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Tag 068Q

Chapter 15: More on Algebra > Section 15.27: Koszul regular sequences

Lemma 15.27.17. Let $R$ be a ring. Let $f_1, \ldots, f_n$ be a Koszul-regular sequence in $R$ such that $(f_1, \ldots, f_r) \not = R$. Consider the faithfully flat, smooth ring map $$ R \longrightarrow S = R[\{t_{ij}\}_{i \leq j}, t_{11}^{-1}, t_{22}^{-1}, \ldots, t_{nn}^{-1}] $$ For $1 \leq i \leq n$ set $$ g_i = \sum\nolimits_{i \leq j} t_{ij} f_j \in S. $$ Then $g_1, \ldots, g_n$ is a regular sequence in $S$ and $(f_1, \ldots, f_n)S = (g_1, \ldots, g_n)$.

Proof. The equality of ideals is obvious as the matrix $$ \left( \begin{matrix} t_{11} & t_{12} & t_{13} & \ldots \\ 0 & t_{22} & t_{23} & \ldots \\ 0 & 0 & t_{33} & \ldots \\ \ldots & \ldots & \ldots & \ldots \end{matrix} \right) $$ is invertible in $S$. Because $f_1, \ldots, f_n$ is a Koszul-regular sequence we see that the kernel of $R \to R^{\oplus n}$, $x \mapsto (xf_1, \ldots, xf_n)$ is zero (as it computes the $n$the Koszul homology of $R$ w.r.t. $f_1, \ldots, f_n$). Hence by Lemma 15.27.16 we see that $g_1 = f_1 t_{11} + \ldots + f_n t_{1n}$ is a nonzerodivisor in $S' = R[t_{11}, t_{12}, \ldots, t_{1n}, t_{11}^{-1}]$. We see that $g_1, f_2, \ldots, f_n$ is a Koszul-sequence in $S'$ by Lemma 15.27.5 and 15.27.15. We conclude that $\overline{f}_2, \ldots, \overline{f}_n$ is a Koszul-regular sequence in $S'/(g_1)$ by Lemma 15.27.14. Hence by induction on $n$ we see that the images $\overline{g}_2, \ldots, \overline{g}_n$ of $g_2, \ldots, g_n$ in $S'/(g_1)[\{t_{ij}\}_{2 \leq i \leq j}, t_{22}^{-1}, \ldots, t_{nn}^{-1}]$ form a regular sequence. This in turn means that $g_1, \ldots, g_n$ forms a regular sequence in $S$. $\square$

    The code snippet corresponding to this tag is a part of the file more-algebra.tex and is located in lines 6375–6390 (see updates for more information).

    \begin{lemma}
    \label{lemma-Koszul-regular-flat-locally-regular}
    Let $R$ be a ring. Let $f_1, \ldots, f_n$ be a Koszul-regular sequence
    in $R$ such that $(f_1, \ldots, f_r) \not = R$.
    Consider the faithfully flat, smooth ring map
    $$
    R \longrightarrow
    S = R[\{t_{ij}\}_{i \leq j}, t_{11}^{-1}, t_{22}^{-1}, \ldots, t_{nn}^{-1}]
    $$
    For $1 \leq i \leq n$ set
    $$
    g_i = \sum\nolimits_{i \leq j} t_{ij} f_j \in S.
    $$
    Then $g_1, \ldots, g_n$ is a regular sequence in $S$ and
    $(f_1, \ldots, f_n)S = (g_1, \ldots, g_n)$.
    \end{lemma}
    
    \begin{proof}
    The equality of ideals is obvious as the matrix
    $$
    \left(
    \begin{matrix}
    t_{11} & t_{12} & t_{13} & \ldots \\
    0 & t_{22} & t_{23} & \ldots \\
    0 & 0 & t_{33} & \ldots \\
    \ldots & \ldots & \ldots & \ldots
    \end{matrix}
    \right)
    $$
    is invertible in $S$.
    Because $f_1, \ldots, f_n$ is a Koszul-regular sequence we see that
    the kernel of
    $R \to R^{\oplus n}$, $x \mapsto (xf_1, \ldots, xf_n)$ is zero (as it
    computes the $n$the Koszul homology of $R$ w.r.t.\ $f_1, \ldots, f_n$).
    Hence by
    Lemma \ref{lemma-make-nonzero-divisor}
    we see that $g_1 = f_1 t_{11} + \ldots + f_n t_{1n}$ is a nonzerodivisor
    in $S' = R[t_{11}, t_{12}, \ldots, t_{1n}, t_{11}^{-1}]$. We see that
    $g_1, f_2, \ldots, f_n$ is a Koszul-sequence in $S'$ by
    Lemma \ref{lemma-koszul-regular-flat-base-change} and
    \ref{lemma-independence-of-generators}.
    We conclude that
    $\overline{f}_2, \ldots, \overline{f}_n$ is a Koszul-regular sequence
    in $S'/(g_1)$ by
    Lemma \ref{lemma-truncate-koszul-regular}.
    Hence by induction on $n$ we see that the images
    $\overline{g}_2, \ldots, \overline{g}_n$ of $g_2, \ldots, g_n$ in
    $S'/(g_1)[\{t_{ij}\}_{2 \leq i \leq j}, t_{22}^{-1}, \ldots, t_{nn}^{-1}]$
    form a regular sequence. This in turn means that
    $g_1, \ldots, g_n$ forms a regular sequence in $S$.
    \end{proof}

    Comments (7)

    Comment #1929 by DB on April 25, 2016 a 1:26 pm UTC

    It seems like the definition of regular sequence (Tag 00LF) is incompatible with the weaker notions of regular sequence (Koszul-, H1-, quasi-). The former definition excludes 1 as a regular sequence wheras the latter don't.

    For instance the lemma above is false if we take f_1 = 1. Similar errors occur elsewhere. For instance in the discussion (Tag 062D) after the definition of Koszul-regular sequence.

    Comment #1930 by Johan (site) on April 25, 2016 a 11:51 pm UTC

    OK, thanks for pointing this out. The correct statement probably should be that the sequence is regular provided that the ideal isn't the whole ring. Yes? I will fix this soonish.

    We tried to be careful with this, so please point out any other mistakes you see with this. In fact I think the discussion following the definition of a Koszul regular sequence (in Section 062D) is correct except maybe that in the very last part we need to assume $f = f_1$ is not a unit; is that what you mean?

    Comment #1932 by DB on April 27, 2016 a 4:19 pm UTC

    Yes, I think the statement in section 062D is true if $f = f_1$ is not a unit. And yes I think the statement is correct if one also assume that the $f_i$ does not generate the whole thing.

    Is changing the definitions an option? It seems a bit strange to me to impose the extra condition that $f_1, \ldots f_r$ should not generate the whole ring in the case for regular sequences but not imposing the condition for the other regularity notions. In for instance EGA IV-1 Ch 0 15.2.2 a regular sequence is allowed to generate the whole ring.

    Comment #1933 by Johan (site) on April 28, 2016 a 12:43 am UTC

    OK, many thanks for the reply.

    About the definition of a regular sequence: The reason for the current choice is that in most of the commutative algebra texts I looked at the notion of a regular sequence was defined in this manner. For example, this is how it is defined in:

    1. Eisenbud's commutative algebra
    2. Kaplansky's Ring Theory
    3. Matsumura Commutative Ring Theory
    4. Matsumura, Commutative Algebra
    5. Kunz, Introduction to Commutative Algebra and Algebraic Geometry
    6. and so on

    The notion of a regular sequence is very iffy. It depends on the order of the sequence even in the local situation when the elements all come from the maximal ideal. So making it a notion which is hard to use, may actually be an advantage.

    Afterthought: it is "well known" that over a Noetherian local ring, the order of the elements in a regular sequence does not matter (Lemma 00LJ). But this statement would be false if we changed the definition!

    Comment #1935 by DB on April 28, 2016 a 12:56 pm UTC

    I don't know which definition is "best", but it makes sense to follow the main textbooks on the subject as you say (perhaps together with a warning that also the other definition is common). My guess is that the choice in EGA is made in order to make regular sequences respect localisations.

    But my point was rather that if one makes one choice for one of the regularity conditions, then it might save the readers (and the writers) some headache if one makes the same choice for the others. For instance, in Matsumura's "Commutative ring theory", a quasi-regular seqence is not allowed to generate the whole ring (p. 124), which is consistent with the definition of regular sequence given in op. cit.

    Comment #1937 by Johan (site) on April 29, 2016 a 1:51 pm UTC

    OK, your point is a good one and there are also plenty of references which define regular sequences without requiring the nonvanishing of the quotient. In fact you are in good company because Burt Totaro suggested the same thing in 2013. So yeah, maybe we'll need to change the definition... Anybody who reads this and agrees, please leave a comment!

    Thanks again for the comment. I just fixed it.

    Comment #2775 by Darij Grinberg (site) on August 19, 2017 a 2:30 am UTC

    The "r" in the 2nd sentence should be an "n".

    There is also 1 comment on Section 15.27: More on Algebra.

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