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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
c_{n}^{n} + c_{n1}^{n1} + ... + 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 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 {_{i}_{j}} Proof: Let
x = a_{i}_{i}.
However, since {_{1},...,_{m}} is a basis of F over E, for each i
a_{i} = b_{ij}_{j} (1 < i < m).
Combining (1) and (2) we see that
x = b_{ij}_{j}_{i}.
Therefore, every element
b_{ij}_{j}_{i} = b_{ij}'_{j}_{i},
c_{ij}_{i}_{j} = 0 where c_{ij} = b_{ij}  b_{ij}'.
But
c_{ij}_{i}_{j} = 0 ( c_{ij}_{j})_{i} = 0
c_{ij}_{j} = 0 (1 < i < n)
since the _{i} are linearly independent over E. But then
c_{ij} = 0 (1 < i < n, 1 < j < m)
since the _{j} are linearly independent over E. Therefore,
b_{ij} = b_{ij}'
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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