Lemma 16.9.2. Let $R \to A \to \Lambda \supset \mathfrak q$ be as in Situation 16.9.1. Let $r \geq 1$ and $\pi _1, \ldots , \pi _ r \in R$ map to elements of $\mathfrak q$. Assume

for $i = 1, \ldots , r$ we have

\[ \text{Ann}_{R/(\pi _1^8, \ldots , \pi _{i - 1}^8)R}(\pi _ i) = \text{Ann}_{R/(\pi _1^8, \ldots , \pi _{i - 1}^8)R}(\pi _ i^2) \]

and

\[ \text{Ann}_{\Lambda /(\pi _1^8, \ldots , \pi _{i - 1}^8)\Lambda }(\pi _ i) = \text{Ann}_{\Lambda /(\pi _1^8, \ldots , \pi _{i - 1}^8)\Lambda }(\pi _ i^2) \]

for $i = 1, \ldots , r$ the element $\pi _ i$ maps to a strictly standard element in $A$ over $R$.

Then, if

\[ R/(\pi _1^8, \ldots , \pi _ r^8)R \to A/(\pi _1^8, \ldots , \pi _ r^8)A \to \Lambda /(\pi _1^8, \ldots , \pi _ r^8)\Lambda \supset \mathfrak q/(\pi _1^8, \ldots , \pi _ r^8)\Lambda \]

can be resolved, so can $R \to A \to \Lambda \supset \mathfrak q$.

**Proof.**
We are going to prove this by induction on $r$.

The case $r = 1$. Here the assumption is that there exists a factorization $A/\pi _1^8 \to \bar C \to \Lambda /\pi _1^8$ which resolves the situation modulo $\pi _1^8$. Conditions (1) and (2) are the assumptions needed to apply Lemma 16.7.3. Thus we can “lift” the resolution $\bar C$ to a resolution of $R \to A \to \Lambda \supset \mathfrak q$.

The case $r > 1$. In this case we apply the induction hypothesis for $r - 1$ to the situation $R/\pi _1^8 \to A/\pi _1^8 \to \Lambda /\pi _1^8 \supset \mathfrak q/\pi _1^8\Lambda $. Note that property (2) is preserved by Lemma 16.2.7.
$\square$

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