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Math Help - Proving Irreducibility

  1. #1
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    Proving Irreducibility

    Let k be a field, let K=k(t) be the function field of k.
    Let \alpha \in K and set \alpha = \tfrac{f(t)}{g(t)} in reduced terms, with \alpha \not \in k.
    Now construct F = k(\alpha).
    Note t\in k(t) solves the polynomial \alpha f(x) - g(x)\in F[x].
    This is a non-zero polynomial because otherwise it would contradict \alpha \not \in k.
    How do we prove this polynomial is irreducible over F?
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    Quote Originally Posted by ThePerfectHacker View Post
    Let k be a field, let K=k(t) be the function field of k.
    Let \alpha \in K and set \alpha = \tfrac{f(t)}{g(t)} in reduced terms, with \alpha \not \in k.
    Now construct F = k(\alpha).
    Note t\in k(t) solves the polynomial \alpha f(x) - g(x)\in F[x].
    This is a non-zero polynomial because otherwise it would contradict \alpha \not \in k.
    How do we prove this polynomial is irreducible over F?
    since k(\alpha) is the field of fractions of k[\alpha], by Gauss's lemma we only need to prove that \alpha f - g is irreducible over k[\alpha].

    suppose that \alpha f(x) - g(x)=a(x)b(x), for some a(x) \in k[\alpha][x], \ b(x) \in k[\alpha][x]. now look at a(x), \ b(x) as elements of

    k[x][\alpha]. see that \alpha is transcendental over k(x). now since, as an element of k[x][\alpha], the degree of \alpha f(x)- g(x) is 1, we

    must have a(x)=\alpha u(x) + v(x), and b(x)=w(x), for some u(x),\ v(x), and w(x) in k[x]. therefore u(x)w(x)=f(x) &

    v(x)w(x)=-g(x). hence w(t) must divide \gcd(f(t),g(t))=1. so w(t) \in k^{\times}. hence b(x)=w(x) \in k^{\times}. \ \ \square
    Last edited by NonCommAlg; May 22nd 2008 at 02:43 AM.
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    Quote Originally Posted by NonCommAlg View Post
    u(x)w(x)=f(x) &

    v(x)w(x)=-g(x).
    I think you mean, \alpha f(x) = u(x)w(x), but your argument still works. Thank you.

    It seems to me that the transcendental property of \alpha was very important over here. And I am guessing that is because: the ring generated by a transcendental element is isomorphic to the polynomial ring in a naturally manner, i.e. k[\alpha]\simeq k[x] by sending \alpha\mapsto x and building the isomorphism out of that. The crucial point is that this mapping is well-defined because if an element of k[\alpha] had two different representations in terms of \alpha then that would contradict the fact that \alpha is transcendental. Am I getting the right idea?

    Off topic question: why did you choose to study non-commutative algebra over field theory? Is it because it is more general?
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    Quote Originally Posted by ThePerfectHacker View Post
    I think you mean, \alpha f(x) = u(x)w(x), but your argument still works. Thank you.
    No! the correct one is what i wrote, i.e. f(x)=u(x)w(x). the reason is that from \alpha f(x)-g(x)=(\alpha u(x) + v(x))w(x)

    we'll get (f(x)-u(x)w(x))\alpha -g(x)-w(x)=0. but since \alpha \notin k(x), we get f(x)-u(x)w(x)=-g(x)-w(x)=0.

    Off topic question: why did you choose to study non-commutative algebra over field theory? Is it because it is more general?
    it's actually noncommutative ring theory not only algebras. why did i choose it? i guess because i think ring theory is interesting

    and the theory of commutative rings is too old and, well, too easy!
    Last edited by NonCommAlg; May 22nd 2008 at 05:50 PM.
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  5. #5
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    Quote Originally Posted by NonCommAlg View Post
    it's actually noncommutative ring theory not only algebras. why did i choose it? i guess because i think ring theory is interesting
    I forgot to ask, since this is related to what you do, is there such a thing as Galois theory over skew fields? More specifically let F\subseteq K be skew fields, is there a way to determine all skew fields E so that F\leq E\leq K?

    My other question is about the formal meaning of k[x][\alpha]. I understand it to mean as all polynomials in \alpha with coefficients in k[x], and this is okay because of transcendentalness. My problem is how do we formally define k[x][\alpha]? I never seen it used before.
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