# $p$ is prime iff $\langle p\rangle$ is maximal (euclidean ring)

Let $\mathbb E$ an euclidian ring and $p\in \mathbb E$. Prove $p$ is prime iff $\langle p\rangle$ is maximal.

Proof:

As $\mathbb E$ is an euclidian ring (euclidean domain), then it is a PID (By Fraleigh theorem).

 Notice that for principal ideals: $\ {contains} = \ {divides}$,  i.e. $(a)\supseteq (b)\iff a\mid b,\,$ thus

$\qquad\quad\begin{eqnarray} (p)\,\text{ is maximal} &\iff&\!\!\ (p)\, \text{ has no proper } \,{{container}}\,\ (d), (d)\neq(1)\\ &\iff&\ p\ \ \text{ has no nonunit proper}\,\ {{divisor}}\,\ d\\ &\iff&\ p\ \ \text{ is irreducible}\\ &\iff&\ p\ \ \text{ is prime}\\\end{eqnarray}$

Note: this proof was based on Bill Dubuque's proof in another post for the case of $\mathbb{Z}$.

Is the proof correct??

• Looks good to me. – quasi Jul 14 '17 at 4:44
• @Weaam it's pretty similar, I actually followed his proof, but my proof has little details that requires for an euclidean ring and I posted it because I wanted to be 100% sure that it is correct. – user441848 Jul 14 '17 at 4:59
• @AnneliseToft Thanks for the edit. – Weaam Jul 14 '17 at 5:03
• Just one thing, you have to prove that in a Euclidean domain $p$ irreducible $\iff$ $p$ prime. – Xam Jul 14 '17 at 5:42
• @Xam how to prove it? – user441848 Jul 14 '17 at 16:47