# Thread: Algebraic number problem. Hard one¡

1. ## 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

2. Originally Posted by osodud
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.