The Markov-Zariski topology of an abelian group
aa r X i v : . [ m a t h . G R ] M a y THE MARKOV-ZARISKI TOPOLOGY OF AN ABELIAN GROUP
DIKRAN DIKRANJAN AND DMITRI SHAKHMATOV
Dedicated to Kenneth Kunen on the occasion of his 65th anniversary
Abstract.
According to Markov [25], a subset of an abelian group G of the form { x ∈ G : nx = a } , for some integer n and some element a ∈ G , is an elementary algebraic set; finite unions ofelementary algebraic sets are called algebraic sets. We prove that a subset of an abelian group G is algebraic if and only if it is closed in every precompact (=totally bounded) Hausdorff grouptopology on G . The family of all algebraic subsets of an abelian group G forms the family of closedsubsets of a unique Noetherian T topology Z G on G called the Zariski , or verbal , topology of G [3]. We investigate the properties of this topology. In particular, we show that the Zariski topologyis always hereditarily separable and Fr´echet-Urysohn.For a countable family F of subsets of an abelian group G of cardinality at most the continuum,we construct a precompact metric group topology T on G such that the T -closure of each memberof F coincides with its Z G -closure. As an application, we provide a characterization of the subsetsof G that are T -dense in some Hausdorff group topology T on G , and we show that such a topology,if it exists, can always be chosen so that it is precompact and metric. This provides a partial answerto a long-standing problem of Markov [25]. We use P and N to denote the sets of all prime numbers and all natural numbers, respectively.In this paper, 0 ∈ N . As usual, Z denotes the group of integers, and Z ( n ) denotes the cyclic groupof order n . We use c to denote the cardinality of the continuum. The symbol ω denotes the firstuncountable cardinal. 1. Introduction
Three topologies on a group.
In 1944, Markov [25] introduced four special families ofsubsets of a group G : Definition 1.1. ([25]) A subset X of a group G is called:(a) elementary algebraic if there exist an integer n >
0, elements a , . . . , a n ∈ G and ε , . . . , ε n ∈{− , } , such that X = { x ∈ G : x ε a x ε a . . . a n − x ε n a n = 1 } ,(b) algebraic if X is an intersection of finite unions of elementary algebraic subsets of G ,(c) unconditionally closed if X is closed in every Hausdorff group topology on G ,(d) potentially dense if G admits some Hausdorff group topology T such that X is dense in( G, T ).The family of all unconditionally closed subsets of G coincides with the family of closed setsof a T topology M G on G , namely the infimum (taken in the lattice of all topologies on G ) ofall Hausdorff group topologies on G . This topology has been introduced in [9, 10] as the Markovtopology of G . Key words and phrases. abelian group, algebraic closure, Zariski topology, verbal topology, Markov topology,Noetherian space, irreducible set, unconditionally closed set, potentially dense set, precompact group, totally boundedgroup, combinatorial dimension.The first named author was partially supported by SRA grants P1-0292-0101 and J1-9643-0101, as well as grantMTM2009-14409-C02-01.The second named author was partially supported by the Grant-in-Aid for Scientific Research (C) No. 19540092of the Japan Society for the Promotion of Science (JSPS). Recall that a Hausdorff group topology T on a group G is called precompact (or totally bounded )provided that ( G, T ) is (isomorphic to) a subgroup of some compact Hausdorff group or, equiva-lently, if the completion of ( G, T ) with respect to the two-sided uniformity is compact. Let P G bethe infimum of all precompact Hausdorff group topologies on G . Clearly, P G is a T topology on G , which we call the precompact Markov topology of G [9].One can easily see that the family of all algebraic subsets of G is closed under finite unionsand arbitrary intersections, and contains G and all finite subsets of G ; thus, it can be taken asthe family of closed sets of a unique T topology Z G on G . Markov [24, 25] defined the algebraicclosure of a subset X of a group G as the intersection of all algebraic subsets of G containing X ,i.e., the smallest algebraic set that contains X . This definition satisfies the conditions necessary forintroducing a topological closure operator on G . Since a topology on a set is uniquely determinedby its closure operator, it is fair to say that Markov was the first to (implicitly) define the topology Z G , though he did not name it. To the best of our knowledge, the first name for this topologyappeared explicitly in print in a 1977 paper by Bryant [3], who called it a verbal topology of G . Ina more recent series of papers beginning with [2], Baumslag, Myasnikov and Remeslennikov havedeveloped algebraic geometry over an abstract group G . In an analogy with the celebrated Zariskitopology from algebraic geometry, they introduced the Zariski topology on the finite powers G n ofa group G . In the particular case when n = 1, this topology coincides with the verbal topology ofBryant. For this reason, the topology Z G is also called the Zariski topology of G in [9, 10].Note that ( G, Z G ), ( G, M G ) and ( G, P G ) are quasi-topological groups, i.e., their inversion andshifts are continuous; see [1]. Fact 1.2. Z G ⊆ M G ⊆ P G for every group G . Proof.
An elementary algebraic subset of G must be closed in every Hausdorff group topology on G , which gives the first inclusion. The second inclusion is obvious. (cid:3) Markov’s “algebraic versus unconditionally closed” problem.
In 1944, Markov [24](see also [25]) posed his celebrated problem: is every unconditionally closed subset of a groupalgebraic ? Using the language of Markov and Zariski topologies, this question can be naturallyreformulated as the problem of coincidence of these topologies: does the equality Z G = M G hold forevery group G ? Markov himself obtained a positive answer in the case when G is countable [25].Moreover, in [25], Markov attributes to Perel’man the fact that Z G = M G for every abelian group G . To the best of our knowledge, the proof of this fact never appeared in print until [10]. In thepresent manuscript, we further strengthen this result from [10] as follows: Theorem A. Z G = M G = P G for every abelian group G . Reformulating this theorem in Markov’s terminology results in the following Corollary B.
Corollary B.
A subset of an abelian group G is algebraic if (and only if ) it is closed in everyprecompact Hausdorff group topology on G . An example of a group G with Z G = M G was found by Hesse [20], who apparently was unawarethat his results solve Markov’s problem in the negative. This problem was later highlighted as anopen problem in the survey [5], and an example of a group G with Z G = M G under the ContinuumHypothesis CH was recently provided in [30].If the group G is infinite, the topology M G is discrete if and only if G is non-topologizable ,i.e., it does not admit a non-discrete Hausdorff group topology. The existence of non-topologizablegroups has been another long-standing open problem of Markov [25]; it was later resolved positivelythrough an example under CH by Shelah [29], an uncountable ZFC example by Hesse [20] and acountable ZFC example by Ol ′ shanskij [28].Observe that the topology P G is discrete if and only if the group G is not maximally almostperiodic , i.e., it does not admit a precompact Hausdorff group topology. A classical example of a group G with discrete P G is the group SL (2 , C ) of all complex 2 × G is topologizable (by its usual topology), one has M G = P G .In view of Fact 1.2, if a group G has discrete Z G , then Z G = M G = P G holds because all threetopologies become discrete. Such groups are extremely rare, but a variety of examples have beenconstructed by Ol ′ shanskij and his school: Example 1.3. (a) Ol ′ shanskij’s example [28] of a countable non-topologizable group has dis-crete Z G .(b) Klyachko and Trofimov [23] constructed a finitely generated torsion-free group G such that Z G is discrete.(c) Trofimov [32] proved that every group H admits an embedding into a group G with discrete Z G .(d) Morris and Obraztsov [26] modified Ol ′ shanskij’s example from item (a) to build, for anysufficiently large prime p , a continuum of pairwise non-isomorphic infinite non-topologizablegroups of exponent p , all of whose proper subgroups are cyclic.1.3. Realization of the Zariski closure by some Hausdorff group topology.
Given a topo-logical space ( Y, T ), we denote the T -closure of a set X ⊆ G by cl T ( X ).Let X be a subset of a group G . Given a Hausdorff group topology T on G , one has Z G ⊆ M G ⊆T , and therefore, cl T ( X ) ⊆ cl M G ( X ) ⊆ cl Z G ( X ). This chain of inclusions naturally leads to thefollowing problem: can one always find a Hausdorff group topology T on G such that cl T ( X ) =cl M G ( X ) = cl Z G ( X )? We shall call this question the realization problem for the Zariski closure ; see[9]. This problem was first considered by Markov in [25], who proved that for every subset X of acountable group G , there exists a metric group topology T on G such that cl T ( X ) = cl Z G ( X ). Wemake the following contribution to this general problem in the abelian case: Theorem C.
For an abelian group G , the following conditions are equivalent: (i) | G | ≤ c , (ii) for every subset X of G , there exists a precompact metric group topology T X on G such that cl T X ( X ) = cl Z G ( X ) , (iii) for every countable family X of subsets of G , one can find a precompact metric grouptopology T X on G such that cl T X ( X ) = cl Z G ( X ) for every X ∈ X . It should be noted that item (iii) cannot be pushed further to accommodate families of size ω ;see Remark 9.6.A counterpart to this theorem, with the word ”metric” removed from both items (ii) and (iii)and the inequality in item (i) relaxed to | G | ≤ c , is proved in our subsequent paper [12].1.4. Characterization of potentially dense subsets of abelian groups of size at most c . The last section of Markov’s paper [25] is exclusively dedicated to the following problem: whichsubsets of a group G are potentially dense in G ? Markov succeeded in proving that every infinitesubset of Z is potentially dense in Z [25]. This was strengthened in [14, Lemma 5.2] by proving thatevery infinite subset of Z is dense in some precompact metric group topology on Z . (The authorsof [25] and [14] were apparently unaware that both these results easily follow from the uniformdistribution theorem of Weyl [33].) Further progress was made by Tkachenko and Yaschenko [31],who proved the following theorem: if an abelian group G of size at most c is either almost torsion-free or has exponent p for some prime p , then every infinite subset of G is potentially dense in G .(According to [31], an abelian group G is almost torsion-free if r p ( G ) is finite for every prime p .)In this manuscript, we obtain the following complete characterization of potentially dense subsetsof abelian groups of size at most c : Theorem D.
Let X be a subset of an abelian group G such that | G | ≤ c . Then the followingconditions are equivalent: (i) X is potentially dense in G , (ii) X is T -dense in G for some precompact metric group topology on G , (iii) cl Z G ( X ) = G . Note that a precompact metric group has size at most c , so item (ii) of Theorem D implies | G | ≤ c . Therefore, this cardinality restriction in Theorem D is necessary. A counterpart of thistheorem, with the word ”metric” removed from item (ii) and the condition on a group G relaxedto | G | ≤ c , is proved in our forthcoming paper [12].Item (iii) of Theorem D highlights the importance of characterizing Zariski dense subsets ofan abelian group G . This is accomplished in Theorem 7.1. Moreover, we explicitly calculate theZariski closure of an arbitrary subset of G ; see Theorem 8.1 and its corollaries in Section 8.1.5. The structure of the Zariski topology Z G of an abelian group. Bryant [3] establishedthat Z G is Noetherian for an abelian-by-finite group G ; for the convenience of the reader, we providea self-contained proof of this fact in the abelian case in Theorem 3.5. In particular, for an abeliangroup G , the space ( G, Z G ) has common properties shared by all Noetherian ( T ) spaces; see Facts2.1 and 2.5. We prove that, for an abelian group G , the space ( G, Z G ) also has some specificproperties not shared by all Noetherian spaces:( α ) Every subspace X of ( G, Z G ) contains a countable subset Y that is dense in X , i.e., ( G, Z G )is hereditarily separable , using topological terminology; see [16].( β ) The subset Y of X from item ( α ) can be chosen so that it has the cofinite topology; it nowfollows that for every point x in Z G -closure of X one can find a sequence { y n : n ∈ N } ofpoints of Y such that the sequence y n converges to x in the Zariski topology Z G . This lastproperty implies that ( G, Z G ) is Fr´echet-Urysohn , using topological terminology; see [16].( γ ) Irreducible components of an elementary algebraic set are disjoint, but irreducible compo-nents of an algebraic set need not be disjoint.Item ( α ) is proved in Corollary 8.8, item ( β ) is proved in Corollary 8.9, and item ( γ ) followsfrom Corollary 4.8(ii) and Example 4.9.1.6. A brief overview of the structure of the paper.
The paper is organized as follows. Sincethe Zariski topology Z G of an abelian group G is Noetherian, we recall basic properties of Noetherianspaces in Section 2. Irreducible sets play a principal role in the geometry of Noetherian spaces, aswitnessed by the fact that every subset X of a Noetherian space is a union of a uniquely determinedfinite family of irreducible sets; these irreducible sets are the so-called irreducible components (thatis, the maximal irreducible subsets) of X . At the end of Section 2, we recall also relevant factsabout combinatorial dimension dim.Section 3 introduces the Zariski topology Z G and establishes its basic properties. For everyabelian group G , the pair ( G, Z G ) is a so-called quasi-topological group, that is, a T -group in theterminology of Kaplansky [22], but it is not a topological group unless G is finite; see Corollary 3.6.Motivated by items ( α ) and ( β ) above, we call a countably infinite subset X of an abelian group G a Z G -atom if X has the cofinite topology (that is, the coarsest T topology) as a subspace of ( G, Z G );see Definition 3.14. An important result demonstrating why Z G -atoms are so useful is Proposition3.15 which says, in particular, that the Zariski closure of a Z G -atom is an irreducible (and thus,an elementary algebraic) set. Furthermore, Z G -atoms are precisely the irreducible one-dimensionalcountably infinite subsets of ( G, Z G ); see Fact 2.9.Since the topology of any subset of ( G, Z G ) is completely determined by its finitely many irre-ducible components, we study in detail Zariski irreducible sets in Section 4. In Theorem 4.6(ii),we prove that the irreducible component of zero and the connected component of zero of the space( G, Z G ) are both equal to G [ n ], where n coincides with the so-called essential order eo ( G ) of thegroup G (defined by Givens and Kunen [18] in the case of a bounded group G ; see Definition 4.3). Section 5 provides the “technical core” of the manuscript. A prominent role in our paper isplayed by the notion of an almost n -torsion set; see Definition 5.1. To assist the reader in betterunderstanding of this notion, we provide its equivalent forms in Lemma 5.6. The equivalent con-dition for n = 0 is particularly clear: A subset of an abelian group G is almost 0-torsion if andonly if its intersection with every coset of the form a + G [ m ], where a ∈ G and m ∈ N \ { } , isfinite; see Corollary 5.8. The general case is similar to this special case, except that the finitenesscondition is imposed only on integers m that are proper divisors of n ; see Lemma 5.6(ii). Propo-sition 5.10 uncovers a connection between the algebraic notion of an almost n -torsion set and thetopological notion of a Z G -atom. The main result of Section 5 is Proposition 5.12 characterizingsets that contain almost n -torsion subsets. An especially simple characterization of sets containingan almost 0-torsion set is given in Proposition 5.14. Corollary 5.13, Corollary 5.15 and Proposition5.16 describe abelian groups that do, or do not, contain almost n -torsion sets, for different integers n . Section 6 reveals the main reason for considering almost n -torsion sets in this manuscript. Indeed,in Theorem 6.2 we prove that every Z G -atom X is a translate of some almost n -torsion set, fora suitable integer n that is uniquely determined by X . This creates an essential bridge betweenalgebra (almost n -torsion sets) and topology ( Z G -atoms). Another crucial result in this section isCorollary 6.5 saying that Zariski irreducible sets are precisely those sets that contain a Z G -dense Z G -atom (or, equivalently, those sets that contain a Z G -dense translate of an almost n -torsion set,for a suitable integer n ). In particular, Zariski closures of Z G -atoms give all closed irreduciblesubsets of ( G, Z G ); see Corollary 6.6. Since every subspace X of ( G, Z G ) is a finite union of itsirreducible components, this shows that Z G -atoms (that is, translates of almost n -torsion sets)completely determine the topology of X .Since a potentially dense set is Zariski dense, as a necessary step towards solving Markov’spotential density problem, we study Zariski dense sets in Section 7. Theorem 7.1 completelydescribes such sets by means of (translates of) almost n -torsion sets. Zariski dense subsets ofunbounded abelian groups are given an especially nice and simple characterization in Theorem 7.4.Building on results from previous sections, in Section 8 we derive a complete description of theZariski closure of an arbitrary set via Z G -atoms or, equivalently, translates of almost n -torsion sets.Theorem C is proven in Section 9, while Theorem A is proven in Section 10. Finally, Theorem Dis proven in Section 11. In Section 12 we collect some open questions about the Zariski and Markovtopologies in the non-abelian case, based on the results in the abelian case.2. Background on Noetherian spaces, irreducible components and combinatorialdimension dimRecall that a topological space is said to be
Noetherian if it satisfies the ascending chain conditionon open sets, or the equivalent descending chain condition on closed sets. Kaplansky [22, Chap. IV,p. 26] calls the Noetherian T -spaces Z -spaces . We summarize here the key (mostly well-known)properties of Noetherian spaces. Fact 2.1. (1) A subspace of a Noetherian space is Noetherian.(2) If f : X → Z is a continuous surjection and X is Noetherian, then so is Z .(3) Every (subspace of a) Noetherian space is compact.(4) Every non-empty family of closed subsets of a Noetherian space has a minimal elementunder set inclusion.(5) Every infinite subspace Y of a Noetherian space X contains an infinite subspace Z suchthat every proper closed subset of Z is finite.(6) No infinite subspace of a Noetherian space can be Hausdorff; in particular, every HausdorffNoetherian space is finite. Proof. (1) and (2) are straightforward.
In view of (1), to show (3) it suffices to prove that every Noetherian space is compact. Thisimmediately follows from the fact that every descending chain of closed sets stabilizes.(4) follows easily from the fact that the Noetherian spaces satisfy the descending chain conditionon closed sets.To prove (5), it suffices to consider the case when Y = X , since according to (1), a subspace ofa Noetherian space is Noetherian. Let F be the family of all infinite closed subsets of X . Since X ∈ F 6 = ∅ , we can use item (4) to find a minimal element Z of F . Clearly, every proper closedsubset of Z is finite.(6) Let X be an infinite Noetherian space. Apply item (5) to Y = X to get an infinite subspace Z of Y = X as in the conclusion of this item. Since Z is an infinite space without proper infiniteclosed subsets, Z cannot be Hausdorff. Since Z is a subspace of X , we conclude that X is notHausdorff either. (cid:3) Bryant [3] discovered a useful technique for building Noetherian spaces:
Fact 2.2. ([3]) Let E be a family of subsets of a set X closed under finite intersections and satisfyingthe descending chain condition. Assume also that X ∈ E . Then the family E ∪ consisting of finiteunions of the members of E forms the family of closed sets of a unique topology T E on X such thatthe space ( X, T E ) is a Noetherian space.From this fact, we can easily show that finite products of Noetherian spaces are Noetherian.Recall that a topological space X is called irreducible ( connected ) provided that for every parti-tion (respectively, disjoint partition) X = F ∪ F of X into closed sets F and F , either F = X or F = X holds. Note that an irreducible space is connected. For an example of a connected T -space that is not irreducible, take the reals or any connected infinite Hausdorff space.An easy induction establishes the following fact. Fact 2.3. If X = ∅ is an irreducible subset of a space Y and F is a finite family of closed subsetsof Y such that X ⊆ S F , then X ⊆ F for some F ∈ F .We omit the easy proof of the following fact. Fact 2.4. (i) A space X is irreducible if and only if every non-empty open subset of X is densein X .(ii) Every continuous map from an irreducible space to a Hausdorff space is constant. Inparticular, the only irreducible Hausdorff spaces are the singletons.(iii) Every space with a dense irreducible subspace is irreducible.(iv) A dense subset Y of a topological space X is irreducible if and only if X is irreducible.If X is a space and x ∈ X , then a maximal element of the family of all connected (irreducible)subsets of X containing x , ordered by set-inclusion, is called the connected component (respectively,an irreducible component ) of x . Note that the connected component of a point is uniquely deter-mined, while the irreducible ones need not be unique. Connected and irreducible components arealways closed. Fact 2.5. ([19]) Let Y be a Noetherian space. Then every subset X of Y admits a unique de-composition X = X ∪ · · · ∪ X n into a finite union of irreducible, relatively closed (in the subspacetopology of X ) subsets X i such that X i \ X j = ∅ for i = j . Moreover, each such X i is an irreduciblecomponent of X . Proof.
Assume that X is not a finite union of closed irreducible subspaces. Then the family Z ofclosed subspaces Z of X that are not a finite union of closed irreducible subspaces is non-empty.Choose a minimal element Z of Z . Then Z cannot be irreducible, so Z = F ∪ F , where F and F are proper closed subspaces of Z . Then, by the minimality of Z , both F and F are finite unionsof closed irreducible subspaces of X , which yields Z
6∈ Z , giving a contradiction. This proves the existence of a decomposition X = X ∪ · · · ∪ X n of X into a finite union of closed irreducible subsets.By deleting some X i , we can assume, without loss of generality, that X i \ X j = ∅ for i = j .To prove the uniqueness of the decomposition, suppose that X = X ′ ∪ · · · ∪ X ′ m is anotherdecomposition of X into a finite union of closed irreducible subsets such that X ′ i \ X ′ j = ∅ for i = j .Fix i = 1 , , . . . , n . From X i ⊆ X ′ ∪ · · · ∪ X ′ m and Fact 2.3, we deduce that X i ⊆ X ′ j for some j .Analogously, from X ′ j ⊆ X ∪ · · · ∪ X m and Fact 2.3, we must have X ′ j ⊆ X k for some k . Now X i ⊆ X k yields k = i , and hence X i = X ′ j . This shows that m = n , and thus, X k = X ′ f ( k ) for all k = 0 , , . . . , n , where f is an appropriate permutation of n .It remains only to be shown that each X i is a maximal irreducible subset of X . Let T be anirreducible subset of X such that X i ⊆ T . Then T ⊆ X = X ∪ · · · ∪ X n , and Fact 2.3 yields that X i ⊆ T ⊆ X j for some j . This implies i = j and T = X i . (cid:3) Fact 2.6.
Let X be a subset of a Noetherian space Y .(i) X has finitely many connected components. Every connected component C k of X is aclopen subset of X , and the finite family of connected components of X forms a (disjoint)partition X = S mk =1 C k .(ii) Every irreducible component X i of X is contained in some connected component C k .Moreover, if Y is a T -space, then the following also holds:(iii) The set D of isolated points of X is finite, and { d } is a connected component of X for every d ∈ D .(iv) If X is infinite, then the set X \ D has no isolated points, its irreducible components areinfinite and coincide with the infinite irreducible components of X . Proof. (i) Assume that X has infinitely many connected components { C n : n ∈ N } . Choose apoint y n ∈ C n for every n ∈ N . By Fact 2.1(5), there exists an infinite subspace Z of the set Y = { y n : n ∈ N } such that all proper closed subsets of Z are finite. Clearly, Z is connected. Since Z intersects with (infinitely many) distinct connected components of Y , we get a contradiction.Therefore, X has finitely many distinct connected components C , . . . , C m .(ii) Let X i be an irreducible component of X . Since X i ⊆ X = S mk =1 C k , from Fact 2.3 weconclude that X i ⊆ C k for some k = 1 , . . . , m .(iii) Let d be an isolated point of X . Since Y is a T -space, the set { d } is both open and closed in X . It follows that { d } is a connected component of X . Applying (i), we conclude that D is finite.(iv) Suppose now that X is infinite. Since Y is a T -space, and D is finite by (iii), X \ D isa non-empty open subset of X . Therefore, an isolated point of X \ D is an isolated point of X as well, contradicting the definition of D . Let X \ D = S ni =1 X i be the decomposition of X \ D into irreducible components of X \ D . Now one can easily see that X = S ni =1 X i ∪ S d ∈ D { d } is thedecomposition of X into irreducible components of X . (cid:3) A space in which all closed irreducible subsets have a dense singleton is known as a sober space.Recall that the spectrum of every commutative ring is a sober space.
Fact 2.7.
A sober Noetherian T -space must be finite. Proof.
Let X be a sober Noetherian T -space, and let X = X ∪ · · · ∪ X n be the decomposition of X into its irreducible components; see Fact 2.5. Since X is sober, each X i contains an element x i such that { x i } is dense in X i . Since X is a T -space, the set { x i } is closed in X , which implies X i = { x i } . Therefore, X = { x , . . . , x n } . (cid:3) Given a topological space X and a natural number n , we write dim X ≥ n if there exists astrictly increasing chain(1) F ⊆ F ⊆ . . . ⊆ F n of non-empty irreducible closed subsets of X . The combinatorial dimension dim X of a space X isthe smallest number n ∈ N satisfying dim X ≤ n , if such a number exists, or ∞ otherwise. Clearly,every Hausdorff space (as well as every anti-discrete space) has combinatorial dimension 0. Recallthat the Krull dimension of a commutative ring coincides with the combinatorial dimension of itsspectrum. Fact 2.8.
Let X be a Noetherian T -space.(a) dim X > X is infinite.(b) If Y ⊆ X , then dim Y ≤ dim X .(c) If S = S kj =1 S j is a decomposition of a subset S of X into irreducible components of S ,then dim S = max ≤ j ≤ k dim S j .(d) If Y, Z ⊆ X , then dim( Y ∪ Z ) = max { dim Y, dim Z } . Proof. (a) follows directly from the definition.(b) If Y is closed, then every chain (1) witnessing dim Y ≥ n will witness dim X ≥ n as well. If Y is dense in X , then every chain (1) witnessing dim Y ≥ n will give rise to a chain F ⊆ F ⊆ . . . ⊆ F n witnessing dim X ≥ n . In the general case, we have dim Y ≤ dim Y ≤ dim X by the preceding twocases.(c) Indeed, the inequality dim S ≥ max ≤ j ≤ k dim S j follows from (b). Let n = dim S , and assumethat (1) is a chain of closed irreducible subsets of S witnessing dim S ≥ n . Since the irreducible set F n is contained in the finite union S kj =1 S j = S of closed sets, we conclude that F n ⊆ S i for some i = 1 , , . . . k . In other words, the whole chain (1) is contained in S i , implying dim S i ≥ n . Hencedim S = n ≤ dim S i ≤ max ≤ j ≤ k dim S j . This proves (c).(d) Let Y = S ni =1 C i and Z = S mj =1 K j be decompositions of Y and Z into irreducible compo-nents; see Fact 2.5. Note that Y ∪ Z = S ni =1 C i ∪ S mj =1 K j need not be the representation of Y ∪ Z as a union of irreducible components, since C i ∪ K j = C i or C i ∪ K j = K j (i.e., K j ⊆ C i or C i ⊆ K j )may occur. Nevertheless, by removing the redundant members in the union S mj =1 K j , one obtainsthe decomposition Y ∪ Z = S ν L ν of Y ∪ Z into irreducible components L ν , and from (c) we getdim( Y ∪ Z ) = max { dim L ν } ≤ max (cid:26) max ≤ i ≤ n dim C i , max ≤ j ≤ m dim K j (cid:27) = max { dim Y, dim Z } . The inverse inequality dim( Y ∪ Z ) ≥ max { dim Y, dim Z } follows from (b). (cid:3) Fact 2.9.
For a subspace X of a Noetherian space Y , the following conditions are equivalent:(a) X is an irreducible T -space of combinatorial dimension 1,(b) X is infinite and carries the cofinite topology { X \ F : F is a finite subset of X } ∪ {∅} ,(c) X is an infinite space with the coarsest T -topology. Proof.
It is easy to see that (b) and (c) are equivalent and imply (a), even without the assumptionthat Y is Noetherian.(a) → (b) Since dim X = 1 >
0, our X must be infinite by Fact 2.8(a). Let F be a non-emptyclosed subset of X , and let F = F ∪ · · · ∪ F n be the decomposition of F into its irreduciblecomponents F i ; see Fact 2.5. Let i ≤ n be a non-negative integer. Choose x i ∈ F i . Since all threesets in the chain { x i } ⊆ F i ⊆ X are irreducible and dim X = 1, at least one of the inclusions cannotbe proper; that is, either F i = X or F i = { x i } . From F = F ∪ · · · ∪ F n , we conclude that either F = X or F is finite. (cid:3) Easy examples show that the implication (a) → (b) fails if X is not Noetherian.3. Properties of algebraic sets and the Zariski topology of an abelian group
For an abelian group G and an integer n , we set G [ n ] = { x ∈ G : nx = 0 } . Clearly, G [ n ] is asubgroup of G , with G [0] = G and G [1] = { } . As usual, for integers m and n , n | m means that n is a divisor of m , and ( m, n ) denotes thegreatest common divisor of m and n , in case at least one of these integers is non-zero. For the sakeof convenience, we set (0 ,
0) = 0.
Lemma 3.1.
Suppose G is an abelian group, a, b ∈ G and n, m ∈ N . Then: (i) a + G [ n ] ⊆ b + G [ m ] if and only if G [ n ] ⊆ G [ m ] and a − b ∈ G [ m ] , (ii) a + G [ n ] = b + G [ m ] if and only if G [ n ] = G [ m ] and a − b ∈ G [ m ] = G [ n ] , (iii) G [ n ] ∩ G [ m ] = G [ d ] for d = ( m, n ) ; in particular, G [ n ] = G [ d ] provided that G [ n ] ⊆ G [ m ] , (iv) if z ∈ ( a + G [ n ]) ∩ ( b + G [ m ]) , then ( a + G [ n ]) ∩ ( b + G [ m ]) = z + G [ d ] , where d = ( n, m ) .Proof. (i) From a + G [ n ] ⊆ b + G [ m ], we deduce G [ n ] ⊆ b − a + G [ m ], so b − a ∈ G [ m ] and G [ n ] ⊆ G [ m ].(ii) follows from (i), and the proof of (iii) is straightforward.(iv) By our hypothesis, z − a ∈ G [ n ] and z − b ∈ G [ m ], so we conclude that z + G [ n ] = a + G [ n ]and z + G [ m ] = b + G [ m ]. Since G [ n ] ∩ G [ m ] = G [ d ] by (iii), we get( a + G [ n ]) ∩ ( b + G [ m ]) = ( z + G [ n ]) ∩ ( z + G [ m ]) = z + ( G [ n ] ∩ G [ m ]) = z + G [ d ] . (cid:3) Notation 3.2.
Let G be an abelian group. Then E G denotes the family of all elementary algebraicsets of G , and A G denotes the family of all finite unions of elementary algebraic sets of G . Forconvenience, we define E { } = {∅ , { }} , even though ∅ is not an elementary algebraic subset of thetrivial group G = { } .The next lemma collects some basic properties of E G . Lemma 3.3.
For every abelian group G , the family E G has the following properties: (i) if E = ∅ , then E ∈ E G if and only if E = a + G [ n ] for some a ∈ G and n ∈ N , (ii) ∅ ∈ E G and G ∈ E G , (iii) E G is closed under the operation x
7→ − x , (iv) E G is closed under taking translations, (v) E , . . . , E n ∈ E G implies E + . . . + E n ∈ E G , (vi) E G is closed with respect to taking finite intersections, (vii) every descending chain in E G stabilizes, so E G is closed with respect to taking arbitraryintersections, (viii) if k ∈ Z and E ∈ E G , then { x ∈ G : kx ∈ E } ∈ E G .Proof. (i) Replacing the multiplicative notation from Definition 1.1(i) with the additive notationand using commutativity, one immediately gets that E ∈ E G if and only if E = { x ∈ G : nx + b = 0 } for a suitable integer n and an element b ∈ G . Replacing b with − b , if necessary, we may assumethat n ∈ N . Choose a ∈ E . Then na + b = 0, and so E = { x ∈ G : n ( x − a ) = 0 } = { x ∈ G : x − a ∈ G [ n ] } = a + G [ n ].(ii) Note that G = G [0] ∈ E G by (i). If G = { } , then ∅ ∈ E G according to Notation 3.2. If G = { } , choose g ∈ G with g = 0 and observe that ∅ = { x ∈ G : 0 x + g = 0 } ∈ E G .(iii) and (iv) follow easily from (i).(v) Without loss of generality, we can assume that n = 2 and that both E and E are non-empty.Applying (i), we conclude that E i = a i + G [ m i ] for some a i ∈ G and m i ∈ N ( i = 1 , m bethe least common multiple of m and m . Then G [ m ] = G [ m ] + G [ m ], and (i) now yields E + E = a + G [ m ] + a + G [ m ] = ( a + a ) + G [ m ] + G [ m ] = ( a + a ) + G [ m ] ∈ E G . (vi) follows from (i) and Lemma 3.1(iv).(vii) Assume that m, n ∈ N , a, b ∈ G , m ≥ a + G [ n ] is a proper subset of b + G [ m ]. Applyingitems (i) and (iii) of Lemma 3.1, we conclude that G [ n ] = G [ d ] for a positive divisor d of m . Since m ≥ E G under arbitrary intersections follows from (vi).(viii) The assertion is true if G = { } , since E G then coincides with the power set of G accordingto Notation 3.2. Suppose now that G = { } , and let E ′ = { x ∈ G : kx ∈ E } . If E ′ = ∅ , then E ′ ∈ E G by (ii). If E ′ = ∅ , choose x ∈ E ′ . Then kx ∈ E = ∅ , so E = a + G [ n ] for some a ∈ G and n ∈ N ; see (i). Hence, kx = a + t for some t ∈ G with nt = 0. Let us check that E ′ = x + G [ kn ] ∈ E G . Assume x ∈ E ′ . Then kx ∈ E , and so kx = a + s for some s ∈ G [ n ]. As kx = a + t with t ∈ G [ n ], we conclude that k ( x − x ) = s − t ∈ G [ n ], and so x − x ∈ G [ kn ]. Thisproves that x ∈ x + G [ kn ]. In the opposite direction, if x ∈ x + G [ kn ], then kn ( x − x ) = 0, sothat k ( x − x ) ∈ G [ n ], and thus kx − kx ∈ G [ n ]. As kx − a ∈ G [ n ], we conclude that kx − a ∈ G [ n ]as well, and so kx ∈ a + G [ n ] = E . (cid:3) Lemma 3.4.
For every abelian group G , the family A G of all algebraic subsets of G has the followingproperties: (i) A G is closed under the operation x
7→ − x , (ii) A G is closed under taking translations, (iii) A , . . . , A n ∈ A G implies A + . . . + A n ∈ A G , (iv) if k ∈ Z and A ∈ A G , then { x ∈ G : kx ∈ A } ∈ A G .Proof. (i) follows from Lemma 3.3(iii); (ii) follows from Lemma 3.3(iv); (iii) follows from Lemma3.3(v); and (iv) follows from Lemma 3.3(viii). (cid:3) Theorem 3.5. ([3])
Let G be an abelian group. Then: (i) ( G, Z G ) is a Noetherian space, (ii) A G coincides with the family of all Z G -closed sets.Proof. In view of (ii), (vi), (vii) of Lemma 3.3, we can apply Fact 2.2 to conclude that A G = E G ∪ is the family of all closed sets of a unique Noetherian topology on G , and one can easily see thatthis topology coincides with Z G . (cid:3) It follows from Theorem 3.5(i) that, for an abelian group G , the space ( G, Z G ) has all the basicproperties of Noetherian spaces described in Facts 2.1 and 2.5. Clearly, infinite groups G with thediscrete Zariski topology Z G (see Example 1.3) are not Noetherian, and for them, all the propertieslisted in Facts 2.1 and 2.5 fail.It follows from Theorem 3.5(ii) that, for a countable abelian group G , the Zariski topology Z G is countable as well. In the non-abelian case, one has a completely different situation. Indeed, let G be a countable group G such that Z G is discrete; see items (a) and (b) of Example 1.3. Then Z G has cardinality c . In particular, every subset of G is algebraic, so G has c -many algebraic sets.Since G has only countably many elementary algebraic sets, not every algebraic set is a finite unionof elementary algebraic sets (compare this with Theorem 3.5(ii)). Corollary 3.6.
For an abelian group G , the following conditions are equivalent: (i) Z G is Hausdorff, (ii) Z G is sober, (iii) G is finite, (iv) ( G, Z G ) is a topological group.Proof. The proof of the implication (i) → (ii) is trivial. Since ( G, Z G ) is a Noetherian space byTheorem 3.5(i), the implication (ii) → (iii) follows from Fact 2.7.To prove the implication (iii) → (iv), note that the topology Z G is T , and since G is finite, weconclude that ( G, Z G ) is discrete. In particular, ( G, Z G ) is a topological group.Finally, the implication (iv) → (i) holds, since ( G, Z G ) is a T -space, and a T topological groupis Hausdorff. (cid:3) Corollary 3.6 is typical for the abelian case but fails for non-commutative groups. Indeed, anyinfinite group G with discrete Z G (see Example 1.3) provides an example in which ( G, Z G ) is aHausdorff (and thus a sober) topological group. Corollary 3.7.
Let G be an abelian group. Then: (i) the inverse operation x
7→ − x is Z G -continuous, (ii) for every a ∈ G , the translation by a , x x + a , is a homeomorphism of ( G, Z G ) ontoitself, (iii) for every k ∈ Z , the map f k : ( G, Z G ) → ( G, Z G ) , defined by f k ( x ) = kx for x ∈ G , iscontinuous.Proof. In view of Theorem 3.5(ii), (i) follows from Lemma 3.4(i); (ii) follows from Lemma 3.4(ii);and (iii) follows from Lemma 3.4(iv). (cid:3)
Let T be a T -topology on an abelian group G . If the “inverse” operation x
7→ − x is T -continuous, and the addition operation ( x, y ) x + y is separately T -continuous, then Kaplanskycalls the pair ( G, T ) a T -group ; see [22, Chap. IV, p. 27]. The same pair is often called a quasi-topological group ; see [1]. Corollary 3.8.
If a group G is abelian, then ( G, Z G ) is a T -group (that is, a quasi-topologicalgroup). According to Kaplansky, a Z -group is a T -group with a Noetherian topology, and a C -group isa T -group such that, for any fixed a ∈ G , the map x x − ax is continuous; see [22, Chap. IV,p. 28]. According to Bryant and Yen [4], a CZ -group is a C -group that is also a Z -group. FromTheorem 3.5(i) and Corollary 3.8, it follows that ( G, Z G ) is a CZ -group.The Zariski topology Z G is always T . The next fact completely describes the abelian groups G for which Z G is the coarsest T topology, i.e., the cofinite topology. Fact 3.9. ([31, Theorem 5.1]). For an abelian group G , the following conditions are equivalent:(i) every proper algebraic subset of G is finite; that is, Z G coincides with the cofinite topologyof G ,(ii) every proper elementary algebraic subset of G is finite,(iii) either G is almost torsion-free, or G has exponent p for some prime p . Proof.
The proofs of (i) ↔ (ii) and (iii) → (ii) are obvious. Let us prove (ii) → (iii). Assume that(ii) holds, and G is not almost torsion-free. Then there exists a prime p such that r p ( G ) is infinite.Then the subgroup G [ p ] of G is infinite. Since G [ p ] is an elementary algebraic subset of G , (ii)implies that G = G [ p ]. (cid:3) As usual, for a subset Y of a topological space ( X, T ), we denote by T ↾ Y the subspace topology { Y ∩ U : U ∈ T } generated by T .The next lemma shows that the Zariski topology behaves well under taking subgroups. Thisis a typical property in the abelian case (for counter-examples in the non-abelian case, see [10]).Although this lemma can be deduced from [10, Lemma 2.2(a), Lemma 3.7 and Corollary 5.7], weprefer to give a direct, short and transparent proof here for the reader’s convenience. Lemma 3.10.
For every subgroup H of an abelian group G , one has Z G ↾ H = Z H .Proof. In view of Theorem 3.5(ii) and the formulae E G ∪ = A G , E H ∪ = A H , it suffices to check that E H = { H ∩ A : A ∈ E G } . Note that(2) h + H [ n ] = H ∩ ( h + G [ n ]) for all n ∈ N and h ∈ H. Assume that ∅ 6 = B ∈ E H . Then B = h + H [ n ] for some h ∈ H and n ∈ N ; see Lemma3.3(i). Since A = h + G [ n ] ∈ E G by Lemma 3.3(i), and B = H ∩ A by (2), we conclude that E H ⊆ { H ∩ A : A ∈ E G } . Let A ∈ E G . Assuming that H ∩ A = ∅ , choose h ∈ H ∩ A . From ∅ 6 = A ∈ E G and Lemma3.3(i), it follows that A = a + G [ n ] for some a ∈ G and n ∈ N . According to Lemma 3.1(ii), onehas A = a + G [ n ] = h + G [ n ], and so H ∩ A = h + H [ n ] ∈ E H by (2) and Lemma 3.3(i). This provesthat { H ∩ A : A ∈ E G } ⊆ E H . (cid:3) Remark 3.11.
Lemma 3.10 on inclusions
H ֒ → G cannot be extended to arbitrary homomorphisms f : H → G . Indeed, define H = Q , G = Q / Z , and let f : H → G be the canonical homomorphism.We claim that f is not continuous when both groups have Zariski topologies. To see this, note thatthe subset F = { } of G is Z G -closed, while Z = f − ( F ) is not Z H -closed, since Z H is the cofinitetopology of H by Fact 3.9. Therefore, Z G is not a functorial topology in Charles sense; see [17, § Theorem 3.12.
Let G be an abelian group. Then: (i) cl Z G ( A + B ) = cl Z G ( A ) + cl Z G ( B ) whenever A ⊆ G and B ⊆ G , (ii) cl Z G ( a + S ) = a + cl Z G ( S ) for each a ∈ G and every subset S of G .Proof. (i) According to Lemma 3.4(iii), the set cl Z G ( A ) + cl Z G ( B ) is Z G -closed and contains A + B ,which implies that(3) cl Z G ( A + B ) ⊆ cl Z G ( A ) + cl Z G ( B ) . Let us prove the inverse inclusion. All translations x x + a ( a ∈ G ) are Z G -continuous byCorollary 3.7(ii), and so(4) cl Z G ( C ) + D ⊆ cl Z G ( C + D ) for every pair of subsets C, D of G. We claim that(5) cl Z G ( A ) + cl Z G ( B ) ⊆ cl Z G ( A + cl Z G ( B )) ⊆ cl Z G (cl Z G ( A + B )) = cl Z G ( A + B ) . Indeed, the first inclusion in (5) is obtained by applying (4) to C = A and D = cl Z G ( B ), and thesecond inclusion in (5) is obtained by applying (4) to C = B and D = A . Combining (3) and (5),we get (i).(ii) Since ( G, Z G ) is a T space, (ii) is a particular case of (i) applied to A = { a } and B = S .Note that (ii) also follows directly from Corollary 3.7(ii). (cid:3) According to Theorem 3.12, the addition function µ : G × G → G defined by µ ( x, y ) = x + y maps the Z G × Z G -closure of a rectangular set A × B to the Z G -closure of the image µ ( A × B ). Thisfact is significant, since this does not hold for arbitrary subsets of G × G , as the following remarkdemonstrates. Remark 3.13.
For an infinite group G , there always exists a set X ⊆ G × G , such that µ mapsthe Z G × Z G -closure of X outside of the Z G closure of µ ( X ) . Indeed, assume that µ maps the Z G × Z G -closure of X in the Z G closure of µ ( X ) for every subset X of G × G . Then the map µ : ( G × G, Z G × Z G ) → ( G, Z G ) must be continuous. That is, the addition ( x, y ) x + y becomes Z G -continuous. Since the inverse operation x
7→ − x is always Z G -continuous by Corollary 3.7(i),it follows that ( G, Z G ) is a topological group. Thus, G must be finite by Corollary 3.6, which is acontradiction.One can easily find an X as in the above example in the case G = Z , or any almost torsion-freeinfinite group. Indeed, let Y be a countably infinite subset of G . Since G carries a cofinite topologyby Fact 3.9, one can easily check that the set X = { ( y, − y ) : y ∈ Y } ⊆ G × G is Z G × Z G -dense,and so G × G = cl Z G × Z G ( X ). On the other hand, µ ( X ) = { } , and so cl Z G ( µ ( X )) = { } . Thisshows that µ (cl Z G × Z G ( X )) \ cl Z G ( µ ( X )) = µ ( G × G ) \ { } = G \ { } 6 = ∅ . Definition 3.14.
A countably infinite subset X of an abelian group G will be called a Z G -atom provided that Z G ↾ X = { X \ F : is a finite subset of X } S {∅} is the cofinite topology of X . We refer the reader to Fact 2.9 for other conditions equivalent to the condition from Definition3.14.Obviously, an infinite subset of a Z G -atom is still a Z G -atom.The Zariski closures of Z G -atoms are precisely the irreducible (and hence, elementary) algebraicsets; one of the implications is proved in item (i) of the next proposition, while the other one isgiven in Corollaries 6.5 and 6.6. Proposition 3.15.
Let X be a Z G -atom of an abelian group G . Then: (i) cl Z G ( X ) is irreducible, (ii) there exists a ∈ G and n ∈ N such that cl Z G ( X ) = a + G [ n ] , (iii) if F is a Z G -closed subset of G , then either F ∩ X is finite or X ⊆ F , (iv) any faithfully enumerated sequence { x k : k ∈ N } of points of X converges to every point x ∈ cl Z G ( X ) .Proof. (i) Clearly, X is irreducible. Therefore, cl Z G ( X ) must be irreducible by Fact 2.4(iii).(ii) Note that cl Z G ( X ) ∈ A G by Theorem 3.5(ii), so cl Z G ( X ) = S F for some finite family F ⊆ E G ⊆ Z G . Since cl Z G ( X ) is irreducible by item (i), applying Fact 2.3, we conclude thatcl Z G ( X ) = F for some F ∈ F . Now (ii) follows from Lemma 3.3(i).(iii) Let F be a Z G -closed subset of G . Then F ∩ X is a closed subset of ( X, Z G ↾ X ). Since X isa Z G -atom, from Fact 2.9(b) one concludes that either F ∩ X is finite or F ∩ X = X (and hence, X ⊆ F ).(iv) Let { x k : k ∈ N } ⊆ X be a faithfully indexed sequence. Fix an arbitrary point x ∈ cl Z G ( X ).Let U be a Z G -open subset of G containing x . Then U ∩ X is a non-empty Z G ↾ X -open subset of X .Since X is a Z G -atom, from Fact 2.9(b) one concludes that U ∩ X = X \ T for some finite subset T of X . In particular, X \ T ⊆ U . It follows that U contains all but finitely many elements of thesequence { x k : k ∈ N } . Since U was taken arbitrarily, we conclude that the sequence { x k : k ∈ N } converges to x . (cid:3) Essential order and Zariski irreducible subsets
For a subset X of an abelian group G and a natural number n , we define nX = { nx : x ∈ X } .The group G is bounded if nG = { } for some n ∈ N \ { } , and G is unbounded otherwise.We say that d ∈ N is a proper divisor of n ∈ N provided that d
6∈ { , n } and dm = n for some m ∈ N . Note that, according to our definition, each d ∈ N \ { } is a proper divisor of 0. Definition 4.1.
Let n ∈ N . An abelian group G is said to be of exponent n if nG = { } but dG = { } for every proper divisor d of n . In this case, we call n the exponent of G .In particular, G [ n ] has exponent n precisely when G [ n ] = G [ d ] for every proper divisor d of n .Note that the only abelian group of exponent 1 is { } , while a group G is of exponent 0 preciselywhen G is unbounded (i.e., nG = { } for every positive integer n ). Lemma 4.2. If G is an abelian group and m ∈ N , then there exists n ∈ N such that G [ m ] = G [ n ] and G [ n ] has exponent n .Proof. If G [ m ] is unbounded, then m = 0 and G [ m ] has exponent 0, so n = m works. If G [ m ] isbounded, then G [ m ] has exponent n for some divisor n of m . Clearly, G [ m ] = G [ n ] in this case. (cid:3) Definition 4.3.
Let G be an abelian group.(i) If G is bounded, then the essential order eo ( G ) of G is the smallest positive integer n suchthat nG is finite.(ii) If G is unbounded, we define eo ( G ) = 0. The notion of the essential order of a bounded abelian group G , as well as the notation eo ( G ),are due to Givens and Kunen [18], although the definition in [18] is different (but equivalent) toours. Lemma 4.4. If G be a bounded abelian group, n = eo ( G ) and mG is finite for some m ∈ N \ { } ,then n divides m . In particular, the essential order eo ( G ) of a bounded group G divides its exponent.Proof. Clearly, m ≥ n by Definition 4.3(i). Assume that n does not divide m . Then there existsunique integers q ∈ N and r satisfying m = qn + r and 0 < r < n . Since r = m − qn , the group rG ⊆ mG + qnG is finite, which contradicts n = eo ( G ). (cid:3) Lemma 4.5.
Let G be an abelian group and n = eo ( G ) . Then: (i) G [ n ] has finite index in G , (ii) eo ( G [ n ]) = eo ( G ) = n .Proof. If n = 0, then G [ n ] = G [0] = G , and both items (i) and (ii) trivially hold. So we now assumethat n ≥
1. Then G is a bounded torsion group.(i) By Definition 4.3(i), nG ∼ = G/G [ n ] must finite. Hence, G [ n ] is a finite index subgroup of G .(ii) Let m = eo ( G [ n ]). Since G [ n ] is bounded, m ≥
1. By Lemma 4.4, m divides n , and so G [ m ] ⊆ G [ n ] and m ≤ n . From (i) we conclude that G [ m ] has a finite index in G [ n ]. Since G [ n ]has a finite index in G , it follows that G [ m ] has a finite index in G . Hence mG ∼ = G/G [ m ] is finite,which implies n = eo ( G ) ≤ m . This proves that m = n . (cid:3) The next theorem describes the connected component and the irreducible component of theidentity of an abelian group in the Zariski topology.
Theorem 4.6.
Let G be an infinite abelian group and n = eo ( G ) . Then: (i) G [ n ] is a Z G -clopen subgroup of G , (ii) G [ n ] is both the connected component and the irreducible component of the identity of thespace ( G, Z G ) .Proof. (i) By Lemma 4.5(i), G [ n ] has a finite index in G . Since each of the finitely many pairwisedisjoint cosets of G [ n ] are Z G -closed, all of them must also be Z G -open. Therefore, G [ n ] is Z G -clopen.(ii) The irreducible component E of 0 must be an elementary algebraic set, and so E = a + G [ m ]for suitable a ∈ G and m ∈ N ; see Lemma 3.3(i). Since 0 ∈ E , we conclude that E = G [ m ]. FromCorollary 3.7, we obtain that a + G [ m ] is the irreducible component of a for every a ∈ G . Since( G, Z G ) is Noetherian by Theorem 3.5(i), the number of irreducible components of ( G, Z G ) mustbe finite according to 2.5. Therefore, G = G [ m ] + F for some finite subset F of G . Since the set mG = mG [ m ] + mF = mF is finite, n = eo ( G ) divides m by Lemma 4.4. Therefore, G [ n ] ⊆ G [ m ].By (i), G [ n ] is a clopen subgroup of G . Since G [ m ] is connected, we must have G [ m ] = G [ n ]. (cid:3) Corollary 4.7.
For an infinite abelian group G , the following conditions are equivalent: (i) ( G, Z G ) is connected, (ii) ( G, Z G ) is irreducible, (iii) G = G [ n ] , where n = eo ( G ) . Corollary 4.8.
Let E be an elementary algebraic set of an infinite abelian group H . Then: (i) if E is connected, then E is irreducible, (ii) the irreducible components of E are disjoint.Proof. Assume that E = ∅ . According to Lemma 3.3(i), E = a + H [ m ] for some a ∈ H and m ∈ N .By Corollary 3.7(ii), all topological properties involved in (i) and (ii) are translation invariant, sowe can assume without loss of generality that E = H [ m ]. Let G = E and n = eo ( G ). Lemma 3.10implies that Z G = Z H ↾ G . Therefore, without loss of generality, we may also assume that H = G .Applying Lemma 4.5(i) and Theorem 4.6 to G , we conclude that G [ n ] is an irreducible Z G -clopen subgroup of G of finite index, and G [ n ] coincides with the connected component of the identity of( G, Z G ).(i) Since G = E is connected, we deduce that E = G [ n ], and so E is irreducible.(ii) Let g + G [ n ] , . . . , g k + G [ n ] be pairwise disjoint cosets of G [ n ] such that G = S ≤ i ≤ k g i + G [ n ].Since G [ n ] is irreducible, each g i + G [ n ] is also irreducible by Corollary 3.7(ii). It now follows that g + G [ n ] , . . . , g k + G [ n ] are precisely the irreducible components of G = E . (cid:3) Example 4.9.
Let G = Z (6) ( ω ) . Applying Theorem 4.6 to the groups E = G [2] and E = G [3],we conclude that E and E are irreducible (and hence, connected) Z G -closed subsets of G . Since0 ∈ E ∩ E = ∅ , the algebraic set F = E ∪ E is a connected non-irreducible set in G . Furthermore, E and E are irreducible components of F . This shows that both items of Corollary 4.8 fail for(non-elementary) algebraic sets. Proposition 4.10.
For an abelian group G and an integer n ∈ N the following conditions areequivalent: (i) eo ( G [ n ]) = n , (ii) G [ n ] is irreducible and has exponent n .Proof. Let H = G [ n ]. By Lemma 3.10, H is an irreducible subset of ( G, Z G ) if and only if ( H, Z H )is irreducible. Hence, we will argue with the group H = H [ n ] instead of G .(i) → (ii) As eo ( H ) = n , Theorem 4.6 implies that ( H, Z H ) is irreducible. Since eo ( H ) = n and nH = 0, it follows that H has exponent n .(ii) → (i) If n = 0, then H is unbounded, as a group of exponent 0. Hence, eo ( H ) = 0 byDefinition 4.3(ii). If n = 1, then H = { } , and so eo ( H ) = 1. Assume now that n >
1. Suppose m ∈ N , 1 ≤ m < n and mH is finite. As mH ∼ = H/H [ m ] is finite, the closed subgroup H [ m ] of H has finite index, so it is a clopen subgroup of H . As H is irreducible, this yields H [ m ] = H .Thus, mH = { } . Since m < n , this contradicts the fact that the group H has exponent n . Thiscontradiction proves that mH is infinite for every m ∈ N satisfying 1 ≤ m < n . Since nH = { } ,we get eo ( H ) = n . (cid:3) Let us give an example illustrating the usefulness of Proposition 4.10.
Example 4.11.
Let G be a divisible abelian group. If p is a prime number, n ∈ N \ { } and G [ p n ] is infinite, then G [ p n ] is an irreducible subgroup of G . Indeed, let B = { x i : i ∈ I } be a minimal setof generators of the subgroup G [ p ]; that is, B is a base of G [ p ] considered as a linear space over thefield Z /p Z . Since G is divisible, for every i ∈ I , there exists y i ∈ G such that p n − y i = x i . Then y i ∈ G [ p n ] for every i ∈ I , the subgroup of G generated by { y i : i ∈ I } coincides with G [ p n ] and G [ p n ] ∼ = L I Z ( p n ). Since G [ p n ] is infinite, I must be infinite as well. This yields eo ( G [ p n ]) = p n ,and the implication (i) → (ii) from Proposition 4.10 shows that G [ p n ] is irreducible.According to Pr¨ufer’s theorem [17, Theorem 17.2], every bounded abelian group G is a directsum of cyclic groups, so that G = M p ∈ P M s ∈ N Z ( p s ) ( κ p,s ) , where only finitely many of the cardinals κ p,s are nonzero; these cardinals are known as Ulm-Kaplansky invariants of G . For every p ∈ P , the κ p,s > s is referred to as the leading Ulm-Kaplansky invariant of G relative to p .If G is a bounded torsion abelian group and n = eo ( G ), then it easily follows from Pr¨ufer’stheorem that there exists a decomposition G = G ⊕ F , where F is finite, eo ( G ) = n , and G hasexponent n .We will now characterize the condition eo ( G [ n ]) = n (for n >
1) that appears in Proposition4.10 in terms of the Ulm-Kaplansky invariants of G . Proposition 4.12.
Given an integer n > and an infinite abelian group G , the following conditionsare equivalent: (i) eo ( G [ n ]) = n , (ii) all leading Ulm-Kaplansky invariants of G [ n ] are infinite, (iii) there exists a monomorphism Z ( n ) ( ω ) ֒ → G [ n ] .Proof. (i) → (ii) If eo ( G [ n ]) = n and p is a prime dividing n , then the leading Ulm-Kaplanskyinvariant relative to p is determined by the p -rank of the subgroup ( n/p ) G [ n ] of exponent p . Now,by our hypothesis, this group is infinite, and hence, the leading Ulm-Kaplansky invariant relativeto p is infinite as well.(ii) → (iii) Let P be the finite set of primes p dividing n . For every p ∈ P , let p k p be the highestpower of p dividing n . By (ii), the leading Ulm-Kaplansky invariant κ p,k p of G [ n ] is infinite. Hence,the p -primary component of G [ n ] has a direct summand of the form Z ( p k p ) ( κ p,kp ) . Therefore, thereexists a monomorphism Z ( p k p ) ( ω ) ֒ → G [ n ]. Since L p ∈ P Z ( p k p ) ( ω ) ∼ = Z ( n ) ( ω ) , this implies (iii).(iii) → (i) Suppose that m ∈ N and 1 ≤ m < n . Then the group mG [ n ] contains the infinitesubgroup m (cid:0) Z ( n ) ( ω ) (cid:1) ∼ = Z ( n/d ) ( ω ) , where d = ( m, n ) is the greatest common divisor of m and n .Since nG [ n ] = { } , this proves that eo ( G [ n ]) = n . (cid:3) Almost n -torsion sets as building blocks for the Zariski topology Definition 5.1. (i) For a given n ∈ N , we say that a countably infinite subset S of an abeliangroup G is almost n -torsion in G if S ⊆ G [ n ] and the set { x ∈ S : dx = g } is finite for each g ∈ G and every proper divisor d of n ; see [8].(ii) For n ∈ N , let T n ( G ) denote the family of all almost n -torsion sets in G .(iii) Define T ( G ) = S { T n ( G ) : n ∈ N } .In order to clarify Definition 5.1 and to facilitate future references, we collect basic properties ofalmost n -torsion sets in our next remark. Remark 5.2.
Let G be an abelian group.(i) T ( G ) = ∅ ; that is, there are no almost 0-torsion sets.(ii) T n ( G ) ∩ T m ( G ) = ∅ for distinct m, n ∈ N .(iii) Each family T n ( G ) is closed under taking infinite subsets, and so T ( G ) has the same prop-erty.(iv) If H is a subgroup of G , then T n ( H ) = { S ∈ T n ( G ) : S ⊆ H } for every n ∈ N ; see [8,Lemma 4.4]. In particular, whether a set S is almost n -torsion in G depends only on thesubgroup of G generated by S . Notation 5.3.
For an abelian group G and S ∈ T ( G ), we use n S to denote the unique integer n ∈ N such that S ∈ T n ( G ). (The uniqueness of such n follows from Remark 5.2(ii).)The notion of almost n -torsion set was introduced first in [15, Definition 3.3] under a differentname and has been even split into two cases; see [8, Remark 4.2] for an extended comparisonbetween this terminology and the one proposed in [15].The almost n -torsion sets were used in [8, 15] to build countably compact group topologies onabelian groups, while they were used in [31] to construct independent group topologies on abeliangroups. But only in the context of the Zariski topology can one fully realize the true power of theseremarkable sets due to their close relation to Z G -atoms; see Theorem 6.2. In fact, this relationpermits us to describe the Zariski topology of the abelian groups in full detail; see Sections 6 and8. Definition 5.4.
For a subset X of an abelian group G , we define(6) M ( X ) = { n ∈ N \ { } : X ∩ ( a + G [ n ]) is infinite for some a ∈ G } . and(7) m ( X ) = (cid:26) min M ( X ) , if M ( X ) = ∅ , if M ( X ) = ∅ . We shall see in Corollary 5.7 below that the function X m ( X ) is an extension of the function S n S from the family T ( G ) to the family of all subsets of G .From Definition 5.4, it immediately follows that m ( X ) = m ( g + X ) for every g ∈ G and each X ⊆ G . Our next lemma shows that in (6), it suffices to consider only those a ∈ G that areelements of X . Lemma 5.5.
For every subset X of an abelian group G , one has (8) M ( X ) = { n ∈ N \ { } : X ∩ ( x + G [ n ]) is infinite for some x ∈ X } . Proof.
Assume that X ∩ ( a + G [ n ]) is infinite for some a ∈ G and n ∈ N . Choose x ∈ X ∩ ( a + G [ n ]).Since x + G [ n ] = a + G [ n ], the intersection X ∩ ( x + G [ n ]) = X ∩ ( a + G [ n ]) must be infinite aswell. This proves the non-trivial inclusion in (8). (cid:3) The following lemma provides two reformulations of the notion of an almost n -torsion set. Lemma 5.6.
Let G be an abelian group and n ∈ N . For a countably infinite set S ⊆ G [ n ] , thefollowing conditions are equivalent: (i) S is almost n -torsion, (ii) S ∩ ( a + G [ d ]) is finite whenever a ∈ G and d is a proper divisor of n , (iii) n = m ( S ) .Proof. The equivalence of (i) and (ii) is proven in [15, Remark 3.4]. Let us now prove that (ii) and(iii) are also equivalent. We consider two cases.
Case 1 . n = 0. Since every integer n ∈ N \ { } is a proper divisor of 0, condition (ii) is equivalentto M ( S ) = ∅ by (6), and the latter condition is equivalent to m ( S ) = 0 by (7). Case 2 . n ∈ N \ { } . Since S ∩ G [ n ] = S is infinite, n ∈ M ( S ) = ∅ by (6).Let us prove that (ii) → (iii). Let m = m ( S ). Since M ( S ) = ∅ , m ≥ m ∈ M ( S ),(6) implies that the set S ′ = S ∩ ( a + G [ m ]) must be infinite for some a ∈ G . Since S ′ ⊆ S ⊆ G [ n ],we have S ′ ⊆ G [ n ] ∩ ( a + G [ m ]) = z + G [ d ], where z ∈ G and d = ( n, m ); see Lemma 3.1(iv).Since m ≥ n ≥
1, we have d ≥
1. Since S ∩ ( z + G [ d ]) contains the infinite set S ′ , d ∈ M ( S ) by (6). Since m = min M ( S ), we conclude that m ≤ d . Since d divides m , we must have d = m . Since d = m ∈ M ( S ), from (ii) it follows that d cannot be a proper divisor of n . Hence, n = d = m = m ( S ). This establishes (iii).The reverse implication (iii) → (ii) is obvious. (cid:3) Corollary 5.7. If G is an abelian group and S ∈ T ( G ) , then m ( S ) = n S . Corollary 5.8.
A countably infinite subset X of an abelian group G is almost -torsion if and onlyif m ( X ) = 0 .Proof. Since X ⊆ G = G [0], the conclusion follows from the equivalence (iii) ↔ (i) of Lemma5.6. (cid:3) Lemma 5.9.
Suppose that m, n ∈ N , S is a almost n -torsion subset of an abelian group G , a ∈ G and S ∩ ( a + G [ m ]) is infinite. Then G [ n ] ⊆ a + G [ m ] .Proof. By Lemma 3.1(iv), S ⊆ G [ n ] ∩ ( a + G [ m ]) = z + G [ d ] for some z ∈ S , where d = ( n, m )is the greatest common divisor of n and m . Since S is infinite, from the implication (i) → (ii) ofLemma 5.6, it follows that d cannot be a proper divisor of n . Hence, d = n . Since z ∈ S ⊆ G [ n ],we have z + G [ d ] = z + G [ n ] = G [ n ]. We have thus proven that G [ n ] ∩ ( a + G [ m ]) = G [ n ], whichyields G [ n ] ⊆ a + G [ m ]. (cid:3) Proposition 5.10.
Given n ∈ N and a countably infinite subset S of an abelian group G , thefollowing conditions are equivalent: (i) S is almost n -torsion, (ii) G [ n ] has exponent n , cl Z G ( S ) = G [ n ] and S is a Z G -atom.Proof. (i) → (ii) Suppose that d ( G [ n ]) = 0 (i.e., G [ n ] = G [ d ]) for some proper divisor d of n . Then { x ∈ S : dx = 0 } = { x ∈ S : nx = 0 } = S , because S ⊆ G [ n ]. Since S is infinite, S is not almost n -torsion, which contradicts (i). This proves that G [ n ] has exponent n .Let F be a Z G -closed set. According to Theorem 3.5(ii), there exist k ∈ N \ { } , a , ..., a k ∈ G and positive integers n , ..., n k such that F = ( a + G [ n ]) ∪ ( a + G [ n ]) ∪ . . . ∪ ( a k + G [ n k ]). If S ∩ F is infinite, then S ∩ ( a i + G [ n i ]) must be infinite for some i ≤ k . Now Lemma 5.9 impliesthat S ⊆ G [ n ] ⊆ a i + G [ n i ] ⊆ F . This shows that S is a Z G -atom satisfying G [ n ] ⊆ cl Z G ( S ). Since S ⊆ G [ n ] and G [ n ] is Z G -closed, we also have the reverse inclusion, namely, cl Z G ( S ) ⊆ G [ n ].(ii) → (i) It suffices to prove that S satisfies the condition (ii) of Lemma 5.6. Let a ∈ G and d be a proper divisor of n . Since G [ n ] has exponent n , the set E = a + G [ d ] is either disjoint from G [ n ] or a proper closed subset of G [ n ]. In the first case, S ∩ E = ∅ . In the second case, since E is a proper closed subset of G [ n ] and cl Z G ( S ) = G [ n ], S ∩ E must be a proper closed subset of S .Since S is a Z G -atom, it follows that S ∩ E must be finite. (cid:3) Lemma 5.11.
Suppose that Y is a subset of an abelian group G and H 6 = ∅ is a countable family ofsubgroups of G such that Y \ S i ≤ k ( a i + H i ) = ∅ whenever k ∈ N , a , . . . , a k ∈ G and H , . . . , H k ∈ H .Then there exists an infinite set S ⊆ Y such that S ∩ ( a + H ) is finite whenever a ∈ G and H ∈ H .Proof. Let H = { H n : n ∈ N } be an enumeration of H . For a finite set F ⊆ G and k ∈ N , define E F,k = S i ≤ k F + H i , and note that Y \ E F,k = ∅ by the assumption of our lemma. Therefore, byinduction on k , we can choose a sequence { y k : k ∈ N } such that y k +1 ∈ Y \ E F k ,k for every k ∈ N ,where F k = { y , . . . , y k } . Indeed, Y = ∅ allows one to choose y ∈ Y . Assuming that y , . . . , y k have already been chosen, select y k +1 ∈ Y \ E F k ,k = ∅ .We claim that S = { y k : k ∈ N } is as required. Clearly, S ⊆ Y . Since H 6 = ∅ , from our choiceof y k +1 , it follows that y k +1 F k + H = { y , . . . , y k } + H . Since 0 ∈ H , we conclude that y k +1
6∈ { y , . . . , y k } . Therefore, S is infinite. Assume now that a ∈ G and H ∈ H . Then thereexists n ∈ N such that H = H n . We must show that the set S ∩ ( a + H ) = S ∩ ( a + H n ) isfinite. If this set is empty, then the proof is complete. Suppose now that y m ∈ a + H n for some m ∈ N . Define k = max( m, n ). Let j > k be an arbitrary integer. Suppose that y j ∈ a + H n . Then y j − y m ∈ H n , and so y j ∈ y m + H n ⊆ F k + H n ⊆ F j − + H n ⊆ E F j − ,j − , which contradicts the choice of y j . This proves that S ∩ ( a + H n ) ⊆ { y , . . . , y k } . (cid:3) Our next proposition is the main result of this section. It provides an important characterizationof sets containing an almost n -torsion set that will be used frequently in subsequent sections. Proposition 5.12.
Suppose that n ∈ N and G is an abelian group. Then for every infinite set Y ⊆ G [ n ] , the following conditions are equivalent: (i) Y contains an almost n -torsion set, (ii) G [ n ] is an irreducible subgroup of G of exponent n , and cl Z G ( Y ) = G [ n ] , (iii) eo ( G [ n ]) = n and cl Z G ( Y ) = G [ n ] .Proof. Since G [1] = { } contains no infinite subsets Y , our proposition trivially holds for n = 1.Therefore, for the rest of the proof, we assume that n ∈ N \ { } .(i) → (ii) Let S be an almost n -torsion subset of Y . Applying the implication (i) → (ii) ofProposition 5.10, we conclude that S is Z G -atom, G [ n ] has exponent n , and G [ n ] = cl Z G ( S ) ⊆ cl Z G ( Y ) ⊆ G [ n ]. In particular, cl Z G ( Y ) = G [ n ]. Since S is a Z G -atom, G [ n ] = cl Z G ( S ) is irreducibleby Proposition 3.15(i).(ii) → (i) Define H = { G [ d ] : d ∈ D } , where D is the set of all proper divisors of n . We claimthat G , Y and H satisfy the assumptions of Lemma 5.11. Indeed, observe that { } = G [1] ∈ H 6 = ∅ ,since 1 is a proper divisor of every n ∈ N \{ } . Furthermore, suppose that k ∈ N , a , . . . , a k ∈ G and H , . . . , H k ∈ H . Assume that Y ⊆ E = S i ≤ k ( a i + H i ). Since E is Z G -closed, G [ n ] = cl Z G ( Y ) ⊆ E .Since G [ n ] is irreducible, G [ n ] ⊆ a i + H i for some i ≤ k by Fact 2.3. As H i = G [ d i ] for some d i ∈ D ,Lemma 3.1(i) yields G [ n ] ⊆ G [ d i ]. From d i ∈ D , it follows that d i is a proper divisor of n , andso G [ d i ] ⊆ G [ n ]. Thus, G [ d i ] = G [ n ], which contradicts our assumption that G [ n ] has exponent n .This shows that Y \ S i ≤ k ( a i + H i ) = ∅ .Let S be a set from the conclusion of Lemma 5.11. Then S is almost n -torsion by implication(ii) → (i) of Lemma 5.6.(ii) ↔ (iii) This follows from Proposition 4.10. (cid:3) Corollary 5.13.
For a given integer n ∈ N , an abelian group G contains an almost n -torsionsubset if and only if eo ( G [ n ]) = n .Proof. Apply Proposition 5.12 to Y = G [ n ]. (cid:3) Our next proposition provides a simple characterization of the sets that contain an almost 0-torsion set. It will be used in Theorem 7.4 to describe the Zariski dense subsets of an unboundedabelian group G . Proposition 5.14.
For a subset Y of an abelian group G , the following conditions are equivalent: (i) Y contains an almost -torsion set, (ii) mY is infinite for every integer m ∈ N \ { } , (iii) Y \ ( G [ m ] + F ) = ∅ for every integer m ∈ N \ { } and each finite subset F of G .Proof. (i) → (ii) Let S be an almost 0-torsion subset of Y . Assume that mS is finite. Then we canfind s ∈ S such that ms = ms for infinitely many s ∈ S . Since m is a proper divisor of 0, thiscontradicts Definition 5.1. This shows that mS (and then mY as well) must be infinite.(ii) → (iii) Assume Y ⊆ G [ m ] + F for some integer m ∈ N \ { } and some finite subset F of G .Then mY ⊆ mF is finite, which contradicts (ii).(iii) → (i) We claim that G , Y and H = { G [ n ] : n ∈ N \ { }} satisfy the assumptions of Lemma5.11. Suppose that k ∈ N , a , . . . , a k ∈ G and H , . . . , H k ∈ H . Then there exist m , . . . , m k ∈ N \ { } such that H i = G [ m i ] for i ≤ k . Let m = m m . . . m k . Then [ i ≤ k ( a i + H i ) = [ i ≤ k ( a i + G [ m i ]) ⊆ [ i ≤ k ( a i + G [ m ]) = { a , . . . , a k } + G [ m ] , and (iii) implies that Y \ S i ≤ k ( a i + H i ) = ∅ .Let S be a set from the conclusion of Lemma 5.11. Since (iii) obviously implies that G isunbounded, G has exponent 0. Since N \ { } is the set of all proper divisors of 0 and S satisfiesLemma 5.11, S is almost 0-torsion by the implication (ii) → (i) of Lemma 5.6. (cid:3) Corollary 5.15.
An abelian group G contains an almost -torsion subset if and only if G isunbounded.Proof. This follows from Proposition 5.14 when Y = G . Alternatively, this also follows fromCorollary 5.13 when n = 0 and Definition 4.3(ii). (cid:3) There is another, direct proof of this corollary. Clearly, bounded groups have no almost 0-torsionsubsets. Now assume that G is unbounded, and choose for every n > s n ∈ G suchthat n ! s n = 0. If G is non-torsion, then any infinite cyclic subgroup of G is an almost 0-torsionsubset, so we shall assume that G is torsion. In this case, S = { s n : n ∈ N } is an almost 0-torsion subset of G . Indeed, consider a non-zero d ∈ N and h ∈ G . It suffices to prove that the equation ds n = h holds only for finitely many n ∈ N . Since G is torsion, n h = 0 for some n >
0. Supposethat n ∈ N and n ≥ n + d . Then n d | n !, and n ! s n = 0 yields n ds n = 0. Since n h = 0, it followsthat ds n = h . Proposition 5.16.
An infinite abelian group G is almost torsion-free if and only if G has no almost n -torsion sets for any integer n > .Proof. Assume that G is almost torsion-free. For every prime p , the p -rank r p ( G ) of G is definedas the dimension of the subgroup G [ p ] over the field Z /p Z . Hence, the subgroup G [ p m ] is finite forevery m ∈ N . Now fix a positive n ∈ N . The subgroup G [ n ] of G is a sum of subgroups of the form G [ p m ], where the prime powers p m run over all divisors of the form p m in the prime factorizationof n ; hence, G [ n ] is finite. This proves that G has no almost n -torsion sets.On the other hand, if p is an arbitrary prime and G has no almost p -torsion sets p , then r p ( G ) < ∞ . Therefore, G is almost torsion-free. (cid:3) Irreducible sets and Z G -atoms Lemma 6.1.
Assume that n ∈ N , S is an almost n -torsion subset of an abelian group G and a ∈ G .Then a + S is almost n -torsion if and only if a ∈ G [ n ] .Proof. Assume that a + S is almost n -torsion. Then a + S ⊆ G [ n ]. Since S ⊆ G [ n ], we concludethat a ∈ G [ n ].Assume now that a ∈ G [ n ]. Since S ⊆ G [ n ], we have a + S ⊆ G [ n ]. Let g ∈ G , and let d be aproper divisor of n . Since S is almost n -torsion, the set { s ∈ S : ds = g − da } = { x ∈ a + S : dx = g } must be finite. Thus, a + S is almost n -torsion as well. (cid:3) Our first theorem in this section shows that Z G -atoms X are precisely the translates of almost m ( X )-torsion sets. (See Definition 5.4 for the number m ( X ).) Theorem 6.2.
For a countably infinite subset X of an abelian group G , the following conditionsare equivalent: (i) X is a Z G -atom, (ii) there exists a ∈ G such that X − a ∈ T ( G ) , (iii) for every x ∈ X , the set X − x is almost m ( X ) -torsion.Proof. (i) → (ii) By Proposition 3.15(ii), cl Z G ( X ) = a + G [ n ] for some a ∈ G and n ∈ N . By Lemma4.2, we may assume that G [ n ] has exponent n . Let S = X − a . By Theorem 3.12(ii), cl Z G ( S ) =cl Z G ( X − a ) = cl Z G ( X ) − a = G [ n ]. Furthermore, since X is a Z G -atom, from Corollary 3.7(ii) weconclude that S must be Z G -atom as well. Applying the implication (ii) → (i) of Proposition 5.10,we obtain that S must be almost n -torsion. That is, S ∈ T n ( G ) ⊆ T ( G ).(ii) → (iii) Let S = X − a and n = n S . By Corollary 5.7, m ( S ) = n . Therefore, m ( X ) = m ( X − a ) = m ( S ) = n since the function m ( − ) is translation invariant. Let x ∈ X . Since x − a ∈ G [ n ], we have a − x ∈ G [ n ], and so a − x + S = X − x is almost n -torsion by Lemma 6.1.(iii) → (i) Let x ∈ X . Since S = X − x is almost m ( X )-torsion, it is a Z G -atom by the implication(i) → (ii) of Proposition 5.10. Then X = x + S is a Z G -atom by Corollary 3.7(ii). (cid:3) Corollary 5.8 shows that the translates in the above theorem are not necessary when m ( X ) = 0.We now characterize all elements b ∈ G such that X − b is an almost m ( X )-torsion set. Proposition 6.3.
Let X be a Z G -atom of an abelian group G , and let n = m ( X ) . Given b ∈ G ,the set X − b is almost n -torsion if and only if b ∈ X + G [ n ] .Proof. Assume that b ∈ X + G [ n ]. Then there exists x ∈ X such that b − x ∈ G [ n ]. By theimplication (i) → (iii) of Theorem 6.2, S x = X − x is an almost n -torsion set. Since x − b ∈ G [ n ],Lemma 6.1 implies that X − b = x − b + S x is an almost n -torsion set. Now assume that X − b is an almost n -torsion set. By implication (ii) → (i) of Theorem 6.2, X is a Z G -atom. Choose any x ∈ X . Then S x = X − x is an almost n -torsion set by the implication(i) → (iii) of Theorem 6.2. Since X − b = x − b + S x is an almost n -torsion set, Lemma 6.1 impliesthat b − x ∈ G [ n ]. Therefore, b ∈ x + G [ n ] ∈ X + G [ n ]. (cid:3) Now we characterize the non-trivial irreducible sets in the Zariski topology.
Theorem 6.4.
Let G be an abelian group and X an infinite subset of G . Then the followingconditions are equivalent: (i) X is an irreducible subset of ( G, Z G ) , (ii) there exist a ∈ G and S ∈ T ( G ) such that a + S is Z G -dense in X , (iii) there exist n ∈ N and a ∈ X such that X − a contains a Z G -dense almost n -torsion subsetof G .Proof. (i) → (iii) Since X is irreducible, by Fact 2.4(iii), cl Z G ( X ) is a Z G -closed irreducible set.Therefore, cl Z G ( X ) = a + G [ n ] for some a ∈ X and n ∈ N . By Lemma 4.2, we may assume that G [ n ] has exponent n . Since translates are Z G -homeomorphisms by Corollary 3.7(ii), Y = X − a is an infinite Z G -dense subset of G [ n ], and G [ n ] is irreducible. From the implication (ii) → (i) ofProposition 5.12, it follows that Y contains an almost n -torsion subset S . Since G [ n ] = cl Z G ( S ) ⊆ cl Z G ( Y ) ⊆ G [ n ] by Proposition 5.10, S is Z G -dense in Y .(iii) → (ii) Let S ⊆ X − a be an almost n -torsion set that is Z G -dense in X − a . Once againapplying the fact that translates are Z G -homeomorphisms, we conclude that a + S must be Z G -densein a + ( X − a ) = X .(ii) → (i) By Theorem 6.2, a + S is a Z G -atom, and so it is irreducible. Since a + S is Z G -densein X , the latter set must be irreducible as well by Fact 2.4(iii). (cid:3) Corollary 6.5.
A subset of an abelian group is irreducible if and only if it contains a Z G -dense Z G -atom.Proof. From Theorem 6.4, we conclude that a subset X of an abelian group G is irreducible if thereexist n ∈ N , a ∈ G and an almost n -torsion set S such that a + S is Z G -dense in X . By Proposition5.10, S is a Z G -atom. Furthermore, a + S is a Z G -atom by Corollary 3.7(ii). (cid:3) Corollary 6.6.
For a subset F of an abelian group G , the following conditions are equivalent: (i) F is a closed, irreducible subset of ( G, Z G ) , (ii) F is a Z G -irreducible elementary algebraic set, (iii) F is a Z G -connected elementary algebraic set, (iv) F coincides with the Z G -closure of some Z G -atom.Proof. The implication (i) → (ii) is trivial. The equivalence (ii) ↔ (iii) follows from Corollary 4.8.Finally, the implications (ii) → (iv) and (iv) → (i) follow from Corollary 6.5. (cid:3) Description of Zariski dense sets
Theorem 7.1.
Let G be an infinite abelian group and n = eo ( G ) . Then for every subset X of G ,the following conditions are equivalent: (i) X is Zariski dense in G , (ii) a + X contains an almost n -torsion set for every a ∈ G , (iii) there exist a finite set F ⊆ G and a family { S x : x ∈ F } of almost n -torsion sets such that F + G [ n ] = G and x + S x ⊆ X for each x ∈ F .Proof. According to Lemma 4.5(i) and Theorem 4.6, G [ n ] is an irreducible, Z G -clopen subgroup of G with a finite index. (i) → (ii). Fix a ∈ G . From (i) and Corollary 3.7(ii), we conclude that a + X is Z G -dense in G .Since G [ n ] is Z G -open, Y = ( a + X ) ∩ G [ n ] must be Z G -dense in G [ n ]. Since G [ n ] is irreducible, wecan apply Proposition 5.12 to conclude that Y must contain an almost n -torsion set.(ii) → (iii) This implication holds because G [ n ] has a finite index in G .(iii) → (i). Let a ∈ F . Since a + X contains an almost n -torsion set, G [ n ] ⊆ cl Z G ( a + X ) byProposition 5.10. Since G [ n ] is Z G -open in G , X a = ( a + X ) ∩ G [ n ] is Z G -dense in G [ n ]. Hence − a + X a is Z G -dense in − a + G [ n ] by Corollary 3.7(ii). Since − a + X a ⊆ X , it follows that X ∩ ( − a + G [ n ]) is Z G -dense in − a + G [ n ].Note that − F + G [ n ] = − ( F + G [ n ]) = G , and so S {− a + G [ n ] : a ∈ F } = G . Since each X ∩ ( − a + G [ n ]) is Z G -dense in − a + G [ n ], we conclude that X is Z G -dense in G . (cid:3) One may wonder if it is possible to strengthen Theorem 7.1(iii) by requiring all S x to be equalto a single almost n -torsion set S . Since S x ∈ F ( x + S x ) = F + S in this case, the modified item(iii) then reads as “ F + S ⊆ X for some almost n -torsion set S of G ”. We will show in Theorem7.3 below that such a replacement is possible precisely when ( G, Z G ) is irreducible. For the proofof this modified theorem, we need the following lemma: Lemma 7.2.
Assume that X is a countably infinite subset of an abelian group G . Then there existinfinite disjoint sets Y , Y ⊆ X such that ( a + Y ) ∩ ( a + Y ) is finite whenever a , a ∈ G .Proof. The smallest subgroup H of G containing X is countable, so we can fix an enumeration H = { h n : n ∈ N } of H . For n ∈ N , define F n = { h , . . . , h n } . Since X is infinite, usinga straightforward recursion on n ∈ N we can choose x n ∈ X \ ( { x , . . . , x n − } + ( F n ∪ − F n )).We claim that Y = { x n : n ∈ N } and Y = { x n +1 : n ∈ N } are as required by Lemma7.2. Clearly, Y and Y are infinite disjoint subsets of X . Suppose that a , a ∈ G . Note that( a + Y ) ∩ ( a + Y ) = a + ( Y ∩ ( a − a + Y )), so we may assume without loss of generality that a = 0. If Y ∩ ( a + Y ) = ∅ , the proof is finished. So assume that Y ∩ ( a + Y ) = ∅ . Then a = 0as Y and Y are disjoint, and a ∈ Y − Y ⊆ X − X ⊆ H . Thus, there exists n ∈ N such that a = h n . We claim that Y ∩ ( a + Y ) ⊆ { x , . . . , x n } . Indeed, suppose that x k ∈ Y ∩ ( a + Y )for some k > n . Then there exists some l ∈ N such that x k = a + x l = h n + x l . Clearly, k = l since a = 0. If k < l , then n < l holds as well, and so x l = x k − h n ∈ { x , . . . , x l − } − F l − ,which is a contradiction. If l < k , then x k = h n + x l ∈ F k − + { x , . . . , x k − } , which is again acontradiction. (cid:3) Theorem 7.3.
Let G be a non-trivial abelian group and n = eo ( G ) . Then the following conditionsare equivalent: (a) ( G, Z G ) is irreducible, (b) for every Z G -dense set X , there exist an almost n -torsion S of G and a finite set F with F + G [ n ] = G and F + S ⊆ X , (c) for every Z G -dense set X , there exist an infinite set Z ⊆ G [ n ] and a subset F of G suchthat F + G [ n ] = G and F + Z ⊆ X .Proof. (a) → (b) Let X be a Z G -dense subset of G . Since G = G [ n ] is irreducible by Corollary 4.7, X contains an almost n -torsion set S by Proposition 5.12. Now (b) holds when F = { } .(b) → (c) follows obviously from the definition of an almost n -torsion set.(c) → (a) Clearly, G is infinite, and so by Corollary 4.7, it suffices to prove that G = G [ n ].Assume, by contradiction, that G [ n ] = G . Thus, we can choose g ∈ G \ G [ n ]. By Lemma 4.5(ii), eo ( G [ n ]) = eo ( G ) = n . Thus, G contains an almost n -torsion set S by Corollary 5.13. Let Y and Y be subsets of S satisfying Lemma 7.2 with S substituted by X . As an infinite subset of analmost n -torsion set, each Y i is almost n -torsion. Thus, both Y and Y are Z G -dense in G [ n ] byProposition 5.10. Hence, g + Y is Z G -dense in g + G [ n ] by Theorem 3.12(ii). It follows that the set(9) X = Y ∪ ( g + Y ) ∪ ( G \ ( G [ n ] ∪ ( g + G [ n ]))) is Z G -dense in G , and so (c) allows us to fix a finite set F ⊆ G and an infinite set Z ⊆ G [ n ]satisfying F + G [ n ] = G and F + Z ⊆ X . Since Y ∪ Y ⊆ S ⊆ G [ n ] and g G [ n ], from (9) we get(10) X ∩ G [ n ] = Y and X ∩ ( g + G [ n ]) = Y . Since 0 ∈ G = F + G [ n ], there exists some f ∈ F ∩ G [ n ]. Since f + Z ⊆ F + Z ⊆ X , we have f + Z = ( f + Z ) ∩ G [ n ] ⊆ X ∩ G [ n ] = Y by (10). Therefore, Z ⊆ − f + Y .Similarly, since g ∈ G = F + G [ n ], there exists some f ∈ F ∩ ( g + G [ n ]). Since f + Z ⊆ F + Z ⊆ X ,we have f + Z = ( f + Z ) ∩ ( g + G [ n ]) ⊆ X ∩ ( g + G [ n ]) = g + Y by (10). Therefore, Z ⊆ g − f + Y .We have proved that Z ⊆ ( − f + Y ) ∩ ( g − f + Y ). Since the latter set is finite by our choiceof Y and Y , we conclude that Z must be finite as well, which is a contradiction. (cid:3) For an unbounded abelian group G , our next theorem provides a very simple characterization ofZariski dense subsets of G . Theorem 7.4.
Let G be an unbounded abelian group. For an infinite subset X of G , the followingconditions are equivalent: (i) X is Z G -dense in G , (ii) mX is infinite for every integer m ∈ N \ { } , (iii) X contains an almost -torsion set.Proof. First, note that G [0] = G .(i) → (iii) By (i), we have cl Z G ( X ) = G = G [0]. Since G is unbounded, eo ( G [0]) = eo ( G ) = 0by Definition 4.3(ii). Applying the implication (iii) → (i) of Proposition 5.12, we conclude that X contains an almost 0-torsion set.(iii) → (i) Let S be an almost 0-torsion subset of X . Now G = G [0] = cl Z G ( S ) ⊆ cl Z G ( X ) ⊆ G by the implication (i) → (ii) of Proposition 5.10. Thus, X is Z G -dense in G .The equivalence (ii) ↔ (iii) is proved in Proposition 5.14. (cid:3) Description of the Zariski closure
Theorem 8.1.
Let X be an infinite subset of an abelian group G , let D be the finite set of Z G -isolated points of X , and let X \ D = S ki =1 X i be the (unique) decomposition of X \ D into irreduciblecomponents. Then there exist a , . . . , a k ∈ G and S , . . . , S k ∈ T ( G ) such that (i) each a i + S i is Z G -dense in X i , and (ii) cl Z G ( X ) = D ∪ k [ i =1 ( a i + cl Z G ( S i )) = D ∪ k [ i =1 ( a i + G [ n S i ]) .Proof. Observe that ( X, Z G ↾ S ) is a Noetherian space by Theorem 3.5(i) and Fact 2.1(1). Hence, D is finite by Fact 2.6. According to the same fact, each X i is infinite, so we can apply Theorem6.4 to find a i ∈ G and S i ∈ T ( G ) such that a i + S i is Z G -dense in X i . This yields (i). To prove (ii),note that(11) cl Z G ( X i ) = cl Z G ( a i + S i ) = a i + cl Z G ( S i ) = a i + G [ n S i ] for every i ≤ k, where the first equation follows from item (i), the second equation follows from Theorem 3.12(ii),and the third equation follows from Proposition 5.10. Since D is a finite subset of a T -space, D isclosed in ( G, Z G ), and socl Z G ( X ) = cl Z G D ∪ k [ i =1 X i ! = D ∪ k [ i =1 cl Z G ( X i ) , which together with (11) yields (ii). (cid:3) Remark 8.2. (i) Let X , G and D be as in the assumption of Theorem 8.1. Define I X = { ( n, a ) ∈ N × G : X − a contains an almost n -torsion set } . We shall see now that the set A = { a + G [ n ] : ( n, a ) ∈ I X } is finite, and(12) cl Z G ( X ) = D ∪ [ ( n,a ) ∈ I X ( a + G [ n ]) = D ∪ [ A . Indeed, let X \ D = S ki =1 X i be the decomposition of X \ D into irreducible components.Let a , . . . , a k ∈ G and S , . . . , S k ∈ T ( G ) be as in the conclusion of Theorem 8.1. Clearly,( n S i , a i ) ∈ I X for all i . This proves the inclusion cl Z G ( X ) ⊆ D ∪ S ( n,a ) ∈ I X ( a + G [ n ]). To provethe reverse inclusion, suppose that ( n, a ) ∈ I X . Then, X − a contains some set S ∈ T n ( G ),and so a + G [ n ] = cl Z G ( a + S ) ⊆ cl Z G ( S ) by Proposition 5.10. Since the elementary algebraicset a + G [ n ] ∈ A contains the dense irreducible set a + S , it is irreducible as well, and theinclusion a + G [ n ] ⊆ cl Z G ( X ) = D ∪ S ki =1 ( a i + G [ n S i ]) now yields a + G [ n ] ⊆ a i + G [ n S i ] forsome i . This proves that each closed set a + G [ n ] ∈ A is contained in one of the finitelymany closed sets a i + G [ n S i ]. Since our space ( G, Z G ) is Noetherian, this yields that A isfinite.(ii) The union (12) may be redundant; here is an easy example to this effect. Let G = Z (4) ( ω ) ,and let S be the canonical base of G . Then S is an almost 4-torsion set. Let X = S ∪ S . Then X is irreducible (since it contains the dense irreducible set S ) and Z G -dense, so cl Z G ( X ) = G = G [4] and the union from Theorem 8.1(ii) has a single member,namely cl Z G ( S ) = cl Z G ( X ). However, A = { G [2] , G [4] } , so the term G [2] in the union G = G [2] ∪ G [4] from (12) is redundant.(iii) The union from Theorem 8.1(ii) is not redundant, since it gives the (unique) decompositionof the closed set cl Z G ( X ) into a union of its irreducible components; see Fact 2.5. For every d ∈ D , the singleton { d } is an irreducible component of cl Z G ( X ), which means that the set D is disjoint from S ki =1 ( a i + G [ n S i ]). In particular, D = ∅ when G is infinite, and X is Z G -dense in G . Remark 8.3.
If at least one of the integers n S i appearing in Theorem 8.1(ii) is equal to , then D = ∅ , k = 1 , and the union from Theorem 8.1(ii) contains only one member, namely, G [0] = G . Indeed, assume that n S i = 0 for some i ≤ k . Then G [ n S i ] = G [0] = G , and so G = a i + G [ n S i ] ⊆ S kj =1 ( a j + G [ n S j ]) ⊆ G . Hence, D = ∅ , and k = i = 1 by Remark 8.2(iii). Remark 8.4.
Let X be a Z G -atom of an abelian group G . Then S = X − x is an almost m ( X )-torsion set for some (in fact, each) x ∈ X ; see Theorem 6.2. Therefore, cl Z G ( X ) = cl Z G ( x + S ) = x + cl Z G ( S ) = x + G [ m ( X )] by Theorem 3.12(ii) and the implication (i) → (ii) of Proposition 5.10.Since the union from Theorem 8.1(ii) is not redundant by Remark 8.2(iii), we conclude that theunique number n S i appearing in Theorem 8.1(ii) coincides with m ( X ). This conclusion fails in amore general case, for example, when X is an irreducible set. Indeed, let us note that for the subset X of the group G from Remark 8.2(ii) one has m ( X ) = 2, while cl Z G ( X ) = G [4]. Example 8.5.
Here we describe the closure of an infinite subset X of G in the case when X is a subgroup .(a) If X is unbounded, then X contains an almost 0-torsion set by Corollary 5.15. Hence, X is Z G -dense.(b) If X is a bounded subgroup, then by Pr¨ufer’s theorem X is a direct sum of finite cyclicsubgroups. Hence, we can write X = F ⊕ X , where F is finite and X ∼ = L ki =1 Z ( n i ) ( α i ) forsome infinite cardinals α i . In particular, X contains a subgroup isomorphic to Z ( n ) ( ω ) ∼ = L ki =1 Z ( n i ) ( ω ) , where n is the least common multiple of all n i . By Proposition 4.12(b), X contains an almost n -torsion set S , and so cl Z G ( X ) = G [ n ] by Proposition 5.10. Being finite, F is Z G -closed, and we conclude from Theorem 3.12(i) that cl Z G ( X ) = cl Z G ( F )+cl Z G ( X ) = F + G [ n ].(c) It follows from (a) and (b) that a subgroup X of G is Z G -dense if and only if either X isunbounded, or both X and G are bounded with eo ( X ) = eo ( G ) = n and nG = nX .(d) From (a) and (b), we deduce that the Zariski closure of a subgroup is always a subgroup.This can also be deduced directly from Theorem 3.12. Since ( G, Z G ) is a quasi-topologicalgroup (see the analysis following Corollary 3.7), the same conclusion can be deduced from[1, Proposition 1.4.13] as well.In the rest of this section, we present the major corollaries of Theorem 8.1. Our first corollarysignificantly strengthens [15, Lemma 3.6]: Corollary 8.6.
Let X be an infinite subset of an abelian group G . Then there exist k ∈ N , a , . . . , a k ∈ G and S , . . . , S k ∈ T ( G ) such that S ki =1 ( a i + S i ) is Z G -dense in X \ D , where D isthe finite set of Z G -isolated points of X . Our next proposition describes the case in which the union S ki =1 ( a i + S i ) in Corollary 8.6 is notonly Z G -dense in X \ D , but actually coincides with X \ D . Proposition 8.7.
Let X be a countably infinite subset of an abelian group G . Then dim X = 1 ifand only if there exist a natural number k and a Z G -atom X i for each i ≤ k , such that S ki =1 X i = X \ D , where D is the finite set of Z G -isolated points of S .Proof. To prove the “if” part, assume that sets X i with the desired properties are given. Accordingto Definition 3.14 and Fact 2.9, dim X i = 1. Then dim ( S mi =1 X i ) = 1 by Fact 2.8(d), and this yieldsdim X = 1, as dim D = 0 by Fact 2.8(a).To prove the “only if” part, assume that dim X = 1. Let D be the finite set of Z G -isolated pointsof X , and let X \ D = S ki =1 X i be the decomposition of X \ D into infinite irreducible components;see Fact 2.6. Fix i = 1 , , . . . , k . By Fact 2.8(b), dim X i ≤ dim X = 1. Since X i is infinite, fromFact 2.8(a) we derive the reverse inequality dim X i ≥
1. It follows that X i satisfies the assumptionsof Fact 2.9(a), and so X i is a Z G -atom by Definition 3.14. (cid:3) Corollary 8.8.
For an abelian group G , the space ( G, Z G ) is hereditarily separable; that is, everysubset X of ( G, Z G ) has a countable dense subset.Proof. If X is finite, then X is its own countable dense subspace. Assume now that X is infinite.Let D , k , a , . . . , a k and S , . . . , S k be as in Theorem 8.1. Then D ∪ S ki =1 ( a i + S i ) is a countabledense subset of X . (cid:3) Corollary 8.9.
For an abelian group G , the space ( G, Z G ) is Fr´echet-Urysohn; that is, if X is asubset of G and x ∈ cl Z G ( X ) , then there exists a sequence of points { x j : j ∈ N } of X convergingto x .Proof. If x ∈ X , by defining x j = x for all j ∈ N , we derive the required sequence. Therefore, fromnow on we shall assume that x X ; in particular X is infinite. Let D , k , X , . . . , X k , a , . . . , a k and S , . . . , S k be as in Theorem 8.1. Since x ∈ cl Z G ( X ) \ X , according to Theorem 8.1(ii), weconclude that x D . Then x ∈ cl Z G ( a i + S i ) for some i ≤ k . By Theorem 6.2, a i + S i is a Z G -atom.Let { y j : j ∈ N } be any faithful enumeration of a i + S i . From Proposition 3.15(iv), we concludethat the sequence { y j : j ∈ N } converges to x . Finally, note that y j ∈ a i + S i ⊆ X i ⊆ X for every j ∈ N . (cid:3) Item (b) of the next example demonstrates that Corollaries 8.8 and 8.9 are specific results aboutthe space ( G, Z G ) for an abelian group G that do not hold in general for Noetherian T spaces. Example 8.10. (a) Let X be a set, and let T i , i = 0 , , . . . , n , be Noetherian topologies on X .Then T = sup i T i is also a Noetherian topology on X . Indeed, the family E = { T ni =0 F i : X \ F i ∈ T i for all i ≤ n } satisfies the descending chain condition, E is closed underfinite intersections, and X ∈ E . Thus, by Fact 2.2 we conclude that the family E ∪ formsthe family of closed sets of a unique topology T E on X such that the space ( X, T E ) is aNoetherian space. It remains only to note that T = T E .(b) We can use (a) to get an example of a Noetherian T -space that is neither separable norFr´echet-Urysohn. Indeed, let α > ω be an ordinal, and let X = α . Let T = {{ x ∈ X : β < x } : β < α } S {∅ , X } be the upper topology of X , and let T be the co-finite topologyof X . Since both topologies are Noetherian, the topology T = sup {T , T } is Noetherianby (a). Since T is a T topology, so is T . Since ( X, T ) is not separable, ( X, T ) is notseparable either. To show that ( X, T ) is not Fr´echet-Urysohn, observe that ω ∈ X is inthe T -closure of the set S = { γ ∈ X : γ < ω } , yet no sequence of points of S converges to ω .Our last corollary of Theorem 8.1 is used in the proof of Theorem 9.5. Corollary 8.11.
Let X , G , D , k and S , . . . , S k ∈ T ( G ) be as in Theorem 8.1. If T is a Hausdorffgroup topology on G such that cl T ( S i ) = G [ n S i ] for every i ≤ k , then cl T ( X ) = cl Z G ( X ) .Proof. By the assumption of our corollary, a i + G [ n S i ] = a i + cl T ( S i ) = cl T ( a i + S i ) ⊆ cl T ( S i ) forevery i ≤ k . Combining this with Theorem 8.1(ii) yieldscl Z G ( X ) = D ∪ k [ i =1 ( a i + G [ n S i ]) ⊆ D ∪ k [ i =1 cl T ( S i ) = cl T ( X ) . The reverse inclusion cl T ( X ) ⊆ cl Z G ( X ) follows from Z G ⊆ M G ⊆ T . (cid:3) A precompact metric group topology realizing the Zariski closure
The main purpose of this section is to prove Theorem 9.5. The proof of Theorem C, provided inthe end of this section, then follows easily from Theorem 9.5.We start with two lemmas from [8] that will be needed in our proofs.
Lemma 9.1. ([8, Lemma 4.10])
Suppose that E is a countable family of subsets of an abelian group G , g ∈ G and g = 0 . Then there exists a group homomorphism h : G → T such that: (i) h ( g ) = 0 , (ii) if n ∈ N , z ∈ T [ n ] , l ∈ N \ { } and E ∈ E ∩ T n ( G ) , then the set { x ∈ E : | h ( x ) − z | < /l } is infinite. Lemma 9.2. ([8, Lemma 3.17])
Let G and H be Abelian groups such that | G | ≤ r ( H ) and | G | ≤ r p ( H ) for each p ∈ P . Suppose also that G ′ a subgroup of G such that r ( G ′ ) < r ( H ) and r p ( G ′ ) Let G be a countable abelian group and S be a countable subfamily of T ( G ) . Thenthere exists a monomorphism h : G → T N satisfying the following property: if S ∈ S , O is an opensubset of T N and O ∩ T [ n S ] N = ∅ , then the set { x ∈ S : h ( x ) ∈ O } is infinite. In particular, h ( S ) is dense in T [ n S ] N for every S ∈ S .Proof. Let S = { S j : j ∈ N } be an enumeration of S . For typographical reasons, let n j = n S j foreach j ∈ N . Let { g j : j ∈ N } be an enumeration of the countable set G \ { } and B be a countablebase of T with ∅ 6∈ B . Define E − = S . By induction on k ∈ N , we construct a countable family E k of subsets of G and a homomorphism h k : G → T with the following properties:(i k ) h k ( g k ) = 0,(ii k ) E k − ⊆ E k , (iii k ) if E ∈ E k − ∩ T ( G ), U ∈ B and U ∩ T [ n E ] = ∅ , then the set S E,U,k = { x ∈ E : h k ( x ) ∈ U } is infinite,(iv k ) if E ∈ E k − , U ∈ B and the set S E,U,k is infinite, then S E,U,k ∈ E k .We apply Lemma 9.1 to find a homomorphism h : G → T satisfying (i ) and (iii ). Define E = E − ∪ { S E,U, : E ∈ E − , U ∈ B , S E,U, is infinite } . Then (ii ) and (iv ) hold trivially. Thiscompletes the basis of induction. Suppose now that k ∈ N \ { } and for every j < k we have alreadyconstructed a countable family E j of subsets of G and a homomorphism h j : G → T satisfying (i j )–(iv j ). Let us define a countable family E k of subsets of G and a homomorphism h k : G → T satisfying (i k )–(iv k ). We apply Lemma 9.1 once again to find a homomorphism h k : G → T satisfying (i k ) and (iii k ). Define E k = E k − ∪ { S E,U,k : E ∈ E k − , U ∈ B , S E,U,k is infinite } and notethat (ii k ) and (iv k ) hold.Define h : G → T N by h ( x ) = { h n ( x ) } n ∈ N for x ∈ G . Clearly, h is a group homomorphism. If g ∈ G \ { } , then g = g k for some k ∈ N , and so h k ( g k ) = 0 by (i k ), which yields h ( g ) = h ( g k ) = 0.Hence h is a monomorphism.Assume that j ∈ N , and O is an open subset of T N such that O ∩ T [ n j ] N = ∅ . We shall provethat the set { x ∈ S j : h ( x ) ∈ O } is infinite. There exists k ∈ N and a sequence U , . . . , U k ofelements of B such that U × · · · × U k × T N \{ ,...,k } ⊆ O and U i ∩ T [ n j ] = ∅ for all i ≤ k . From(iii ) and our definition of E − , it follows that E = S S j ,U , is infinite, and (iv ) gives E ∈ E . Asan infinite subset of the almost n j -torsion set S j , the set E is also almost n j -torsion by Remark5.2(iii), and so (iii ) implies that E = S E ,U , is infinite. Thus, E ∈ E by (iv ). Continuingthis argument, we conclude that each set E i = S E i − ,U i ,i is infinite for i ≤ k . By construction, E k ⊆ E k − ⊆ · · · ⊆ E ⊆ S j . Finally, note that E k is an infinite subset of S j such that h i ( x ) ∈ U i whenever x ∈ E k and i ≤ k . Therefore, h ( E k ) ⊆ U × · · · × U k × T N \{ ,...,k } ⊆ O , which implies E k ⊆ { x ∈ S j : h ( x ) ∈ O } . Since the former set is infinite, so is the latter. (cid:3) Corollary 9.4. A countably infinite Abelian group G is isomorphic to a dense subgroup of T N ifand only if G is unbounded.Proof. Let j : G → T N be a monomorphism such that j ( G ) is a dense subgroup of T N . Then nj ( G )must be dense in T N = n T N , so nG = { } for every positive n ∈ N . Therefore, G is unbounded.If G is unbounded, then G contains an almost 0-torsion set S ; see Corollary 5.15. By Lemma9.3 applied to S = { S } , there exists a monomorphism j : G → T N such that j ( S ) is dense in T N .So j ( G ) is dense in T N as well. (cid:3) Theorem 9.5. Let G be an abelian group with | G | ≤ c , and let X be a countable family of subsetsof G . Then there exists a precompact metric group topology T on G such that the T -closure of each X ∈ X coincides with its Z G -closure.Proof. Applying Corollary 8.11, for every X ∈ X we can fix a finite family S X ⊆ T ( G ) with thefollowing property:( † X ) if T is a Hausdorff group topology on G such that cl T ( S ) = G [ n S ] for each S ∈ S X , thencl T ( X ) = cl Z G ( X ).Define S = S { S X : X ∈ X } . Let G ′ be the countable subgroup of G generated by S S .Since S ⊆ T ( G ) and S ⊆ G ′ for each S ∈ S , from Remark 5.2(iv) we conclude that S ⊆ T ( G ′ ).Therefore, we can apply Lemma 9.3 to find a monomorphism h : G ′ → T N such that h ( S ) is dense in T [ n S ] N for every S ∈ S . Since | T N | = r ( T N ) = c , r p ( T N ) = c for all p ∈ P ([17]; see also [7, Lemma4.1]), | G | ≤ c , | G ′ | = | N | < c , and T N is divisible, Lemma 9.2 allows us to find a monomorphism π : G → T N extending h . Denote by T the precompact metric group topology induced on π ( G ) bythe usual topology of T N .Let X ∈ X . Choose S ∈ S X arbitrarily. Since π ( G )[ n S ] = π ( G ) ∩ T [ n S ] N and π ( S ) = h ( S ) isdense in T [ n S ] N , it follows that π ( S ) is T -dense in π ( G )[ n S ]. Identifying G with π ( G ), we conclude that S is T -dense in G [ n S ]. In particular, cl T ( S ) = cl T ( G [ n S ]). Since G [ n S ] is Z G -closed in G and Z G ⊆ T , it follows that cl T ( G [ n S ]) = G [ n S ]. We proved that cl T ( S ) = G [ n S ] for every S ∈ S X .Now ( † X ) yields cl T ( X ) = cl Z G ( X ). (cid:3) Remark 9.6. Let G be an infinite abelian group.(i) There exists a family X of subsets of G such that | X | = c , and for every Hausdorff grouptopology T on G , the T -closure of some X ∈ X differs from its Z G -closure. Indeed, let H be a countably infinite subgroup of G and X be the family of all subsets of H . Clearly, | X | = c . Assume, by contradiction, that T is a Hausdorff group topology on G such thatcl T ( X ) = cl Z G ( X ) for every X ∈ X . For X ∈ X , we have cl T ↾ H ( X ) = H ∩ cl T ( X ) = H ∩ cl Z G ( X ) = cl Z H ( X ) by Lemma 3.10, and since X is the family of all subsets of H , weconclude that T ↾ H = Z H . Since T is a Hausdorff topology on G , the subspace topology T ↾ H is Hausdorff as well. This contradicts Corollary 3.6.(ii) Item (i) shows that the conclusion of Theorem 9.5 may fail for families X of size c . Underthe Continuum Hypothesis, the conclusion of Theorem 9.5 would then fail for families X ofsize ω . Thus, in general, a single Hausdorff group topology on an abelian group G cannotrealize the Zariski closure of uncountably many subsets of G , that is, the assumption ofcountability for the family X in Theorem 9.5 is necessary. Proof of Theorem C. The implication (i) → (iii) is proved in Theorem 9.5. The implication(iii) → (ii) is trivial. Finally, the implication (ii) → (i) holds because a precompact metric grouphas size at most c (this follows from the well known fact that infinite compact metrizable grouphas size c [1, Corollary 5.2.7 b)]). (cid:3) Proof of the equality Z G = M G = P G for an abelian group G For an abelian group H we denote by H ∗ the closed subgroup of T H consisting of all homomor-phisms χ : H → T ; in particular, H ∗ is compact in the subspace topology inherited from T H . Asubgroup N of H ∗ is called point separating if for every x ∈ H \ { } , there exists χ ∈ N such that χ ( x ) = 0.For every subgroup N of H ∗ , let e N,H : H → T N be the map defined by e N,H ( x )( χ ) = χ ( x ) for χ ∈ N and each x ∈ H , and let T N,H denote the coarsest topology on H with respect to which e N,H becomes continuous. The straightforward proof of the following fact is left to the reader. Fact 10.1. For an abelian group H and a subgroup N of H ∗ , the following conditions are equivalent:(i) N is point separating;(ii) T N,H is a precompact Hausdorff group topology on H ;(iii) e N,H is a monomorphism;(iv) the map e N,H : ( H, T N,H ) ֒ → T N is a topological isomorphism between ( H, T N,H ) and thesubgroup e N,H ( H ) of T H .The relevant fact that the correspondence N T N,H between point separating subgroups of H ∗ and precompact Hausdorff group topologies on H is a bijection was pointed out by Comfort andRoss [6].Let H be a normal subgroup of a group G and T ′ a Hausdorff group topology on H . In general,it is impossible to find a Hausdorff group topology T on G that induces the original topology T ′ on H . There are severe obstacles to the extension of group topologies even when the subgroup H is abelian [10, 13]. However, in the case when the subgroup H is central (in particular, when thegroup G itself is abelian), the extension problem becomes trivial. One can simply take the familyof all T ′ -neighborhoods of 0 as a base of the family of T -neighborhoods of 0 of a group topology T on G . This topology T is Hausdorff and obviously induces the original topology T ′ on H . Notethat there may exist other group topologies T ′′ on G with T ′′ ↾ H = T ′ . The extension of precompact Hausdorff group topologies from a subgroup of an abelian group tothe whole group is a bit more delicate. Theorem 10.2. Let H be a subgroup of an abelian group G and T ′ a precompact Hausdorff grouptopology on H . Then there exists a precompact Hausdorff group topology T on G that induces T ′ on H ; that is, T ↾ H = T ′ holds.Proof. Let ρ : G ∗ → H ∗ be the restriction homomorphism defined by ρ ( χ ) = χ ↾ H for every χ ∈ G ∗ .Since T is a divisible group, for every χ ∈ H ∗ , there exists an extension e χ ∈ G ∗ of χ . Therefore, ρ is a surjection. It follows from Peter-Weyl’s theorem that the subgroup N of H ∗ consisting of all T ′ -continuous characters is point separating and T ′ = T N,H ; see [6].We claim that A = ρ − ( N ) is a point separating subgroup of G ∗ . Indeed, let g ∈ G \ { } . Weneed to find ϕ ∈ A such that ϕ ( g ) = 0. If g ∈ H , then χ ( g ) = 0 for some χ ∈ N , as N is apoint separating subgroup of H ∗ . Since ρ is a surjection, there exists ϕ ∈ G ∗ such that ρ ( ϕ ) = χ .In particular, ϕ ∈ ρ − ( N ) = A . Furthermore, ϕ ( g ) = ϕ ↾ H ( g ) = χ ( g ) = 0. Suppose now that g ∈ G \ H . Let π : G → G/H be the canonical quotient homomorphism. There exists a character ψ : G/H → T such that ψ ( π ( g )) = 0. Clearly, ϕ = ψ ◦ π ∈ G ∗ and ρ ( ϕ ) = ϕ ↾ H = 0 ∈ N , whichyields ϕ ∈ ρ − ( N ) = A . Finally, note that ϕ ( g ) = 0.Since A is a point separating subgroup of G , it follows from Fact 10.1 that T = T A,G is aprecompact Hausdorff group topology on G . It remains only to show that T ↾ H = T ′ . In otherwords, we aim to show that the inclusion map j : ( H, T N,H ) ֒ → ( G, T A,G ) is a topological groupembedding.For a set Y and an element y ∈ Y , let p Y,y : T Y → T be the canonical projection definedby p Y,y ( f ) = f ( y ) for every y ∈ Y . Since A = ρ − ( N ) and ρ is a surjection, ρ ↾ A : A → N issurjective as well. Therefore, the the map ι : T N → T A defined by ι ( f ) = f ◦ ρ ↾ A for f ∈ T N ,is a monomorphism. Furthermore, the map ι is continuous, as p A,χ ◦ ι = p N,ρ ( χ ) is continuous forevery χ ∈ A . By the compactness of T N , we conclude that ι : T N ֒ → T A is a topological groupembedding.By Fact 10.1, both maps e A,G : ( G, T A,G ) ֒ → T A and e N,H : ( H, T N,H ) ֒ → T N are topologicalgroup embeddings. Since the diagram ( G, T A,G ) (cid:31) (cid:127) e A,G / / T A ( H, T N,H ) (cid:31) (cid:127) e N,H / / ?(cid:31) j O O T N ?(cid:31) ι O O is commutative, we conclude that j is a topological group embedding as well. (cid:3) Remark 10.3. (i) In connection with Fact 10.1, one should mention the following importantresult of Comfort and Ross [6]: Given an abelian group H , a subgroup N of H ∗ is pointseparating if and only if N is dense in H ∗ .(ii) With the help of the equivalence in (i), one can offer an alternative argument showing thatthe subgroup A of G ∗ in the proof of Theorem 10.2 is point separating. Indeed, since G ∗ isa compact group and ρ : G ∗ → H ∗ is a continuous surjective homomorphism, ρ is an openmap. Since N is dense in H ∗ by item (i), this implies that A = ρ − ( N ) is dense in G ∗ .Applying (i) once again, we conclude that A is a point separating subgroup of G ∗ . Corollary 10.4. If H is a subgroup of an abelian group G , then P G ↾ H ⊆ P H .Proof. Let X be a subset of H that is not P H -closed. There exists a precompact Hausdorff grouptopology T ′ on H such that X is not T ′ -closed. By Theorem 10.2, there exists a Hausdorff grouptopology T on G such that T ↾ H = T ′ . Then X is not T -closed, and so X cannot be P G -closedeither. Since X ⊆ H , we conclude that X is not P G ↾ H -closed. (cid:3) Proof of Theorem A. Since Z G ⊆ M G ⊆ P G , it suffices to prove that P G ⊆ Z G . Let us note thatthis follows from Theorem 9.5 when | G | ≤ c . Indeed, let F be a subset of G that is not Z G -closed.Applying Theorem 9.5 with X = { F } , we can find a precompact metric group topology T on G such that cl T ( F ) = cl Z G ( F ) = F ; that is, F is not T -closed. Hence, F cannot be P G -closed either.To prove that P G ⊆ Z G in the general case, it suffices to pick an arbitrary subset F of G andcheck that cl Z G ( F ) ⊆ cl P G ( F ). According to Corollary 8.8, there exists a countable subset X of F such that F ⊆ cl Z G ( X ). In particular, cl Z G ( F ) ⊆ cl Z G ( X ). Since cl P G ( X ) ⊆ cl P G ( F ), it remains tocheck that cl Z G ( X ) ⊆ cl P G ( X ).Fix an arbitrary x ∈ cl Z G ( X ), and let H be the (countable) subgroup of G generated by X and x . By Lemma 3.10, x ∈ H ∩ cl Z G ( X ) = cl Z G ↾ H ( X ) = cl Z H ( X ). By the initial part of theargument, Z H = P H , as H is countable. So x ∈ cl Z H ( X ) = cl P H ( X ). Since X ⊆ H , Corollary10.4 yields cl P H ( X ) ⊆ cl P G ↾ H ( X ) = H ∩ cl P G ( X ). Therefore, x ∈ cl P G ( X ). This proves thatcl Z G ( X ) ⊆ cl P G ( X ). (cid:3) Corollary 10.5. For an arbitrary subset X of an abelian group G , the following conditions areequivalent: (a) X is unconditionally closed, (b) X is closed in every precompact Hausdorff group topology on G , (c) X is algebraic. The equivalence of (a) and (c) was attributed by Markov [25] to Perel ′ man, though the proofnever appeared in print. Recently, a proof of this equivalence was provided in [10]; see also [31,Theorem 3.13] for almost torsion-free abelian groups and [31, Proposition 4.6] for abelian groupsof prime exponent. The group topologies involved in both results in [31] are not precompact, sothese results do not include also the equivalence with (b) even in those two particular cases. Remark 10.6. According to Lemma 3.10 and Corollary 10.4, one might first study the spaces( G, Z G ) and ( G, P G ) for divisible groups G and then use them to obtain information on theirsubgroups. The advantage of having a divisible group G is that, for every prime number p and eachinteger n ∈ N , the subgroup G [ p n ] of G is always irreducible whenever it is infinite; see Example4.11. 11. A partial solution to a problem of Markov Proof of Theorem D. (i) → (iii) Assume that X is potentially dense in G , and let T be aHausdorff group topology on G such that X is T -dense in G . Then G = cl T ( X ) ⊆ cl Z G ( X ) ⊆ G ,which yields G = cl Z G ( X ).(iii) → (ii) This follows from Theorem 9.5.(ii) → (i) This is trivial. (cid:3) Combining Theorems D and 7.1, we obtain a partial solution to Markov’s problem regardingpotentially dense sets for all abelian groups of size at most continuum: Corollary 11.1. Let X be an infinite subset of be an abelian group G of size ≤ c . Define n = eo ( G ) .Then the following conditions are equivalent: (i) X is potentially dense in G , (ii) a + X contains an almost n -torsion set for every a ∈ G , (iii) there exist a finite set F ⊆ G and a family { S x : x ∈ F } of almost n -torsion sets such that F + G [ n ] = G and x + S x ⊆ X for each x ∈ F . Recall that every infinite subset of an abelian group G is Zariski dense if and only if G is eitheralmost torsion-free or has a prime exponent (see Fact 3.9). Corollary 11.2. For an abelian group G , consider the following conditions: (a) every infinite subset of G is potentially dense in G , (b) every proper unconditionally closed subset of G is finite (that is, M G coincides with thecofinite topology of G ), (c) every proper algebraic subset of G is finite (that is, Z G coincides with the cofinite topologyof G ).Then (a) → (b) ↔ (c). Moreover, all three conditions are equivalent whenever | G | ≤ c .Proof. To show (a) → (b) assume that X is an infinite unconditionally closed subset of G . Then X must be simultaneously closed and dense in some Hausdorff group topology on G . Thus, X = G .The equivalence (b) ↔ (c) follows from Corollary 10.5.The last assertion is an obvious consequence of Theorem D. (cid:3) Remark 11.3. Since item (a) of Corollary 11.2 yields | G | ≤ c , it is not possible to invert theimplication (a) → (b) in Corollary 11.2 if | G | > c . In our forthcoming paper [12], we invert it forgroups of size at most 2 c .From Theorems D and 7.4 we obtain the following: Corollary 11.4. Let G be an unbounded abelian group such that | G | ≤ c . Then an infinite subset X of G is potentially dense in G if and only if mX is infinite for every m ∈ N \ { } . The potentially dense subsets of almost torsion-free or divisible abelian groups of arbitrary sizeare described in [11]. 12. Open questions Since our results provide a sufficiently clear picture in the abelian case, we include a list ofquestions concerning the possibilities to extend some of them in the non-abelian case.Theorem A leaves the following question open. Question 12.1. Which of the equalities Z G = M G = P G from Theorem A remain true for nilpotentgroups?According to Bryant’s theorem, Z G is Noetherian for every abelian group G . This fails to betrue in general, e.g., there exist infinite (necessarily non-abelian) groups G with discrete Z G ; seeExample 1.3. Nevertheless, there is a huge gap between Noetherian and discrete topologies. Infact, being Noetherian is a much stronger property than compactness: it is easy to see that a spaceis Noetherian if and only if all its subspaces are compact. This justifies the following question. Question 12.2. Let G be a group. If Z G is compact, must Z G be necessarily Noetherian?According to Corollary 3.6, the Zariski topology Z G of an infinite abelian group is never Hausdorff,while it is Noetherian by Bryant’s theorem. This motivates the following question. Question 12.3. Does there exist an infinite group G such that its Zariski topology Z G is bothcompact and Hausdorff?Let us note that it was necessary to relax “Noetherian” to “compact” in the above question,since an infinite Noetherian space is never Hausdorff; see Fact 2.1. In particular, a positive answerto this question would provide a negative answer to Question 12.2.Finally, let us recall two questions from [9]. Question 12.4. Let G be a group of size at most 2 c and E be a countable family of subsets of G .Can one find a Hausdorff group topology T E on G such that the T E -closure of every E ∈ E coincideswith its M G -closure? For an Abelian group G with | G | ≤ c the answer is positive, and in fact the topology T E in thiscase can be chosen to be precompact by Theorem 9.5.The counterpart of Question 12.4 for Z G instead of M G has a consistent negative answer; see thecomment in [9].Let us consider now the counterpart of Question 12.4 for P G instead of M G . Question 12.5. Let G be a group of size at most 2 c having at least one precompact Hausdorff grouptopology, and let E be a countable family of subsets of G . Can one find a precompact Hausdorffgroup topology T E on G such that the T E -closure of every E ∈ E coincides with its P G -closure?Again, for an Abelian group G with | G | ≤ c , the answer is positive by Theorem 9.5. Historic remark. 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