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Let F be a field, E an extension of F, S a subset of E. If S = S1 union S2, then F(S) = F(S1)F(2). That is, F(S) can be obtained by adjoining the elements of S2 to F(S1). F(alpha)/F is finite if and only if alpha is algebraic over F. Let V be a vector space. If V has a finite set of generators, then we say that V is finite-dimensional and we define its dimension, denoted dimFV, to be the minimum possible number of elements in such a set. If V does not have a finite set of generators, then we say the V is infinite-dimensional, and we define dimFV to be infinity S is said to be linearly dependent if there exist distinct elements s1, ..., sn belonging to S and alpha1, ..., alphan belonging to F, not all 0 such that alpha1s1 + ...+ alphansn = 0. If S is not linearly dependent, then we say that S is linearly independent. Let V be a vector space over F. A basis of V is a subset {ei} of V (finite or infinite) with the property that every element v of V can be uniquely written in the form v = sum over ialphaiei. We say that alpha is algebraic over F if there exists a nonzero polynomial f element of F[X] such that f(alpha) = 0. If alpha is not algebraic over F, then we say that alpha is transcendental over F.

Algebraic Extensions

Let F be a field and let E be an extension of F. We say that E is an algebraic extension of F if every alpha element of E is algebraic over F. One of the main results of this section asserts that if alpha is algebraic over F, then F(alpha) is an algebraic extension of F.

Proposition 1: Let E/F be finite. Then E is algebraic over F.

Proof: Let alpha element of E, n = deg(E/F). Then the n + 1 elements

1, alpha, ..., alphan

of E must be linearly dependent over F, since dimF(E) = n. Therefore, there exist ci element of F (0 < i < n), ci not all 0, such that

cnalphan + cn-1alphan-1 + ... + c0 = 0.

Therefore, alpha is algebraic over F.

The next two results are elementary, but of critical importance.

Proposition 2: deg(E/F) = 1 if and only if E = F.

Proof: backward implication Obvious.

forward implicationSuppose that deg(E/F) = 1. Then we can choose a basis of E over F consisting of one element alpha, and every element beta element of E is of the form aalpha for some a element of F. In particular, 1 = a0alpha for some a0 element of F, so that we have alpha = a0-1 element of F. Therefore, aalpha element of F implies E subset of F. Thus since F subset of E, we have E = F.

Theorem 3: Let E subset of F subset of G be three fields. Assume that deg(F/E) and deg(G/F) are finite. Then deg(G/E) is finite and

deg(G/E) = deg(G/F) · deg(F/E).

Further, if {alpha1,...,alphan} is a basis of G over F and {beta1,...,betam} is a basis of F over E, then {alphaibetaj} (1 < i < n, 1 < j < m) is a basis for G over E.

Proof: Let x element of G. Then there exist elements ai element of F (1 < i < n) such that

(1)
x = sum over i aialphai.

However, since {beta1,...,betam} is a basis of F over E, for each i (1 < i < n), there exist elements bij element of E (1 < j < m) such that

(2)
ai = sum over j bijbetaj   (1 < i < m).

Combining (1) and (2) we see that

x = sum over isum over j bijbetajalphai.

Therefore, every element x element of G is expressible as a linear combination of the elements alphaibetaj. Moreover this expression is unique, since

sum over isum over j bijbetajalphai = sum over isum over j bij'betajalphai,
impliessum over isum over j cijalphaibetaj = 0    where cij = bij - bij'.

But

sum over isum over j cijalphaibetaj = 0 implies sum over i(sum over j cijbetaj)alphai = 0
implies sum over j cijbetaj = 0    (1 < i < n)

since the alphai are linearly independent over E. But then

cij = 0    (1 < i < n, 1 < j < m)

since the betaj are linearly independent over E. Therefore,

bij = bij'

for all i, j.

In what follows, let the elements alpha1,alpha2,...,alphan,beta be drawn from some extension E of F.

Corollary 4: If alpha1 and alpha2 are algebraic over F, then F(alpha1,alpha2) is of finite degree over F and is therefore algebraic over F.

Proof: By Theorem 5 of the section on algebraic elements, F(alpha1) is finite over F. Moreover, alpha2 algebraic over F implies that alpha2 is algebraic over F(alpha1). [For if alpha2 is the zero of a polynomial f element of F[X], then it certainly is the zero of some polynomial in F(alpha1)[X], namely f.] Therefore, by Proposition 1of the section on concepts, F(alpha1)(alpha2) = F(alpha1,alpha2) is finite over F(alpha1). Thus, by Theorem 3, F(alpha1,alpha2) is finite over F, so that by Proposition 1, F(alpha1,alpha2) is algebraic over F.

Corollary 5: If alpha1,...,alphan are algebraic over F, then F(alpha1,...,alphan) is of finite degree over F and is therefore algebraic over F.

Proof: Apply Corollary 4 and induction on n.

Corollary 6: Let alpha and beta be algebraic over F, then alpha ± beta, alpha · beta, alpha/beta (beta not equal 0) are algebraic over F.

Proof: F(alpha,beta) is algebraic over F, and alpha ± beta, alpha · beta, alpha/beta (beta not equal 0) belong to F(alpha,beta).