## 10.68 Regular sequences

In this section we develop some basic properties of regular sequences.

Definition 10.68.1. Let $R$ be a ring. Let $M$ be an $R$-module. A sequence of elements $f_1, \ldots , f_ r$ of $R$ is called an *$M$-regular sequence* if the following conditions hold:

$f_ i$ is a nonzerodivisor on $M/(f_1, \ldots , f_{i - 1})M$ for each $i = 1, \ldots , r$, and

the module $M/(f_1, \ldots , f_ r)M$ is not zero.

If $I$ is an ideal of $R$ and $f_1, \ldots , f_ r \in I$ then we call $f_1, \ldots , f_ r$ a *$M$-regular sequence in $I$*. If $M = R$, we call $f_1, \ldots , f_ r$ simply a *regular sequence* (in $I$).

Please pay attention to the fact that the definition depends on the order of the elements $f_1, \ldots , f_ r$ (see examples below). Some papers/books drop the requirement that the module $M/(f_1, \ldots , f_ r)M$ is nonzero. This has the advantage that being a regular sequence is preserved under localization. However, we will use this definition mainly to define the depth of a module in case $R$ is local; in that case the $f_ i$ are required to be in the maximal ideal – a condition which is not preserved under going from $R$ to a localization $R_\mathfrak p$.

Example 10.68.2. Let $k$ be a field. In the ring $k[x, y, z]$ the sequence $x, y(1-x), z(1-x)$ is regular but the sequence $y(1-x), z(1-x), x$ is not.

Example 10.68.3. Let $k$ be a field. Consider the ring $k[x, y, w_0, w_1, w_2, \ldots ]/I$ where $I$ is generated by $yw_ i$, $i = 0, 1, 2, \ldots $ and $w_ i - xw_{i + 1}$, $i = 0, 1, 2, \ldots $. The sequence $x, y$ is regular, but $y$ is a zerodivisor. Moreover you can localize at the maximal ideal $(x, y, w_ i)$ and still get an example.

Lemma 10.68.4. Let $R$ be a local Noetherian ring. Let $M$ be a finite $R$-module. Let $x_1, \ldots , x_ c$ be an $M$-regular sequence. Then any permutation of the $x_ i$ is a regular sequence as well.

**Proof.**
First we do the case $c = 2$. Consider $K \subset M$ the kernel of $x_2 : M \to M$. For any $z \in K$ we know that $z = x_1 z'$ for some $z' \in M$ because $x_2$ is a nonzerodivisor on $M/x_1M$. Because $x_1$ is a nonzerodivisor on $M$ we see that $x_2 z' = 0$ as well. Hence $x_1 : K \to K$ is surjective. Thus $K = 0$ by Nakayama's Lemma 10.20.1. Next, consider multiplication by $x_1$ on $M/x_2M$. If $z \in M$ maps to an element $\overline{z} \in M/x_2M$ in the kernel of this map, then $x_1 z = x_2 y$ for some $y \in M$. But then since $x_1, x_2$ is a regular sequence we see that $y = x_1 y'$ for some $y' \in M$. Hence $x_1 ( z - x_2 y' ) =0$ and hence $z = x_2 y'$ and hence $\overline{z} = 0$ as desired.

For the general case, observe that any permutation is a composition of transpositions of adjacent indices. Hence it suffices to prove that

\[ x_1, \ldots , x_{i-2}, x_ i, x_{i-1}, x_{i + 1}, \ldots , x_ c \]

is an $M$-regular sequence. This follows from the case we just did applied to the module $M/(x_1, \ldots , x_{i-2})$ and the length $2$ regular sequence $x_{i-1}, x_ i$.
$\square$

Lemma 10.68.5. Let $R, S$ be local rings. Let $R \to S$ be a flat local ring homomorphism. Let $x_1, \ldots , x_ r$ be a sequence in $R$. Let $M$ be an $R$-module. The following are equivalent

$x_1, \ldots , x_ r$ is an $M$-regular sequence in $R$, and

the images of $x_1, \ldots , x_ r$ in $S$ form a $M \otimes _ R S$-regular sequence.

**Proof.**
This is so because $R \to S$ is faithfully flat by Lemma 10.39.17.
$\square$

Lemma 10.68.6. Let $R$ be a Noetherian ring. Let $M$ be a finite $R$-module. Let $\mathfrak p$ be a prime. Let $x_1, \ldots , x_ r$ be a sequence in $R$ whose image in $R_{\mathfrak p}$ forms an $M_{\mathfrak p}$-regular sequence. Then there exists a $g \in R$, $g \not\in \mathfrak p$ such that the image of $x_1, \ldots , x_ r$ in $R_ g$ forms an $M_ g$-regular sequence.

**Proof.**
Set

\[ K_ i = \mathop{\mathrm{Ker}}\left(x_ i : M/(x_1, \ldots , x_{i - 1})M \to M/(x_1, \ldots , x_{i - 1})M\right). \]

This is a finite $R$-module whose localization at $\mathfrak p$ is zero by assumption. Hence there exists a $g \in R$, $g \not\in \mathfrak p$ such that $(K_ i)_ g = 0$ for all $i = 1, \ldots , r$. This $g$ works.
$\square$

Lemma 10.68.7. Let $A$ be a ring. Let $I$ be an ideal generated by a regular sequence $f_1, \ldots , f_ n$ in $A$. Let $g_1, \ldots , g_ m \in A$ be elements whose images $\overline{g}_1, \ldots , \overline{g}_ m$ form a regular sequence in $A/I$. Then $f_1, \ldots , f_ n, g_1, \ldots , g_ m$ is a regular sequence in $A$.

**Proof.**
This follows immediately from the definitions.
$\square$

Lemma 10.68.8. Let $R$ be a ring. Let $0 \to M_1 \to M_2 \to M_3 \to 0$ be a short exact sequence of $R$-modules. Let $f_1, \ldots , f_ r \in R$. If $f_1, \ldots , f_ r$ is $M_1$-regular and $M_3$-regular, then $f_1, \ldots , f_ r$ is $M_2$-regular.

**Proof.**
By Lemma 10.4.1, if $f_1 : M_1 \to M_1$ and $f_1 : M_3 \to M_3$ are injective, then so is $f_1 : M_2 \to M_2$ and we obtain a short exact sequence

\[ 0 \to M_1/f_1M_1 \to M_2/f_1M_2 \to M_3/f_1M_3 \to 0 \]

The lemma follows from this and induction on $r$. Some details omitted.
$\square$

Lemma 10.68.9. Let $R$ be a ring. Let $M$ be an $R$-module. Let $f_1, \ldots , f_ r \in R$ and $e_1, \ldots , e_ r > 0$ integers. Then $f_1, \ldots , f_ r$ is an $M$-regular sequence if and only if $f_1^{e_1}, \ldots , f_ r^{e_ r}$ is an $M$-regular sequence.

**Proof.**
We will prove this by induction on $r$. If $r = 1$ this follows from the following two easy facts: (a) a power of a nonzerodivisor on $M$ is a nonzerodivisor on $M$ and (b) a divisor of a nonzerodivisor on $M$ is a nonzerodivisor on $M$. If $r > 1$, then by induction applied to $M/f_1M$ we have that $f_1, f_2, \ldots , f_ r$ is an $M$-regular sequence if and only if $f_1, f_2^{e_2}, \ldots , f_ r^{e_ r}$ is an $M$-regular sequence. Thus it suffices to show, given $e > 0$, that $f_1^ e, f_2, \ldots , f_ r$ is an $M$-regular sequence if and only if $f_1, \ldots , f_ r$ is an $M$-regular sequence. We will prove this by induction on $e$. The case $e = 1$ is trivial. Since $f_1$ is a nonzerodivisor under both assumptions (by the case $r = 1$) we have a short exact sequence

\[ 0 \to M/f_1M \xrightarrow {f_1^{e - 1}} M/f_1^ eM \to M/f_1^{e - 1}M \to 0 \]

Suppose that $f_1, f_2, \ldots , f_ r$ is an $M$-regular sequence. Then by induction the elements $f_2, \ldots , f_ r$ are $M/f_1M$ and $M/f_1^{e - 1}M$-regular sequences. By Lemma 10.68.8 $f_2, \ldots , f_ r$ is $M/f_1^ eM$-regular. Hence $f_1^ e, f_2, \ldots , f_ r$ is $M$-regular. Conversely, suppose that $f_1^ e, f_2, \ldots , f_ r$ is an $M$-regular sequence. Then $f_2 : M/f_1^ eM \to M/f_1^ eM$ is injective, hence $f_2 : M/f_1M \to M/f_1M$ is injective, hence by induction(!) $f_2 : M/f_1^{e - 1}M \to M/f_1^{e - 1}M$ is injective, hence

\[ 0 \to M/(f_1, f_2)M \xrightarrow {f_1^{e - 1}} M/(f_1^ e, f_2)M \to M/(f_1^{e - 1}, f_2)M \to 0 \]

is a short exact sequence by Lemma 10.4.1. This proves the converse for $r = 2$. If $r > 2$, then we have $f_3 : M/(f_1^ e, f_2)M \to M/(f_1^ e, f_2)M$ is injective, hence $f_3 : M/(f_1, f_2)M \to M/(f_1, f_2)M$ is injective, and so on. Some details omitted.
$\square$

Lemma 10.68.10. Let $R$ be a ring. Let $f_1, \ldots , f_ r \in R$ which do not generate the unit ideal. The following are equivalent:

any permutation of $f_1, \ldots , f_ r$ is a regular sequence,

any subsequence of $f_1, \ldots , f_ r$ (in the given order) is a regular sequence, and

$f_1x_1, \ldots , f_ rx_ r$ is a regular sequence in the polynomial ring $R[x_1, \ldots , x_ r]$.

**Proof.**
It is clear that (1) implies (2). We prove (2) implies (1) by induction on $r$. The case $r = 1$ is trivial. The case $r = 2$ says that if $a, b \in R$ are a regular sequence and $b$ is a nonzerodivisor, then $b, a$ is a regular sequence. This is clear because the kernel of $a : R/(b) \to R/(b)$ is isomorphic to the kernel of $b : R/(a) \to R/(a)$ if both $a$ and $b$ are nonzerodivisors. The case $r > 2$. Assume (2) holds and say we want to prove $f_{\sigma (1)}, \ldots , f_{\sigma (r)}$ is a regular sequence for some permutation $\sigma $. We already know that $f_{\sigma (1)}, \ldots , f_{\sigma (r - 1)}$ is a regular sequence by induction. Hence it suffices to show that $f_ s$ where $s = \sigma (r)$ is a nonzerodivisor modulo $f_1, \ldots , \hat f_ s, \ldots , f_ r$. If $s = r$ we are done. If $s < r$, then note that $f_ s$ and $f_ r$ are both nonzerodivisors in the ring $R/(f_1, \ldots , \hat f_ s, \ldots , f_{r - 1})$ (by induction hypothesis again). Since we know $f_ s, f_ r$ is a regular sequence in that ring we conclude by the case of sequence of length $2$ that $f_ r, f_ s$ is too.

Note that $R[x_1, \ldots , x_ r]/(f_1x_1, \ldots , f_ ix_ i)$ as an $R$-module is a direct sum of the modules

\[ R/I_ E \cdot x_1^{e_1} \ldots x_ r^{e_ r} \]

indexed by multi-indices $E = (e_1, \ldots , e_ r)$ where $I_ E$ is the ideal generated by $f_ j$ for $1 \leq j \leq i$ with $e_ j > 0$. Hence $f_{i + 1}x_ i$ is a nonzerodivisor on this if and only if $f_{i + 1}$ is a nonzerodivisor on $R/I_ E$ for all $E$. Taking $E$ with all positive entries, we see that $f_{i + 1}$ is a nonzerodivisor on $R/(f_1, \ldots , f_ i)$. Thus (3) implies (2). Conversely, if (2) holds, then any subsequence of $f_1, \ldots , f_ i, f_{i + 1}$ is a regular sequence in particular $f_{i + 1}$ is a nonzerodivisor on all $R/I_ E$. In this way we see that (2) implies (3).
$\square$

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