## 15.30 Koszul regular sequences

Please take a look at Algebra, Sections 10.68, 10.69, and 10.72 before looking at this one.

Definition 15.30.1. Let $R$ be a ring. Let $r \geq 0$ and let $f_1, \ldots , f_ r \in R$ be a sequence of elements. Let $M$ be an $R$-module. The sequence $f_1, \ldots , f_ r$ is called

1. $M$-Koszul-regular if $H_ i(K_\bullet (f_1, \ldots , f_ r) \otimes _ R M) = 0$ for all $i \not= 0$,

2. $M$-$H_1$-regular if $H_1(K_\bullet (f_1, \ldots , f_ r) \otimes _ R M) = 0$,

3. Koszul-regular if $H_ i(K_\bullet (f_1, \ldots , f_ r)) = 0$ for all $i \not= 0$, and

4. $H_1$-regular if $H_1(K_\bullet (f_1, \ldots , f_ r)) = 0$.

We will see in Lemmas 15.30.2, 15.30.3, and 15.30.6 that for elements $f_1, \ldots , f_ r$ of a ring $R$ we have the following implications

\begin{align*} f_1, \ldots , f_ r\text{ is a regular sequence} & \Rightarrow f_1, \ldots , f_ r\text{ is a Koszul-regular sequence} \\ & \Rightarrow f_1, \ldots , f_ r\text{ is an }H_1\text{-regular sequence} \\ & \Rightarrow f_1, \ldots , f_ r\text{ is a quasi-regular sequence.} \end{align*}

In general none of these implications can be reversed, but if $R$ is a Noetherian local ring and $f_1, \ldots , f_ r \in \mathfrak m_ R$, then the four conditions are all equivalent (Lemma 15.30.7). If $f = f_1 \in R$ is a length $1$ sequence and $f$ is not a unit of $R$ then it is clear that the following are all equivalent

1. $f$ is a regular sequence of length one,

2. $f$ is a Koszul-regular sequence of length one, and

3. $f$ is a $H_1$-regular sequence of length one.

It is also clear that these imply that $f$ is a quasi-regular sequence of length one. But there do exist quasi-regular sequences of length $1$ which are not regular sequences. Namely, let

$R = k[x, y_0, y_1, \ldots ]/(xy_0, xy_1 - y_0, xy_2 - y_1, \ldots )$

and let $f$ be the image of $x$ in $R$. Then $f$ is a zerodivisor, but $\bigoplus _{n \geq 0} (f^ n)/(f^{n + 1}) \cong k[x]$ is a polynomial ring.

Lemma 15.30.2. An $M$-regular sequence is $M$-Koszul-regular. A regular sequence is Koszul-regular.

Proof. Let $R$ be a ring and let $M$ be an $R$-module. It is immediate that an $M$-regular sequence of length $1$ is $M$-Koszul-regular. Let $f_1, \ldots , f_ r$ be an $M$-regular sequence. Then $f_1$ is a nonzerodivisor on $M$. Hence

$0 \to K_\bullet (f_2, \ldots , f_ r) \otimes M \xrightarrow {f_1} K_\bullet (f_2, \ldots , f_ r) \otimes M \to K_\bullet (\overline{f}_2, \ldots , \overline{f}_ r) \otimes M/f_1M \to 0$

is a short exact sequence of complexes where $\overline{f}_ i$ is the image of $f_ i$ in $R/(f_1)$. By Lemma 15.28.8 the complex $K_\bullet (R, f_1, \ldots , f_ r)$ is isomorphic to the cone of multiplication by $f_1$ on $K_\bullet (f_2, \ldots , f_ r)$. Thus $K_\bullet (R, f_1, \ldots , f_ r) \otimes M$ is isomorphic to the cone on the first map. Hence $K_\bullet (\overline{f}_2, \ldots , \overline{f}_ r) \otimes M/f_1M$ is quasi-isomorphic to $K_\bullet (f_1, \ldots , f_ r) \otimes M$. As $\overline{f}_2, \ldots , \overline{f}_ r$ is an $M/f_1M$-regular sequence in $R/(f_1)$ the result follows from the case $r = 1$ and induction. $\square$

Lemma 15.30.3. A $M$-Koszul-regular sequence is $M$-$H_1$-regular. A Koszul-regular sequence is $H_1$-regular.

Proof. This is immediate from the definition. $\square$

Lemma 15.30.4. Let $f_1, \ldots , f_{r - 1} \in R$ be a sequence and $f, g \in R$. Let $M$ be an $R$-module.

1. If $f_1, \ldots , f_{r - 1}, f$ and $f_1, \ldots , f_{r - 1}, g$ are $M$-$H_1$-regular then $f_1, \ldots , f_{r - 1}, fg$ is $M$-$H_1$-regular too.

2. If $f_1, \ldots , f_{r - 1}, f$ and $f_1, \ldots , f_{r - 1}, g$ are $M$-Koszul-regular then $f_1, \ldots , f_{r - 1}, fg$ is $M$-Koszul-regular too.

Proof. By Lemma 15.28.11 we have exact sequences

$H_ i(K_\bullet (f_1, \ldots , f_{r - 1}, f) \otimes M) \to H_ i(K_\bullet (f_1, \ldots , f_{r - 1}, fg) \otimes M) \to H_ i(K_\bullet (f_1, \ldots , f_{r - 1}, g) \otimes M)$

for all $i$. $\square$

Lemma 15.30.5. Let $\varphi : R \to S$ be a flat ring map. Let $f_1, \ldots , f_ r \in R$. Let $M$ be an $R$-module and set $N = M \otimes _ R S$.

1. If $f_1, \ldots , f_ r$ in $R$ is an $M$-$H_1$-regular sequence, then $\varphi (f_1), \ldots , \varphi (f_ r)$ is an $N$-$H_1$-regular sequence in $S$.

2. If $f_1, \ldots , f_ r$ is an $M$-Koszul-regular sequence in $R$, then $\varphi (f_1), \ldots , \varphi (f_ r)$ is an $N$-Koszul-regular sequence in $S$.

Proof. This is true because $K_\bullet (f_1, \ldots , f_ r) \otimes _ R S = K_\bullet (\varphi (f_1), \ldots , \varphi (f_ r))$ and therefore $(K_\bullet (f_1, \ldots , f_ r) \otimes _ R M) \otimes _ R S = K_\bullet (\varphi (f_1), \ldots , \varphi (f_ r)) \otimes _ S N$. $\square$

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. $\square$

For nonzero finite modules over Noetherian local rings all of the types of regular sequences introduced so far are equivalent.

Lemma 15.30.7. Let $(R, \mathfrak m)$ be a Noetherian local ring. Let $M$ be a nonzero finite $R$-module. Let $f_1, \ldots , f_ r \in \mathfrak m$. The following are equivalent

1. $f_1, \ldots , f_ r$ is an $M$-regular sequence,

2. $f_1, \ldots , f_ r$ is a $M$-Koszul-regular sequence,

3. $f_1, \ldots , f_ r$ is an $M$-$H_1$-regular sequence,

4. $f_1, \ldots , f_ r$ is an $M$-quasi-regular sequence.

In particular the sequence $f_1, \ldots , f_ r$ is a regular sequence in $R$ if and only if it is a Koszul regular sequence, if and only if it is a $H_1$-regular sequence, if and only if it is a quasi-regular sequence.

Proof. The implication (1) $\Rightarrow$ (2) is Lemma 15.30.2. The implication (2) $\Rightarrow$ (3) is Lemma 15.30.3. The implication (3) $\Rightarrow$ (4) is Lemma 15.30.6. The implication (4) $\Rightarrow$ (1) is Algebra, Lemma 10.69.6. $\square$

Lemma 15.30.8. Let $A$ be a ring. Let $I \subset A$ be an ideal. Let $g_1, \ldots , g_ m$ be a sequence in $A$ whose image in $A/I$ is $H_1$-regular. Then $I \cap (g_1, \ldots , g_ m) = I(g_1, \ldots , g_ m)$.

Proof. Consider the exact sequence of complexes

$0 \to I \otimes _ A K_\bullet (A, g_1, \ldots , g_ m) \to K_\bullet (A, g_1, \ldots , g_ m) \to K_\bullet (A/I, g_1, \ldots , g_ m) \to 0$

Since the complex on the right has $H_1 = 0$ by assumption we see that

$\mathop{\mathrm{Coker}}(I^{\oplus m} \to I) \longrightarrow \mathop{\mathrm{Coker}}(A^{\oplus m} \to A)$

is injective. This is equivalent to the assertion of the lemma. $\square$

Lemma 15.30.9. Let $A$ be a ring. Let $I \subset J \subset A$ be ideals. Assume that $J/I \subset A/I$ is generated by an $H_1$-regular sequence. Then $I \cap J^2 = IJ$.

Proof. To prove this choose $g_1, \ldots , g_ m \in J$ whose images in $A/I$ form a $H_1$-regular sequence which generates $J/I$. In particular $J = I + (g_1, \ldots , g_ m)$. Suppose that $x \in I \cap J^2$. Because $x \in J^2$ can write

$x = \sum a_{ij} g_ ig_ j + \sum a_ j g_ j + a$

with $a_{ij} \in A$, $a_ j \in I$ and $a \in I^2$. Then $\sum a_{ij}g_ ig_ j \in I \cap (g_1, \ldots , g_ m)$ hence by Lemma 15.30.8 we see that $\sum a_{ij}g_ ig_ j \in I(g_1, \ldots , g_ m)$. Thus $x \in IJ$ as desired. $\square$

Lemma 15.30.10. Let $A$ be a ring. Let $I$ be an ideal generated by a quasi-regular sequence $f_1, \ldots , f_ n$ in $A$. Let $g_1, \ldots , g_ m \in A$ be elements whose images $\overline{g}_1, \ldots , \overline{g}_ m$ form an $H_1$-regular sequence in $A/I$. Then $f_1, \ldots , f_ n, g_1, \ldots , g_ m$ is a quasi-regular sequence in $A$.

Proof. We claim that $g_1, \ldots , g_ m$ forms an $H_1$-regular sequence in $A/I^ d$ for every $d$. By induction assume that this holds in $A/I^{d - 1}$. We have a short exact sequence of complexes

$0 \to K_\bullet (A, g_\bullet ) \otimes _ A I^{d - 1}/I^ d \to K_\bullet (A/I^ d, g_\bullet ) \to K_\bullet (A/I^{d - 1}, g_\bullet ) \to 0$

Since $f_1, \ldots , f_ n$ is quasi-regular we see that the first complex is a direct sum of copies of $K_\bullet (A/I, g_1, \ldots , g_ m)$ hence acyclic in degree $1$. By induction hypothesis the last complex is acyclic in degree $1$. Hence also the middle complex is. In particular, the sequence $g_1, \ldots , g_ m$ forms a quasi-regular sequence in $A/I^ d$ for every $d \geq 1$, see Lemma 15.30.6. Now we are ready to prove that $f_1, \ldots , f_ n, g_1, \ldots , g_ m$ is a quasi-regular sequence in $A$. Namely, set $J = (f_1, \ldots , f_ n, g_1, \ldots , g_ m)$ and suppose that (with multinomial notation)

$\sum \nolimits _{|N| + |M| = d} a_{N, M} f^ N g^ M \in J^{d + 1}$

for some $a_{N, M} \in A$. We have to show that $a_{N, M} \in J$ for all $N, M$. Let $e \in \{ 0, 1, \ldots , d\}$. Then

$\sum \nolimits _{|N| = d - e, \ |M| = e} a_{N, M} f^ N g^ M \in (g_1, \ldots , g_ m)^{e + 1} + I^{d - e + 1}$

Because $g_1, \ldots , g_ m$ is a quasi-regular sequence in $A/I^{d - e + 1}$ we deduce

$\sum \nolimits _{|N| = d - e} a_{N, M} f^ N \in (g_1, \ldots , g_ m) + I^{d - e + 1}$

for each $M$ with $|M| = e$. By Lemma 15.30.8 applied to $I^{d - e}/I^{d - e + 1}$ in the ring $A/I^{d - e + 1}$ this implies $\sum _{|N| = d - e} a_{N, M} f^ N \in I^{d - e}(g_1, \ldots , g_ m)$. Since $f_1, \ldots , f_ n$ is quasi-regular in $A$ this implies that $a_{N, M} \in J$ for each $N, M$ with $|N| = d - e$ and $|M| = e$. This proves the lemma. $\square$

Lemma 15.30.11. Let $A$ be a ring. Let $I$ be an ideal generated by an $H_1$-regular sequence $f_1, \ldots , f_ n$ in $A$. Let $g_1, \ldots , g_ m \in A$ be elements whose images $\overline{g}_1, \ldots , \overline{g}_ m$ form an $H_1$-regular sequence in $A/I$. Then $f_1, \ldots , f_ n, g_1, \ldots , g_ m$ is an $H_1$-regular sequence in $A$.

Proof. We have to show that $H_1(A, f_1, \ldots , f_ n, g_1, \ldots , g_ m) = 0$. To do this consider the commutative diagram

$\xymatrix{ \wedge ^2(A^{\oplus n + m}) \ar[r] \ar[d] & A^{\oplus n + m} \ar[r] \ar[d] & A \ar[r] \ar[d] & 0 \\ \wedge ^2(A/I^{\oplus m}) \ar[r] & A/I^{\oplus m} \ar[r] & A/I \ar[r] & 0 }$

Consider an element $(a_1, \ldots , a_{n + m}) \in A^{\oplus n + m}$ which maps to zero in $A$. Because $\overline{g}_1, \ldots , \overline{g}_ m$ form an $H_1$-regular sequence in $A/I$ we see that $(\overline{a}_{n + 1}, \ldots , \overline{a}_{n + m})$ is the image of some element $\overline{\alpha }$ of $\wedge ^2(A/I^{\oplus m})$. We can lift $\overline{\alpha }$ to an element $\alpha \in \wedge ^2(A^{\oplus n + m})$ and substract the image of it in $A^{\oplus n + m}$ from our element $(a_1, \ldots , a_{n + m})$. Thus we may assume that $a_{n + 1}, \ldots , a_{n + m} \in I$. Since $I = (f_1, \ldots , f_ n)$ we can modify our element $(a_1, \ldots , a_{n + m})$ by linear combinations of the elements

$(0, \ldots , g_ j, 0, \ldots , 0, f_ i, 0, \ldots , 0)$

in the image of the top left horizontal arrow to reduce to the case that $a_{n + 1}, \ldots , a_{n + m}$ are zero. In this case $(a_1, \ldots , a_ n, 0, \ldots , 0)$ defines an element of $H_1(A, f_1, \ldots , f_ n)$ which we assumed to be zero. $\square$

Lemma 15.30.12. Let $A$ be a ring. Let $f_1, \ldots , f_ n, g_1, \ldots , g_ m \in A$ be an $H_1$-regular sequence. Then the images $\overline{g}_1, \ldots , \overline{g}_ m$ in $A/(f_1, \ldots , f_ n)$ form an $H_1$-regular sequence.

Proof. Set $I = (f_1, \ldots , f_ n)$. We have to show that any relation $\sum _{j = 1, \ldots , m} \overline{a}_ j \overline{g}_ j$ in $A/I$ is a linear combination of trivial relations. Because $I = (f_1, \ldots , f_ n)$ we can lift this relation to a relation

$\sum \nolimits _{j = 1, \ldots , m} a_ j g_ j + \sum \nolimits _{i = 1, \ldots , n} b_ if_ i = 0$

in $A$. By assumption this relation in $A$ is a linear combination of trivial relations. Taking the image in $A/I$ we obtain what we want. $\square$

Lemma 15.30.13. Let $A$ be a ring. Let $I$ be an ideal generated by a Koszul-regular sequence $f_1, \ldots , f_ n$ in $A$. Let $g_1, \ldots , g_ m \in A$ be elements whose images $\overline{g}_1, \ldots , \overline{g}_ m$ form a Koszul-regular sequence in $A/I$. Then $f_1, \ldots , f_ n, g_1, \ldots , g_ m$ is a Koszul-regular sequence in $A$.

Proof. Our assumptions say that $K_\bullet (A, f_1, \ldots , f_ n)$ is a finite free resolution of $A/I$ and $K_\bullet (A/I, \overline{g}_1, \ldots , \overline{g}_ m)$ is a finite free resolution of $A/(f_ i, g_ j)$ over $A/I$. Then

\begin{align*} K_\bullet (A, f_1, \ldots , f_ n, g_1, \ldots , g_ m) & = \text{Tot}(K_\bullet (A, f_1, \ldots , f_ n) \otimes _ A K_\bullet (A, g_1, \ldots , g_ m)) \\ & \cong A/I \otimes _ A K_\bullet (A, g_1, \ldots , g_ m) \\ & = K_\bullet (A/I, \overline{g}_1, \ldots , \overline{g}_ m) \\ & \cong A/(f_ i, g_ j) \end{align*}

The first equality by Lemma 15.28.12. The first quasi-isomorphism $\cong$ by (the dual of) Homology, Lemma 12.25.4 as the $q$th row of the double complex $K_\bullet (A, f_1, \ldots , f_ n) \otimes _ A K_\bullet (A, g_1, \ldots , g_ m)$ is a resolution of $A/I \otimes _ A K_ q(A, g_1, \ldots , g_ m)$. The second equality is clear. The last quasi-isomorphism by assumption. Hence we win. $\square$

To conclude in the following lemma it is necessary to assume that both $f_1, \ldots , f_ n$ and $f_1, \ldots , f_ n, g_1, \ldots , g_ m$ are Koszul-regular. A counter example to dropping the assumption that $f_1, \ldots , f_ n$ is Koszul-regular is Examples, Lemma 109.14.1.

Lemma 15.30.14. Let $A$ be a ring. Let $f_1, \ldots , f_ n, g_1, \ldots , g_ m \in A$. If both $f_1, \ldots , f_ n$ and $f_1, \ldots , f_ n, g_1, \ldots , g_ m$ are Koszul-regular sequences in $A$, then $\overline{g}_1, \ldots , \overline{g}_ m$ in $A/(f_1, \ldots , f_ n)$ form a Koszul-regular sequence.

Proof. Set $I = (f_1, \ldots , f_ n)$. Our assumptions say that $K_\bullet (A, f_1, \ldots , f_ n)$ is a finite free resolution of $A/I$ and $K_\bullet (A, f_1, \ldots , f_ n, g_1, \ldots , g_ m)$ is a finite free resolution of $A/(f_ i, g_ j)$ over $A$. Then

\begin{align*} A/(f_ i, g_ j) & \cong K_\bullet (A, f_1, \ldots , f_ n, g_1, \ldots , g_ m) \\ & = \text{Tot}(K_\bullet (A, f_1, \ldots , f_ n) \otimes _ A K_\bullet (A, g_1, \ldots , g_ m)) \\ & \cong A/I \otimes _ A K_\bullet (A, g_1, \ldots , g_ m) \\ & = K_\bullet (A/I, \overline{g}_1, \ldots , \overline{g}_ m) \end{align*}

The first quasi-isomorphism $\cong$ by assumption. The first equality by Lemma 15.28.12. The second quasi-isomorphism by (the dual of) Homology, Lemma 12.25.4 as the $q$th row of the double complex $K_\bullet (A, f_1, \ldots , f_ n) \otimes _ A K_\bullet (A, g_1, \ldots , g_ m)$ is a resolution of $A/I \otimes _ A K_ q(A, g_1, \ldots , g_ m)$. The second equality is clear. Hence we win. $\square$

Lemma 15.30.15. Let $R$ be a ring. Let $I$ be an ideal generated by $f_1, \ldots , f_ r \in R$.

1. If $I$ can be generated by a quasi-regular sequence of length $r$, then $f_1, \ldots , f_ r$ is a quasi-regular sequence.

2. If $I$ can be generated by an $H_1$-regular sequence of length $r$, then $f_1, \ldots , f_ r$ is an $H_1$-regular sequence.

3. If $I$ can be generated by a Koszul-regular sequence of length $r$, then $f_1, \ldots , f_ r$ is a Koszul-regular sequence.

Proof. If $I$ can be generated by a quasi-regular sequence of length $r$, then $I/I^2$ is free of rank $r$ over $R/I$. Since $f_1, \ldots , f_ r$ generate by assumption we see that the images $\overline{f}_ i$ form a basis of $I/I^2$ over $R/I$. It follows that $f_1, \ldots , f_ r$ is a quasi-regular sequence as all this means, besides the freeness of $I/I^2$, is that the maps $\text{Sym}^ n_{R/I}(I/I^2) \to I^ n/I^{n + 1}$ are isomorphisms.

We continue to assume that $I$ can be generated by a quasi-regular sequence, say $g_1, \ldots , g_ r$. Write $g_ j = \sum a_{ij}f_ i$. As $f_1, \ldots , f_ r$ is quasi-regular according to the previous paragraph, we see that $\det (a_{ij})$ is invertible mod $I$. The matrix $a_{ij}$ gives a map $R^{\oplus r} \to R^{\oplus r}$ which induces a map of Koszul complexes $\alpha : K_\bullet (R, f_1, \ldots , f_ r) \to K_\bullet (R, g_1, \ldots , g_ r)$, see Lemma 15.28.3. This map becomes an isomorphism on inverting $\det (a_{ij})$. Since the cohomology modules of both $K_\bullet (R, f_1, \ldots , f_ r)$ and $K_\bullet (R, g_1, \ldots , g_ r)$ are annihilated by $I$, see Lemma 15.28.6, we see that $\alpha$ is a quasi-isomorphism.

Now assume that $g_1, \ldots , g_ r$ is a $H_1$-regular sequence generating $I$. Then $g_1, \ldots , g_ r$ is a quasi-regular sequence by Lemma 15.30.6. By the previous paragraph we conclude that $f_1, \ldots , f_ r$ is a $H_1$-regular sequence. Similarly for Koszul-regular sequences. $\square$

Lemma 15.30.16. Let $R$ be a ring. Let $a_1, \ldots , a_ n \in R$ be elements such that $R \to R^{\oplus n}$, $x \mapsto (xa_1, \ldots , xa_ n)$ is injective. Then the element $\sum a_ i t_ i$ of the polynomial ring $R[t_1, \ldots , t_ n]$ is a nonzerodivisor.

Proof. If one of the $a_ i$ is a unit this is just the statement that any element of the form $t_1 + a_2 t_2 + \ldots + a_ n t_ n$ is a nonzerodivisor in the polynomial ring over $R$.

Case I: $R$ is Noetherian. Let $\mathfrak q_ j$, $j = 1, \ldots , m$ be the associated primes of $R$. We have to show that each of the maps

$\sum a_ i t_ i : \text{Sym}^ d(R^{\oplus n}) \longrightarrow \text{Sym}^{d + 1}(R^{\oplus n})$

is injective. As $\text{Sym}^ d(R^{\oplus n})$ is a free $R$-module its associated primes are $\mathfrak q_ j$, $j = 1, \ldots , m$. For each $j$ there exists an $i = i(j)$ such that $a_ i \not\in \mathfrak q_ j$ because there exists an $x \in R$ with $\mathfrak q_ jx = 0$ but $a_ i x \not= 0$ for some $i$ by assumption. Hence $a_ i$ is a unit in $R_{\mathfrak q_ j}$ and the map is injective after localizing at $\mathfrak q_ j$. Thus the map is injective, see Algebra, Lemma 10.63.19.

Case II: $R$ general. We can write $R$ as the union of Noetherian rings $R_\lambda$ with $a_1, \ldots , a_ n \in R_\lambda$. For each $R_\lambda$ the result holds, hence the result holds for $R$. $\square$

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 $S$. $\square$

Comment #6250 by Eric Jovinelly on

The proof of lemma 062I has a typo: g_i + x_r g_r should be g_i + x_i g_r

Also, it's not clear to me why the result holds for the multi-index (0,...,0,n). I'm not sure how the M-H_1 regularity of f_1,...,f_r^n allows us to isolate this single multi-index. If it does, since we may reorder the f_i, why not just reorder them and raise them to different powers? I must be missing something

Comment #7392 by Elie Studnia on

In the discussion between Definition 062E and Lemma 062F, I believe that the equivalence is not entirely correct, because elements acting invertibly on $M$ will automatically be $M$-Kozsul-regular but not $M$-regular. But such elements need not be units of $R$: for instance, let $R=k[x]$ a polynomial ring over a field, $f=x$, $M=R/(x-1)$.

Comment #7393 by on

@#7392. In the discussion between Definition 062E and Lemma 062F there is no $M$, so I think the discussion is fine.

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