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
M-Koszul-regular if H_ i(K_\bullet (f_1, \ldots , f_ r) \otimes _ R M) = 0 for all i \not= 0,
M-H_1-regular if H_1(K_\bullet (f_1, \ldots , f_ r) \otimes _ R M) = 0,
Koszul-regular if H_ i(K_\bullet (f_1, \ldots , f_ r)) = 0 for all i \not= 0, and
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
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 15.30.2. 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.
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}, 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.
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 15.30.6. 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. 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
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 15.30.2. 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 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
(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 qth 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 110.15.1.
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 qth 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.
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 15.28.3. 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.reference 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 nthe 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
Comments (6)
Comment #2777 by Darij Grinberg on
Comment #2885 by Johan on
Comment #6250 by Eric Jovinelly on
Comment #6252 by Johan on
Comment #7392 by Elie Studnia on
Comment #7393 by Johan on