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*}

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 (4)

Comment #918 by JuanPablo on

In the last paragraph of the proof the ideal of should be generated by (not ) where and . The equivalence between (2) and (3) is then immediate.

Comment #922 by on

Hi! I do not understand your comment. For example if and then what you say would imply that the degree 2 part of is annihilated by and it is not.

Comment #925 by JuanPablo on

In the last paragraph of the proof the ideal of should be generated by (not ) where and . The equivalence between (2) and (3) is then immediate.

Hi. In the example it seems that anihilates the degree part because (it is the ideal generated by not the algebra).

Comment #928 by on

Ah, yes, of course. Don't know what I was thinking, sorry! The fix is here.


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 07DW. Beware of the difference between the letter 'O' and the digit '0'.