abelian group is divisible if and only if it is an injective object
Proposition. Abelian group
A is divisible if and only if A is an injective object in the category of abelian groups.
Proof. ,,⇐” Assume that A is not divisible, i.e. there exists a∈A and n∈ℕ such that the equation nx=a has no solution in A. Let B=<a> be a cyclic subgroup generated by a and i:B→A the canonical inclusion. Now there are two possibilities: either B is finite or infinite.
If B is infinite, then let H=ℤ and let f:B→H be defined on generator by f(a)=n. Now A is injective
, thus there exists h:H→A such that h∘f=i. Thus
n⋅h(1)=h(1)+⋯+h(1)=h(1+⋯+1)=h(n)=h(f(a))=i(a)=a. |
Contradiction with definition of n∈ℕ and a∈A.
If B is finite, then let k=|B| (note that n does not divide k) and let H=ℤn⋅k. Furtheremore define f:B→H on generator by f(a)=n (note that in this case f is a well defined homomorphism). Again injectivity of A implies existence of h:H→A such that h∘f=i. Similarly we get contradiction:
n⋅h(1)=h(1)+⋯+h(1)=h(1+⋯+1)=h(n)=h(f(a))=i(a)=a. |
This completes first implication
.
,,⇒” This implication is proven here (http://planetmath.org/ExampleOfInjectiveModule). □
Remark. It is clear that in the category of abelian groups 𝒜ℬ, a group A is projective if and only if A is free. This is since 𝒜ℬ is equivalent to the category
of ℤ-modules and projective modules
are direct summands
of free modules
. Since ℤ is a principal ideal domain
, then every submodule of a free module is free, thus projective ℤ-modules are free.
Title | abelian group is divisible if and only if it is an injective object |
---|---|
Canonical name | AbelianGroupIsDivisibleIfAndOnlyIfItIsAnInjectiveObject |
Date of creation | 2013-03-22 18:48:15 |
Last modified on | 2013-03-22 18:48:15 |
Owner | joking (16130) |
Last modified by | joking (16130) |
Numerical id | 8 |
Author | joking (16130) |
Entry type | Theorem |
Classification | msc 20K99 |