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Robert Gilmer, in his paper "Commutative rings in which each prime ideal is principal", says:

Some well known theorems indicate that certain ideal-theoretic structure properties of a commutative ring $R$ are determined by the set of prime ideals of $R$. For example,
$R$ is Noetherian if and only if each prime ideal of $R$ is finitely generated (1, 2),
$R$ is a multiplication ring if and only if each prime ideal of $R$ is a multiplication ideal, and
if $R$ contains an identity, each nonzero ideal of $R$ is invertible if and only if each nonzero prime ideal of $R$ is invertible.

I would like to have a list of theorems that a property of a commutative ring $R$ (or $R$-modules) is determined by the prime ideals of $R$.

I will gradually list the few things I remember here, and I would appreciate if you complete it:

  • Baer's criterion for injective modules has been refined in many ways including a result that for a commutative Noetherian ring, it suffices to consider only prime ideals instead of all ideals.
  • when $R$ is a commutative Noetherian local ring with maximal ideal $m$, global dimension of the ring $R$ can be alternatively defined as the projective dimension of the residue field $R/m$.
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    $\begingroup$ "determined by the set of prime ideals" is a bit ambiguous. What structure do you retain on this set? For instance, you can consider this set as an ordered set. E.g., what "be a multiplication ideal" retains? Without further information I think that any ring property can be disguised into a property in terms of prime ideals. $\endgroup$
    – YCor
    Commented Sep 19, 2022 at 14:13
  • $\begingroup$ properties like the ones listed in the question. $\endgroup$
    – user 1
    Commented Sep 19, 2022 at 18:54
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    $\begingroup$ Each prime ideal being finitely generated is not a property of the set of prime ideals. $\endgroup$
    – Fernando Muro
    Commented Jan 16 at 8:29
  • $\begingroup$ thank you Fernando Muro . edited. If the question seems dumb, please see the text I quoted from Gilmer. I mean examples like ones in that text; Also, what I have written in the bounty text. $\endgroup$
    – user 1
    Commented Jan 16 at 8:37
  • $\begingroup$ Noetherianness is certainly not determined by the (partially ordered) set of prime ideals. Every valuation ring of rank 1 has the same poset of prime ideals, but there are non-Noetherian valuation rings of rank 1. $\endgroup$
    – Z. M
    Commented Jan 18 at 19:07

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Proposition 1.2.10(a) in "Cohen–Macaulay Rings" by W. Bruns, H.J. Herzog: if $R$ is a Noetherian ring, $I$ an ideal of $R$, and $M$ a finite $R$-module, then $$\operatorname{grade} (I,M)= \inf \{\operatorname{depth} M_p \mid p \in V(I)\}.$$

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