Lemma 16.10.2. Let $\Lambda$ be a Noetherian ring. Let $I \subset \Lambda$ be an ideal. Let $I \subset \mathfrak q$ be a prime. Let $n, e$ be positive integers Assume that $\mathfrak q^ n\Lambda _\mathfrak q \subset I\Lambda _\mathfrak q$ and that $\Lambda _\mathfrak q$ is a regular local ring of dimension $d$. Then there exists an $n > 0$ and $\pi _1, \ldots , \pi _ d \in \Lambda$ such that

1. $(\pi _1, \ldots , \pi _ d)\Lambda _\mathfrak q = \mathfrak q\Lambda _\mathfrak q$,

2. $\pi _1^ n, \ldots , \pi _ d^ n \in I$, and

3. for $i = 1, \ldots , d$ we have

$\text{Ann}_{\Lambda /(\pi _1^ e, \ldots , \pi _{i - 1}^ e)\Lambda }(\pi _ i) = \text{Ann}_{\Lambda /(\pi _1^ e, \ldots , \pi _{i - 1}^ e)\Lambda }(\pi _ i^2).$

Proof. Set $S = \Lambda \setminus \mathfrak q$ so that $\Lambda _\mathfrak q = S^{-1}\Lambda$. First pick $\pi _1, \ldots , \pi _ d$ with (1) which is possible as $\Lambda _\mathfrak q$ is regular. By assumption $\pi _ i^ n \in I\Lambda _\mathfrak q$. Thus we can find $s_1, \ldots , s_ d \in S$ such that $s_ i\pi _ i^ n \in I$. Replacing $\pi _ i$ by $s_ i\pi _ i$ we get (2). Note that (1) and (2) are preserved by further multiplying by elements of $S$. Suppose that (3) holds for $i = 1, \ldots , t$ for some $t \in \{ 0, \ldots , d\}$. Note that $\pi _1, \ldots , \pi _ d$ is a regular sequence in $S^{-1}\Lambda$, see Algebra, Lemma 10.106.3. In particular $\pi _1^ e, \ldots , \pi _ t^ e, \pi _{t + 1}$ is a regular sequence in $S^{-1}\Lambda = \Lambda _\mathfrak q$ by Algebra, Lemma 10.68.9. Hence we see that

$\text{Ann}_{S^{-1}\Lambda /(\pi _1^ e, \ldots , \pi _{i - 1}^ e)}(\pi _ i) = \text{Ann}_{S^{-1}\Lambda /(\pi _1^ e, \ldots , \pi _{i - 1}^ e)}(\pi _ i^2).$

Thus we get (3) for $i = t + 1$ after replacing $\pi _{t + 1}$ by $s\pi _{t + 1}$ for some $s \in S$ by Lemma 16.10.1. By induction on $t$ this produces a sequence satisfying (1), (2), and (3). $\square$

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