Baer-Specker group is not free abelian: Difference between revisions
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==Facts used == | ==Facts used == | ||
# A slight variant of [[pure subgroup implies direct factor in torsion-free abelian group that is finitely generated as a module over the ring of integers localized at a set of primes]] (we need to relax the condition of finite generation | # A slight variant of [[pure subgroup implies direct factor in torsion-free abelian group that is finitely generated as a module over the ring of integers localized at a set of primes]] (we need to relax the condition of finite generation) | ||
# [[uses::Homomorphism from Baer-Specker group to group of integers that is zero on restricted direct product is zero]] | # [[uses::Homomorphism from Baer-Specker group to group of integers that is zero on restricted direct product is zero]] | ||
Latest revision as of 23:02, 1 July 2017
Statement
The Baer-Specker group, which is defined as the external direct product of countably many copies of the group of integers, is not a free abelian group.
Facts used
- A slight variant of pure subgroup implies direct factor in torsion-free abelian group that is finitely generated as a module over the ring of integers localized at a set of primes (we need to relax the condition of finite generation)
- Homomorphism from Baer-Specker group to group of integers that is zero on restricted direct product is zero
Proof
Proof follows by combining (1) and (2).