1

Question: Let $D$ be an integral domain and let $c\in D$ be irreducible in $D$. Show the ideal $(x,c)$ in $D[x]$ is not principal.

Thoughts: Since $c$ is irreducible in $D$, $c$ is noninvertible in $D$. Suppose $(x,c)$ is principal. Then, there exists some $d\in D$ such that $(x,c)=(d)$, and so $d\vert c$ and $d\vert x$. Since $c$ is irreducible, $d$ is either a unit, or $d=ux$, for some $u\in D$. I am trying to, eventually, get a contradiction, and I am assuming that I will be contradicting that $c$ is invertible by eventually getting that $c$ must be a unit, but I am having trouble getting there.

User7238
  • 2,474
  • trick: replace $D$ with $\mathbb{Z}$ and then try to translate the proof :) – Peter Müller Dec 30 '21 at 20:48
  • If $(x,c)$ is principal, you can find a $d\in D[x]$ (not in $D$!) such that $(x,c)=(d)$. It is true that necessarily $d\in D\subseteq D[x]$, but this needs an argument. Only then it makes sense to consider that $d\in c$ in $D$ and use the irreducibility hypothesis. – Thorgott Dec 30 '21 at 20:49
  • Apply the Lemma in the linked dupe (and recall that irreducibles are nonzero nonunits by definition). – Bill Dubuque Dec 31 '21 at 02:14
  • It suffices for $c$ to be a nonzero nonunit (not necessarily irreducible). In particular, if $D[x]$ is a PID, then $D$ must be a field. – Geoffrey Trang Dec 31 '21 at 04:00
  • @Geoffrey Yes, that is what is proved in the Lemma in the linked dupe. – Bill Dubuque Dec 31 '21 at 09:32

0 Answers0