Theorem 1

Suppose that H is noetherian.

a) Every chain of graded prime ideals of H, saturated as a chain of graded prime ideals, is saturated as a chain of prime ideals.

b) For every graded prime ideal p of H, one has htgr(p) = ht(S^{-1}p) = ht(p).

c) One has dimgr(H) = dim(S^{-1}H) = dim(H).

To prove a), it is enough to prove that, if p and q are two distinct graded prime ideals of H such that q ⊂ p and every graded prime ideal contained between q and p is equal either to p or to q, then ht(p/q) = 1. Passing to the quotient by q, one is reduced to the case where q = {0}. It is therefore a matter of proving that, if H is an integral domain, and if p is an ideal of H, minimal among the graded prime ideals ≠ {0}, one has ht(p) = 1. Now let a be a nonzero homogeneous element of p, and let r be a prime ideal of H such that a ∈ r ⊂ p and minimal for these properties (II, § 2, No. 6, Lemma 2). Since r^{gr} is prime (Lemma 1, a)) and nonzero, one has r^{gr} = p, hence p = r. Since H is an integral domain and noetherian, p has height 1 (§ 3, No. 1, Prop. 1), whence a).

Let us prove b). Let p be a graded prime ideal of H. One has

(Prop. 2); let us prove the inequality ht(p) ≤ htgr(p) by induction on htgr(p). If htgr(p) = 0, p is minimal among the graded prime ideals, hence minimal (Lemma 1, c)), and one has ht(p) = 0. Suppose that one has htgr(p) > 0 and let us prove the inequality ht(q) ≤ htgr(p) − 1 for every prime ideal q contained in p and distinct from p. Let us distinguish two cases. If q is graded, one concludes by the induction hypothesis. If q is not graded, then one has q^{gr} ≠ p, hence ht(q^{gr}) ≤ htgr(q^{gr}) by the induction hypothesis, whence ht(q) ≤ htgr(q^{gr}) + 1 by Prop. 3; it remains to prove the inequality htgr(q^{gr}) ≤ htgr(p) − 2; but if one had htgr(q^{gr}) = htgr(p) − 1, the chain q^{gr} ⊂ p would be saturated by a), which it is not since q^{gr} ≠ q ≠ p.

Finally let us prove c). One has dimgr(H) ≤ dim(S^{-1}H) ≤ dim(H) (Prop. 2), and it remains to prove dim(H) ≤ dimgr(H), or again ht(p) ≤ dimgr(H) for every prime ideal p of H. Let therefore p be a prime ideal of H. If p is graded, one has ht(p) = htgr(p) ≤ dimgr(H). If p is not graded, one has ht(p) = htgr(p^{gr}) + 1 by Prop. 3; let m be a maximal graded ideal of H containing p^{gr}; by Lemma 1, b), m is maximal, hence distinct from p^{gr}, and one has htgr(p^{gr}) + 1 ≤ htgr(m) ≤ dimgr(H), whence again ht(p) ≤ dimgr(H). This completes the proof.