Let $R$ be a PID , let $a,b \in R$ be co-prime elements , i.e. $aR+bR=R$ . Then obviously $R[X]/(aX-b)$ is an integral domain . How to show that $R[X]/(aX-b) \cong R[1/a]$ ?
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From this answer we learn that $R[X]/(aX-b) \simeq R[b/a]$. But in your case $R[b/a]=R[1/a]$. – user26857 Sep 04 '17 at 16:52
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Every overring $E$ of fractions of a $\rm\color{#c00}{Bezout}$ domain is a localization because $,b/a\in E,\Rightarrow, 1/a\in E,,$ which follows by dividing $\ \color{#c00}{a j+ b k = 1}\ $ by $,a.,$ See this answer for more. – Bill Dubuque Sep 04 '17 at 20:41
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1Possible duplicate of If $a,b$ is an $R$-sequence, then $ax-b$ is prime (Eisenbud, Exercise 10.4) – Ken Duna Sep 05 '17 at 17:13
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Hint:
Consider a Bézout's relation: $\;ua+vb=1,\quad (u,v\in R)$. In $R[1/a]$, you have: $$\frac 1a=u+\frac{vb}a.$$
Bernard
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Okay ... but I am still not getting it ... how to get a surjective homomorphism from $R[X] \to R[1/a]$ with kernel $(aX-b)$ ? Or are you suggesting any other method ..? – Sep 04 '17 at 12:32
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You map $X$ to $b/a$,thus defining a homomorphism from $\mathbf R[X]$ to $R[1/a]$. As $aX-b$ lies in the kernel, it passes to the quotient $R[X]/(aX-b)$, and it's surjective because of the relation I mention in my answer. – Bernard Sep 04 '17 at 13:41
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It's not hard to see it's a prime ideal. If the kernel strictly contained the ideal $(aX-b)$, it would be a maximal ideal of $R[X]$, which would have a non-zero intersection $R$, and the canonical morphism $R\to R[1/a]$ would not be injective. – Bernard Sep 04 '17 at 16:09
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@Bernard You are using too much that $R$ is a PID in a question that obviously holds in a more general context. – user26857 Sep 05 '17 at 09:18
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@Bernard I won't be so sure, at least not when it comes to show that the ideal $(aX-b)$ is prime. – user26857 Sep 05 '17 at 09:21
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Oh! this point… Yes, I use the fact the structure of prime ideals in a polynomial ring over a P.I.D. is known. – Bernard Sep 05 '17 at 09:23