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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
c n n + c n-1 n-1 + ... + c 0 = 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 { i j} (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 =  a i i.
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)
a i =  b ij j (1 < i < m).
Combining (1) and (2) we see that
Therefore, every element x G is expressible as a linear combination of the elements i j. Moreover this expression is unique, since
   c ij i j = 0 where c ij = b ij - b ij'.
But
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( , ).
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