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Math Help - Field Theory - Nicholson - Section 6.2 - Exercise 31

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    Super Member Bernhard's Avatar
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    Field Theory - Nicholson - Section 6.2 - Exercise 31

    In Section 6.2 of Nicholson: Introduction to Abstract Algebra, Exercise 31 reads as follows:

    Let  E \supseteq F be fields and let  u \in E be transcendental over F.

    (a) Show that  F(u) = \{ f(u){g(u)}^{-1} \ | \ f,g \in F[x] ; g(x) \ne 0 \}

    (b) Show that  F(u) \cong F(x) where F(x) is the field of quotients of the integral domain F[x].

    (c) Show that every element  w \in F(u), w \notin F , is transcendental over F.

    Can someone help me approach this problem.

    Peter
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    Re: Field Theory - Nicholson - Section 6.2 - Exercise 31

    Since E is a field extension of F, there exists a function f_0(x) \in F[x] such that f_0(u) = 0. Let f_1(x) = f_0(x)+1. Let f_u(x) = x. Let f_k(x) = kf_1(x) for any k \in F. Essentially, what I am trying to do is find all elements of the form a+bu. I would think this is the way to start.
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    Super Member Bernhard's Avatar
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    Re: Field Theory - Nicholson - Section 6.2 - Exercise 31

    Thanks SlipEternal

    One quick question:

    You write: "Since E is a field extension of F, there exists a function  f_0(x) \in F[x] such that  f_0(u) = 0 ."

    But  u \in E is transcendental over F, so doesn't this mean that there is no polynomial  f_0(x) in F[x] such that  f_0(u) = 0 .

    Can you clarify?

    Thanks.

    Peter
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    Re: Field Theory - Nicholson - Section 6.2 - Exercise 31

    Right, I am getting tired. This should be much easier than that. So f_{a+bu}(x) = a+bx for any a,b \in F. Now, f_{a+bu}(u) = a+bu. Now through products of functions, you should be able to get any element of F(u)
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