The Krull-Akizuki situation: proofs

D. J. Bernstein

19951124

Proof of Theorem F1. Let x be a nonzero element of R. Define a k-module homomorphism φ : R → R by φ(y) = xy. R is a domain, so φ is injective, so dim_{k} R = dim_{k} φ(R), so dim_{k}(R/φ(R)) = 0, so φ(R) = R. In particular 1 ∈ φ(R), so x is a unit.

Proof of Theorem L1. Say I = a_{1}R + a_{2}R + ⋯ + a_{n}R. Consider the surjective map φ : R^{n} → I defined by sending (r_{1}, r_{2}, …, r_{n}) to a_{1}r_{1} + a_{2}r_{2} + ⋯ + a_{n}r_{n}. If each r_{i} ∈ P then ∑ a_{i}r_{i} ∈ P I; thus φ induces a surjective map (R/P)^{n} → I/P I. Hence len_{R}(I/P I) ≤ len_{R}((R/P)^{n}) = n len_{R}(R/P) = n.

Proof of Theorem L2. We have len_{R}(R/P^{n}) = len_{R}(R/P) + len_{R}(P/P^{2}) + ⋯ + len_{R}(P^{n−1}/P^{n}). Each term len_{R}(P^{k}/P^{k+1}) is finite by Theorem L1, since P^{k} is finitely generated.

Proof of Theorem A1. P and P′ are maximal ideals. Distinct maximal ideals are coprime.

Proof of Theorem A2. By Noetherian factorization (Exercise 17) there are prime ideals P_{1}, P_{2}, …, P_{n} such that 0R = P_{1}P_{2} … P_{n}. If Q is any other prime ideal then Q is coprime to each P_{i} by Theorem A1, so Q is coprime to 0R, contradiction.

Define e(P) as the number of i such that P_{i} = P; then 0R = ∏_{P} P^{e(P)}. By Theorem A1, the factors P^{e(P)} are coprime. Thus, by the Chinese Remainder Theorem, R/0R is isomorphic to ∏_{P} (R/P^{e(P)}).

Proof of Theorem A3. By Theorem A2, R is isomorphic to ∏_{P} (R/P^{e(P)}). Any prime ideal P of R is finitely generated since R is Noetherian; thus len_{R}(R/P^{e(P)}) is finite by Theorem L2. Hence len_{R} R = ∑_{P} len_{R}(R/P^{e(P)}) is finite.

Proof of Theorem C1. Let Q be a prime ideal of R/I. Write P = Q + I; note that P is nonzero. Now R/P is isomorphic to (R/I)/Q, which is a domain, so P is prime. But dim R ≤ 1, so P is maximal, so R/P is a field, so Q is maximal.

Proof of Theorem V1. By Theorem C1, R/a is Artin. By Theorem A3, len_{R}(R/a) = len_{R/a}(R/a) is finite.

Write n = dim_{F} MF. Select an F -basis for MF consisting of n vectors v_{1}, …, v_{n} ∈ M. Write T = Rv_{1} ⊕ ⋯ ⊕ Rv_{n} ⊆ M. Find a common denominator d ∈ R − {0} for the generators of M, so that dM ⊆ T.

Observe that len_{R}(T/aT) = n len_{R}(R/a) is finite. Similarly, len_{R}(d^{−1}T/T) is finite. Thus len_{R}(d^{−1}T/aT) is finite. A fortiori len_{R}(M/aT) is finite.

Finally len(M/aM) + len(aM/aT) = len(M/aT) = len(M/T) + len(T/aT). All quantities here are finite, and len(M/T) = len(aM/aT), so len(M/aM) = len(T/aT) = n len_{R}(R/a).

Proof of Theorem V2. Write t = (dim_{F} V) len_{R}(R/a). We will show that any finitely generated Rsubmodule of M/aM has length at most t. Then, by Exercise 46, M/aM has length at most t.

Any finitely generated R-submodule of M/aM is of the form N/aM, where N is a finitely generated Rsubmodule of M. By Theorem V1, len_{R}(N/aN) = (dim_{F} NF) len_{R}(R/a) ≤ t; N/aN maps onto N/aM, so len_{R}(N/aM) ≤ t.

Proof of Theorem V3. Pick a nonzero element a ∈ I. Then M/aM maps onto M/IM, so len_{R}(M/IM) ≤ len_{R}(M/aM). By Theorem V2, len_{R}(M/aM) is finite.

Proof of Theorem K1. Apply Theorem V3 with V = K and M = A.

Proof of Theorem K2. Pick any nonzero α ∈ J. K is a finitely generated F -module, so α is integral over F. Let f = x^{d} + f_{d−1}x^{d−1} + ⋯ + f_{0} ∈ F [x] be a monic polynomial of smallest degree such that f(α) = 0. If f_{0} = 0 then f/x is a monic polynomial of smaller degree with α as a root, contradiction; hence f_{0} ≠ 0. Clear denominators: r_{d}α^{d} + r_{d−1}α^{d−1} + ⋯ + r_{0} = 0, with r_{0} ≠ 0. Finally r_{0} = −r_{d}α^{d} − ⋯ − r_{1}α ∈ J.

Proof of Theorem K3. Let C be any nonempty collection of ideals of A. We will show that C has a maximal element. If C = {0A} then 0A is maximal. Otherwise C contains some nonzero ideal J. Define D as the collection of ideals in C containing J; it is enough to show that D has a maximal element.

Define I = J ∩ R. By Theorem K2, I is nonzero. By Theorem K1, A/IA is a finite-length R-module, hence a finitely generated R-module, hence a Noetherian R-module. Define E = {T/IA : T ∈ D}; then E is a nonempty collection of R-submodules of A/IA, so E has some maximal element T/IA.

Finally T is a maximal element of D. Indeed, if T ⊆ U ∈ D then T/IA ⊆ U/IA so T/IA = U/IA; but T and U contain IA by definition of D, so T = U.

Proof of Theorem K4. Write P = Q ∩ R. By Theorem K2, P is nonzero. If ab ∈ P with a, b ∈ R then ab ∈ Q, so a ∈ Q or b ∈ Q, so a ∈ P or b ∈ P.

Proof of Theorem K5. Let Q be a nonzero prime ideal of A. Define P = Q ∩ R. By Theorem K4, P is a nonzero prime ideal of R, so P is maximal, so R/P is a field. Consider the R/P -vector space A/P A; by Theorem K1, dim_{R/P} (A/P A) = len_{R}(A/P A) is finite. Thus dim_{R/P} (A/Q) is finite. A/Q is a domain, so by Theorem F1 it is a field. Hence Q is maximal.