If every chain in a poset has an upper bound then the poset has a maximal element.
Definition (Partially ordered set)
A partially ordered set, or poset, is a set with a relation , i.e., for all , either or . The relation must satisfy
(Reflexivity) For all , .
(Antisymmetry) For all , if and then .
(Transitivity) For all , if and then .
Example: powerset
If is a set then , the powerset of , under inclusion is a poset.
Definition (Chain)
A chain in a poset is a totally ordered subset , i.e., for all then or (all elements are comparable).
Definition (Upper bound)
In a poset , is an upper bound for a subset if for all .
Definition (Maximal element)
In a poset , is maximal if for implies .
Exercise: Every vector space has a basis
Let be a vector space. Prove that has a basis.
Proof
We use Zorn’s lemma. Take under inclusion as the poset. We then check that the hypotheses of Zorn’s lemma are satisfied, and therefore a maximal element of is a basis (this needs to be checked).
Injective modules
Recall that there are two equivalent definitions of injective modules:
Definition 1 (Injective modules)
A module is injective if is exact.
Definition 2 (Injective modules)
A modules is injective if for all injective homomorphisms , and homomorphisms , there exists a homomorphism such that the following diagram commutes:
Theorem (Baer’s criterion)
Let be an algebra and be an -module. Then is injective if and only if for all left ideals of and homomorphisms , there exists such that the following diagram commutes:
Definition (Left ideal)
A left ideal of is a submodule of as a left -module.
Example:
For , only ideals are for . WLOG, we can assume as the case is trivial. We consider the diagram
So we only need to be “divisible by ” for all . For example, is an injective abelian group, but as are , , , etc.
#TODO2
Assume that satisfies Baer’s criterionBaer’s criterion. Let and be as in the definitiondefinition of injective (but we do not assume is injective). WLOG, we can assume that is a submodule by identifying with . Define the poset
with the partial order
i.e. is an extension of .
Every chain has an upper bound
Let be a chain. Let and let
Q: Why is is a module? A: Consider . Then there exist such that and , . Since is a chain, then either or . WLOG, assume that , and so , and . Closure under scalar multiplication is similar.
We can check that is well-defined #TODO (exercise). Now, if , then , therefore every chain has an upper bound.
Maximal element is an extension of
By Zorn’s lemma, has a maximal element. Let be a maximal element. If , then we’re done. Suppose . Then there exists . Fix this choice of and let , the submodule of generated by and . Explicitly,
Now consider the diagram
where is multiplication by on the right, defined , and
Our goal is to find an extension of to , or equivalently, to find . The question is what are the constraints? In fact, is the source of our constraints, for if , and therefore , we have
where is already known.
Claim: For any -module , there is a natural bijection
i.e., a map is equivalent to a way to map out of , , and a way to map out of , , such that the following diagram commutes: