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
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