Lemma 10.69.3. Let $R \to R'$ be a flat ring map. Let $M$ be an $R$-module. Suppose that $f_1, \ldots , f_ r \in R$ form an $M$-quasi-regular sequence. Then the images of $f_1, \ldots , f_ r$ in $R'$ form a $M \otimes _ R R'$-quasi-regular sequence.

Proof. Set $J = (f_1, \ldots , f_ r)$, $J' = JR'$ and $M' = M \otimes _ R R'$. We have to show the canonical map $\mu : R'/J'[X_1, \ldots X_ r] \otimes _{R'/J'} M'/J'M' \to \bigoplus (J')^ nM'/(J')^{n + 1}M'$ is an isomorphism. Because $R \to R'$ is flat the sequences $0 \to J^ nM \to M$ and $0 \to J^{n + 1}M \to J^ nM \to J^ nM/J^{n + 1}M \to 0$ remain exact on tensoring with $R'$. This first implies that $J^ nM \otimes _ R R' = (J')^ nM'$ and then that $(J')^ nM'/(J')^{n + 1}M' = J^ nM/J^{n + 1}M \otimes _ R R'$. Thus $\mu$ is the tensor product of (10.69.0.1), which is an isomorphism by assumption, with $\text{id}_{R'}$ and we conclude. $\square$

Comment #920 by JuanPablo on

I think it wasn't entirely clear how the flatness hypothesis is used. Here flatness hypothesis is used for the following:

If $I$ is an ideal of $R$ and $I' = IR'$ then $IM \otimes_R R'=I'M'$. This is seen by tensoring the exact sequence $0 \rightarrow IM \rightarrow M$. In this lemma for $I=J^n$. The lemma follows tensoring the equation in the definition of quasi-regular sequences.

Comment #4179 by Nils Waßmuth on

I think there is a typo in the definition of the map $\mu$: There are $n$ variables but our sequence only has $r$ entries.

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