This blog post will be used in a later one. Please skip ahead to the next one.<\/p>\n
Let R be a regular ring. Let R —> S = R\/J be a quotient. Assume we have f_1, …, f_c in J and derivations D_1, …, D_c : R —> R as well as an element z’ in R such that z’ J is contained in (f_1, …, f_c) + J^2. Let B be the Koszul algebra on f_1, …, f_c over R and let<\/p>\n
z = det(D_i(f_j))<\/p><\/blockquote>\n
be as in the previous blog post.<\/p>\n
We can extend B —> S to a Tate resolution. Thus we may assume we have<\/p>\n
R —> B —> A —-> S<\/p><\/blockquote>\n
where A is gotten from B by adjoining variables and extending the differential. In particular A —> S is a quasi-isomorphism and A is free over R and over B as a graded module and B —> A is the inclusion of a direct summand (as a graded B-module).<\/p>\n
Lemma:<\/strong> There exists an n >= 1 such that (zz’)^n annihilates Cone(B ⊗ S —> A ⊗ S) in D(S). (Tensor products over R.)<\/p>\n
Proof:<\/strong> After inverting zz’ the immersion is regular by Tag 0GEE<\/a>. This uses that R is regular! For J = (f_1, …, f_c) and f_1, …, f_c a regular sequence the map is a quasi-isomorphism as both sides compute Tor^R_*(S, S). EndProof<\/strong><\/p>\n
Remark:<\/strong> For a while I tried to see if n = 1 is sufficient. I haven’t yet found a counter example. I think stuff in the literature may say that this is true if S is CM.<\/p>\n
Lemma:<\/strong> Let M’ in D(S). Let M be a dg B-module which is graded free and which comes with a quasi-isomorphism M —> M’ of dg B-modules. The cone of the map M ⊗_B S —> M’ is annihilated by (zz’)^n<\/p>\n
Proof.<\/strong> By standard things the module M ⊗_B S is quasi-isomorphic to M’ ⊗_B A. Then we can replace B by B’ = B ⊗_R S and A by A’ = A ⊗_R S. Thus we have to show that the cone on M’ ⊗_{B’} A’ —> M’ is annihilated by (zz’)^n. By the lemma above we know that the cone C’ of the map B’ —> A’ is annihilated by (zz’)^n. Now we have the short exact sequence of B’-modules<\/p>\n
0 —> M’ —> M’ ⊗_{B’} A’ —> M’ ⊗_{B’} C’ —> 0<\/p><\/blockquote>\n
whose first arrow is a splitting to the arrow M’ ⊗_{B’} A’ —> M’. Anyway, this shows that the cone we are looking at is isomorphic to a shift of M’ ⊗_{B’} C’ which proves what we want. EndProof<\/strong><\/p>\n
Denote i_* : D(S) —> D(R) the pushforward and denote i* : D(R) —> D(S) the pullback. Let M be an object of D(S). We have a counit map<\/p>\n
e : i*i_*M —> M<\/p><\/blockquote>\n
Lemma:<\/strong> For any S-module M’ there is a map s : M’ —> i*i_*M’ whose composition with e is equal to multiplication by z^{2n + 1} (z’)^{2n}.<\/p>\n
Proof.<\/strong> Denote a_* : D(S) —> D(B), a* : D(B) —> D(S), b_* : D(B) —> D(R), b* : D(R) —> D(B) the usual functors. By the previous lemma we see that there is a map a*M —> M’ whose cone is annihilated by (zz’)^n. Then it suffices to prove the lemma for a*M but with the multiplier being z (in this argument we get the 2n powers of z and z’ in the statement). So we want to construct a map<\/p>\n
a*M —> i*i_*a*M = a*b*b_*a_*a*M<\/p><\/blockquote>\n
For this we can first use the unit for a to get a*b*b_*M —> a*b*b_*a_*a*M. Then we can use the construction of the previous post to get a*M —> a*b*b_*M and this is where z comes in! EndProof<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"
This blog post will be used in a later one. Please skip ahead to the next one. Let R be a regular ring. Let R —> S = R\/J be a quotient. Assume we have f_1, …, f_c in J … Continue reading