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Math Help - Transcendental extension fields

  1. #1
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    Transcendental extension fields

    Hi, I'm having problems with the second part of this problem.

    Let F be an extension of K and a \in F transcendental over K.
    Let f \in K[x] with \text{deg } f >0. Show:
    a) f(a) is transcendental over K.
    b) If b \in F and f(b)=a, then b is transcendental over K.

    For part a, I assumed otherwise, so there exists a g \in K[x] such that g(f(a))=0, but g(f(x))=h(x) \in K[x] and h(a)=0 which contradicts a being transcendental over K.

    For part b, I'm not sure. I tried doing it by contradiction also, but I'm not getting anywhere. I feel like it's relatively simple, and I'm just forgetting something obvious. Any help would be appreciated! Thanks!
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  2. #2
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    Re: Transcendental extension fields

    Tell me if you have covered such a result or not but finite sums and products of algebraic elements are algebraic. So we can prove the contrapositive. Let b be algebraic. We need to show f(b) \neq a. To this end, f(b) is just a finite sum of products of b and the coefficients of f(x), that is f(b) is algebraic and cannot be transcendental, i.e. f(b) \neq a.
    Last edited by Vlasev; October 19th 2012 at 01:32 AM.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Re: Transcendental extension fields

    Another way. If b is algebraic over K, then [K(b):K] is finite and as a consequence K(b) is an algebraic extension of K. As a\in K(b), a would be algebraic over K (contradiction).
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    Re: Transcendental extension fields

    I have not encountered that result, although it seems to make sense (it would be frightening to think of transcendentals being the result of finite sums and/or products of algebraic elements).

    I like Fernando's proof though, and it's something I should have realized, it's just that the problem got my mind set on using functions.

    This was my solution:
    (Preliminary stuff)
    Notation: aK=\{ak:k \in K\} and a+K=\{a+k:k \in K\}
    If a \in K then a( F \setminus K) = F \setminus K
    Proof: Since a \in K, aK = K. Now, aF=F but aF = a(K \cup (F \setminus K)) = aK \cup a(F \setminus K) = K \cup a(F \setminus K), thus a( F \setminus K) = F \setminus K.
    Similarly, If a \in K then a + ( F \setminus K) = F \setminus K.
    Thus, if b \in F \setminus K, then bK=F \setminus K and b+K=F \setminus K.

    Let f(x)=\sum_{i=0}^n f_i x^i \in K[x], so f(b)=a.
    Now f(b) - f_0 = b \sum_{i=1}^n f_i b^{i-1} = a - f_0.

    Note that: a \in F \setminus K, b \in F \setminus K (because if it is in K then f(b) \in K which means a \in K which contradicts a being transcendental).
    So, using the preliminary stuff, a - f_0 \in F \setminus K implies that b \sum_{i=1}^n f_i b^{i-1} \in F \setminus K and since b \in F \setminus K, we have that \sum_{i=1}^n f_i b^{i-1} \in K.
    So, h(x) = (x - f_0)(\sum_{i=1}^n f_i b^{i-1})^{-1} \in K[x] and h(a)=b. Thus, by part (a), b is transcendental.

    Is there anything wrong with what I've done?
    What I think I basically did is:
    If \exists f \in K[x] such that f(b)=a, then \exists g \in K[x] such that g(a)=b.
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