Fields Institute Communications Volume 00, 0000

Left valuation rings, left cones, and a question of Frege’s

Hans-Heinrich Brungs

Department of Mathematical Sciences University of Alberta Edmonton, Alberta Canada T6G 2G1 hbrungs @ math.ualberta.ca

Günter Törner

Gerhard-Mercator-Universität 47048 Duisburg Germany g.toerner @ t-online.de

Abstract. A left cone P in a group G is a submonoid of G so that the lattice of left ideals of P is totally ordered and that G is generated by P. A subring R of a skew field F is a left valuation ring of F if and only if R^{∗} = R\{0} is a left cone in F^{∗}. The ideal theory of these left cones is investigated, left cones in a wreath product of groups and in free groups are constructed which are not right cones. Such examples, which give an answer to a question of Frege’s, were encountered in the mid seventies during the construction of left valuation rings. The final section deals with the question whether the existence of left cones in G implies the existence of certain cones in G.

1 Introduction

Let G be a group. A non-empty subset P of G is called a left cone of G if (i) P P ⊆ P, (ii) P P^{−1} = P ∪ P^{−1}, and (iii) P^{−1}P = G. This is equivalent with the condition that P is a submonoid of G that generates G and for any two elements a, b in P either P a ⊇ P b or P b ⊃ P a. A non-empty subset I of P is a left ideal of P if P I ⊆ I and it follows that the lattice of left ideals of a left cone P is totally ordered. The intersection P ∩ P^{−1} = U(P) is the subgroup of units of P.

G. Frege [6] at around 1900 is concerned with, among other things, the construction of real numbers. He essentially asks whether a left cone P of a group G with U(P) = {e} must also be a right cone. A negative answer is given in [1] where this problem is discussed in greater detail. An example of a left cone P with

\begin{matrix}c \\ ◯ \end{matrix}0000 American Mathematical Society

1991 Mathematics Subject Classification. 00F15, 20F60.

Key words and phrases. Left valuation ring, right ordered group, left cone, Frege.

The first author is supported in part by NSERC.

trivial unit group which is not a right cone was obtained earlier, see [2], during the construction of left valuation rings. See Section 3 where we also give further examples of such cones in free groups. The ideal theory of left cones is discussed in Section 2.

A left valuation ring of a skew field F is a subring R of F so that R\{0} is a left cone in the multiplicative group F^{∗} = (F \{0}, ·) of F. A subset I ⊃ {0} of R is a left ideal of the ring R if and only if I\{0} is a left ideal of the left cone R\{0} : If a and b are elements in I with b = ca for c in R, then a − b = (1 − c)a. Such rings have interesting ring theoretical properties, see [8], [3]. W. Klingenberg in [9] considers projective planes over local rings R. The condition that R is a right valuation ring is equivalent with the geometrical condition that there is a line through any two distinct points in the plane. If, in addition, the lattice of left ideals is totally ordered, then R is the coordinate ring of a desarguesian projective plane: any two lines will intersect in at least one point.

If a group G contains a left cone P with U(P) = {e}, it contains a cone Q with U(Q) = {e}, i.e. G is then right ordered, see [10]. In the last section some related results are considered.

2 Ideal theory

Let P be a left cone in a group G. A left ideal of P is a non-empty subset I of P with P I ⊆ I. Right ideals and (two-sided) ideals are defined similarly. An ideal I ≠ P of P is a prime ideal if AB ⊆ I implies A ⊆ I or B ⊆ I for ideals A, B of P, and the ideal I ≠ P is completely prime if ab ∈ I implies a ∈ I or b ∈ I for elements a, b in P.

Definition 2.1 A prime segment Q_{1} ⊃ Q_{2} of the left cone P consists of two completely prime ideals Q_{1} ≠ Q_{2} of P so that no further completely prime ideal exists between Q_{1} and Q_{2}. If Q_{1} is a minimal completely prime ideal of P, then Q_{1} ⊃ Q_{2} with Q_{2} = ∅ is also a prime segment.

If P ≠ G is a left cone of G with U = U(P) as its subgroup of units, then J(P) = P \U is its unique maximal ideal which is also completely prime. The next result shows how further completely prime ideals of P can be obtained; see [4] for the next two results.

Proposition 2.2 Let P be a left cone of a group G and I ≠ P an ideal of P. Then Q = ⋂_{n∈ℕ}I^{n} is either the empty set or a completely prime ideal of P. In

particular, idempotent ideals I = I^{2} ≠ P of P are completely prime.

Prime segments Q_{1} ⊃ Q_{2} of the left cone P fall into one of three categories. Proposition 2.2 plays an essential role in the proof of this result.

Theorem 2.3 Let P be a left cone of a group G and let Q_{1} ⊃ Q_{2} a be a prime segment of P. Then exactly one of the following alternatives occurs:

(a): The prime segment Q_{1} ⊃ Q_{2} is left invariant; i.e. aQ_{1} ⊆ Q_{1}a for all

a ∈ Q_{1}\Q_{2}. (b): The prime segment Q_{1} ⊃ Q_{2} is simple; i.e. there are no further ideals of

P between Q_{1} and Q_{2}.

(c): The prime segment Q_{1} ⊃ Q_{2} is exceptional; i.e. there exists a prime ideal Q of P with Q_{1} ⊃ Q ⊃ Q_{2}. Then there are no further ideals between Q_{1} and Q and ⋂_{n∈ℕ}Q^{n} is equal to Q_{2}.

This theorem provides useful information about the structure of the lattice of ideals of a left cone P.

More precise results can be obtained if P is a rank one left cone which is the case where J(P) is the only completely prime ideal of P. The only segment J(P) ⊃ ∅ of P is then left invariant if and only if P is archimedean in the following sense: For a, b ∈ J(P) there exists a natural number n with P b ⊇ P a^{n}. In that case, P is left invariant, i.e. aP ⊆ P a for all a in P. Then U = U(P) satisfies aU ⊆ Ua for all a in P and the quotient structure P/U consisting of the elements Ua, a ∈ P can be formed; the operation in P/U is given by Ua · Ub = Uab. The condition aU ⊆ Ua for all a in P is not sufficient to guarantee that the semigroup P/U is embeddable into a group. This however follows from the assumption that P is a rank one left invariant cone: We show that left cancellation holds in P/U. Hence, assume that UaUb = Uab = UaUc = Uac for a, b, c in P and, say, c = db for d in P. Then Ua = Uad. We want to prove that this implies d ∈ U and hence Uc = Ub.

If d ∈ J(P) and ad = d′a by the left invariance of P for some d′ in P, then Uad = Ud′a = Ua implies d′ in U. Hence, P a = P d′a = P d^{′n}a = P ad^{n} ⊆ P d^{n} for any n, which contradicts the assumption that P is archimedean. With Hölder’s Theorem [7] we obtain the following result:

Theorem 2.4 Let P be a left cone in G so that J(P) is the only completely prime ideal of P and that J(P) ⊃ ∅ is a left invariant prime segment. Then P is left invariant, P/U(P) = \overline{P} exists and is order isomorphic to a subsemigroup of (ℝ, +).

There are no similarly complete results for rank one left cones P for which the prime segment J(P) ⊃ ∅ is simple or exceptional. For examples and further results see [4], see also Example 4.2.

3 Left cones that are not cones

The authors in [1] rephrase a problem that Frege left unresolved in [6], Band 2, p. 172, as follows: Does there exist a group G with a non-empty subset H so that the following three conditions

α): p, q ∈ H implies pq ∈ H;

β): 1 ∉ H;

γ): p, q ∈ H implies p = q or pq^{−1} ∈ H or qp^{−1} ∈ H;

are satisfied, but the following fourth condition δ) is not satisfied? δ): p, q ∈ H implies p = q or q^{−1}p ∈ H or p^{−1}q ∈ H.

They construct such a subset H in the free group of rank 2 by considering regularly acting automorphisms of a non-linear upper semilinear set, which in turn is a set of finite subsets of the free group of rank 2.

We establish first a relationship between subsets H of a group G with α), β) and γ) and left cones P and G.

Proposition 3.1 Let G ≠ {1} be a group. Then: a): Let P be a left cone of G with U(P) = {1}. Then H = J(P) generates G and satisfies α), β), γ); condition δ) is satisfied by H if and only if P is a right cone.

b): If H is a subset of G that satisfies α), β), γ), and generates G, then P =

H ∪ {1} is a left cone of G with U(P) = {1}. c): If P is a left cone of G and Q is a right and left cone of U(P) with U(Q) = {1}, then H = J(P) ∪ J(Q) generates G and satisfies α), β), and γ). If P is not a right cone, H does not satisfy δ).

Proof The proofs of a) and b) follow from the definitions. The statement c) follows from a) if U(P) = {1}. Otherwise, J(Q) generates U(P) and H = J(P) ∪ J(Q) generates G. To show that α) is satisfied by H it is sufficient to consider p ∈ J(P) and q ∈ J(Q) ⊆ U(P). Then pq, qp ∈ J(P). To prove that condition γ) is true for H let p ≠ q be elements in H. If both p and q are in J(Q), then either pq^{−1} or qp^{−1} is in J(Q). If p is in J(P) and q is in J(Q), then pq^{−1} is in J(P) ⊆ H. Finally, if both p and q are in J(P), then either p = cq or q = cp for c in J(P) or p = uq for u ∈ U(P) = Q. In this last case it follows that u or u^{−1} is in J(Q) which proves γ) for H. If P is not a right cone, there exist p, p′ ∈ P with p^{−1}p′, p^{′−1}p ∉ P. It follows that p ≠ p′ in J(P) ⊆ H and H does not satisfy δ).

We will now give several examples of subsets H in groups G that satisfy α), β), γ) and not δ). The first of these examples for the wreath product of two infinite cyclic groups was obtained during the construction of left valuation rings in [2].

Example 3.2 Let B = ∑_{i∈ℤ}⟨t_{i}⟩ be the direct sum of infinite cyclic groups

⟨t_{i}⟩. Then B can be ordered lexicographically, contains the subgroup B_{0} = ∑_{i=0}^{∞}⟨t_{i}⟩,

and admits the automorphism σ with σ(t_{i}^{r_{i}})_{i∈ℤ} = (t_{i+1}^{r_{i}})_{i∈ℤ}. The group G is the semidirect product G = B ⋊ ⟨y⟩ of (B, σ) with an infinite cyclic group ⟨y⟩; hence, the elements of G have the form by^{n} for b ∈ B, n ∈ ℤ, and yb = σ(b)y.

The subset H = {by^{n} ∈ G|n > 0 and b ∈ B_{0}, or n = 0 and e < b ∈ B_{0}} satisfies the conditions α), β), and γ), where we observe that p = p′q for p, p′, q ∈ P if and only if p = b_{1}y^{n_{1}}, q = b_{2}y^{n_{2}} and either n_{1} > n_{2} or n_{1} = n_{2} and b_{1} > b_{2} in B_{0}.

In addition, pq = (b_{1}y^{n_{1}})(b_{2}y^{n_{2}}) = b_{1}σ^{n_{1}}(b_{2})σ^{n_{2}}(b_{1})^{−1}y^{n_{2}}b_{1}y^{n_{1}} = q′p for any p, q ∈ P and q′ = b_{1}σ^{n_{1}}(b_{2})σ^{n_{2}}(b_{1})^{−1}y^{n_{2}} ∈ P.

On the other hand, p = y and q = t_{0}y (with t_{0} = (t_{i}^{r_{i}})_{i∈ℤ}, r_{0} = 1, r_{i} = 0 otherwise) are elements in P, but y ≠ t_{0}yp′ and t_{0}y ≠ yp′ for all p′ ∈ H, i.e. δ) does not holds for H.

Given a group G generated by a subset H that satisfies the conditions α), β), and γ). Then a binary relation < can be defined on G by the rule that a < b for a, b in G if and only if b = pa for some p ∈ H. It follows immediately from the definition that a < b implies ag < bg for any g in G. The relation < is called limp, i.e. left invariant under multiplication by positive elements, if a < b for a, b in G and q in H implies qa < qb. Finally, the relation < is called dense if H does not have a smallest element.

The relation defined in this way in Example 3.2 is not dense, since t_{0} is the smallest element, but it is limp: a < b for a, b in G implies b = pa for p in H, and qb = qpa = p′qa for q in H and a suitable q′ in H.

To obtain a group G with a subset H that satisfies α), β), γ) but not δ) so that the corresponding relation < is limp and dense, we modify the above example.

Example 3.3 The group B = ∑_{i∈ℚ}⟨t_{i}⟩ with ⟨t_{i}⟩ infinite cyclic for all i can again be ordered lexicographically. B contains the subgroup B_{0} = ∑_{0<i∈ℚ}⟨t_{i}⟩ and admits the automorphism σ with σ(t_{i}^{r_{i}}) = (t_{i+1}^{r_{i}}). Then G and H are defined as in Example 3.2.

In order to obtain subsets H in free groups F that satisfy conditions α), β), γ), but not δ), we combine the construction of left cones P in F given by P.M. Cohn in [5], p. 534/5, with Proposition 3.1.c).

Example 3.4 Let X = {x_{α}|α < τ} be a set of free generators of the free group F, where τ is an ordinal. Let U be the subgroup of F generated by the elements of the form

U_{βα_{1}α_{2}...α_{r}} = x_{α_{r}} … x_{α_{2}}x_{α_{1}}x_{β}x_{α_{1}}^{−1} … x_{α_{r}}^{−1} with β < α_{1} ≥ α_{2} ≥ ⋯ ≥ α_{r}.

Then P, the submonoid of F generated by U and X, is a left cone that is not a right cone if |X| ≥ 2. Every element p in P can be written uniquely in the form p = ux_{α_{r}} … x_{α_{2}}x_{α_{1}} with u ∈ U = U(P), and α_{r} ≤ ⋯ ≤ α_{2} ≤ α_{1}. For u in U and x_{α} there exists u′ in U with x_{α}u = u′x_{α}, and x_{α}x_{β} = (x_{α}x_{β}x_{α}^{−1}))x_{α} = u_{1}x_{α} for β < α and u_{1} = x_{α}x_{β}x_{α}^{−1} ∈ U. There does not exist an element h in P with x_{1}x_{2} = x_{2}h or x_{2} = x_{1}x_{2}h. Since F can be linearly ordered, so can the subgroup U of F. If Π = {u ∈ U = U(P)|u > e}, then H = J(P) ∪ Π is a subset of F that satisfies α, β, and γ, but not δ).

4 Cones from left cones

A group G which has a right and left cone P with U(P) = {e}, i.e. a right ordered group, is in general not linearly ordered, it does not have a cone P′ with U(P′) = {e} and gP′g^{−1} = P′ for all g in G. However, the following arguments, see [10], show that a group G with left cone P and U(P) = {e} is right orderable: If x_{1}, …, x_{n} are elements ≠ e in G, then x_{i} = q^{−1}p_{i} for some elements q, p_{i} in P. Hence, there exists an index i_{0} with P p_{i_{0}} ⊆ P p_{i} for i = 1, …, n, and it follows that for h = p_{i_{0}} the elements h and hx_{i}^{−1} are in P for all i. Then, either P h ⊆ P hx_{i}^{−1} and hx_{i}h^{−1} ∈ P, or P hx_{i}^{−1} ⊆ P h and hx_{i}^{−1}h^{−1} ∈ P. It follows that there exist ε_{i} ∈ {1, −1} so that hx_{i}^{ε_{i}}h^{−1} ∈ J(P) and that the sub-semigroup of G generated by the hx_{i}^{ε_{i}}h^{−1} does not contain e. Hence, the sub-semigroup generated by the x_{i}^{ε_{i}} does not contain e, and, by Conrad’s criterion, G contains a right and left cone with {e} as group of units −G is right orderable.

We obtain:

Proposition 4.1 If P is a left cone in the group G so that U = U(P) is a normal subgroup of G, then G contains a right and left cone Q with U(Q) = U.

To prove this we apply the above results to the factor group G/U.

However, there exist cones P in groups G so that U(P) is not normal in G.

Example 4.2 Let G = ⟨x⟩ ⋉_{σ} {t^{r}|r ∈ ℝ} be the semidirect product of the infinite cyclic group generated by x and the group {t^{r}|r ∈ ℝ} isomorphic to (ℝ, +) with t^{r}x = xt^{2r}, i.e. σ(t^{r}) = t^{2r}. Then P = {x^{n}t^{r}|n ∈ ℤ, r ≥ 0} is a cone of G with U(P) = ⟨x⟩. However, txt^{−1} = xt ∉ U(P) shows that U is not normal in G.

Let F be a skew field and R a left valuation ring of F. Then there is the following easy but somewhat surprising result:

Proposition 4.3 Let R be a left valuation ring of the skew field F. Then R is an invariant valuation ring of F if and only if U(R) is a normal subgroup of F^{∗}.

Proof If R is an invariant valuation ring of F, then dRd^{−1} = R and dU(R)d^{−1} = U(R) follows for d ∈ F^{∗}; hence, U(R) is normal in F^{∗}. Conversely, assume that U(R) is normal in F^{∗}. An element r in R is either in U(R) or in J(R). In the last case, 1 + r ∈ U(R), hence, d(1 + r)d^{−1} = 1 + drd^{−1} ∈ U(R), and drd^{−1} ∈ R. It follows that dRd^{−1} ⊆ R for all d in F^{∗} and dRd^{−1} = R.

This last result is not true for left cones P in groups G or even cones P in G : The cone P, with U(P) = {e} normal in G, of a right orderable group that is not linearly orderable, can serve as an example.

We consider one more case where the existence of a left cone P in a group G implies the existence of a cone P̃ in G; however U(P̃) is the normal closure of U(P) in this case. Let P be a left cone in the group G so that every left ideal of P is an ideal; we say that P is left invariant. In this case aP ⊆ P a for all a in P and P ⊆ a^{−1}P a ⊆ a^{−1}b^{−1}P ba for a, b in P.

Lemma 4.4 (i) Assume that P is a left invariant left cone of a group G. Then P̃ = ∪a^{−1}P a, a ∈ P, is an invariant cone of G with Ũ = U(P̃) the normal closure of U(P). (ii) Assume that R is a left invariant left valuation ring of a skew field F. Then ∪d^{−1}Rd = R̃, d ∈ R^{∗}, is an invariant valuation ring of F with Γ = F^{∗}/Ũ as valuation group, where Ũ = ∪d^{−1}U(R)d, d ∈ R^{∗}.

Proof It is enough to prove (i). It follows from the observation before the lemma that the left cones a^{−1}P a, a ∈ P, form a chain and P̃ is therefore a left cone of G. If p̃ is any element in P̃, then p̃ ∈ a^{−1}P a for some a in P and d^{−1}p̃d, dp̃d^{−1} ∈ P̃ follows for d in P. Since P generates G we have g^{−1}P̃g = P̃ for all g in G and all right ideals of P are ideals; P̃ is a cone. Any unit of P̃ is a unit in some a^{−1}P a, hence contained in some a^{−1}Ua for a in P.

It appears to be a difficult problem to construct for a group G with left cone P a skew field F and a left valuation ring R of F so that G ⊆ F^{∗}, R ∩ G = P, and for every element 0 ≠ r in R exist u ∈ U(R) and p ∈ P with r = up. Every principal left ideal ≠ (0) of R has then the form Rp for some p ∈ P and Rp_{1} = Rp_{2}, for p_{1}, p_{2} ∈ P, only if P p_{1} = P p_{2}. It follows that then there is a correspondence between the left ideal ≠ (0) of R and the left ideals of P. We say that R and P are associated if in this correspondence ideals, completely prime ideals and prime ideals correspond to ideals, completely prime ideals, and prime ideals respectively.

References

[1] Adeleke, S.A.; Dummett, M.A.E.; Neumann, P.H., On a question of Frege’s about rightordered groups, Bull. London Math. Soc. 19 (1987), 513-521.

[2] Botto-Mura, R.T.; Brungs, H.H.; Fisher, J.L., Chain rings and valuation semigroups, Comm. Algebra 5 (1977), no. 14, 1529-1547.

[3] Brungs, H.H.; Törner, G., Right chain domains with prescribed value holoids, J. Algebra 176 (1995), 346-355.

[4] H.H. Brungs, and G. Törner, Ideal theory of right cones and associated rings, J. Algebra 210 (1998), 145-164.

[5] Cohn, P.M., Free rings and their relations. Second edition. London Mathematical Society Monographs, 19, Academic Press, Inc. London-New York, 1985.

[6] Frege, G., Grundesetze der Arithmetik. Bände I, II (reprinted), Hildesheim: Georg Olms Verlagsbuchhandlung, 1962.

[7] Fuchs, L. Teilweise geordnete algebraische Strukturen, Vandenhoek and Ruprecht, Göttingen, 1966.

[8] Jategaonkar, A.V., A counter-example in ring theory and homological algebra, J. Algebra 12 (1969), 418-440.

[9] Klingenberg, W., Projective Geometrien mit Homomorphismus, Math. Annalen 132 (1956), 180-200.

[10] Kopytov, V.M, Medvedev, N.Y., Right ordered groups, Siberian School of Algebra and Logic, New York, London, Moscow, 1996.

[11] Skornyakov, L.A., Left valuation semigroups, Siberian Math. J. 11 (1970) 168-182