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Algebraic ExtensionsLet F be a field and let E be an extension of F. We say that E is an algebraic extension of F if every Proposition 1: Let E/F be finite. Then E is algebraic over F. Proof: Let
1, , ..., n
of E must be linearly dependent over F, since
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 Proof: Obvious.
Suppose that
Theorem 3: Let
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} Proof: Let
x = aii.
However, since {1,...,m} is a basis of F over E, for each i
ai = bijj (1 < i < m).
Combining (1) and (2) we see that
x = bijji.
Therefore, every element
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 Proof: By Theorem 5 of the section on algebraic elements,
Corollary 5: If 1,...,n are algebraic over F, then Proof: Apply Corollary 4 and induction on n.
Corollary 6: Let and be algebraic over F, then Proof: |
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