Proof. Let $R$ be a ring and let $M$ be an $R$-module. Let $f_1, \ldots , f_ r$ be an $M$-$H_1$-regular sequence. Denote $J = (f_1, \ldots , f_ r)$. The assumption means that we have an exact sequence

$\wedge ^2(R^ r) \otimes M \to R^{\oplus r} \otimes M \to JM \to 0$

where the first arrow is given by $e_ i \wedge e_ j \otimes m \mapsto (f_ ie_ j - f_ je_ i) \otimes m$. Tensoring the sequence with $R/J$ we see that

$JM/J^2M = (R/J)^{\oplus r} \otimes _ R M = (M/JM)^{\oplus r}$

is a finite free module. To finish the proof we have to prove for every $n \geq 2$ the following: if

$\xi = \sum \nolimits _{|I| = n, I = (i_1, \ldots , i_ r)} m_ I f_1^{i_1} \ldots f_ r^{i_ r} \in J^{n + 1}M$

then $m_ I \in JM$ for all $I$. In the next paragraph, we prove $m_ I \in JM$ for $I = (0, \ldots , 0, n)$ and in the last paragraph we deduce the general case from this special case.

Let $I = (0, \ldots , 0, n)$. Let $\xi$ be as above. We can write $\xi = m_1 f_1 + \ldots + m_{r - 1}f_{r - 1} + m_ I f_ r^ n$. As we have assumed $\xi \in J^{n + 1}M$, we can also write $\xi = \sum _{1 \leq i \leq j \leq r - 1} m_{ij}f_ if_ j + \sum _{1 \leq i \leq r - 1}m'_ i f_ if_ r^ n + m'' f_ r^{n + 1}$. Then we see that

$\begin{matrix} (m_1 - m_{11}f_1 - m'_1f_ r^ n)f_1 + \\ (m_2 - m_{12}f_1 - m_{22}f_2 - m'_2f_ r^ n)f_2 + \\ \ldots + \\ (m_{r - 1} - m_{1 r - 1}f_1 - \ldots - m_{r - 1 r - 1}f_{r - 1} - m'_{r - 1}f_ r^ n)f_{r - 1} + \\ (m_ I - m'' f_ r)f_ r^ n = 0 \end{matrix}$

Since $f_1, \ldots , f_{r - 1}, f_ r^ n$ is $M$-$H_1$-regular by Lemma 15.30.4 we see that $m_ I - m'' f_ r$ is in the submodule $f_1M + \ldots + f_{r - 1}M + f_ r^ nM$. Thus $m_ I \in f_1M + \ldots + f_ rM$.

Let $S = R[x_1, x_2, \ldots , x_ r, 1/x_ r]$. The ring map $R \to S$ is faithfully flat, hence $f_1, \ldots , f_ r$ is an $M$-$H_1$-regular sequence in $S$, see Lemma 15.30.5. By Lemma 15.28.4 we see that

$g_1 = f_1 - \frac{x_1}{x_ r} f_ r, \ \ldots , \ g_{r - 1} = f_{r - 1} - \frac{x_{r - 1}}{x_ r} f_ r, \ g_ r = \frac{1}{x_ r}f_ r$

is an $M$-$H_1$-regular sequence in $S$. Finally, note that our element $\xi$ can be rewritten

$\xi = \sum \nolimits _{|I| = n, I = (i_1, \ldots , i_ r)} m_ I (g_1 + x_ i g_ r)^{i_1} \ldots (g_{r - 1} + x_ i g_ r)^{i_{r - 1}} (x_ rg_ r)^{i_ r}$

and the coefficient of $g_ r^ n$ in this expression is

$\sum m_ I x_1^{i_1} \ldots x_ r^{i_ r}$

By the case discussed in the previous paragraph this sum is in $J(M \otimes _ R S)$. Since the monomials $x_1^{i_1} \ldots x_ r^{i_ r}$ form part of an $R$-basis of $S$ over $R$ we conclude that $m_ I \in J$ for all $I$ as desired. $\square$

Comment #6253 by on

The question asked was: why do we have $m_I \in JM$ for $I = (0, \ldots, 0, n)$? Well, we can write $\xi = m_1 f_1 + \ldots + m_{r - 1}f_{r - 1} + m_I f_r^n$ and we know it is in $J^{n + 1}M$ so we can also write $\xi = \sum_{1 \leq i \leq j \leq r - 1} m_{ij}f_if_j + \sum_{1 \leq i \leq r - 1}m'_i f_if_r^n + m'' f_r^{n + 1}$. Then we see that By definition of $H_1$-regular we see that $m_I - m'' f_r$ is in the submodule $f_1M + \ldots + f_{r - 1}M + f_r^nM$. OK? I could not see how this implies the same for the other coefficients, because I don't know how to do the same algebra trick for the other coefficients.

I shall add some text to the proof the next time I go through all the comments.

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• 4 comment(s) on Section 15.30: Koszul regular sequences

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