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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 a1,...,an in F then k(a1,...,an) denotes the smallest field extension of k containing all the ai.

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)=xn+b1xn-1+...+b0. Now I claim that 1,a,a2,...,an-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 ar-nm(a)=0 we see that ar=-b1ar+n-1+...+b0ar-n. That is we can express ar in terms of lower powers of a. By induction we only need to use the spanning set indicated.

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