Betti numbers of modules over Noetherian rings with ... - IPM

Chapter 3: Growth of the Betti sequence of the canonical module 48
Let E = E(k) and ΩE denote the first syzygy module of E. Then minimality implies
that ΩE ⊆ m.R d and so m.ΩE = 0. That is ΩE is a k-vector space of length
ℓ(ΩE) = ℓ(R d ) − ℓ(E(k)) = dℓ(R) − ℓ(R) = (d − 1)ℓ(R) = (d − 1)(d + 1) = d 2 − 1.
Hence for all i > 0, Tor R i+2(E, E) ∼ = Tor R i (k, k) (d2−1) 2
. We claim that Tor R i (k, k) ∼ = kdi. For this note that since R ⊇ k d , continuing the construction of the given exact
sequence, we end up with the minimal free resolution of k over R as
Thus Tor R i+2(E, E) ∼ = k di (d 2 −1) 2
· · · → R d3
→ R d2
→ R d → R → k → 0.
to check the non-vanishing of Tor R 2 (E, E) and Tor R 1 (E, E).
Note that since β 1 R (E) = d2 − 1,
which is again non-zero.
Finally, the sequence
for all i > 0, which is of course non-zero. It remains
Tor R 2 (E, E) ∼ = Tor R 1 (k d2 −1 , E) ∼ = Tor R 1 (k, E) d2 −1 ∼ = k (d 2 −1) 2
0 → Tor R 1 (E, E) → E ⊗R k d2 −1 → E d → E ⊗R E → 0
is exact. Since E ⊗R k d2 −1 ∼ = (E/mE) d 2 −1 and E d is of length (d + 1)d, a simple length
comparison shows that
d(d 2 − 1) > (d + 1)d ⇐⇒ d > 2,
hence for d > 2, the homomorphism E ⊗R k d2 −1 → E d has non-zero kernel, that is
Tor R 1 (E, E) = 0 if d > 2.
But since d can only take values larger than 1, we just need to consider the case
d = 2. Then d(d 2 − 1) = d 2 + d = 6 and so Tor R
1 (E, E) = 0 ⇐⇒ E ⊗R E = 0. In