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By Siegfried Bosch, Werner Lutkebohmert, Michel Raynaud

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Then, we see that H(K, n) + H(N, n) = H(N0 , n), and so we find that ∆d−1 H(N0 , n) − H(N, n) = ∆d−1 H(K, n). Now, if deg H(K, n) = d−1, then ∆d−1 H(K, n) > 0, since the polynomial takes only positive values, for large enough n. If the degree is lower, then ∆d−1 H(K, n) = 0. In either case, we find that ∆d−1 H(N, n) ≤ ∆d−1 H(N0 , n) = l(N0 ). 32 2. GRADED RINGS AND MODULES II: FILTRATIONS AND HILBERT FUNCTIONS Now, we proceed to the proof of the equivalences. It’s easy to see that (3) ⇒ (2) ⇒ (1), using the example above.

14). 24. The degree of the Samuel polynomial χqM depends only on the finite set Supp M ∩ V (q) and the module M . Proof. Consider M as an S-module instead, where S = R/ ann(M ), and replace q and q by qS and q S. In this case, we have ann(M ) = 0, and so V (q) = V (q ), which implies that rad(q) = rad(q ). Hence, we can find n ∈ N such that qn ⊂ q ; and so χqM (mn − 1) ≥ χqM (m − 1), for all m ∈ N. This shows that deg χqM ≥ deg χqM . We get the other inequality by symmetry. As we discussed in our introduction to I-adic filtrations and the associated graded ring, we have a natural surjective map (M/qM )[t0 , .

6) ⇒ (1): Observe that H • (C) = 0 if and only if H • (C)⊗R M = 0. 2), this will do. As for the final statement, it’s clear from characterization (2), that, whenever M is faithfully flat, we have Supp M = Spec R, since faithfully flat modules are in particular non-zero. 5. Let (R, m) be a local ring, and let M be a finitely generated R-module. Then the following are equivalent: (1) M is free. 52 3. FLATNESS (2) M is flat. (3) M is faithfully flat. Proof. 7), and (3) ⇒ (2) follows by definition.

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