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 $e_ i \wedge e_ j \otimes m \mapsto (f_ ie_ j - f_ je_ i) \otimes m$. In particular this implies that

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

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 $I$. Note that $f_1, \ldots , f_{r - 1}, f_ r^ n$ is an $M$-$H_1$-regular sequence by Lemma 15.30.4. Hence we see that the required result holds for the multi-index $I = (0, \ldots , 0, n)$. It turns out that we can reduce the general case to this case as follows.

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

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

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

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

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

$\sum m_ I x_1^{i_1} \ldots x_ r^{i_ r} \in J(M \otimes _ R S).$

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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