The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

10.67 Regular sequences

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

Definition 10.67.1. 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:

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

  2. 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 10.67.2. 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 10.67.3. 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 10.67.4. 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 10.19.1. 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 10.67.5. 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

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

  2. 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 10.38.17. $\square$

Lemma 10.67.6. 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 10.67.7. 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 10.67.8. Let $R$ be a ring. Let $M$ be an $R$-module. Let $f_1, \ldots , f_ r \in R$ be an $M$-regular sequence. Then for $e_1, \ldots , e_ r > 0$ the sequence $f_1^{e_1}, \ldots , f_ r^{e_ r}$ is $M$-regular too.

Proof. We will prove this by induction on $r$. If $r = 1$ this follows from the fact that a power of a nonzerodivisor on $M$ is a nonzerodivisor on $M$. If $r > 1$, then by induction applied to $M/f_1M$ we have that $f_1, f_2^{e_2}, \ldots , f_ r^{e_ r}$ is an $M$-regular sequence. Thus it suffices to show that $f_1^ e, f_2, \ldots , f_ r$ is an $M$-regular sequence 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 we have a short exact sequence

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

By induction the elements $f_2, \ldots , f_ r$ are $M/f_1M$ and $M/f_1^{e - 1}M$-regular sequences. It follows from the snake lemma that they are also $M/f_1^ eM$-regular sequences. $\square$

Lemma 10.67.9. Let $R$ be a ring. Let $f_1, \ldots , f_ r \in R$ which do not generate the unit ideal. The following are equivalent:

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

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

  3. $f_1x_1, \ldots , f_ rx_ r$ 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 by Lemma 10.67.8, i.e., hence $f_{i + 1}$ is a nonzerodivisor on all $R/I_ E$. In this way we see that (2) implies (3). $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0AUH. Beware of the difference between the letter 'O' and the digit '0'.