1998/99 GCW
Suppose that 
is a subset of a field 
. We can consider all the elements got from by repeated addition, subtraction, multiplication and division. These will constitute a subfield of , the smallest subfield containing . It is the intersection of all the subfields of that contain . Each element of this subfield can be expressed as a rational function in elements of , i.e. in the form
where 
and 
are polynomials in 
variables with integer coefficients and 
are elements of and is some positive integer. We say that the subfield is generated by .
If 
is a subfield of a field 
and 
is a set of elements of then we denote by
the subfield of 
generated by 
. If 
for some 
, we say that 
is a simple extension and that 
is a primitive element for this extension.
Suppose that is a subfield of a field and that is an element of . We say that is algebraic over if there is a nonzero polynomial 
, i.e. with coefficients in , such that
If is not algebraic over we call it transcendental over .
The element gives us an intermediate extension
It determines a homomorphism of rings 
taking a polynomial 
to 
, whose kernel consists of all the polynomials with coefficients in 
having as a root. Since 
is a principal ideal domain, this kernel has the form 
where 
is a polynomial in 
of least degree having as a root. If we specify to be monic, i.e. have leading coefficient equal to 1, then is unique. We call it the minimal polynomial of over . It must be irreducible over because if we had 
, so that 
, then either 
or 
contradicting the minimality of the degree of . We say that is algebraic of degree 
if its minimal polynomial has degree 
.
Proposition: Suppose that is a subfield of a field and that is an element of . Then is algebraic over if and only if 
is finite.
Proof: (
) Suppose that 
is finite. Then the infinite sequence of elements
cannot be linearly independent over 
. So there exists an integer and elements 
in 
such that
(
) Let be the minimal polynomial over of a, and let 
be such that 
. Then does not divide 
and since is irreducible over it must be coprime to 
. Hence, by the Euclidean Algorithm, we can find 
such that
It follows, by substituting for 
, that 
. In other words, any nonzero polynomial expression in with coefficients in has an inverse which is also a polynomial expression in with coefficients in . So any element of 
can be expressed in this form. Furthermore, if has degree , we may choose the polynomial expression to have degree less than , because, by the remainder theorem for polynomials, any polynomial 
can be written in the form
where either 
or the degree of 
is less than . Substituting 
for 
gives
Since the remainder is unique, we see that 
is a basis for 
as a vector space over .
Corollary: is algebraic over of degree if and only if 
.
Corollary: If 
is algebraic over with its minimal polynomial over then the homomorphism of rings 
which substitutes for 
induces an isomorphism
If is transcendental over then the homomorphism of rings 
which substitutes for 
induces an isomorphism
where 
denotes the field of fractions of 
.
We call an extension of fields 
algebraic if each element of is algebraic over . This does not mean that 
is finite. It means that 
is finite for each 
.
We say that an extension of fields 
is finite if 
is finite. Evidently, finite extensions are algebraic, but the converse is not true.
Theorem: A composite of algebraic extensions is algebraic.
Proof: Let 
and 
be algebraic extensions. Each element 
of 
is algebraic over 
. Let 
be the coefficients of the minimal polynomial over of 
. If we can prove that the extension
is finite, it will follow that its intermediate extension 
is finite. We may prove by induction on 
that 
is finite, as follows: first 
is finite, since 
is algebraic. For the inductive step, we note that
is finite, because the minimal polynomial of 
over 
has degree less than or equal to the minimal polynomial of 
over , and that the composite extension
is finite because of the inductive hypothesis and the multiplication theorem.
A field is called algebraically closed if every polynomial in 
of degree greater than zero can be written as a product of polynomials of degree one (i.e. linear factors).
Theorem: The following are equivalent for a field :
is algebraically closed.
of degree greater than zero has a root in .
is an isomorphism.
and element 
, if 
then 
is transcendental over .
is irreducible if and only if it has degree one.
Proof: (i) 
(ii) is trivial. (ii) 
(v) because an irreducible polynomial cannot have a root unless it has degree one. (v) 
(iii) because the minimal polynomials of elements of 
have to have degree one. (iii) 
(iv) is trivial. (v) 
(i) because every polynomial can be expressed as a product of irreducible polynomials.
We say that an extension 
is an algebraic closure of if
is algebraically closed, and
with 
algebraically closed has 
as an intermediate extension.
Theorem: Every field has an algebraic closure.
Proof: Let be a field. Let be a set in bijective correspondence with the set of all monic irreducible polynomials of degree at least two in 
. For each 
let 
be the corresponding irreducible polynomial. Let 
be the ring of polynomials with coefficients in in a finite number from a set of variables 
. We have a ring homomorphism 
sending an element of to the corresponding constant polynomial. Let 
be the ideal of generated by the set of elements 
. The axiom of choice asserts that there is a maximal ideal 
of containing 
. Let 
be the quotient ring 
. It is a field because 
is a maximal ideal of . We have an extension of fields 
given by the composite homomorphism 
. Let us denote by 
the image of 
in . Then 
is a root in of 
, since 
. As every polynomial of degree greater than zero in 
is a product of irreducible polynomials, and as every irreducible polynomial in 
acquires a root in we see that every polynomial in 
of degree greater than zero acquires a root in . Of course, is likely to have new polynomials that do not have any roots in . So we repeat this construction, and obtain recursively a sequence of extensions
where 
and 
. Every polynomial in 
of degree greater than zero has a root in 
. We have a directed family of fields, so we may define 
to be their union. Suppose we have a polynomial 
of degree greater than zero in 
. Each coefficient belongs to some subfield 
of 
. Hence there is an index 
such that 
belongs to 
and so 
has a root in the subfield 
of 
. So 
is algebraically closed.
Suppose we have an extension of fields with algebraically closed. For each 
we may choose a root 
of 
. Then we have a ring homomorphism 
defined by the condition that for each 
gets taken to 
. The kernel of this homomorphism contains 
. We choose the maximal ideal 
so that it contains this kernel. In consequence we get 
as an intermediate extension 
. Repeating this construction gives factorizations 
, and so a factorization 
.
The field 
of complex numbers is algebraically closed. The subring of algebraic numbers (those complex numbers which are roots of polynomials with rational coefficients) is the algebraic closure of 
, the field of rational numbers.
The point about algebraic closure is that every field can be considered a subfield of an algebraically closed field, and polynomials can always be split up into linear factors over that. So if 
is a monic polynomial of degree we know that there are elements 
such that 
in 
. We call
a splitting field of 
. It is unique up to isomorphism.