Abstract
Abstract. We will prove that if G and H are modules over a principal ideal domain R such that the endomorphism rings End R
Introduction
The Baer-Kaplansky Theorem, [6, Theorem 108 .1], states that two primary abelian groups with isomorphic endomorphism rings are necessarily isomorphic. This statement was extended to various classes of modules (abelian groups), e.g. in [8] , [14] , [16] , [20] , [21] . However straightforward examples show that in order to obtain such extensions we need to impose restrictions on these classes. For instance the endomorphism rings of the Prüfer group Z(p ∞ ) and of the group of p-adic integers Z p are both isomorphic to the ring J p of p-adic integers. This fact suggests that we need to restrict to some good classes of modules in order to obtain a Baer-Kaplansky type theorem. Such a result (valid for torsion-free modules over valuation domains) was proved in [21] . It is well known that Baer-Kaplansky Theorem cannot be extended to torsion-free groups (of rank 1) since there are infinitely many pairwise non-isomorphic torsion-free groups of rank 1 whose endomorphism rings are isomorphic to Z, [1] . However, similar results to Baer-Kaplansky Theorem hold for some special classes of torsion-free groups, see [2] . In the setting of modules over complete valuation domains W. May proved a theorem, [15, Theorem 1] , for reduced modules which are neither torsion nor torsion-free and have a nice subgroup B such that M/B is totally projective: if M is such a module and N is an arbitrary module such that they have isomorphic endomorphism rings then
The main aim of this note is to prove a Baer-Kaplansky theorem for arbitrarily modules over principal ideal domains (Theorem 2): if R is (commutative) principal ideal domain then the correspondence (from the class of R-modules to the class of rings) Φ : G → End R (R ⊕ G) reflects isomorphisms of endomorphism rings. Moreover, this property characterizes principal ideal domains in the class of Dedekind domains: if R is a Dedekind domain such that the correspondence Φ reflects isomorphisms then R is a PID. The restriction to Dedekind domains is motivated by the fact that these domains have the cancellation property, i.e. the endofunctor R ⊕ − : Mod-R → Mod-R on the category of all R-modules reflects isomorphisms:
We need this property in order to obtain that Φ reflects isomorphisms (cf. Remark 6). However, in order to obtain such a correspondence which reflects isomorphisms the cancellation property is not enough, as it is proved in Proposition 7 (in contrast with the similar problem for subgroup lattices, approached in [3, Lemma 2]).
A Baer-Kaplansky theorem
The main result proved in this note is the following Theorem 2. Let R be a Dedekind domain. The following are equivalent:
(1) The ring R is a principal ideal domain;
Proof. (1)⇒(2) Let e and f be the idempotents in End R (G ′ ) which are induced by the direct decomposition G ′ = R ⊕ G. Using the version for principal ideal domains of [6, Theorem 106.1], we observe that there are isomorphisms
is an isomorphism then the idempotents e = ϕ(e) and f = ϕ(f ) induce a direct decomposition H ′ = B ⊕ K, where B = e(H ′ ) and
there is an isomorphism End R (B) ∼ = R. Moreover, as before, we have the isomorphisms (of R-modules)
and
We claim that B ∼ = R. Using this and Theorem 1 we obtain H ∼ = K, and we have
In order to prove our claim, suppose that B ≇ R. Let α : B → R be an Rhomomorphism. Since R is a PID it follows that Im(α) ∼ = R, hence Ker(α) = 0. Moreover, Im(α) is a projective module, hence we have a direct decomposition B ∼ = Ker(α) ⊕ Im(α). But End(B) ∼ = R has no non-trivial idempotents, hence B is indecomposable. It follows that Im(α) = 0, hence Hom R (B, R) = 0.
If we consider the direct decomposition H ′ = R⊕H and the canonical projection π R : H ′ → R, it follows that B is contained in H, the kernel of π R . From H ′ = B⊕K we obtain H = (H ∩ K) ⊕ B. Using the equalities
we deduce that K ∼ = R⊕(H ∩K) (as complements for the direct summand B), hence K has a direct summand isomorphic to R. Therefore Hom R (G, R) ∼ = Hom R (K, B) has a direct summand isomorphic to B. Since R is commutative, Hom R (G, R) is an R-module which can be embedded as a submodule in the direct product R G of copies of R (here we view R G as the set of all maps G → R, endowed with pointwise addition and scalar multiplication; see [5, Exercise 43.1] ). Therefore we can embed B in R G . Since B = 0 it follows that we can find a projection π : R G → R such that π(B) = 0. This implies Hom R (B, R) = 0, a contradiction, and it follows that B ∼ = R.
(2)⇒(1) Let I be a non-zero ideal in R. Since R is noetherian and integrally closed we can apply [7, Theorem I.3.7] to conclude that End R (I) ∼ = R. Moreover, since I is invertible, we can use Steinitz isomorphism formula, [7, p.165] . Therefore, for every positive integer n we have an isomorphism (⊕ n−1 k=1 R) ⊕ I n ∼ = ⊕ n k=1 I, hence there are ring isomorphisms
Using again the cancellation property of R, Theorem 1, we conclude that I n is principal for all n ≥ 2. If C(R) is the ideal class group associated to R and [I] is the class of I in this group, it follows that [I] n = 1 for all n ≥ 2, hence [I] = 1. Then I is principal and the proof is complete. As a consequence of Theorem 2 we obtain that locally free modules over principal ideal domains are determined by their endomorphism rings. This is also a consequence of [21, Theorem A].
Remark 3. From the above proof it follows that if R is a principal ideal domain then every ring isomorphism ϕ : End
R (R ⊕ G) → End R (R ⊕ H) induces a direct decomposition R⊕ H = B ⊕ K with B = ϕ(e)(R⊕ H) ∼ = R and (1 − ϕ(e))(R⊕ H) = K ∼ = G,
Corollary 5. If two locally free modules over a principal ideal domain have isomorphic endomorphism rings then they are isomorphic.
Remark 6. In the proof of Theorem 2 we used the cancellation property of the regular module R. If R has not this property (e.g. there are Dedekind-like domains without cancellation property, [11] ) then there are two R-modules G ≇ H such that R ⊕ G ∼ = R ⊕ H, hence End R (R ⊕ G) ∼ = End R (R ⊕ H). If we write these endomorphism rings as matrix rings
we observe that the (2, 1)-blocks in these representations are isomorphic to G, respectively to H. These two blocks are not isomorphic even the corresponding matrix rings are isomorphic. It is obvious that in this case Theorem 2 is not valid.
We will prove that we cannot replace in the implication (1) ⇒ (2) of Theorem 2 the direct summand R by an arbitrary module wich have the cancellation property. The following proposition shows that the property of the regular module R stated in Theorem 2 is more stronger than the usual cancellation property (see [18, Theorem B] ).
Proposition 7.
The following are equivalent for an indecomposable torsion-free abelian group F = 0 of finite rank:
(1) If G and H are abelian groups such that End(
Proof. (1)⇒(2) If F is not isomorphic to Z then F ∼ = Q or F is a reduced abelian group which has no free direct summands.
For the case F ∼ = Q, we can choose G and H two non-isomorphic subgroups of Q such that End(G) = End(H) = Z. It is not hard to see that both endomorphism rings End(F ⊕ G) and End(F ⊕ H) are isomorphic to the matrix ring Q 0 Q Z , so F does not verify the condition (1). If F is a reduced abelian group which has no free direct summands then we can construct, using [18, Theorem] , two (reduced) finite rank torsion-free groups G and H of the same rank such that Hom(F, G) = Hom(F, H) = Hom(G, F ) = Hom(H, F ) = 0 and End(G) = End(H) = Z.
In this case both endomorphism rings End(F ⊕ G) and End(F ⊕ H) are isomorphic to the ring End(F ) × Z, so F does not verify the condition (1).
(2)⇒(1) This is a consequence of Theorem 2.
Remark 8. There are also versions for the Baer-Kaplansky theorem proved for automorphism groups, Jacobson radicals or for ring anti-isomorphisms, [4] , [10] , [13] , [17] . It would be nice to know if Theorem 2 is still true if we consider only automorphism groups or Jacobson radicals.

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