Prove that $\beta$ is a root of a $n$ degree polynomial of K [x] with leading coefficient $1$.

62 Views Asked by At

Prove that $\beta$ is a root of a $n$ degree polynomial of K [x] with leading coefficient $1$. Here $K$ is UFD, $Q$ is its field of fractions, $Q(\beta)$ is algebraic extension of $n>1$ degree of the field of fractions, obtained by adjoining integral over $K$ element $\beta$.

1

There are 1 best solutions below

3
On

It should be a consequence of Gauss' Lemma.


That is, consider the monic polynomial $f$ of minimal degree in $K[x]$ which has $\beta$ as a root. This is clearly irreducible in $K[x]$, and thus, by Gauss' Lemma, also in $Q[x]$. So $f$ is the minimal polynomial of $\beta$ over $Q$, and thus $n = \lvert Q(\beta) : Q \rvert = \operatorname{degree(f)}$.