The algebraic extension condition is important, otherwise algebraic closures aren’t unique, for example
Existence
Algebraic closures exist
Prerequisite lemma
Let be field.
There exists an extension such that every (non-constant) polynomial over has a root in .
(Note, this doesn’t say that all roots are in , just that at least one is).
Proof
Let be the set of monic, non-constant polynomials.
Form the polynomial ring
This is a HUGE ring.
It is a polynomial ring with infinite indeterminants that can be indexed based on polynomials in .
Moreover, it is an integral domain.
In symbols, this means we can write some polynomial
Then we can “plug in” the indeterminant .
Claim: is a proper ideal.
Assume that is not a proper ideal, in other words so
Let be an extension which contains roots of .
How to build this extension
It’s important to know that we can build this extension .
Consider the case for a single root:
We want to construct a field such that and is a root of .
We might as well assume is irreducible otherwise works just fine.
We can consider the field .
Note that is a perfectly good element, and
so is a root of .
This process can be done times to give the desired extension.
Then we can look at the equation
as living in as the identity.
We then plug in to give which is a contradiction.
Hence, is not a proper ideal, so by Zorn’s lemma, there exists a maximal ideal such that .
We can now construct the field
Thus, for any monic, non-constant polynomial , then .
Also and so in .
Therefore, the polynomial has a root (the image of ) in .
Proof of existence
The lemma means that we can always build a “bigger” field that has a root for every polynomial in .
The problem is, we cannot be sure that every polynomial in this bigger field has roots in .
Therefore, we can iterate the construction:
Take as base field, and apply the lemma to get .
This new field, may not have roots for all its polynomials, so do it again, to create .
Repeat.
Define the field
This field is algebraically closed since for any that is non-constant, then .
Hence, has a root in .
Lastly, we now know is algebraically closed, however it may not be an algebraic extension of .
Thus, we take the field
This is an algebraic closure of .
Uniqueness
Algebraic closures are unique up to isomorphism (but not unique isomorphism).
Necessary lemma
Let be an algebraic closure.
Let and be algebraic extensions.
There exists compatible -linear embeddings
such that the following diagram commutes
Intuition
This means that algebraic closures are “maximal” among algebraic extensions.
All other algebraic extensions can be embedded in them.
Proof
By definition is algebraic and is algebraic so by transitivity of algebraic extensions, is algebraic.
So the middle downward arrow is a direct consequence of this, and we may as well assume that .
So it suffices to construct an extension of to .
Consider the set
First, we can see that since .
Poset structure: We define the partial ordering if and only if and the following diagram commutes.
In order to use Zorn’s lemma, we must prove that is a poset for which every chain has an upper bound.
Consider a chain
then the element
for note that for then , so .
It is trivial that this is an upper bound for the chain.
Therefore, by Zorn’s lemma there is a maximal element to , we may call this element .
Claim:.
Note, by assumption since .
Let .
Since is algebraic, then is algebraic (by transitivity).
Let be the minimal polynomial for .
Consider
We know that since is an algebraic closure of then it is also an algebraic closure of .
So has a root in , call it .
Consider the homomorphism