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Math Help - Field of Fractions and Extensions

  1. #1
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    Field of Fractions and Extensions

    Hello, I'd like some help regarding field extensions.
    Let L:K be a field extension. \alpha \in L is algebraic if and only if K[\alpha] = K(\alpha) \simeq F[X]/(f) where f is the minimal polynomial of alpha in K.
    Now before I even get to the F[X]/(f) I'd like some clarification about the equality between the polynomials in alpha and the rational functions in alpha.
    I know one inclusion is trivial; the other inclusion (showing that all non-zero elements in K[\alpha] are invertible) is fine except the proof I've seen states 'it is enough to show that the inverse of \alpha is in the ring'. I don't understand why this is enough? Generally K[\alpha] is not of the form k+t\alpha with k,t in K (right?) so how is showing \alpha ^{-1} is in the ring enough to show all the other inverses are in the ring?
    For the isomorphism I'm not sure what map to use..
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  2. #2
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    Quote Originally Posted by bleys View Post
    Hello, I'd like some help regarding field extensions.
    Let L:K be a field extension. \alpha \in L is algebraic if and only if K[\alpha] = K(\alpha) \simeq F[X]/(f) where f is the minimal polynomial of alpha in K.
    Now before I even get to the F[X]/(f) I'd like some clarification about the equality between the polynomials in alpha and the rational functions in alpha.
    I know one inclusion is trivial; the other inclusion (showing that all non-zero elements in K[\alpha] are invertible) is fine except the proof I've seen states 'it is enough to show that the inverse of \alpha is in the ring'. I don't understand why this is enough? Generally K[\alpha] is not of the form k+t\alpha with k,t in K (right?) so how is showing \alpha ^{-1} is in the ring enough to show all the other inverses are in the ring?
    For the isomorphism I'm not sure what map to use..

    It is enough because clearly K[\alpha] is an integral domain...

    Tonio
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  3. #3
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    Call me stupid, but I still don't understand
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  4. #4
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    Quote Originally Posted by bleys View Post
    Call me stupid, but I still don't understand
    Not stupid, but perhaps al little careless: what's missing an integral domain to be a field? Well then, what must be proved in order to show some int. domain is actually a field?

    Tonio
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  5. #5
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    Well, for any non-zero a in the integral domain, you must show a^{-1} is also in the integral domain. We've shown it for \alpha, what about the rest? How is knowing \alpha ^{-1} is in K[\alpha] imply any polynomial in \alpha is invertible?
    I'm not really sure where you're going; the best I could do is use induction on the degree of polynomials in K[\alpha], but I don't think that's the kind of proof you're (or the book is) thinking about, even though it is quite simple and short.
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  6. #6
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    Quote Originally Posted by bleys View Post
    Well, for any non-zero a in the integral domain, you must show a^{-1} is also in the integral domain. We've shown it for \alpha, what about the rest? How is knowing \alpha ^{-1} is in K[\alpha] imply any polynomial in \alpha is invertible?
    I'm not really sure where you're going; the best I could do is use induction on the degree of polynomials in K[\alpha], but I don't think that's the kind of proof you're (or the book is) thinking about, even though it is quite simple and short.

    I think it's simpler than that, but deeper from an abstract algebra point of view: look at the ring homom. \phi: K[x]\rightarrow K[\alpha]

    defined by \phi(f(x)):=f(\alpha) . Since K[\alpha] in an int. domain, <f(x)>:=\ker\phi is a prime ideal, but since

    K[x] is an Euclidean domain this prime ideal is in fact maximal (and thus f(x) is irreducible: the irr.

    pol. of \alpha , and thus K[\alpha] is a field, and since K[\alpha]\subset K(\alpha) , then...

    Tonio
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  7. #7
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    Alright, I see. It didn't occur to me to use the substitution homomorphism...
    Anyway, thanks for the help, Tonio!
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