Saturated sets and A-algebras [duplicate]

Consider $S$, the localisation of $\mathbb{Z}[x]$ at the multiplicative set $\{ 1, x^3, x^6, \ldots \}$. Now, $x^2/x^3$ is an element of $S$, and it seems to me that this behaves the same as $1/x$. Therefore, it seems we get the same result as if we had localised at $\{ 1,x,x^2,x^3,\ldots \}$ instead.

Am I correct that these two localisations are "the same"? If so, given a ring $R$, and multiplicative subsets $M, N \subseteq R$, when are $\mathrm{Loc}(R,M)$ and $\mathrm{Loc}(R,N)$ "the same" / canonically isomorphic?

Solution 1:

Yes, these two localizations are the same, and your logic is completely fine.

To answer your larger question, we define the saturation $\hat{S}$ of a multiplicatively closed subset $S\subset R$ to be the set of elements $r\in R$ so that there exists an $r'\in R$ so that $rr'\in S$. It's not so difficult to see that $S^{-1}R$ is isomorphic to $\hat{S}^{-1}R$ as an $R$-algebra: clearly $R\to \hat{S}^{-1}R$ factors uniquely through $R\to S^{-1}R$ since every element in $S$ is invertible in $\hat{S}^{-1}R$; conversely, every element in $\hat{S}$ is invertible in $S^{-1}R$ as $r\cdot r'/(rr')=1$ so $R\to S^{-1}R$ factors uniquely through $R\to \hat{S}^{-1}R$.

This establishes that $S^{-1}R$ is always isomorphic as an $R$-algebra to $\hat{S}^{-1}R$, so if $S,T$ are two multiplicatively closed subsets having the same saturation, then $S^{-1}R\cong T^{-1}R$ as a $R$-algebras. Conversely, if $S$ and $T$ are two multiplicatively closed subsets with different saturations, then $S^{-1}R$ and $T^{-1}R$ are not isomorphic as $R$-algebras with their standard $R$-algebra structures: we might as well assume that $S$ and $T$ are saturated and that there's some element $s\in S$ not in $T$, and then $s$ has invertible image in $S^{-1}R$ but not $T^{-1}R$.