Some properties of positive derivations on f-rings II

Theorem 10(b)(i) in the article “Some properties of positive derivations on f -rings” by Henriksen and Smith asserts that if D is a positive derivation on a reduced f -ring and if x ∈ ker D, then {x}⊥⊥ ⊆ ker D. A counterexample is provided to show that this assertion is false, and correct proofs are given for some results in the paper by Henriksen and Smith that use Theorem 10(b)(i) in their proofs.


Introduction
Recall that a lattice-ordered ring ( -ring) is a ring (R, +, ·) with a lattice order ≥ that is compatible with the operations in the sense that if a ≥ b in R and c ∈ R, then a + c ≥ b + c, and if x, y ≥ 0 in R, then xy ≥ 0. For an -ring (R, +, ·, ≥), an endomorphism D of (R, +) is positive if D(R + ) ⊆ R + , where R + = {x ∈ R x ≥ 0} is the positive cone of R, and a derivation if D(ab) = aD(b) + D(a)b for all a, b ∈ R. An f -ring is an -ring R such that if x ∧ y = 0 and z ∈ R + , then xz ∧ y = 0 = zx ∧ y. Recall that if x ∧ y = 0 in an f -ring, then xy = 0 [1, 9.1.10(iv)]. Colville, Davis, and Keimel initiated the study of positive derivations on f -rings in [2]. In [4], Henriksen and Smith extended the work of [2] and provided a direct and elementary proof to its main result.
Throughout the sequel, A denotes an f -ring and D(A) denotes the set of positive derivations on A. For notation and terminology left undefined, we refer the reader to [5].
The result of [4] with which we are concerned is Theorem 10. Recall that an element e ∈ A is regular if x = 0 whenever ex = 0 or xe = 0, that a band of A is a convex sublattice subgroup B of A such that if X ⊆ B and X ∈ A, then X ∈ B, that rad A denotes the set of all nilpotent elements of (a) If e is regular and ex ∈ ker D, then x ∈ ker D.
(b) If A is reduced, then: (iii) ker D is a band; (iv) D n = 0 implies that D = 0; (v) e 2 = e ∈ A implies that e ∈ ker D.

(c) If A has an identity element and U (A) is the smallest band containing the units of
The proof of Theorem 10(b)(i) in [4] used the incorrect inclusion: While the first inclusion is correct, the second inclusion is not (see [1, 3.2.2] and Example 2.1 below); it should be In The proofs of several results in [4] used Theorem 10(b)(i). In Section 3, we provide correct proofs for some of the these and prove special cases for others.
For use in the sequel, recall that an -ring L is Archimedean if for all nonzero x and y in L + , there exists a positive integer n such that nx ≤ y, that an -ideal of L is an ideal of the ring (L, +, ·) that is also a subgroup of (L, +) and a convex sublattice of (L, ≥), and that an -ideal I of L is -prime provided that I = L and if JK ⊆ I for -ideals J and K, either where π : L → L/P is the usual projection π(a) = a+P . The projection π is said to preserve infinite sups if for any subset X of L, whenever X exists in L, π(X) exists in L/P and π(X) = π( X). It is easy to see that if L is the direct sum, or direct product, of totally ordered rings, then the projections onto the factors preserve infinite sups (see also Corollary 3.8 below).
Finally recall that if R is a ring with unit element and D is a derivation on R, then D(1) = 0 because

The Examples
Our first example shows that both Theorem 10(b)(i) and the second inclusion in (1) are false.
Example 2.1 Let R be the totally ordered field of real numbers and R = R[x] be the polynomial ring over R. Define a polynomial in R positive if the coefficient of its highest power is positive. Then R becomes a totally ordered domain which is not Archimedean, and the usual derivative D( Then ker D is the set of constant polynomials, and for any 0 = a ∈ ker D, {a} ⊥ = {0} and hence We next note that an Archimedean f -ring may have a derivation D for which ker D is not a band. The f -ring we construct is not reduced and has a separating set of minimal -prime -ideals whose projections preserve infinite sups. Pick an integer κ > 1 and define a multiplication on L by letting It is easy to check that (L, +, ·, ≥) is a commutative Archimedean f -ring. For j ≥ 1, let b j ∈ L is the element Then b κ is a nonzero element of L such that b 2 κ = 0, and thus L is not reduced. For each integer k ≥ 1, let P k = {f ∈ L f (k) = 0}. It is easy to see that for each f ∈ P k , there exists g ∈ P k such that f g ∈ P k , and hence that P k is a minimal -prime -ideal by [5,Theorem 3.2.22]. It is obvious that {P k } is separating, and it is easy to check that each projection π k : L → L/P k , defined by π k (r) = r + P k , preserves infinite sups.
Define D : L −→ L by letting It is easy to check that D is a positive endomorphism of (L, +, ≥) with kernel ker D = ∞ n=1 R. Furthermore, by the definitions given above, for all f, g ∈ L, D(f g) = 0 and f D(g) + D(f )g = 0 + 0 = 0 as well so that D is a derivation on L. Certainly each b j is in ∞ n=1 R, and since ∞ j=1 b j is the function that is constantly 1,

Lemma 3.1 Suppose that D ∈ D(A) and z ∈ A. If A is reduced, then |D(z)| = D(|z|).
and thus D n−2 (D(x) 2 x n−2 ) = 0. Applying this argument sufficiently many times, we conclude that D(x) n = 0, and thus since R is reduced, that D(x) = 0. Since this is true for all x ∈ R + , D = 0. The other results in [4] whose proofs use Theorem 10(b)(i) are Theorem 10(b)(iii) and the first part of Theorem 10(c) (that U (A) ⊂ ker D). We conclude by showing that these assertions are indeed true for totally ordered rings (Theorem 3.6) and for reduced f -rings that possess a separating set of minimal -prime -ideals whose projections preserve infinite sups (Theorem 3.7). Proof. Suppose that X ⊆ ker D and that x = X in T . For any a ∈ X, x ≥ a, and since D is a positive derivation on T , D(x) ≥ D(a) = 0. Suppose that 0 = z ∈ X. Then x − |z| < x and thus x − |z| is not an upper bound for X. So since T is totally ordered, there exists an element w ∈ X such that x − |z| ≤ w. It follows that D(x) − D(|z|) ≤ D(w) = 0. But since T is totally ordered, either D(|z|) = D(z) = 0 or D(|z|) = D(−z) = 0, and hence D(x) ≤ 0. So 0 ≤ D(x) ≤ 0, i.e., x ∈ ker D, and thus ker D is a band. If T has an identity element, then U (T ) ⊆ ker D by Lemma 3.5.

Lemma 3.5 Suppose that A has a unit element and that D ∈ D(A). If u is a unit of
Corollary 3.8 Suppose that {T α } is a collection of reduced totally ordered rings and recall that α T α is a reduced f -ring with respect to the coordinatewise operations and order. If R is an -subring of α T α that contains α T α and D ∈ D(R), then ker D is a band, and if as well R has an identity element, then U (R) ⊆ ker D.
Proof. We will show that each projection π α : R → T α preserves infinite sups. Suppose by way of contradiction that the projection π γ does not preserve infinite sups. Then there exists {x i } ⊆ R such that i x i exists in R but π γ ( i x i ) = i π γ (x i ). Then since T γ is totally ordered and π γ ( i x i ) > π γ (x i ) for all i, there exists 0 < t ∈ T γ such that π γ ( i x i ) − t ≥ π γ (x i ) for all i. Since α T α ⊆ R, the following element t exists in R: x i for all i, a contradiction. It follows that each π α preserves infinite sups and hence by Theorem 3.7 that ker D is a band and if R has an identity element as well, that U (R) ⊆ ker D.
Example 2.2 shows that Corollary 3.8 may fail if R is not reduced.