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Extension Field : Kronecker's Theorem

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For your information,
Theorem 29.3 (Kronecker's Theorem):
Let F be a field and let f(x) be a non-constant polynomial in F[x]. Then there exists an extention field E of F and an alpha in E such that f(alpha) = 0

Theorem 29.18:
Let E be a simple extension F(alpha) of a field F, and let alpha be algebraic over F. Let the degree of irr(alpha, F) be n >= (greater or equal) 1. Then every element Beta of E = F(alpha) can be uniquely expressed in the form
Beta = b(sub)0 + (b(sub)1) (alpha) + ..... + b(sub) (n-1)) (alpha^(n-1))
where b(sub)i are in F

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Extension Fields are investigated using Kronecker's Theorem. The solution is detailed and well presented. The response received a rating of "5" from the student who originally posted the question.

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, then we note that , , . So is irreducible in .
a) Suppose has a root , ...

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