A field K
is called algebraically closed
if every polynomial
of positive degree
has a zero in K
. (Equivalently, such a polynomial must split into a product of linear
For example the fundamental theorem of algebra says that
C is algebraically closed. (Any proof of this
eventually boils down to the completeness of C.
proof I know uses Liouville's theorem. If f(z) is a
polynomial function on the complex numbers which vanishes nowhere
then its reciprocal must be bounded and holomorphic on the plane,
hence constant by Liouville.)
Despite its name this result is not especially important for algebra.
However, it is important to know that given a field K we can construct
an algebraic closure of K. So let me explain what this is.
For a field K an algebraic closure of K is a algebraic
field extension L of K that is an algebraically closed field.
That leaves two natural questions: can we construct algebraic
closures and are they unique? Here's the answer.
Given a field K there exists an algebraic closure L
of K. If M is also an algebraic closure of K
then there is an isomorphism of rings
f:L-->M such that f(a)=a for all a in K.
I'm going to show the construction of an algebraic closure.
First we need
If K<=L is a field extension then the set F of
elements of L which are algebraic over K form
a field extension of K.
Proof. Choose a,b in F.
By the field extension writeup [K(a):K]
is finite. Since b is algebraic over K it
is a fortiori algebraic over K(a). Thus again
we have that [K(a,b):K(b)] is finite.
By the lemma about dimensions of towers of field extensions in
the proof of the uniqueness of splitting fields we see that
[K(a,b):K] is also finite. It follows that every
element of this extension is algebraic over K. For take
any such element c. Then 1,c,c2,... cannot
be linearly independent over K. This gives a nonzero
polynomial in K[x] with c as a root.
It follows that a-b and ab and (for a not zero)
a-1 are all algebraic over K. Hence the result.
Proof that algebraic closures exist.
First we show that K is a subfield of an algebraically closed field.
The argument I give is due to Emil Artin.
For each polynomial f in K[x] of positive degree we associate
a variable Xf. Then form the polynomial
ring in all these variables R. (Note there will be infinitely
many variables here.) Now consider the ideal of R generated by
all f(Xf), where f runs through the polynomials
in K[x] of positive degree. I claim this ideal is proper.
If not we have an equation
1 = f(Xf1)g1 +...+ f(Xfn)gn
Now we can form L
a finite dimensional field extension of K
in which each fi
has a zero (see splitting field
. There are finitely many variables
involved in the gi
is not one of the fi
then we just put
At this point we can for substitute the variables in the equation with the
various ah. When we do that obviously the right hand side vanishes
and we get 1=0. This contradiction shows that the ideal generated
by the f(Xf) is proper. Hence there exists a maximal ideal
I of R that contains it. Thus M1=R/I is a field
and we have a natural injective ring homomorphism K-->M1.
By construction, in the field extension M1
of K we have that
each polynomial f in K[x] of positive degree has a zero.
Iterating we can form a chain of fields
M1 < M2 < ...
so that each polynomial in
of positive degree
has a zero in Mi+1
. Obviously the union M
these fields is itself a field extension of K
. Further, by
construction it is algebraically closed.
Finally, we apply the previous lemma to see that the collection F
of elements in M that are algebraic over K is an field
extension of K. It is also algebraically closed. For
suppose that f in F[x] is a polynomial
of positive degree. Then f has a root a in M, since
M is algebraically closed. But since this means
a will be algebraic over F and hence over K
we see that a is in F, as was needed to show F
See also proof that the algebraic closure of a field is unique