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Few Definitions:
Let $R$ be a nonzero commutative ring which does not necessarily contain identity. Let $S$ be a multiplicatively closed subset of $R$ such that $S$ does not contain the zero element and zero divisors.

Define an Equivalence relation $\sim$ on $R\times S$ by setting $(r, s) \sim (r',s')$ if $rs' = sr'$. and denote the equivalnece class of $(r, s)$ by $\frac{r}{s}=\{(a,b)|a\in R, b\in S \enspace \text{and}\ rb=as\}$.

Now, the set $S^{-1}R$ is the set of equivalence classes under $\sim$.

Define multiplication and addition on $S^{-1}R$ by $$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd} \enspace \text{and}\ \frac{a}{b}\times \frac{c}{d}=\frac{ac}{bd}$$

My Question:

Let $R$ be a commutative ring with $1\neq 0$. Let $s$ be a nonzero divisor of $R$ and $S$ denote the multiplicatively closed subset $\{1,s,s^2,...\}$.Then prove that $$S^{-1}R\cong \frac{R[x]}{(sx-1)}$$

My Idea:
Define $$\varphi:R[x]\rightarrow S^{-1}R\\r\mapsto r \enspace \forall \enspace r\in R\\x\mapsto \frac{1}{s}\enspace \forall \enspace s\in S$$ Now, I need to show that this map is surjective ring Homomorphism and need to determine its kernel. Then, I will use First Isomorphism Theorem for Rings to get the result.

It is easy to note that $\varphi$ is a ring homomorphism under the above definition.

Now, any element $\alpha \in S^{-1}R$ is of the form $\frac{a}{s^m}$ for some $m\in \mathbb{N}\cup\{0\}$, where $a\in R$ and $s^m\in S$.
Now, $$\frac{a}{s^m}=a\times \frac{1}{s^m}= a\times (\frac{1}{s})^m\\=\varphi(a)\times (\varphi(x))^m\\=\varphi(a)\times \varphi(x^m) =\varphi(ax^m)$$ Therefore, $\varphi$ is a surjective ring homomorphism.

But I don't know how to compute its Kernel.

Can Anyone please help me and show me a proof?

Bill Dubuque
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Kumar
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  • @BillDubuque The question looks similar to it but it's a different question. I need my proof to be verified and also, need an extension to it(a simple way of finding its Kernel). I would request you to please open it. Thanks. – Kumar Jun 30 '19 at 17:18
  • @Kunar Please chase the links, I've added more. This is a FAQ discussed here many times. If you have a specific question not answered in any of those pages then please elaborate in the question and we can reopen it. – Bill Dubuque Jun 30 '19 at 17:19
  • @BillDubuque Thanks for adding more links. Now, it more clear. and I satisfied with your duplicate marking. :) – Kumar Jun 30 '19 at 17:28
  • Glad I could help. There are probably many more links but hopefully those will give you the general idea. – Bill Dubuque Jun 30 '19 at 17:30

1 Answers1

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You get one inclusion from the fact that $\varphi(sx-1) = s\cdot \frac{1}{s}-1 = 0 \in S^{-1}R$, such that $(sx-1) \subset \text{ker}(\varphi)$. Now take an element from the kernel and try to show the other inclusion.

Con
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  • After Posting the above question I figured out this inclusion. But Thanks. But I really want to see the other inclusion. – Kumar Jun 30 '19 at 17:03