The Stacks project

Proof. Let $R$, $M$, $f_1, \ldots , f_ r$ be as in the first sentence of the lemma. If $r = 1$, it is immediate that $f_1$ is $M$-Koszul-regular. Assume $r > 1$. Since $f_1$ is a nonzerodivisor on $M$, we obtain a short exact sequence of complexes:

Here $\overline{f}_ i$ is the image of $f_ i$ in $R/(f_1)$. By Lemma 15.28.8 the complex $K_\bullet (f_1, \ldots , f_ r)$ 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 $R/(f_1)$, $M/f_1M$, $\overline{f}_2, \ldots , \overline{f}_ r$ satisfy the conditions of the lemma, by induction we conclude this complex is acyclic in postive degrees. This finishes the proof of the first statement. The second statement immediately follows from the first. $\square$

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 $S$. $\square$