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Let $\mathbb{F}[x]$ be a polynomial ring over a field $\mathbb{F}$. Let $I \subset \mathbb{F}[x]$ be an ideal, that is,

$\forall s,t \in I: s-t \in I$ and $\forall r \in \mathbb{F}[x]: rs,sr \in I$.

Prove that $\exists c(x) \in \mathbb{F}[x] : I = \{c(x)g(x): g(x) \in \mathbb{F}[x]\}$.

Any hints as to which $c(x) \in \mathbb{F}[x]$ I should pick and how I should proceed? Also my notes highlighted this result as particularly important... why is this? Thanks!

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    Hint: what can you say about the degree of $c(x)$ vs $c(x)g(x)$. – Michael Burr Dec 31 '19 at 10:15
  • So I should pick $c \in I$ with least degree? –  Dec 31 '19 at 10:16
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    That should work (depending on how you define degree for $0$). – Michael Burr Dec 31 '19 at 10:17
  • Yes, so $c$ of least non-zero degree? –  Dec 31 '19 at 10:21
  • Yes, that's a good choice. Further, use that the remainder has less degree than the dividend. – Berci Dec 31 '19 at 10:26
  • Ah nice, so it's just double inclusion + euclid's algorithm, thanks! Also, do you know why we care about this result in the first place? –  Dec 31 '19 at 10:27
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    @user736948 This result is quite important: it states that $\mathbb{F}[x]$ is what is called a "principal ideal domain" (PID).Finite type modules over PIDs have a neat structure theorem (this gives, for instance, a classification of finite type abelian groups when applied to $\mathbb{Z}$, but also "Frobenius Reduction" when applied to $\mathbb{F}[x]$). –  Dec 31 '19 at 10:42

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