# The Stacks Project

## Tag: 00LN

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Lemma 9.68.2. 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 $I' = (i_1, \ldots, i_{d-1}, 0)$. 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 $J' = (f_1, \ldots, f_{c-1})$. We observe that $\sum\nolimits_{|I'| = n - l} a_{I', l} f^{I'}$ is mapped into $J'$ by $f_c^{l}$ and hence (because $f_c$ is not a zerodivisor on $R/J'$) it is in $J'$. By induction hypotheses (for the induction on $c$), we see 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 $$\sum\nolimits_{|I'| = n - l + 1} (a_{I', l-1} + f_c b_{I', l - 1}) f^{I'})f_c^{l-1} + \sum\nolimits_{e = 0}^{l - 2} \left( \sum\nolimits_{|I'| = n - e} a_{I', e} f^{I'} \right) f_c^e = 0$$ 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. $\square$

\begin{lemma}
\label{lemma-regular-quasi-regular}
Let $R$ be a ring.
\begin{enumerate}
\item A regular sequence $f_1, \ldots, f_c$ of $R$ is a quasi-regular
sequence.
\item 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.
\end{enumerate}
\end{lemma}

\begin{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
$I' = (i_1, \ldots, i_{d-1}, 0)$. 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 $J' = (f_1, \ldots, f_{c-1})$.
We observe that $\sum\nolimits_{|I'| = n - l} a_{I', l} f^{I'}$
is mapped into $J'$ by $f_c^{l}$ and hence
(because $f_c$ is not a zerodivisor on $R/J'$) it is in $J'$.
By induction hypotheses (for the induction on $c$),
we see 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
$$\sum\nolimits_{|I'| = n - l + 1} (a_{I', l-1} + f_c b_{I', l - 1}) f^{I'})f_c^{l-1} + \sum\nolimits_{e = 0}^{l - 2} \left( \sum\nolimits_{|I'| = n - e} a_{I', e} f^{I'} \right) f_c^e = 0$$
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.

\medskip\noindent
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.
\end{proof}


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