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Thread: Algebraic number problem. Hard one°

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

    Let $\displaystyle x$ be an algebraic number.

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

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

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



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

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

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

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



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

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

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

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

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

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

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

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