Proposition 6

Let A be a ring. a) If a is an ideal of A, one has dim(A/a) $\leq$ dim(A). b) If S is a multiplicative subset of A, one has dim(S$^{-1}$A) $\leq$ dim(A). c) One has dim(A) = sup dim(A/p), where p runs through the set of minimal prime ideals of A. d) If A has only a finite number of minimal prime ideals (for example if A is noetherian (II, § 4, No. 3, cor. 3 to prop. 14)) and if p is a prime ideal of A, one has

where q runs through the set of minimal prime ideals of A contained in p. e) Let a be an ideal of A which is contained in no minimal prime ideal of A; then one has dim(A) $\geq$ dim(A/a) + 1. In particular, if A is integral, one has dim(A) $\geq$ dim(A/a) + 1 for every nonzero ideal a of A.

By the remark of II, § 4, No. 3, if a is an ideal of A, the topological space Spec(A/a) is homeomorphic to the closed subspace V(a) of Spec(A). Assertion a) results from this and from prop. 1, a) of No. 1. Assertion b) results from loc. cit., corollary to prop. 13. By loc. cit., cor. 2 to prop. 14, the irreducible components of Spec(A) are homeomorphic to the spaces Spec(A/p), where p is a minimal prime ideal of A, and assertion c) results from the corollary of prop. 1 of No. 1. Under the assumption of d), the space Spec(A) has only a finite number of irreducible components; the irreducible components of Spec(A) containing p are the sets V(q), where q is a minimal prime ideal contained in p. Assertion d) then results from cor., b) of prop. 1 of No. 1.

Let us prove e) finally. It is a question of proving that, for every chain $p_0 \subset ... \subset p_n$ of prime ideals of A such that $a \subset p_0$, one has dim(A) $\geq n + 1$. In view of the assumption made on a, there exists a prime ideal $p_{-1}$ of A contained in $p_0$, distinct from $p_0$, and $p_{-1} \subset p_0 \subset ... \subset p_n$ is a chain of prime ideals of A, of length $n + 1$.