## 10.69 Quasi-regular sequences

There is a notion of 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

10.69.0.1
\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 (10.69.0.1) 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 (10.69.0.1) 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 (10.69.0.1), 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 (10.69.0.1). 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 $\bigoplus _{n \geq 0} J^ nM/J^{n + 1} \cong M/JM[X_1, \ldots , X_ c]$ 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 (Koszul regular sequences). In the paper [Kabele] the author introduces 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 $H_1(K_{\bullet }(R, x_{\bullet })) = 0$. If $R$ is a local ring (possibly non-Noetherian) and the sequence consists of elements of the maximal ideal, then 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. 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 $J^ nM/J^{n + 1}M \cong \overline{J}^ n\overline{M}/\overline{J}^{n + 1}\overline{M}$. $\square$

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