Let F be a field, E an extension of F, S a subset of E. If S = S1 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()/F is finite if and only if 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 S is said to be linearly dependent if there exist distinct elements s1, ..., sn belonging to S and 1, ..., n belonging to F, not all 0 such that 1s1 + ...+ nsn = 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 = iei. We say that is algebraic over F if there exists a nonzero polynomial f F[X] such that f() = 0. If is not algebraic over F, then we say that 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 E is algebraic over F. One of the main results of this section asserts that if is algebraic over F, then F() is an algebraic extension of F.

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

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

1, , ..., n

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

cnn + cn-1n-1 + ... + c0 = 0.

Therefore, 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: Obvious.

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

Theorem 3: Let E F 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 {1,...,n} is a basis of G over F and {1,...,m} is a basis of F over E, then {ij} (1 < i < n, 1 < j < m) is a basis for G over E.

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

(1)
x = aii.

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

(2)
ai = bijj   (1 < i < m).

Combining (1) and (2) we see that

x = bijji.

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

bijji = bij'ji,
cijij = 0    where cij = bij - bij'.

But

cijij = 0 ( cijj)i = 0
cijj = 0    (1 < i < n)

since the i are linearly independent over E. But then

cij = 0    (1 < i < n, 1 < j < m)

since the j are linearly independent over E. Therefore,

bij = bij'

for all i, j.

In what follows, let the elements 1,2,...,n, be drawn from some extension E of F.

Corollary 4: If 1 and 2 are algebraic over F, then F(1,2) is of finite degree over F and is therefore algebraic over F.

Proof: By Theorem 5 of the section on algebraic elements, F(1) is finite over F. Moreover, 2 algebraic over F implies that 2 is algebraic over F(1). [For if 2 is the zero of a polynomial f F[X], then it certainly is the zero of some polynomial in F(1)[X], namely f.] Therefore, by Proposition 1of the section on concepts, F(1)(2) = F(1,2) is finite over F(1). Thus, by Theorem 3, F(1,2) is finite over F, so that by Proposition 1, F(1,2) is algebraic over F.

Corollary 5: If 1,...,n are algebraic over F, then F(1,...,n) is of finite degree over F and is therefore algebraic over F.

Proof: Apply Corollary 4 and induction on n.

Corollary 6: Let and be algebraic over F, then ± , · , / ( 0) are algebraic over F.

Proof: F(,) is algebraic over F, and ± , · , / ( 0) belong to F(,).