1

$I$ is irreducible ideal of $R$ when $R$ is a PID , I want to show it's a prime ideal.

The definition I am familiar with: $I$ a proper ideal of $R$ is an irreducible ideal if for any two ideals $J,K$ such that $I\subseteq J$, $I\subseteq K$ , then $I\neq J\cap K$

Well denoting $I=\langle a\rangle$ , Assuming $I$ is not a primes, means there are $x,y\in R$ , such that $xy\in\langle a\rangle$ such that $x\notin\langle a \rangle$ , $y\notin\langle a \rangle$.

Taking $\langle x \rangle \cap \langle y \rangle$ , we get an ideal such that $a\in \langle x \rangle \cap \langle y \rangle$ because $a=xy\in \langle x \rangle$ , $a=xy\in \langle y \rangle$. So $\langle a \rangle \subset\langle x \rangle\cap \langle y \rangle$.

For $r\in \langle x \rangle\cap \langle y \rangle$ I tried to show $r\in \langle a \rangle$ but didn't succeed, I gotthe next results:

  • $r = xd , r=ye \Rightarrow r^2=xyed=a\cdot ed\Rightarrow r^2 \in \langle a \rangle$
  • $r=xd , r=ye \Rightarrow ry=xyd=ad$ and$ rx=yxe=ae\Rightarrow rx,ry\in \langle a \rangle$.
user5721565
  • 1,390
  • 11
  • 23
  • The question in the link talks about $a\in$PID is irreducible $\Leftrightarrow$ $a$ is prime. Yet I am asking about an irreducible Ideal, not about an irreducible element. Can someone please explain the equivalence (if exists)? – user5721565 Jan 25 '19 at 04:37

0 Answers0