Lemma 10.69.5. Let $R$ be a ring. Let $M$ be an $R$-module. Let $f_1, \ldots , f_ c \in R$ be an $M$-quasi-regular sequence. For any $i$ the sequence $\overline{f}_{i + 1}, \ldots , \overline{f}_ c$ of $\overline{R} = R/(f_1, \ldots , f_ i)$ is an $\overline{M} = M/(f_1, \ldots , f_ i)M$-quasi-regular sequence.

Proof. It suffices to prove this for $i = 1$. Set $\overline{J} = (\overline{f}_2, \ldots , \overline{f}_ c) \subset \overline{R}$. Then

\begin{align*} \overline{J}^ n\overline{M}/\overline{J}^{n + 1}\overline{M} & = (J^ nM + f_1M)/(J^{n + 1}M + f_1M) \\ & = J^ nM / (J^{n + 1}M + J^ nM \cap f_1M). \end{align*}

Thus, in order to prove the lemma it suffices to show that $J^{n + 1}M + J^ nM \cap f_1M = J^{n + 1}M + f_1J^{n - 1}M$ because that will show that $\bigoplus _{n \geq 0} \overline{J}^ n\overline{M}/\overline{J}^{n + 1}\overline{M}$ is the quotient of $\bigoplus _{n \geq 0} J^ nM/J^{n + 1}M \cong M/JM[X_1, \ldots , X_ c]$ by $X_1$. Actually, we have $J^ nM \cap f_1M = f_1J^{n - 1}M$. Namely, if $m \not\in J^{n - 1}M$, then $f_1m \not\in J^ nM$ because $\bigoplus J^ nM/J^{n + 1}M$ is the polynomial algebra $M/J[X_1, \ldots , X_ c]$ by assumption. $\square$

## Comments (2)

Comment #8809 by on

Typo: in the iso after "is the quotient of", there's a missing $M$ in the denominator of the LHS.

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• 4 comment(s) on Section 10.69: Quasi-regular sequences

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