Unique factorization in Dedekind domains: proofs

D. J. Bernstein

19950918

Proof of Theorem G1. If I ∉ N then R = R ÷ I so I = IR = I(R ÷ I) = R, contradiction.

Proof of Theorem G2. Since P ∈ N, there is some z ∉ R such that zP ⊆ R. We must check two things:

(1) P ≠ R; (2) if ab ∈ P then a ∈ P or b ∈ P.

If P = R then 1 ∈ P so z ∈ zP ⊆ R, contradiction; hence P ≠ R.

Say ab ∈ P, a, b ∈ R. Either za ∈ R or za ∉ R. If za ∈ R then z(P + aR) = zP + zaR ⊆ R so P + aR ∈ N. But P is maximal in N so P = P + aR so a ∈ P as desired.

If za ∉ R then za(P + bR) = azP + zabR ⊆ aR + zP R ⊆ R so P + bR ∈ N. But P is maximal in N so P = P + bR so b ∈ P as desired.

Proof of Theorem G3. Say I ⊈ P. Pick a ∈ I − P. For any b ∈ J we have ab ∈ IJ ⊆ P, but a ∉ P, so b ∈ P. Hence J ⊆ P.

Proof of Theorem M1. Using Theorem G3 inductively we see that P_{i} ⊆ P for some i. Now P_{i} is nonzero and dim R ≤ 1 so P_{i} is a maximal ideal. If P_{i} ⊂ P then P = R, contradiction; thus P_{i} = P.

Proof of Theorem I1. If I = I(R ÷ J) then R ⊆ R ÷ J ⊆ I ÷ I = R so R ÷ J = R.

Proof of Theorem I2. By Theorem I1, I ⊂ I(R ÷ I) ⊆ R; I is a maximal ideal, so I(R ÷ I) = R.

Proof of Theorem D1. Let S be the collection of all nonzero ideals J such that (1) J ∈ N; (2) I ⊆ J. By Theorem G1, I ∈ N, so I ∈ S; hence S is nonempty. R is Noetherian so S has some maximal element J. By construction J ∈ N, and if J ⊂ K then K ∉ S and I ⊆ K so K ∉ N; hence J is a maximal element of N. By Theorem G2, J is prime. Furthermore I is nonzero so J is nonzero. Since dim R ≤ 1, J is a maximal ideal. By Theorem I2, J is invertible.

Proof of Theorem D2. We apply Noetherian induction. If I = R then I is the empty product. Now assume I ≠ R. By Theorem D1, I is contained in some invertible maximal ideal J. Consider the ideal I(R ÷ J). By Theorem I1, I ⊂ I(R ÷ J). Certainly I(R ÷ J) is a nonzero invertible ideal. By induction, I(R ÷ J) is a product of invertible maximal ideals. Hence I = I(R ÷ J)J is also a product of invertible maximal ideals.

Proof of Theorem D3. Select a nonzero x ∈ P. Then xR is a nonzero invertible ideal. By Theorem D2, xR is a product of invertible maximal ideals; say xR = P_{1}P_{2} … P_{r}. Then P_{1}P_{2} … P_{r} ⊆ P. By Theorem M1, P = P_{i} for some i; since P_{i} is invertible, P is invertible.

Proof of Theorem D4. We apply Noetherian induction. If I = R then I is invertible. Now assume I ≠ R. Let P be a maximal ideal containing I. Then P is invertible by Theorem D3, so I ⊂ I(R ÷ P) by Theorem I1. By Noetherian induction I(R ÷ P) is invertible. Hence I = I(R ÷ P)P is invertible.