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Let $R$ be a ring and let $S=\{1,s,s^2,s^3,\dots\}$ be a multiplicative system of $R$. Consider the canonical map $R\rightarrow S^{-1}R$. Is $S^{-1}R$ a finitely generated algebra over $R$? It looks like $\frac{1}{s}$ will generate $S^{-1}R$ over $R$.

  1. If $P$ is a prime ideal of $R$, is $R_{P}$ a finitely generated algebra over $R$?

  2. If instead of $S=\{1,s,s^2,s^3,\dots\}$ we take any arbitrary multiplicative system, is $S^{-1}R$ a finitely generated algebra over $R$?

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Babai
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    You are correct in the first assumption. For the second, think of localizing a prime ideal in the integers. – Tobias Kildetoft Dec 10 '14 at 14:46
  • So are you claiming $\mathbb{Z}{P}$ is not finitely generated over $\mathbb{Z}$? I guess $\mathbb{Z}{P}$ will have infinitely many elements of the form $\frac{a}{b}$ where p doesn't divide $p$. Now with ploynomial combinations of finitely many of them one cannot generate the whole of $\mathbb{Z}_{P}$. Is the argument correct? – Babai Dec 10 '14 at 14:58
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    Right (one way to think of it is that given a finite subset, you can only get a finite number of distinct prime divisors in the denominators). – Tobias Kildetoft Dec 10 '14 at 15:02
  • @TobiasKildetoft That should really be posted as an answer to the question. – Seth Dec 10 '14 at 15:30
  • @TobiasKildetoft Instead of $S={1,s,s^2,s^3,\cdots}$ if we take any multiplicative system, is the localization still finitely generated? – Babai Dec 10 '14 at 18:00

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If $S$ is finitely generated as a monoid, then $S^{-1} R$ is finitely generated as an $R$-algebra. In fact, if $S$ is generated by $s_1,\dotsc,s_n$, then $S^{-1} R$ is generated by $\frac{1}{s_1},\dotsc,\frac{1}{s_n}$. If $S$ is not finitely generated, then usually $S^{-1} R$ is not finitely generated. This applies in particular to the case that $S=R \setminus \mathfrak{p}$ for some prime ideal $\mathfrak{p}$. For instance, $\mathbb{Q}$ is not a finitely generated $\mathbb{Z}$-algebra.

Martin Brandenburg
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  • Can it happen that $S$ is not finitely generated but $S^{-1}R$ is still finitely generated as an $R$-algebra? I can't think of an example offhand. – Tobias Kildetoft Dec 10 '14 at 18:28
  • @TobiasKildetoft: Yes, there are lots of trivial examples because if $0 \in S$ then $S^{-1} R = 0$. – Martin Brandenburg Dec 10 '14 at 19:20
  • But usually we disallow that. Are there examples without this? My guess would be that at least one would need to invert some zero divisors. – Tobias Kildetoft Dec 10 '14 at 19:25
  • "usually we disallow that" - why? who? – Martin Brandenburg Dec 10 '14 at 19:31
  • Hmm, for some reason I remembered it as if that was part of the definition of a multiplicative set. Anyway, one can make similar "trivial" examples by only inverting units. I wonder if there are non-trivial examples (i.e. where the multiplicative set does not contain $0$ and the only unit it contains is $1$). – Tobias Kildetoft Dec 11 '14 at 11:32