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Let $R$ be a commutative ring with $1$. (If required, assume also that $R$ is an integral domain.) Consider the localization $R_f$ at $0\neq f \in R$ where the multiplicative set is $S=\{f^n\}_{n \geq 0}$.

Is there any relation between this ring $R_f$ and the polynomial ring $R\big[\frac{1}{f}\big]$? Relation as in (say) some sort of correspondence between the prime ideals of the two rings.

My primary motivation to ask this question comes from trying to understand all the prime ideals of $\mathbb{Z}\big[\frac{1}{n}\big]$. I found this link with the same question as mine. But I don't see why it is marked as a "duplicate" to another question.

Can someone perhaps clarify the relation between $R_f$ and $R\big[\frac{1}{f}\big]$?

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$R[1/f] = R[X]/(Xf-1)$ is isomorphic to $R_f$ via the natural map sending $X$ to $1/f$.

To see this, we can show that $R[1/f]$ has the universal property of a localization: if $\varphi:R\to S$ is a morphism that inverts $f$, then we can extend to a morphism $R[X]$ sending $X$ to $\varphi(f)^{-1}$. This factors through the quotient, i.e. $R[X]/(Xf-1)\to S$, and we can show that this map is the unique one compatible with $\varphi$.

Andrew Dudzik
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    In particular the primes in $\mathbb Z[1/n]$ are generated by those prime numbers, that dont come up in the prime factorization of $n$. And of couse the zero ideal. – MooS Mar 01 '16 at 07:07
  • Can I ask a slightly related question here: If two rings $A$ and $B$ are isomorphic, is it necessarily true that their prime spectra $Spec(A)$ and $Spec(B)$ are bijective? –  Mar 01 '16 at 10:43
  • @monomorphic Yes. An isomorphism $\varphi$ of rings is a bijection such that $\varphi^{-1}(I)$ is prime if and only if $I$ is prime. – Andrew Dudzik Mar 01 '16 at 16:36
  • @Slade I have a difficulty in understanding your first equation. I know it is easy, but I would be pleased if you could suggest some sources on $``localization"$. Thanks in advance. – Tedebbur Mar 29 '20 at 15:56
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    @Tedebbur Any text that covers the theory of rings and modules. Dummit and Foote is one example. – Andrew Dudzik Mar 29 '20 at 16:10