1

$\textbf{Theorem}$: If $R$ is a principal ideal domain, then $p$ is prime $\iff$ $p$ is irreducible.

$Proof:$ $(\Longleftarrow)$ If $p$ is irreducible, then $(p)$ is maximal in the set of all proper principal ideals of $R$. Since $R$ is an integral domain, $(p)$ is prime. Then, $p$ is prime.

However, it seems that I did not use PID, could anyone help me check what's my problem, thanks in advance.

$\textbf{Edit:}$ In an integral domain $R$, $c$ is irreducible $\iff$ $(c)$ is maximal in the set $S$ of all proper principal ideals of $R$.

$Proof$:$(\Longrightarrow)$ If $c$ is irreducible, then $(c)$ is a proper ideal in $R$. If $(c) \subset (d)$, then $c=dx, x \in R$. $(Case1):$ If $d$ is a unit, then $(d)=R$. $(Case2):$ If $x$ is a unit, then $c$ and $d$ are associates. which means $(a)=(b)$.

$(\Longleftarrow):$ If $(c)$ is maximal in the set $S$ of all proper principal ideals, then $c$ is a nonzero nonunit in $R$.If $c=ab$, then $(c)\subset (a) \implies (c)=(a)$ or $(a)=R$. $(Case1):$ If $(a)=R$, then $a$ is a unit. $(Case2):$ If $(c)=(a)$, then $a=cr_1, r_1 \in R.$ $c(r_1b-1_R)=0 \implies r_1 b=1_R$, $i.e.$ $b$ is a unit in $R$

Gerry Myerson
  • 185,413
Bowei Tang
  • 3,763
  • https://math.stackexchange.com/questions/770731/irreducible-elements-in-a-pid-are-prime – Yathi May 19 '24 at 12:01
  • @Yathi I have read it, but I don't know why we need PID since without PID (only integral domain), $(p)$ is still maximal in the set of all proper principle ideals. – Bowei Tang May 19 '24 at 12:08
  • If $R$ is PID, then $\langle p \rangle$ would be a maximal ideal of $R$ and hence a prime ideal (Every maximal ideal is prime). – Yathi May 19 '24 at 12:13
  • @Yathi But in the book, ($c$ is irreducible $\iff$ $(c)$ is maximal in the set $S$ of all proper principle ideals of $R$) this is also true when $R$ is an integral domain – Bowei Tang May 19 '24 at 12:17
  • @Yathi I have edited my problem and provide a proof for that. – Bowei Tang May 19 '24 at 12:20
  • 2
    Yes. But you cannot say that if $I$ is maximal among the proper principal ideals of $R$, then$I$ is a prime ideal. You can say that if $I$ maximal among all the proper ideals, then $I$ is prime. – Yathi May 19 '24 at 12:20
  • @Yathi Thank you for your patience, I get it! – Bowei Tang May 19 '24 at 12:32
  • Glad that it helped. – Yathi May 19 '24 at 14:02

0 Answers0