Lemma 38.7.1. Let R \to S be a ring map. Let N be a S-module. Assume
R is a local ring with maximal ideal \mathfrak m,
\overline{S} = S/\mathfrak m S is Noetherian, and
\overline{N} = N/\mathfrak m_ R N is a finite \overline{S}-module.
Let \Sigma \subset S be the multiplicative subset of elements which are not a zerodivisor on \overline{N}. Then \Sigma ^{-1}S is a semi-local ring whose spectrum consists of primes \mathfrak q \subset S contained in an element of \text{Ass}_ S(\overline{N}). Moreover, any maximal ideal of \Sigma ^{-1}S corresponds to an associated prime of \overline{N} over \overline{S}.
Proof.
Note that \text{Ass}_ S(\overline{N}) = \text{Ass}_{\overline{S}}(\overline{N}), see Algebra, Lemma 10.63.14. This is a finite set by Algebra, Lemma 10.63.5. Say \{ \mathfrak q_1, \ldots , \mathfrak q_ r\} = \text{Ass}_ S(\overline{N}). We have \Sigma = S \setminus (\bigcup \mathfrak q_ i) by Algebra, Lemma 10.63.9. By the description of \mathop{\mathrm{Spec}}(\Sigma ^{-1}S) in Algebra, Lemma 10.17.5 and by Algebra, Lemma 10.15.2 we see that the primes of \Sigma ^{-1}S correspond to the primes of S contained in one of the \mathfrak q_ i. Hence the maximal ideals of \Sigma ^{-1}S correspond one-to-one with the maximal (w.r.t. inclusion) elements of the set \{ \mathfrak q_1, \ldots , \mathfrak q_ r\} . This proves the lemma.
\square
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