An elementary characterisation of Krull dimension

Thierry Coquand (^{∗}) Henri Lombardi (^{†}), Marie-Françoise Roy (^{‡}),

April 2004

Abstract

We give an elementary characterisation of Krull dimension for distributive lattices and commutative rings. This follows the following geometrical intuition: an algebraic variety is of dimension ≤ k if and only if each subvariety has a boundary of dimension < k. Since our results hold for distributive lattices, they hold, by Stone duality [11], for any spectral spaces.

MSC 2000: 13C15, 03F65, 13A15, 13E05

Key words: Krull dimension, Boundary of a subvariety, Constructive Mathematics.

Boundaries of an element in a distributive lattice

By distributive lattice we mean a lattice with a minimum and a maximum (so that all finite parts have a supremum and an infimum) which is distributive.

Let L be a distributive lattice. An ideal of L is a subset I ⊆ L such that

0 ∈ I
x, y ∈ I ⟹ x ∨ y ∈ I
x ∈ I, z ∈ L ⟹ x ∧ z ∈ I

The last property can be written (x ∈ I, y ≤ x) ⇒ y ∈ I.

The dual notion is the notion of filter. A filter F is a subset of L such that

1 ∈ F
x, y ∈ F ⟹ x ∧ y ∈ F
x ∈ F, z ∈ L ⟹ x ∨ z ∈ F

A prime ideal is an ideal I such that 1 ∉ I and

x ∧ y ∈ I ⇒ [x ∈ I or y ∈ I]

and dually a prime filter is a filter F such that 0 ∉ F and

x ∨ y ∈ F ⇒ [x ∈ F or y ∈ F]

Notice that an ideal (resp. a filter) is prime if and only if its complement is a filter (resp. an ideal).

∗ Chalmers, University of Göteborg, Sweden, email: coquand@cs.chalmers.se

† Equipe de Mathématiques, CNRS UMR 6623, UFR des Sciences et Techniques, Université de Franche-Comté, 25 030 BESANCON cedex, FRANCE, email: lombardi@math.univ-fcomte.fr, partiellement financé par le réseau européen RAAG CT-2001-00271

‡ IRMAR (UMR CNRS 6625), Université de Rennes 1, Campus de Beaulieu 35042 Rennes cedex FRANCE, email:

marie-francoise.roy@math.univ-rennes1.fr, partiellement financée par le réseau européen RAAG CT-2001-00271

If x ∈ L we denote by D(x) the set of prime ideals I such that x ∉ I. We have D(0) = ∅ and D(x) ∩ D(y) = D(x ∧ y). The set of all prime ideals of a distributive lattice L has a natural structure of topological space, called the spectrum Sp(L) of L. We take for basic open sets the sets D(x), x ∈ L. It can be shown that, each D(x) is compact, and that the compact open sets of Sp(L) are exactly the subsets of the form D(x), x ∈ L [11].

The spaces (homeomorphic to spaces) of the form Sp(L) are called spectral spaces and it is possible to characterise directly these spaces by topological properties [11, 7]. Most topological spaces used in commutative algebra, Zariski spectrum of a ring, spaces of valuations of a field, …, are spectral spaces.

The set Sp(L) is ordered by inclusion, and the Krull dimension of L is defined as the upper bound of the length of chains of prime ideals (or equivalently chains of prime filters).

If x ∈ L we define the boundary ideal of x as being the ideal generated by x and the elements y ∈ L such that x ∧ y = 0. Dually, we define the boundary filter of x as being the filter generated by x and the elements y ∈ L such that x ∨ y = 1.

Definition 1 The upper boundary of x ∈ L in the distributive lattice L is the distributive lattice L^{{x}} quotient of L by the boundary ideal of x. Thus it is the lattice L, ∧, ∨ with the order

a ≤^{x} b  ⟺  ∃y ∈ L ( x ∧ y = 0 & a ≤ x ∨ y ∨ b)

When L is implicative the definition becomes a ≤ x ∨ ¬x ∨ b.

By considering the dual lattice, one defines the lower boundary L_{{x}}, which is the distributive lattice quotient of L by the boundary filter of x. Thus it is the lattice L, ∧, ∨ with the order

a ≤_{x} b  ⟺  ∃y ∈ L ( x ∨ y = 1 & a ∧ x ∧ y ≤ b)

It can be checked that the boundary of the open D(x), viewed as a subspace of Sp(L), is a spectral space (as a closed set in a spectral space) and corresponds by Stone duality to the distributive lattice L^{{x}}.

Krull dimension of a distributive lattice

The duality between distributive lattice and spectral spaces relies on classical logic and the axiom of choice. From a constructive point of view, this duality is seen as a way to develop the theory of spectral spaces, using distributive lattices as a point-free presentation of these spaces [8]. One is thus led to look for direct definitions of topological notions in term of distributive lattices, and for instance, a direct definition of the Krull dimension.

A first constructive definition of Krull dimension was sketched in [1]. This definition was analysed in the work [6]. The author gave an elementary characterisation of the Krull dimension of a lattice L in term of the Boolean algebra generated by L. In [3], following the idea in [1], the two first authors proved the following result, which gives yet another a concrete characterization of Krull dimension.

Theorem 2 Let L be a distributive lattice generated by a subset S and ℓ a nonnegative integer. The following are equivalent

(1) L has Krull dimension ≤ ℓ

(2) For all x_{0}, …, x_{ℓ} ∈ S there exist a_{0}, …, a_{ℓ} ∈ L such that

a_{0} ∧ x_{0} ≤ 0 , a_{1} ∧ x_{1} ≤ a_{0} ∨ x_{0} , …, a_{ℓ} ∧ x_{ℓ} ≤ a_{ℓ−1} ∨ x_{ℓ−1} , 1 ≤ a_{ℓ} ∨ x_{ℓ}.

In particular a distributive lattice L is of dimension ≤ 0 if and only if L is a Boolean algebra (any element has a complement).

The goal of this paper is to present a simpler inductive characterisation of Krull dimension, which provides also a simple proof of the equivalence between (1) and (2) in the previous theorem. This inductive characterisation corresponds to the following geometrical intuition: a variety is of dimension ≤ k if and only if any subvariety has a boundary of dimension < k. (the induction begins with dimension −1 which defines the trivial lattice).

Theorem 3 Let L be a distributive lattice generated by a subset S and ℓ a nonnegative integer. The following are equivalent

(1) L has Krull dimension ≤ ℓ

(2) For all x ∈ S the boundary L^{{x}} is of Krull dimension ≤ ℓ − 1.

(3) For all x ∈ S the boundary L_{{x}} is of Krull dimension ≤ ℓ − 1.

Proof.

(1) ⇔ (2): We show first that any maximal filter F of L becomes trivial in L^{{x}}, i.e. it contains 0. This means that one can find a ∈ F such that a ≤^{x} 0. If x ∈ F this holds since x ≤^{x} 0. If x ∉ F there exists z ∈ F such that x ∧ z = 0 (since the filter generated by F and x is trivial) and we have then z ≤^{x} 0. This shows that the Krull dimension of L^{{x}} becomes one less than the one of L (if it is finite). Next, we show that if F′ ⊂ F, F maximal and x ∈ F \ F′ then F′ does not become trivial in L^{{x}} (which shows that dim L^{{x}} is dim L − 1 with a good choice of x). Indeed, we would get otherwise z ∈ F′ such that z ∧ x = 0, which is impossible since both z and x are in F. We finally notice that if F′ ⊂ F are distinct prime filters and S generates L one can find x ∈ S such that x ∈ F \ F′.

(1) ⇔ (3) is a consequence of (1) ⇔ (2) by duality. □

By Stone duality [11], we get the following result.

Theorem 4 A spectral space X is of Krull dimension ≤ k if and only if any open compact of X has a boundary of dimension < k.

Since any spectral space can be viewed as the spectrum of a commutative ring, it is natural to define directly boundaries for commutative rings.

The two boundaries of an element in a commutative ring

Let R be a commutative ring. We write ⟨J⟩ for the ideal of R generated by the subset J ⊆ R. We write M(U) for the monoid (a monoid will always be multiplicative) generated by the subset U ⊆ R. Given a commutative ring R the Zariski lattice Zar(R) has for elements the radicals of finitely generated ideals. The order relation is the inclusion and we get

\sqrt{I} ∧ \sqrt{J} = \sqrt{IJ}, \sqrt{I} ∨ \sqrt{J} = \sqrt{I + J}.

We shall write ã for \sqrt{⟨a⟩}. We have

ã_{1} ∨ ⋯ ∨ ã_{m} = \sqrt{⟨a_{1}, …, a_{m}⟩} and ã_{1} ∧ ⋯ ∧ ã_{m} = { \atop a_{1} ⋯ a_{m}}.

Let U and J be two finite subsets of R, we have

⋀_{u∈U} ũ ≤_{Zar(R)} ⋁_{a∈J} ã ⟺ ∏_{u∈U} u ∈ \sqrt{⟨J⟩} ⟺ M(U) ∩ ⟨J⟩ ≠ ∅

This describes completely the lattice Zar(R). More precisely ([3]) we have:

Proposition 5 The lattice Zar(R) of a commutative ring R is (up to isomorphism) the lattice generated by symbols D(x), x ∈ R with the relations

D(0) = 0, D(1) = 1, D(fg) = D(f) ∧ D(g), D(f + g) ≤ D(f) ∨ D(g).

The spectrum of the distributive lattice Zar(R) is naturally isomorphic to the Zariski spectrum of the ring R. So the Krull dimension of a commutative ring R is the same as the Krull dimension of its Zariski lattice Zar(R).

Definition 6 Let R be a commutative ring and x ∈ R.

(1) The boundary R^{\{x\}} of x in R is the quotient ring R/I^{\{x\}} where I^{\{x\}} = xR + (\sqrt{0} : x).

(2) The boundary R_{{x}} of x in R is the localized ring R_{S_{{x}}} where S_{{x}} = x^{ℕ}(1 + xR).

The next proposition is easy.

Proposition 7 Let L = Zar(R) and x ∈ R. Then L^{{x̃}} is naturally isomorphic to Zar(R^{{x}}) and L_{{x̃}} is naturally isomorphic to Zar(R_{{x}}).

We get an elementary inductive characterization of Krull dimension of commutative rings. Recall that a ring R has Krull dimension −1 if and only if it is trivial (i.e., 1_{R} = 0_{R}).

Theorem 8 Let R be a commutative ring and ℓ ≥ 0 an integer. The following are equivalent

(1) The Krull dimension of R is ≤ ℓ.

(2) For all x ∈ R the Krull dimension of R^{{x}} is ≤ ℓ − 1.

(3) For all x ∈ R the Krull dimension of R_{{x}} is ≤ ℓ − 1.

These equivalences are immediate consequences of Theorem 3 and Proposition 7.

Corollary 9 (cf. [3, 10]) Let ℓ be a nonnegative integer. The Krull dimension of R is ≤ ℓ if and only if for all x_{0}, …, x_{ℓ} in R there exists a_{0}, …, a_{ℓ} ∈ R and m_{0}, …, m_{ℓ} ∈ ℕ such that

x_{0}^{m_{0}}(⋯ (x_{ℓ}^{m_{ℓ}}(1 + a_{ℓ}x_{ℓ}) + ⋯) + a_{0}x_{0}) = 0  (1)

Proof. Since dimension −1 corresponds to the trivial ring the equivalence for the case ℓ = 0 is clear. Assume the equivalence has been established for all integers < ℓ and all R. We deduce that the dimension of a localization S^{−1}R is < ℓ if and only if for all x_{0}, …, x_{ℓ−1} ∈ R there exist a_{0}, …, a_{ℓ−1} ∈ R, s ∈ S and m_{0}, …, m_{ℓ−1} ∈ ℕ such that

x_{0}^{m_{0}}(x_{1}^{m_{1}} ⋯ (x_{ℓ−1}^{m_{ℓ−1}}(s + a_{ℓ−1}x_{ℓ−1}) + ⋯ + a_{1}x_{1}) + a_{0}x_{0}) = 0 .  (2)

Notice that s replaces 1 in the similar equality (1) with R instead of S^{−1}R. It remains only to replace s by an arbitrary element in S_{{x_{ℓ}}}, i.e., an element x_{ℓ}^{m_{ℓ}}(1 + a_{ℓ}x_{ℓ}). □

The advantage of this definition is, besides its elementary character, to allow simple proofs by induction on the dimension. We can for instance prove directly in this way the following non-Noetherian version of Bass’ stable range theorem.

Theorem 10 If the dimension of R is < n and 1 = D(a, b_{1}, …, b_{n}) there exists x_{1}, …, x_{n} such that 1 = D(b_{1} + ax_{1}, …, b_{n} + ax_{n}).

Examples If A = ℤ and n ≠ 0, 1, −1 then ℤ^{{n}} = ℤ/nℤ and ℤ_{{n}} = ℚ. These are two 0-dimensional rings. For n = 0, 1 or −1 the two boundaries are trivial. Thus the Krull dimension of ℤ is 1. Let 𝕂 be a field contained in an algebraically closed field 𝕃, and J be a finitely generated ideal of 𝕂[X_{1}, …, X_{n}] and A = 𝕂[X_{1}, …, X_{n}]/J. If V is the algebraic variety corresponding to J in 𝕃^{n}, if f ∈ A defines the subvariety W of V and if B is the boundary of W in V, defined as the intersection of W with the Zariski cloture of its complement in V, then the affine variety B corresponds to the ring A^{{f}}.

Acknowledgments: we thank the referee for carefull rereading and valuable comments.

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