Definition A field K is called algebraically closed if every polynomial in K[x] of positive degree has a zero in K. (Equivalently, such a polynomial must split into a product of linear factors.)

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. The simplest 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.

Definition 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.

Theorem 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

Lemma 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), say afi. There are finitely many variables Xh involved in the gi above. If h is not one of the fi then we just put ah=0.

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 Mi[x] of positive degree has a zero in Mi+1. Obviously the union M of all 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 algebraically closed.