## 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 6329–6344 (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}
```

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