Lemma 15.31.1. Let R be a ring. Let f_1, \ldots , f_ r \in R be an Koszul-regular sequence. Then the extended alternating Čech complex R \to \bigoplus \nolimits _{i_0} R_{f_{i_0}} \to \bigoplus \nolimits _{i_0 < i_1} R_{f_{i_0}f_{i_1}} \to \ldots \to R_{f_1\ldots f_ r} from Section 15.29 only has cohomology in degree r.
15.31 More on Koszul regular sequences
We continue the discussion from Section 15.30.
Proof. By Lemma 15.30.4 and induction the sequence f_1, \ldots , f_{r - 1}, f_ r^ n is Koszul regular for all n \geq 1. By Lemma 15.28.4 any permutation of a Koszul regular sequence is a Koszul regular sequence. Hence we see that we may replace any (or all) f_ i by its nth power and still have a Koszul regular sequence. Thus K_\bullet (R, f_1^ n, \ldots , f_ r^ n) has nonzero cohomology only in homological degree 0. This implies what we want by Lemma 15.29.6. \square
Lemma 15.31.2. Let a, a_2, \ldots , a_ r be an H_1-regular sequence in a ring R (for example a Koszul regular sequence or a regular sequence, see Lemmas 15.30.2 and 15.30.3). With I = (a, a_2, \ldots , a_ r) the blowup algebra R' = R[\frac{I}{a}] is isomorphic to R'' = R[y_2, \ldots , y_ r]/(a y_ i - a_ i).
Proof. By Algebra, Lemma 10.70.6 it suffices to show that R'' is a-torsion free.
We claim a, ay_2 - a_2, \ldots , ay_ n - a_ r is a H_1-regular sequence in R[y_2, \ldots , y_ r]. Namely, the map
(a, ay_2 - a_2, \ldots , ay_ n - a_ r) : R[y_2, \ldots , y_ r]^{\oplus r} \longrightarrow R[y_2, \ldots , y_ r]
used to define the Koszul complex on a, ay_2 - a_2, \ldots , ay_ n - a_ r is isomorphic to the map
(a, a_2, \ldots , a_ r) : R[y_2, \ldots , y_ r]^{\oplus r} \longrightarrow R[y_2, \ldots , y_ r]
used to the define the Koszul complex on a, a_2, \ldots , a_ r via the isomorphism
R[y_2, \ldots , y_ r]^{\oplus r} \longrightarrow R[y_2, \ldots , y_ r]^{\oplus r}
sending (b_1, \ldots , b_ r) to (b_1 - b_2y_2 \ldots - b_ ry_ r, -b_2, \ldots , - b_ r). By Lemma 15.28.3 these Koszul complexes are isomorphic. By Lemma 15.30.5 applied to the flat ring map R \to R[y_2, \ldots , y_ r] we conclude our claim is true. By Lemma 15.28.8 we see that the Koszul complex K on a, ay_2 - a_2, \ldots , ay_ n - a_ r is the cone on a : L \to L where L is the Koszul complex on ay_2 - a_2, \ldots , ay_ n - a_ r. Since H_1(K) = 0 by the claim, we conclude that a : H_0(L) \to H_0(L) is injective, in other words that R'' = R[y_2, \ldots , y_ r]/(a y_ i - a_ i) has no nonzero a-torsion elements as desired. \square
Lemma 15.31.3. Let A \to B be a ring map. Let f_1, \ldots , f_ r be a sequence in B such that B/(f_1, \ldots , f_ r) is A-flat. Let A \to A' be a ring map. Then the canonical map
H_1(K_\bullet (B, f_1, \ldots , f_ r)) \otimes _ A A' \longrightarrow H_1(K_\bullet (B', f'_1, \ldots , f'_ r))
is surjective. Here B' = B \otimes _ A A' and f_ i' \in B' is the image of f_ i.
Proof. The sequence
\wedge ^2(B^{\oplus r}) \to B^{\oplus r} \to B \to B/J \to 0
is a complex of A-modules with B/J flat over A and cohomology group H_1 = H_1(K_\bullet (B, f_1, \ldots , f_ r)) in the spot B^{\oplus r}. If we tensor this with A' we obtain a complex
\wedge ^2((B')^{\oplus r}) \to (B')^{\oplus r} \to B' \to B'/J' \to 0
which is exact at B' and B'/J'. In order to compute its cohomology group H'_1 = H_1(K_\bullet (B', f'_1, \ldots , f'_ r)) at (B')^{\oplus r} we split the first sequence above into the exact sequences 0 \to J \to B \to B/J \to 0, 0 \to K \to B^{\oplus r} \to J \to 0, and \wedge ^2(B^{\oplus r}) \to K \to H_1 \to 0. Tensoring over A with A' we obtain the exact sequences
\begin{matrix} 0 \to J \otimes _ A A' \to B \otimes _ A A' \to (B/J) \otimes _ A A' \to 0
\\ K \otimes _ A A' \to B^{\oplus r} \otimes _ A A' \to J \otimes _ A A' \to 0
\\ \wedge ^2(B^{\oplus r}) \otimes _ A A' \to K \otimes _ A A' \to H_1 \otimes _ A A' \to 0
\end{matrix}
where the first one is exact as B/J is flat over A, see Algebra, Lemma 10.39.12. We conclude that J' = J \otimes _ A A' \subset B' and that K \otimes _ A A' \to \mathop{\mathrm{Ker}}((B')^{\oplus r} \to B') is surjective. Thus
\begin{align*} H_1 \otimes _ A A' & = \mathop{\mathrm{Coker}}\left(\wedge ^2(B^{\oplus r}) \otimes _ A A' \to K \otimes _ A A'\right) \\ & \to \mathop{\mathrm{Coker}}\left( \wedge ^2((B')^{\oplus r}) \to \mathop{\mathrm{Ker}}((B')^{\oplus r} \to B') \right) = H'_1 \end{align*}
is surjective too. \square
Lemma 15.31.4. Let A \to B and A \to A' be ring maps. Set B' = B \otimes _ A A'. Let f_1, \ldots , f_ r \in B. Assume B/(f_1, \ldots , f_ r)B is flat over A
If f_1, \ldots , f_ r is a quasi-regular sequence, then the image in B' is a quasi-regular sequence.
If f_1, \ldots , f_ r is a H_1-regular sequence, then the image in B' is a H_1-regular sequence.
Proof. Assume f_1, \ldots , f_ r is quasi-regular. Set J = (f_1, \ldots , f_ r). By assumption J^ n/J^{n + 1} is isomorphic to a direct sum of copies of B/J hence flat over A. By induction and Algebra, Lemma 10.39.13 we conclude that B/J^ n is flat over A. The ideal (J')^ n is equal to J^ n \otimes _ A A', see Algebra, Lemma 10.39.12. Hence (J')^ n/(J')^{n + 1} = J^ n/J^{n + 1} \otimes _ A A' which clearly implies that f_1, \ldots , f_ r is a quasi-regular sequence in B'.
Assume f_1, \ldots , f_ r is H_1-regular. By Lemma 15.31.3 the vanishing of the Koszul homology group H_1(K_\bullet (B, f_1, \ldots , f_ r)) implies the vanishing of H_1(K_\bullet (B', f'_1, \ldots , f'_ r)) and we win. \square
Lemma 15.31.5. Let A' \to B' be a ring map. Let I \subset A' be an ideal. Set A = A'/I and B = B'/IB'. Let f'_1, \ldots , f'_ r \in B'. Assume
A' \to B' is flat and of finite presentation,
I is locally nilpotent,
the images f_1, \ldots , f_ r \in B form a quasi-regular sequence,
B/(f_1, \ldots , f_ r) is flat over A.
Then B'/(f'_1, \ldots , f'_ r) is flat over A'.
Proof. Set C' = B'/(f'_1, \ldots , f'_ r). We have to show A' \to C' is flat. Let \mathfrak r' \subset C' be a prime ideal lying over \mathfrak p' \subset A'. We let \mathfrak q' \subset B' be the inverse image of \mathfrak r'. By Algebra, Lemma 10.39.18 it suffices to show that A'_{\mathfrak p'} \to C'_{\mathfrak q'} is flat. Algebra, Lemma 10.128.6 tells us it suffices to show that f'_1, \ldots , f'_ r map to a regular sequence in
B'_{\mathfrak q'}/\mathfrak p'B'_{\mathfrak q'} = B_\mathfrak q/\mathfrak p B_\mathfrak q = (B \otimes _ A \kappa (\mathfrak p))_\mathfrak q
with obvious notation. What we know is that f_1, \ldots , f_ r is a quasi-regular sequence in B and that B/(f_1, \ldots , f_ r) is flat over A. By Lemma 15.31.4 the images \overline{f}_1, \ldots , \overline{f}_ r of f'_1, \ldots , f'_ r in B \otimes _ A \kappa (\mathfrak p) form a quasi-regular sequence. Since (B \otimes _ A \kappa (\mathfrak p))_\mathfrak q is a Noetherian local ring, we conclude by Lemma 15.30.7. \square
Lemma 15.31.6. Let A' \to B' be a ring map. Let I \subset A' be an ideal. Set A = A'/I and B = B'/IB'. Let f'_1, \ldots , f'_ r \in B'. Assume
A' \to B' is flat and of finite presentation (for example smooth),
I is locally nilpotent,
the images f_1, \ldots , f_ r \in B form a quasi-regular sequence,
B/(f_1, \ldots , f_ r) is smooth over A.
Then B'/(f'_1, \ldots , f'_ r) is smooth over A'.
Proof. Set C' = B'/(f'_1, \ldots , f'_ r) and C = B/(f_1, \ldots , f_ r). Then A' \to C' is of finite presentation. By Lemma 15.31.5 we see that A' \to C' is flat. The fibre rings of A' \to C' are equal to the fibre rings of A \to C and hence smooth by assumption (4). It follows that A' \to C' is smooth by Algebra, Lemma 10.137.17. \square
Comments (0)