abelian group is divisible if and only if it is an injective object
Proof. ,,” Assume that is not divisible, i.e. there exists and such that the equation has no solution in . Let be a cyclic subgroup generated by and the canonical inclusion. Now there are two possibilities: either is finite or infinite.
If is finite, then let (note that does not divide ) and let . Furtheremore define on generator by (note that in this case is a well defined homomorphism). Again injectivity of implies existence of such that . Similarly we get contradiction:
,,” This implication is proven here (http://planetmath.org/ExampleOfInjectiveModule).
Remark. It is clear that in the category of abelian groups , a group is projective if and only if is free. This is since is equivalent to the category of -modules and projective modules are direct summands of free modules. Since is a principal ideal domain, then every submodule of a free module is free, thus projective -modules are free.
|Title||abelian group is divisible if and only if it is an injective object|
|Date of creation||2013-03-22 18:48:15|
|Last modified on||2013-03-22 18:48:15|
|Last modified by||joking (16130)|