## Tag `062D`

## 15.27. Koszul regular sequences

Please take a look at Algebra, Sections 10.67, 10.68, and 10.71 before looking at this one.

Definition 15.27.1. 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

$M$-Koszul-regularif $H_i(K_\bullet(f_1, \ldots, f_r) \otimes_R M) = 0$ for all $i \not = 0$,$M$-$H_1$-regularif $H_1(K_\bullet(f_1, \ldots, f_r) \otimes_R M) = 0$,Koszul-regularif $H_i(K_\bullet(f_1, \ldots, f_r)) = 0$ for all $i \not = 0$, and$H_1$-Koszul-regularif $H_1(K_\bullet(f_1, \ldots, f_r)) = 0$.

We will see in Lemmas 15.27.2, 15.27.3, and 15.27.6 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 15.27.7). 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

- $f$ is a regular sequence of length one,
- $f$ is a Koszul-regular sequence of length one, and
- $f$ 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 15.27.2. 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 15.26.8 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 15.27.3. 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 15.27.4. Let $f_1, \ldots, f_{r - 1} \in R$ be a sequence and $f, g \in R$. Let $M$ be an $R$-module.

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

Proof.By Lemma 15.26.11 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 15.27.5. 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 $N = M \otimes_R S$.

- 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 $S$.
- 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 $(K_\bullet(f_1, \ldots, f_r) \otimes_R M) \otimes_R S = K_\bullet(\varphi(f_1), \ldots, \varphi(f_r)) \otimes_S N$. $\square$Lemma 15.27.6. An $M$-$H_1$-regular sequence is $M$-quasi-regular.

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 15.27.4. 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 15.27.5. By Lemma 15.26.4 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. $\square$

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

Lemma 15.27.7. 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

- $f_1, \ldots, f_r$ is an $M$-regular sequence,
- $f_1, \ldots, f_r$ is a $M$-Koszul-regular sequence,
- $f_1, \ldots, f_r$ is an $M$-$H_1$-regular sequence,
- $f_1, \ldots, f_r$ 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 15.27.2. The implication (2) $\Rightarrow$ (3) is Lemma 15.27.3. The implication (3) $\Rightarrow$ (4) is Lemma 15.27.6. The implication (4) $\Rightarrow$ (1) is Algebra, Lemma 10.68.6. $\square$Lemma 15.27.8. 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 $$ \mathop{\rm Coker}(I^{\oplus m} \to I) \longrightarrow \mathop{\rm Coker}(A^{\oplus m} \to A) $$ is injective. This is equivalent to the assertion of the lemma. $\square$Lemma 15.27.9. 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 15.27.8 we see that $\sum a_{ij}g_ig_j \in I(g_1, \ldots, g_m)$. Thus $x \in IJ$ as desired. $\square$Lemma 15.27.10. 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 15.27.6. 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 15.27.8 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 15.27.11. 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 15.27.12. 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 15.27.13. 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 15.26.12. The first quasi-isomorphism $\cong$ by (the dual of) Homology, Lemma 12.22.7 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 100.13.1.

Lemma 15.27.14. 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 15.26.12. The second quasi-isomorphism by (the dual of) Homology, Lemma 12.22.7 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 15.27.15. Let $R$ be a ring. Let $I$ be an ideal generated by $f_1, \ldots, f_r \in R$.

- If $I$ can be generated by a quasi-regular sequence of length $r$, then $f_1, \ldots, f_r$ is a quasi-regular sequence.
- 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.
- 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 15.26.3. 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 15.26.6, 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 15.27.6. 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 15.27.16. 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 10.62.19.

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 15.27.17. 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 5812–6432 (see updates for more information).

```
\section{Koszul regular sequences}
\label{section-koszul-regular}
\noindent
Please take a look at
Algebra, Sections \ref{algebra-section-regular-sequences},
\ref{algebra-section-quasi-regular}, and
\ref{algebra-section-depth}
before looking at this one.
\begin{definition}
\label{definition-koszul-regular-sequence}
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
\begin{enumerate}
\item {\it $M$-Koszul-regular} if
$H_i(K_\bullet(f_1, \ldots, f_r) \otimes_R M) = 0$ for
all $i \not = 0$,
\item {\it $M$-$H_1$-regular} if
$H_1(K_\bullet(f_1, \ldots, f_r) \otimes_R M) = 0$,
\item {\it Koszul-regular} if $H_i(K_\bullet(f_1, \ldots, f_r)) = 0$ for
all $i \not = 0$, and
\item {\it $H_1$-Koszul-regular} if $H_1(K_\bullet(f_1, \ldots, f_r)) = 0$.
\end{enumerate}
\end{definition}
\noindent
We will see in Lemmas \ref{lemma-regular-koszul-regular},
\ref{lemma-koszul-regular-H1-regular}, and
\ref{lemma-H1-regular-quasi-regular} 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 \ref{lemma-noetherian-finite-all-equivalent}).
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
\begin{enumerate}
\item $f$ is a regular sequence of length one,
\item $f$ is a Koszul-regular sequence of length one, and
\item $f$ is a $H_1$-regular sequence of length one.
\end{enumerate}
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.
\begin{lemma}
\label{lemma-regular-koszul-regular}
An $M$-regular sequence is $M$-Koszul-regular.
A regular sequence is Koszul-regular.
\end{lemma}
\begin{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 \ref{lemma-cone-koszul}
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.
\end{proof}
\begin{lemma}
\label{lemma-koszul-regular-H1-regular}
A $M$-Koszul-regular sequence is $M$-$H_1$-regular.
A Koszul-regular sequence is $H_1$-regular.
\end{lemma}
\begin{proof}
This is immediate from the definition.
\end{proof}
\begin{lemma}
\label{lemma-mult-koszul-regular}
Let $f_1, \ldots, f_{r - 1} \in R$ be a sequence and $f, g \in R$.
Let $M$ be an $R$-module.
\begin{enumerate}
\item 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.
\item If $f_1, \ldots, f_{r - 1}, f$ and $f_1, \ldots, f_{r - 1}, f$ are
$M$-Koszul-regular then $f_1, \ldots, f_{r - 1}, fg$ is $M$-Koszul-regular
too.
\end{enumerate}
\end{lemma}
\begin{proof}
By
Lemma \ref{lemma-koszul-mult}
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 $i$.
\end{proof}
\begin{lemma}
\label{lemma-koszul-regular-flat-base-change}
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 $N = M \otimes_R S$.
\begin{enumerate}
\item 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 $S$.
\item 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 $S$.
\end{enumerate}
\end{lemma}
\begin{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
$(K_\bullet(f_1, \ldots, f_r) \otimes_R M) \otimes_R S =
K_\bullet(\varphi(f_1), \ldots, \varphi(f_r)) \otimes_S N$.
\end{proof}
\begin{lemma}
\label{lemma-H1-regular-quasi-regular}
An $M$-$H_1$-regular sequence is $M$-quasi-regular.
\end{lemma}
\begin{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 \ref{lemma-mult-koszul-regular}.
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.
\medskip\noindent
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 \ref{lemma-koszul-regular-flat-base-change}.
By
Lemma \ref{lemma-change-basis}
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.
\end{proof}
\noindent
For nonzero finite modules over Noetherian local rings all of the types of
regular sequences introduced so far are equivalent.
\begin{lemma}
\label{lemma-noetherian-finite-all-equivalent}
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
\begin{enumerate}
\item $f_1, \ldots, f_r$ is an $M$-regular sequence,
\item $f_1, \ldots, f_r$ is a $M$-Koszul-regular sequence,
\item $f_1, \ldots, f_r$ is an $M$-$H_1$-regular sequence,
\item $f_1, \ldots, f_r$ is an $M$-quasi-regular sequence.
\end{enumerate}
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.
\end{lemma}
\begin{proof}
The implication (1) $\Rightarrow$ (2) is
Lemma \ref{lemma-regular-koszul-regular}.
The implication (2) $\Rightarrow$ (3) is
Lemma \ref{lemma-koszul-regular-H1-regular}.
The implication (3) $\Rightarrow$ (4) is
Lemma \ref{lemma-H1-regular-quasi-regular}.
The implication (4) $\Rightarrow$ (1) is
Algebra, Lemma \ref{algebra-lemma-quasi-regular-regular}.
\end{proof}
\begin{lemma}
\label{lemma-H1-regular-in-quotient}
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 $I \cap (g_1, \ldots, g_m) =
I(g_1, \ldots, g_m)$.
\end{lemma}
\begin{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
$$
\Coker(I^{\oplus m} \to I)
\longrightarrow
\Coker(A^{\oplus m} \to A)
$$
is injective. This is equivalent to the assertion of the lemma.
\end{proof}
\begin{lemma}
\label{lemma-conormal-sequence-H1-regular}
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 $I \cap J^2 = IJ$.
\end{lemma}
\begin{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 \ref{lemma-H1-regular-in-quotient}
we see that $\sum a_{ij}g_ig_j \in I(g_1, \ldots, g_m)$.
Thus $x \in IJ$ as desired.
\end{proof}
\begin{lemma}
\label{lemma-join-quasi-regular-H1-regular}
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 $A$.
\end{lemma}
\begin{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 \ref{lemma-H1-regular-quasi-regular}.
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 \ref{lemma-H1-regular-in-quotient}
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.
\end{proof}
\begin{lemma}
\label{lemma-join-H1-regular-sequences}
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 $A$.
\end{lemma}
\begin{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.
\end{proof}
\begin{lemma}
\label{lemma-truncate-H1-regular}
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.
\end{lemma}
\begin{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.
\end{proof}
\begin{lemma}
\label{lemma-join-koszul-regular-sequences}
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 $A$.
\end{lemma}
\begin{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 \ref{lemma-join-sequences-koszul-complex}.
The first quasi-isomorphism $\cong$ by (the dual of)
Homology, Lemma \ref{homology-lemma-double-complex-gives-resolution}
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.
\end{proof}
\noindent
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 \ref{examples-lemma-strange-regular-sequence}.
\begin{lemma}
\label{lemma-truncate-koszul-regular}
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.
\end{lemma}
\begin{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 \ref{lemma-join-sequences-koszul-complex}.
The second quasi-isomorphism by (the dual of)
Homology, Lemma \ref{homology-lemma-double-complex-gives-resolution}
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.
\end{proof}
\begin{lemma}
\label{lemma-independence-of-generators}
Let $R$ be a ring. Let $I$ be an ideal generated by $f_1, \ldots, f_r \in R$.
\begin{enumerate}
\item If $I$ can be generated by a quasi-regular sequence of length $r$,
then $f_1, \ldots, f_r$ is a quasi-regular sequence.
\item 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.
\item If $I$ can be generated by a Koszul-regular sequence of length $r$,
then $f_1, \ldots, f_r$ is a Koszul-regular sequence.
\end{enumerate}
\end{lemma}
\begin{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.
\medskip\noindent
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 \ref{lemma-functorial}.
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 \ref{lemma-homotopy-koszul},
we see that $\alpha$ is a quasi-isomorphism.
\medskip\noindent
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 \ref{lemma-H1-regular-quasi-regular}. By the previous paragraph
we conclude that $f_1, \ldots, f_r$ is a $H_1$-regular sequence.
Similarly for Koszul-regular sequences.
\end{proof}
\begin{lemma}
\label{lemma-make-nonzero-divisor}
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.
\end{lemma}
\begin{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 $R$.
\medskip\noindent
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 \ref{algebra-lemma-zero-at-ass-zero}.
\medskip\noindent
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 $R$.
\end{proof}
\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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