not required. – Arturo Magidin Sep 12 '24 at 04:09
  • It is actually easier to map the other way, by using the usual "fraction of fraction" formula: given an element $(r/s)/(t/u)$ of $T^{-1}(S^{-1}R)$, map to $ru/st\in (S_)^{-1}R$: note $st\in S_$ because $st/su = t/u\in T$, so $st\in S_*$. – Arturo Magidin Sep 12 '24 at 04:29
  • There are two proofs here. Though stated with identity because of the question, neither proof uses that $R$ has an identity. – Arturo Magidin Sep 12 '24 at 04:31
  • @ArturoMagidin I see.I really appreciate your help! – hhh Sep 12 '24 at 13:19

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