UCLA ALGEBRA QUALIFYING EXAM Solutions
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چکیده
Proof. Let L/F be an algebraic extension. Let f : L −→ L be a homomorphism fixing F . Recall that field homomorphisms are always injective, it remains to show that it is surjective. Let a ∈ L. As L/F is algebraic, there exists a1, . . . , ad ∈ F such that a satisfy p(x) = x + a1xd−1 + . . .+ ad. Let S = {s ∈ L : p(s) = 0}. As f is a homomorphism fixing the coefficients of the polynomial p(x), if s ∈ S, then f(s) ∈ S as well. Thus we may consider f : S −→ S as a set map. Since it is still injective, as |S| ≤ d is finite, it is also surjective. Thus a is in the image of f , but as a is arbitrary, we are done. ¤
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