Algebraic closure

Let be a field. An algebraic closure of is an algebraically closed field such that is an algebraic extension. #m/def/field Assuming AC,¹ an algebraic closure always exists and is unique up to isomorphism of field extensions, so one often speaks of the algebraic closure.

Proof of existence and uniqueness

The proof of existence and uniqueness requires enough lemmata to warrant a section of this Zettel.² We invoke Zorn's lemma.

Existence

Let be a field. There exists an extension such that every nonconstant polynomial has at least one root in .

Proof due to Emil Artin

Let denote the set of nonconstant monic polynomials in , and let be the corresponding polynomial ring, potentially in infinitely many indeterminates. Consider the ideal

which we will show must be proper. Suppose towards contradiction that , so

for some and . We can then construct an extension where the polynomials have roots , by iteratively Adjoining a root to a field. If we evaluate

which is a contradiction. Since is proper, invoking Zorn it is contained in a maximal ideal , giving the the field extension

where by construction every nonconstant monic (and thus nonconstant general) polynomial has a root .

To guarantee the existence of all roots we iterate this process ad infinitum, so not only does have a root , but has a root , &c. This yields a chain of extensions

Let be the union or limit of this chain. is an algebraically closed field, and thus is an algebraic closure of .

Proof

For every we have for some , so we can just work within whatever is necessary, since the result is independent. Thus is a field. If , then for some , and thus it has a root in .

Uniqueness

See Embedding an algebraic extension into an algebraically closed field.

Suppose and are algebraic closures of . Then there exists an isomorphism of field extensions .