Lemma 15.30.17. Let $R$ be a ring. Let $f_1, \ldots , f_ n$ be a Koszul-regular sequence in $R$ such that $(f_1, \ldots , f_ n) \not= R$. 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 15.30.16 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 15.30.5 and 15.30.15. We conclude that $\overline{f}_2, \ldots , \overline{f}_ n$ is a Koszul-regular sequence in $S'/(g_1)$ by Lemma 15.30.14. 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 $S$.
$\square$

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