Let

*k* be a

field. If

*F* has

*k* as a

subfield
we say that

*F* is a

field extension of

*k*.

For some examples see the subfield writeup but here is one
that is important in Galois theory.
Suppose that we have that *F* is a field extension of *k*
and that *a* is an element of *F*. Then we can form the
smallest field extension of *k* (inside *F*) that contains
*a*. This is written *k(a)*. Another description of this
field extension is

*k(a)={ f(a)/g(a): f(x),g(x)* are polynomials over *k* and *g(a)* is nonzero }

More generally, if *a*_{1},...,a_{n} in
*F* then *k(a*_{1},...,a_{n}) denotes the
smallest field extension of *k* containing all the *a*_{i}.

**Theorem**
Suppose that *a* is algebraic over *k* with minimal polynomial *m(x)*. Then

**Proof:**
We define a map from the polynomial ring

*G:k[x]->k(a)*
by G(f(x))=f(a). It's easy to see that

*G* is a

ring homomorphism.
As

*a* is

algebraic over

*k* then

*G* will be

surjective.
For if

*h(a)* is nonzero then

*h(x)* has no common factor
with

*m(x)*. Since

*h(x)* and

*m(x)* are coprime
this means that by

Euclid's algorithm we can find polynomials

*r(x),s(x)* such that

*1=m(x)r(x)+h(x)s(x)*.
Now evaluate this equation at

*a* and we get

*1=h(a)s(a)*. Thus the typical element of

*k(a)* which is

*f(a)/g(a)* can be written as

*f(a)s(a)*
which is in the image of

*G*.

The kernel of *h* will
be the ideal of *k[x]* generated by *m(x)*.
For if *f(a)=0* then *f(x)* is a multiple of *m(x)*
(see the minimal polynomial writeup for details).
By the first isomorphism theorem this says that *k(a)* is
isomorphic to *k[x]/(m(x))*.

Write *m(x)=x*^{n}+b_{1}x^{n-1}+...+b_{0}. Now I claim
that *1,a,a*^{2},...,a^{n-1} are a basis
for *k(a)* as a *k*-vector space.
It follows from this
that the dimension of *k(a)* as a *k*-vector space
is the degree of *m(x)*. So let's prove the claim. Firstly
these powers of *a* are linearly independent. For any linear
dependence between them wil give us a nonzero polynomial over
*k* which has smaller degree then *m* but has *a*
as a root. This contradicts the definition of minimal polynomial.
So we have to show that these powers span. Take *r>n-1*.
Then since *a*^{r-n}m(a)=0 we see that
*a*^{r}=-b_{1}a^{r+n-1}+...+b_{0}a^{r-n}. That
is we can express *a*^{r} in terms of lower powers of
*a*. By induction we only need to use the spanning set indicated.