A Nilregular Element Property

An element a of a commutative ring R is nilregular if and only if x is nilpotent whenever ax is nilpotent. More generally, an ideal I of R is nilregular if and only if x is nilpotent whenever ax is nilpotent for all a ∈ I. We give a direct proof that if R is Noetherian, then every nilregular ideal contains a nilregular element. In constructive mathematics, this proof can then be seen as an algorithm to produce nilregular elements of nilregular ideals whenever R is coherent, Noetherian, and discrete. As an application, we give a constructive proof of the Eisenbud–Evans–Storch theorem that every algebraic set in n–dimensional affine space is the intersection of n hypersurfaces. AMS Classification: 03F65 (14M10) 1. The nilregular element property Let R be a commutative ring with unit and N its nilradical, i.e. the ideal consisting of its nilpotent elements. We define an element a (respectively, an ideal I) of R to be nilregular if and only if x ∈ N whenever ax ∈ N (respectively, ax ∈ N for all a ∈ I). So an ideal I is nilregular precisely when the transporter ideal (N : I) = {x ∈ R : xI ⊆ N} is contained in N . We present a method to find nilregular elements of nilregular ideals when R is Noetherian. For this, we interpret first the property of being nilregular in a topological way. As usual, let D(a) be the set of prime ideals p of R such that a / ∈ p, and let D(a1, . . . , an) stand for the union of D(a1), . . . , D(an). The intersection of D(a) and D(b) is D(ab), and D(a) is a subset of D(a1, . . . , an) if and only if a belongs to the radical of the ideal (a1, . . . , an) generated by a1, . . . , an. In particular, D(a) = ∅ precisely when a ∈ N . Lemma 1.1. We have D(a + b, ab) = D(a, b) for all a, b ∈ R. If, in particular, D(a) and D(b) are disjoint, then D(a + b) = D(a, b). Proof. Both a = a(a + b)− ab and b = (a + b)b− ab belong to (a + b, ab). It is well–known that the D(a) with a ∈ R form a basis of opens for the Zariski topology on the prime spectrum (the set of all prime ideals) of R. It follows that a ∈ R is nilregular if and only if D(a) is dense for the Zariski topology. Remark 1.2. D(a1, . . . , an) is dense if and only if (a1, . . . , an) is a nilregular ideal. Theorem 1.3. Let R be Noetherian, and a1, . . . , an ∈ R. If D(a1, . . . , an) is dense, then the ideal (a1, . . . , an) contains a nilregular element. Proof. If D(x) 6= ∅, then there exists i such that D(xai) 6= ∅, because D(a1, . . . , an) is dense. Hence if the ring is nontrivial, then we can inductively build a sequence b0, b1, . . . 1 of elements of R in the following way: b0 is one ai such that D(b0) 6= ∅; if D(b0, . . . , bk) is not dense, then bk+1 is a multiple of one aj such that D(bk+1) 6= ∅ and D(bk+1) is disjoint from D(b0, . . . , bk). Since R is Noetherian, this procedure has to stop, and we eventually find p such that D(b0, . . . , bp) is dense and D(bi) ∩D(bj) = ∅ whenever i 6= j. By Lemma 1.1, we have D(b0, . . . , bp) = D(b0 + · · ·+ bp) and b0 + · · ·+ bp is a nilregular element in (a1, . . . , an). As in [1] we define the ideal boundary Na of a ∈ R to be the ideal generated by a and the elements x of R such that ax is nilpotent; in other words, Na = aR + (N : a). Lemma 1.4. Every ideal boundary is a nilregular ideal. Proof. Fix a ∈ R, and assume that bx is nilpotent for all b ∈ Na. Then x is nilpotent. Indeed, ax is nilpotent because a ∈ Na; whence x ∈ Na and thus x is nilpotent. Corollary 1.5. If R is Noetherian, then every ideal boundary contains a nilregular element. Throughout this section we could only have required that the topological space Spec(R) rather than the ring R be Noetherian. 2. Constructive interpretation We interpret the previous argument in the framework of constructive mathematics [6, 7]. Let L(R) be the lattice of radically finitely generated ideals of R: that is, the radicals of finitely generated ideals [2]. Following Joyal [5], the lattice L(R), with inclusion as ordering, can also be defined as the distributive lattice generated by the symbols D(a) with a ∈ R, and equipped with the relations D(0) = 0 , D(1) = 1 , D(ab) = D(a) ∧D(b) , D(a + b) ≤ D(a) ∨D(b) for a, b ∈ A. Writing D(a1, . . . , am) for D(a1) ∨ · · · ∨D(am), it can be shown [2] that D(b1) ∧ · · · ∧D(bn) ≤ D(a1, . . . , am) if and only if the monoid generated by b1, . . . , bn meets the ideal generated by a1, . . . , am. So D(a1, . . . , am) can indeed be identified with the radical of the ideal (a1, . . . , am), and D (a) = 0 precisely when a is nilpotent. Lemma 2.1. If R is coherent, Noetherian, and discrete, then one can decide whether a given element of R is nilpotent. Proof. Let a ∈ R. Every annihilator (0 : a) is a finitely generated ideal with (0 : a) ⊆ (0 : a). Since R is Noetherian, there exists n such that (0 : a) = (0 : a). We even have (0 : a) = ( 0 : a ) for all k. (Indeed, if ab = 0, then ab annihilates a and thus also a, so that ab = 0.) Hence a is nilpotent if and only if a = 0. Corollary 2.2. If R is coherent, Noetherian, and discrete, then equality to 0 is decidable in L (R). 2 We recall that a lattice is a Heyting algebra if and only if one can assign to every pair (u, v) of elements another element u → v such that u ∧ x ≤ v if and only if x ≤ u → v. In a Heyting algebra, one writes ¬u for u → 0. If R is coherent and Noetherian, then L(R) is a Heyting algebra [2]. A direct argument shows that a ∈ R is nilregular if and only if ¬D(a) = 0. Remark 2.3. ¬D(a1, . . . , an) = 0 if and only if (a1, . . . , an) is a nilregular ideal. Lemma 2.4. If R is coherent, Noetherian, and discrete, for given b0, . . . , bk ∈ R we can decide whether ¬D(b0, . . . , bk) = 0; if indeed ¬D(b0, . . . , bk) 6= 0, then we can compute bk+1 ∈ R such that D(bk+1) 6= 0 and D(bk+1) ∧D(b0, . . . , bk) = 0. Proof. Write ¬D(b0, . . . , bk) = D (c1, . . . , cm), and apply Lemma 2.1 successively to the cj. If cj / ∈ N for some j, then bk+1 = cj is as desired. Corollary 2.5. If R is coherent, Noetherian and discrete, then we can decide whether an element b of R is nilregular, and if this is not the case, then we can compute an element x / ∈ N such that bx ∈ N . In this context, ¬D(b0, . . . , bk) = 0 precisely when D(b0, . . . , bk) is dense. Reasoning as in the previous section (Theorem 1.3), we can now conclude. Theorem 2.6. Let R be coherent, Noetherian, and discrete, and a1, . . . , an ∈ R. If ¬D(a1, . . . , an) = 0, then the ideal (a1, . . . , an) contains a nilregular element. This result seems closely connected to the regular element property proved constructively in [7]. The hypothesis is a little weaker (we don’t assume the ring to contain an infinite field), but the statement is a priori different unless the ring is reduced (we use ‘nilregular’ instead of ‘regular’). In view of Lemma 1.4, Corollary 1.5 can be rephrased as follows. Corollary 2.7. If R is coherent, Noetherian, and discrete, then every ideal boundary contains a nilregular element. In terms of L (R), this means that for every a ∈ R there is s ∈ R with ¬D(s) = 0 and D(s) ≤ D(a) ∨ ¬D(a); observe that D(Na) = D(a) ∨ ¬D(a).