The Stacks project

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

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. $\square$

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

Proof. By Lemma 15.28.11 we have exact sequences

for all $i$. $\square$

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

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

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

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

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

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

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

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. $\square$

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. $\square$

Proof. Consider the exact sequence of complexes

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

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

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

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. $\square$

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

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)

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

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

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. $\square$

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

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 subtract 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

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. $\square$

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

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. $\square$

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

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. $\square$

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

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. $\square$

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. $\square$

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

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 $R$. $\square$

Proof. The equality of ideals is obvious as the matrix

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 $S$. $\square$