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Thread: Proving Irreducibility

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    Proving Irreducibility

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

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

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

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

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

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

    It seems to me that the transcendental property of $\displaystyle \alpha$ 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. $\displaystyle k[\alpha]\simeq k[x]$ by sending $\displaystyle \alpha\mapsto x$ and building the isomorphism out of that. The crucial point is that this mapping is well-defined because if an element of $\displaystyle k[\alpha]$ had two different representations in terms of $\displaystyle \alpha$ then that would contradict the fact that $\displaystyle \alpha$ 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, $\displaystyle \alpha f(x) = u(x)w(x)$, but your argument still works. Thank you.
    No! the correct one is what i wrote, i.e. $\displaystyle f(x)=u(x)w(x).$ the reason is that from $\displaystyle \alpha f(x)-g(x)=(\alpha u(x) + v(x))w(x)$

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

    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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    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 $\displaystyle F\subseteq K$ be skew fields, is there a way to determine all skew fields $\displaystyle E$ so that $\displaystyle F\leq E\leq K$?

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