The Stacks project

10.69 Quasi-regular sequences

We introduce the notion of quasi-regular sequence which is slightly weaker than that of a regular sequence and easier to use. Let $R$ be a ring and let $f_1, \ldots , f_ c \in R$. Set $J = (f_1, \ldots , f_ c)$. Let $M$ be an $R$-module. Then there is a canonical map
\begin{equation} \label{algebra-equation-quasi-regular} M/JM \otimes _{R/J} R/J[X_1, \ldots , X_ c] \longrightarrow \bigoplus \nolimits _{n \geq 0} J^ nM/J^{n + 1}M \end{equation}

of graded $R/J[X_1, \ldots , X_ c]$-modules defined by the rule

\[ \overline{m} \otimes X_1^{e_1} \ldots X_ c^{e_ c} \longmapsto f_1^{e_1} \ldots f_ c^{e_ c} m \bmod J^{e_1 + \ldots + e_ c + 1}M. \]

Note that ( is always surjective.

Definition 10.69.1. Let $R$ be a ring. Let $M$ be an $R$-module. A sequence of elements $f_1, \ldots , f_ c$ of $R$ is called $M$-quasi-regular if ( is an isomorphism. If $M = R$, we call $f_1, \ldots , f_ c$ simply a quasi-regular sequence.

So if $f_1, \ldots , f_ c$ is a quasi-regular sequence, then

\[ R/J[X_1, \ldots , X_ c] = \bigoplus \nolimits _{n \geq 0} J^ n/J^{n + 1} \]

where $J = (f_1, \ldots , f_ c)$. It is clear that being a quasi-regular sequence is independent of the order of $f_1, \ldots , f_ c$.

Lemma 10.69.2. Let $R$ be a ring.

  1. A regular sequence $f_1, \ldots , f_ c$ of $R$ is a quasi-regular sequence.

  2. Suppose that $M$ is an $R$-module and that $f_1, \ldots , f_ c$ is an $M$-regular sequence. Then $f_1, \ldots , f_ c$ is an $M$-quasi-regular sequence.

Proof. Set $J = (f_1, \ldots , f_ c)$. We prove the first assertion by induction on $c$. We have to show that given any relation $\sum _{|I| = n} a_ I f^ I \in J^{n + 1}$ with $a_ I \in R$ we actually have $a_ I \in J$ for all multi-indices $I$. Since any element of $J^{n + 1}$ is of the form $\sum _{|I| = n} b_ I f^ I$ with $b_ I \in J$ we may assume, after replacing $a_ I$ by $a_ I - b_ I$, the relation reads $\sum _{|I| = n} a_ I f^ I = 0$. We can rewrite this as

\[ \sum \nolimits _{e = 0}^ n \left( \sum \nolimits _{|I'| = n - e} a_{I', e} f^{I'} \right) f_ c^ e = 0 \]

Here and below the “primed” multi-indices $I'$ are required to be of the form $I' = (i_1, \ldots , i_{c - 1}, 0)$. We will show by descending induction on $l \in \{ 0, \ldots , n\} $ that if we have a relation

\[ \sum \nolimits _{e = 0}^ l \left( \sum \nolimits _{|I'| = n - e} a_{I', e} f^{I'} \right) f_ c^ e = 0 \]

then $a_{I', e} \in J$ for all $I', e$. Namely, set $J' = (f_1, \ldots , f_{c-1})$. Observe that $\sum \nolimits _{|I'| = n - l} a_{I', l} f^{I'}$ is mapped into $(J')^{n - l + 1}$ by $f_ c^{l}$. By induction hypothesis (for the induction on $c$) we see that $f_ c^ l a_{I', l} \in J'$. Because $f_ c$ is not a zerodivisor on $R/J'$ (as $f_1, \ldots , f_ c$ is a regular sequence) we conclude that $a_{I', l} \in J'$. This allows us to rewrite the term $(\sum \nolimits _{|I'| = n - l} a_{I', l} f^{I'})f_ c^ l$ in the form $(\sum \nolimits _{|I'| = n - l + 1} f_ c b_{I', l - 1} f^{I'})f_ c^{l-1}$. This gives a new relation of the form

\[ \left(\sum \nolimits _{|I'| = n - l + 1} (a_{I', l-1} + f_ c b_{I', l - 1}) f^{I'}\right)f_ c^{l-1} + \sum \nolimits _{e = 0}^{l - 2} \left( \sum \nolimits _{|I'| = n - e} a_{I', e} f^{I'} \right) f_ c^ e = 0 \]

Now by the induction hypothesis (on $l$ this time) we see that all $a_{I', l-1} + f_ c b_{I', l - 1} \in J$ and all $a_{I', e} \in J$ for $e \leq l - 2$. This, combined with $a_{I', l} \in J' \subset J$ seen above, finishes the proof of the induction step.

The second assertion means that given any formal expression $F = \sum _{|I| = n} m_ I X^ I$, $m_ I \in M$ with $\sum m_ I f^ I \in J^{n + 1}M$, then all the coefficients $m_ I$ are in $J$. This is proved in exactly the same way as we prove the corresponding result for the first assertion above. $\square$

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 (, which is an isomorphism by assumption, with $\text{id}_{R'}$ and we conclude. $\square$

Lemma 10.69.4. Let $R$ be a Noetherian ring. Let $M$ be a finite $R$-module. Let $\mathfrak p$ be a prime. Let $x_1, \ldots , x_ c$ be a sequence in $R$ whose image in $R_{\mathfrak p}$ forms an $M_{\mathfrak p}$-quasi-regular sequence. Then there exists a $g \in R$, $g \not\in \mathfrak p$ such that the image of $x_1, \ldots , x_ c$ in $R_ g$ forms an $M_ g$-quasi-regular sequence.

Proof. Consider the kernel $K$ of the map ( As $M/JM \otimes _{R/J} R/J[X_1, \ldots , X_ c]$ is a finite $R/J[X_1, \ldots , X_ c]$-module and as $R/J[X_1, \ldots , X_ c]$ is Noetherian, we see that $K$ is also a finite $R/J[X_1, \ldots , X_ c]$-module. Pick homogeneous generators $k_1, \ldots , k_ t \in K$. By assumption for each $i = 1, \ldots , t$ there exists a $g_ i \in R$, $g_ i \not\in \mathfrak p$ such that $g_ i k_ i = 0$. Hence $g = g_1 \ldots g_ t$ works. $\square$

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. $\square$

Lemma 10.69.6. Let $(R, \mathfrak m)$ be a local Noetherian ring. Let $M$ be a nonzero finite $R$-module. Let $f_1, \ldots , f_ c \in \mathfrak m$ be an $M$-quasi-regular sequence. Then $f_1, \ldots , f_ c$ is an $M$-regular sequence.

Proof. Set $J = (f_1, \ldots , f_ c)$. Let us show that $f_1$ is a nonzerodivisor on $M$. Suppose $x \in M$ is not zero. By Krull's intersection theorem there exists an integer $r$ such that $x \in J^ rM$ but $x \not\in J^{r + 1}M$, see Lemma 10.51.4. Then $f_1 x \in J^{r + 1}M$ is an element whose class in $J^{r + 1}M/J^{r + 2}M$ is nonzero by the assumed structure of $\bigoplus J^ nM/J^{n + 1}M$. Whence $f_1x \not= 0$.

Now we can finish the proof by induction on $c$ using Lemma 10.69.5. $\square$

Remark 10.69.7 (Other types of regular sequences). In the paper [Kabele] the author discusses two more regularity conditions for sequences $x_1, \ldots , x_ r$ of elements of a ring $R$. Namely, we say the sequence is Koszul-regular if $H_ i(K_{\bullet }(R, x_{\bullet })) = 0$ for $i \geq 1$ where $K_{\bullet }(R, x_{\bullet })$ is the Koszul complex. The sequence is called $H_1$-regular if $H_1(K_{\bullet }(R, x_{\bullet })) = 0$. One has the implications regular $\Rightarrow $ Koszul-regular $\Rightarrow $ $H_1$-regular $\Rightarrow $ quasi-regular. By examples the author shows that these implications cannot be reversed in general even if $R$ is a (non-Noetherian) local ring and the sequence generates the maximal ideal of $R$. We introduce these notions in more detail in More on Algebra, Section 15.30.

Remark 10.69.8. Let $k$ be a field. Consider the ring

\[ A = k[x, y, w, z_0, z_1, z_2, \ldots ]/ (y^2z_0 - wx, z_0 - yz_1, z_1 - yz_2, \ldots ) \]

In this ring $x$ is a nonzerodivisor and the image of $y$ in $A/xA$ gives a quasi-regular sequence. But it is not true that $x, y$ is a quasi-regular sequence in $A$ because $(x, y)/(x, y)^2$ isn't free of rank two over $A/(x, y)$ due to the fact that $wx = 0$ in $(x, y)/(x, y)^2$ but $w$ isn't zero in $A/(x, y)$. Hence the analogue of Lemma 10.68.7 does not hold for quasi-regular sequences.

Lemma 10.69.9. Let $R$ be a ring. Let $J = (f_1, \ldots , f_ r)$ be an ideal of $R$. Let $M$ be an $R$-module. Set $\overline{R} = R/\bigcap _{n \geq 0} J^ n$, $\overline{M} = M/\bigcap _{n \geq 0} J^ nM$, and denote $\overline{f}_ i$ the image of $f_ i$ in $\overline{R}$. Then $f_1, \ldots , f_ r$ is $M$-quasi-regular if and only if $\overline{f}_1, \ldots , \overline{f}_ r$ is $\overline{M}$-quasi-regular.

Proof. This is true because $J^ nM/J^{n + 1}M \cong \overline{J}^ n\overline{M}/\overline{J}^{n + 1}\overline{M}$. $\square$

Comments (4)

Comment #7891 by Laurent Berger on

In the first sentence "There is a notion of regular sequence which is slightly weaker than that of a regular sequence" the word "quasi" is presumably missing.

Comment #8824 by Ryo Suzuki on

It might be good to add this lemma:

Let be a ring and let . Set . Let be an -module. Suppose that, for any and a homogeneous polynomial of degree , if then the coefficients of is in . Then is a -quasi-regular sequence.

It is essentially proved in the proof of Lemma 00LN. It is also appear in the proof of Lemma 062I and its proof might become cleaner by using this lemma.

Comment #9266 by on

This is essentially just the definition of a quasi-regular sequence and I think we're just using this directly in the places you mention. Going to leave as is for now.

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