I am getting a little confused about the relationship (if any) between two definitions/uses of the term “saturation.” Let $R$ be a commutative ring with 1 and let $I\subset R$ be an ideal.

Atiyah-MacDonald gives one characterization. If $S\subset R$ is a multiplicatively closed subset, then the saturation of $I$ with respect to $S$, denoted $S(I)$, is the contraction of the ideal $S^{-1}I\subset S^{-1}R$. More explicitly, if $\iota:R\to S^{-1}R$ is the natural map to the localization, then $S(I) = \iota^{-1}(\iota(I)S^{-1}R)$, i.e., the contraction of the extension of $I$ along $\iota$.

On the other hand, Eisenbud, in the chapter on Gröbner bases, defines the saturation of $I$ with respect to an ideal $J$ to be $(I:J^\infty) = \bigcup_{n=1}^\infty (I:J^n)$.

I think if $J=(x)$ is principal then taking $S=\{1,x,x^2,\dots\}$, the definitions agree, because $S(I) = \bigcup_{s\in S}(I:s)$. But if $J$ is not principal or $S$ is some arbitrary multiplicatively closed set, I was wondering if there is some relationship between the two definitions that would justify the use of the same terminology.

This question is in part motivated by an exercise from Eisenbud that involves working with the ideal $(I:(x,y)^\infty)$ for some $x,y\in R$. I would like to be able to characterize this ideal in terms of the contraction of the extension of $I$ through some localization—the intuitive guess would be at the multiplicatively closed set $S=\{1,x,y,x^2,xy,y^2,\dots\}$—but it’s not clear to me why this should be possible: I can see that $(I:(x,y)^\infty)\subset S(I)$, but I can’t see any reason to expect the reverse inclusion to hold. (My hope is that I’m missing something, and that the similar terminology between these definitions indicates some connection I’m missing.)

  • 1
    $\begingroup$ I was thinking about this exact situation some time back and came across the following post: math.stackexchange.com/q/2506016/588038. I hope you find it useful. $\endgroup$
    – cqfd
    Feb 28, 2023 at 1:56
  • $\begingroup$ That does seem quite helpful—thanks for the link! $\endgroup$ Mar 1, 2023 at 1:03
  • 1
    $\begingroup$ Hello, if you are still interested, I wrote an answer here. $\endgroup$
    – cqfd
    Mar 16, 2023 at 16:39


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