aa r X i v : . [ m a t h . L O ] O c t Walker’s cancellation theorem
Robert Lubarsky and Fred RichmanFlorida Atlantic University22 August 2012
Abstract
Walker’s cancellation theorem says that if B ⊕ Z is isomorphic to C ⊕ Z in the category of abelian groups, then B is isomorphic to C . Weconstruct an example in a diagram category of abelian groups wherethe theorem fails. As a consequence, the original theorem does nothave a constructive proof even if B and C are subgroups of the freeabelian group on two generators. Both of these results contrast witha group whose endomorphism ring has stable range one, which allowsa constructive proof of cancellation and also a proof in any diagramcategory. An object G in an additive category is cancellable if whenever B ⊕ G isisomorphic to C ⊕ G , then B is isomorphic to C . Elbert Walker, in hisdissertation [7], and P. M. Cohn in [3], independently answered a questionof Irving Kaplansky by showing that finitely generated abelian groups arecancellable in the category of abelian groups. The most interesting case isthat of Z , the additive group of integers. That’s because finitely generatedgroups are direct sums of copies of Z and of cyclic groups of prime powerorder, and a cyclic group of prime power order has a local endomorphismring, hence is cancellable by a theorem of Azumaya [2].It is somewhat anomalous that Z is cancellable. A rank-one torsion-freegroup A is cancellable if and only if A ∼ = Z or the endomorphism ring of A has stable range one [1, Theorem 8.12],[4]. (A ring R has stable range ne if whenever aR + bR = R , then a + bR contains a unit of R .) Thus forrank-one torsion-free groups, the endomorphism ring tells the whole story—except for Z . It turns out that an object is cancellable if its endomorphismring has stable range one. The proof of this in [6, Theorem 4.4] is constructiveand works for any abelian category. It is also true, [6], that semilocal ringshave stable range one, so Azumaya’s theorem is a special case of this. Infact, that the endomorphism ring of A has stable range one is equivalent to A being substitutable, a stronger condition than cancellation [6, Theorem4.4]. We say that A is substitutable if any two summands a group, withcomplements that are isomorphic to A , have a common complement. Thegroup Z is not substitutable: Consider the subgroups of Z generated by(1 , , , , a, b ) with a = ± a − b = ± Z is cancellable in the (abelian)category D T ( Ab ) of diagrams of abelian groups based on a fixed finite poset T with a least element. There is a natural embedding of Ab into D T ( Ab )given by taking a group into the constant diagram on T with identity mapsbetween the groups on the nodes. In particular, we can identify the group ofintegers as an object of D T ( Ab ). As the endomorphism ring of any group G is the same as that of its avatar in D T ( Ab ), a substitutable group issubstitutable viewed as an object in D T ( Ab ). However it turns out that Z is not cancellable in D T ( Ab ) where T is the linearly ordered set { , , } .This result has repercussions for the constructive theory of abelian groups.Because of it, we can conclude that Walker’s theorem does not admit a con-structive proof. In fact, it is not even provable when B and C are restrictedto be subgroups of Z . It was the question of whether Walker’s theoremhad a constructive proof that initiated our investigation. You can think of aconstructive proof as being a proof within the context of intuitionistic logic.Such proofs are normally constructive in the usual informal sense. Most anyproof of Azumaya’s theorem is constructive, so a constructive proof of thecancellability of Z would show that you can cancel finite direct sums of finiteand infinite cyclic groups.As any homomorphism from an abelian group onto Z splits, Walker’stheorem can be phrased as follows: If A is an abelian group, and f, g : A → Z are epimorphisms, then ker f ∼ = ker g . The following theorem gets us partway to a proof of Walker’s theorem. 2 heorem 1 Let A be an abelian group and f, g : A → Z be epimorphisms.Then f (ker g ) = g (ker f ) so that ker g ker f ∩ ker g ∼ = f (ker g ) = g (ker f ) ∼ = ker f ker f ∩ ker g Proof.
Consider the image I of the map A → Z ⊕ Z induced by f and g . As f and g are epimorphisms, I is a subdirect product. Note that f (ker g ) = I ∩ ( Z ⊕
0) when the latter is viewed as a subgroup of Z , and similarly g (ker f ) = I ∩ (0 ⊕ Z ). To finish the proof we show that if ( x, ∈ I ,then (0 , x ) ∈ I . As I is a subdirect product, there exists n ∈ Z such that( n, ∈ I . Thus (0 , x ) = x ( n, − n ( x, ∈ I .Thus we get the desired isomorphism ker f ∼ = ker g if ker f ∩ ker g = 0or if f (ker g ) is projective. Classically, every subgroup of Z is projective, sothis constitutes a classical proof. Indeed, it is a classical proof that in thecategory of modules over a Dedekind domain D , the module D is cancellable[5]. Our example lives in the category D T ( Ab ) of diagrams of abelian groupsbased on the linearly ordered set T = { , , } . The example shows that youcan’t cancel Z in D T ( Ab ).The groups on the nodes will be subgroups A ⊂ A ⊂ A = Z definedby generators: A = (1 , , , , A = (1 , , − , , , , A = (1 , , , , , , , , , (8 , , ∈ A . The maps between these groups are inclu-sions. Define the maps f, g : Z → Z by f ( a, b, c ) = a and g ( a, b, c ) = b .The maps f and g each induce maps from these three groups into Z whichgive two maps from the diagram into the constant diagram Z . We denotethe kernel of the map f restricted to A i by ker i f and similarly for g . Thesekernels admit the following generators:ker f = (0 , ,
0) ker f = (0 , , , ,
64) ker f = (0 , , , , g = (8 , ,
0) ker g = (1 , , − , ,
64) ker g = (1 , , , , B = ker f and C = ker g are clearly each embeddable inthe diagram Z ⊕ Z . That B ⊕ Z is isomorphic to C ⊕ Z follows from the factthat the diagram A can be written as an internal direct sum B ⊕ Z and alsoas an internal direct sum C ⊕ Z . The generator of Z in the first case is theelement (1 , , , , Theorem 2
There is no isomorphism between ker f and ker g in D T ( Ab ) . Proof.
Suppose we had an isomorphism ϕ : ker f → ker g . Looking at theisomorphisms at 0 and 2, there exist e, e ′ = ± x ∈ Z so that ϕ (0 , ,
0) = (8 e, ,
0) and ϕ (0 , ,
1) = ( x, , e ′ )Thus ϕ (0 , ,
8) = ( e + 8 x, , e ′ ). For ( e + 8 x, , e ′ ) to be in ker g , we musthave 8 e ′ +24 ( e + 8 x ) divisible by 64. But 8 e ′ +24 ( e + 8 x ) is equal to 8 e ′ +24 e modulo 64, and this is not divisible by 64.The following result shows that we can’t get an example that is a subob-ject of the diagram Z n using the linearly ordered set T = { , } . Theorem 3
Let T = { , } . In the category D T ( Ab ) , if A and B aresubobjects of Z n , and A ⊕ Z is isomorphic to B ⊕ Z , then A is isomorphic to B . Proof.
Write A ⊆ Z n as A ⊆ A . As A is a finite-rank free abeliangroup, the situation A ⊆ A can be represented by an integer matrix whoserows generate A . Using elementary row and column operations, we candiagonalize this matrix so that each entry on the diagonal divides the next(Smith normal form). Thus A is isomorphic to B exactly when the ranksof the free abelian groups A and B are equal, and A /A ∼ = B /B . If C = A ⊕ Z is isomorphic to D = B ⊕ Z , then the rank of C = A ⊕ Z isequal to the rank of D = B ⊕ Z , so the rank of A is equal to the rank of B , and A /A ∼ = C /C ∼ = D /D ∼ = B /B , so A is isomorphic to B .This theorem leaves open the question of whether there is an counterex-ample of this sort using the poset that looks like a “V”.4 The Brouwerian counterexample
A Brouwerian example is an object depending on a finite family of proposi-tions. The idea is that if a certain statement holds about that object, thensome relation holds among the propositions. Thus a Brouwerian example ispiece of reverse mathematics : the derivation of a propositional formula froma mathematical statement. For example, there may be just one proposition P and if the statement holds for that object, then P ∨ ¬ P holds. Thusfrom the general truth of the statement we could derive the law of excludedmiddle, from which we would conclude that the statement does not admit aconstructive proof. Our Brouwerian counterexample to Walker’s theorem isbased on the diagram of groups of the previous section.Let P and Q be propositions. Let A = (cid:8) x ∈ Z : x ∈ A or P ∧ x ∈ A or P ∧ Q (cid:9) where A and A are defined in the preceding section. The maps f, g : Z → Z are defined as before by f ( a, b, c ) = a and g ( a, b, c ) = b .Note that A is a discrete group (any two elements are either equal ordistinct) as it is a subgroup of the discrete group Z . Theorem 4
The groups ker f and ker g are isomorphic if and only if P ∨ P ⇒ ( Q ∨ ¬ Q ) . Proof.
As before, we denote A i ∩ ker f by ker i f .If P holds, then the isomorphism is induced by ϕ (0 , ,
0) = (1 , , − ϕ (0 , ,
1) = (0 , , P ⇒ ( Q ∨ ¬ Q ) holds. Define ϕ on ker f by ϕ (0 , ,
0) = (0 , , x thatis not in ker f . If x ∈ ker f , and x / ∈ ker f , then P holds, hence either Q or ¬ Q holds. If Q holds, then the isomorphism is induced by ϕ (0 , ,
0) =(1 , ,
0) and ϕ (0 , ,
1) = (0 , , ¬ Q holds, the isomorphism is inducedby ϕ (0 , ,
8) = (3 , , −
8) and ϕ (0 , ,
0) = (8 , , ϕ is an isomorphism. If ϕ (0 , , = ( ± , , P holds, so we may assume that ϕ (0 , ,
0) = (8 , , P ⇒ Q ∨ ¬ Q , suppose P holds. If ϕ (ker f ) = ker g , then Q holds. If ϕ (ker f ) =ker g , then Q cannot hold because that would give an isomorphism in thediagram category contrary to Theorem 2.5o if we could find a constructive proof that ker f and ker g were isomor-phic, then we would have a constructive proof of the propositional form P ∨ P ⇒ ( Q ∨ ¬ Q ) .That means that this form would be a theorem in the intuitionistic proposi-tional calculus. But then by the disjunction property, either P is a theorem,which it is not, or P ⇒ ( Q ∨ ¬ Q ) is a theorem. In the latter case, substi-tuting ⊤ for P gives Q ∨ ¬ Q , the law of excluded middle, which is not atheorem.The diagram example of the preceding section can itself be thought of asan object in a model of intuitionistic abelian group theory, and in this waydirectly shows that Walker’s theorem does not admit a constructive proof,even for subgroups of Z . We have seen that we can’t cancel Z with respect to certain subgroups of Z ⊕ Z . It is natural to ask what the situation is with respect to subgroupsof Z . We give a constructive proof of the following theorem. Theorem 5
Let B be an abelian group such that every nontrivial homomor-phism from B to Z is one-to-one. If f is a homomorphism from B ⊕ Z onto Z , then ker f is isomorphic to mB for some positive integer m . Hence if B is torsion free, then ker f is isomorphic to B . Proof.
Let s = f (0 ,
1) and f the restriction of f to B . As f is onto, wehave f ( B ) + s Z = Z . If s = 0, then f maps B isomorphically onto Z , and0 ⊕ Z is ker f , in which case we can set m = 1. So we may suppose that s >
0. We will show that ker f is isomorphic to sB .When is ( b, n ) in ker f ? As f ( b, n ) = f ( b ) + sn , we see that a necessaryand sufficient condition is that f ( b ) ∈ s Z and n = − f ( b ) /s . Thus ker f isisomorphic to f − ( s Z ). As f ( B ) + s Z = Z , and f ( B ) and s Z are ideals of Z , it follows that f ( B ) ∩ s Z = f ( B ) s Z = sf ( B ). Thus f − ( s Z ) = f − ( f ( B ) ∩ s Z ) = f − ( sf ( B ))6learly f − ( sf ( B )) ⊇ sB . Conversely, if f ( b ) ∈ sf ( B ) = f ( sB ), then f ( b ) = f ( sb ′ ) so b = sb ′ ∈ sB .Note that any torsion-free group B of rank at most one satisfies thehypothesis of the theorem. Also any group with no nontrival maps into Z . Classically, this latter condition simply says that B has no proper Z summands.What other groups B allow cancellation of Z ? It suffices that B be finitelygenerated. To see this, look at Theorem 1. If ker f is finitely generated, then g (ker f ) is a finitely generated subgroup of Z , hence is projective. From thisargument it suffices that any image of B in Z be finitely generated. Noticethat subgroups of Z need not have this property.What about a direct sum of two groups that allow cancellation of Z , suchas a direct sum of two subgroups of Z ? References [1]
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