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Math Help - Algebraic number problem. Hard one°

  1. #1
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    Exclamation Algebraic number problem. Hard one°

    Let x be an algebraic number.

    Show that there is a single polynomial for x in Q[x] of minimal order that has x as a root. This polynomial is called minimal polynomial of x

    Show that every polynomial that has  x as a root, is divisible by the minimal polynomial of x

    Show that the minimal polynomial of x is irreducible in Q[x]



    Thank you very much in advance
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  2. #2
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    Quote Originally Posted by osodud View Post
    Let \color{red}a be an algebraic number.

    Show that there is a single polynomial for \color{red}a in Q[x] of minimal order that has \color{red}a as a root. This polynomial is called minimal polynomial of \color{red}a

    Show that every polynomial that has \color{red}a as a root, is divisible by the minimal polynomial of \color{red}a

    Show that the minimal polynomial of \color{red}a is irreducible in Q[x]



    Thank you very much in advance
    you shouldn't give the same name to your algebraic number and the indeterminate x. i changed the name of your algebraic number to a. since a is algebraic over \mathbb{Q}, there exists a polynomial over

    \mathbb{Q} which has a as a root. among all polynomials in \mathbb{Q}[x] which have a as root choose one with smallest degree. call it p(x). we may assume that p(x) is monic, i.e. the leading coefficient of p(x) is

    1, because, if not, then we could replace p(x) by \frac{1}{c}p(x), where c is the leading coefficient of p(x). now if q(x) is another polynomial which has a as a root, then by Euclidean algorithm there exist

    s(x), r(x) in \mathbb{Q}[x] such that either r(x)=0 or \deg r(x) < \deg p(x) and q(x)=s(x)p(x)+r(x) but then r(a)=q(a)-s(a)p(a)=0. so we can't have that \deg r(x) < \deg p(x), because we

    chose p(x) to be a polynomial with smallest degree satisfying the condition p(a)=0. so there's only one option left for r(x) which is r(x)=0. thus q(x)=s(x)p(x), i.e. p(x) \mid q(x).

    finally p(x) is unique, i.e. if q(x) \in \mathbb{Q}[x] is monic, q(a)=0, and \deg p(x)=\deg q(x), then q(x)=p(x). this is because, by what we just proved, we must have q(x)=s(x)p(x), for some

    s(x) \in \mathbb{Q}[x]. but since \deg p(x)= \deg q(x), we must have \deg s(x)=0, i.e. s(x)=k, a constant. so p(x)=kq(x). but then k=1 because both p(x), q(x) are monic. now we can give p(x) a

    name: we call it " the minimal polynomial of a over \mathbb{Q} ". note that the same result is true if we replace \mathbb{Q} by any field.
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