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
(\pi _1, \ldots , \pi _ d)\Lambda _\mathfrak q = \mathfrak q\Lambda _\mathfrak q,
\pi _1^ n, \ldots , \pi _ d^ n \in I, and
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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