Polynomial ring over a field/One variable/Principal ideal domain/2/Fact/Proof

Proof

Let ${}I$ denote an ideal in ${}K[X]$ different from ${}0$. We consider the non-empty set of natural numbers

${\left\{\operatorname {deg} \,(P)\mid P\in I,\,P\neq 0\right\}}.$

This set has a minimum ${}m\in \mathbb {N}$. This number arises from an element ${}F\in I$, ${}F\neq 0$, ${}m=\operatorname {deg} \,(F)$. We claim that ${}I=(F)$.

The inclusion ${}\supseteq$ is clear. To prove the other inclusion ${}\subseteq$, let ${}P\in I$ be given. Due to fact, we have

$P=FQ+R{\text{ and with }}\operatorname {deg} \,(R)<\operatorname {deg} \,(F){\text{ or }}R=0.$

Because of ${}R\in I$ and the minimality of ${}\operatorname {deg} \,(F)$, the first case can not happen. Therefore, ${}R=0$, and this means that ${}P$ is a multiple of ${}F$.

To fact