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.
$\square$
Comments (0)