I know I need to prove both ways since it is an if and only if statement. So if I say p is a prime element if p is an irreducible element. Can is say if p is a prime element I know p is an integral domain. It is finite so i can use the fact that all finite integral domains are fields, thus must be irreducible?

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    $\begingroup$ Is it ok to assume finite? $\endgroup$ – Nicole Mar 14 '15 at 21:56
  • $\begingroup$ what do you take as the definition of prime and irreducible elements? $\endgroup$ – Eoin Mar 14 '15 at 22:04
  • $\begingroup$ I think I was looking at it as a prime IDEAL so my logic is incorrect. Irreducible to me means there exists no 0,and thus is a field. Prime element is Z3, an example? (Thats Z superscript 3) $\endgroup$ – Nicole Mar 14 '15 at 22:07
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    $\begingroup$ You aren't totally off base. An element $p$ is prime $\iff$ $(p)$ is a prime ideal. An irreducible element, $q\neq 0$ and $q$ not a unit, is one such that whenever $q=ab$ then one of $a,b$ is a unit. The latter is usually taken to be the definition of an irreducible element. $\endgroup$ – Eoin Mar 14 '15 at 22:10
  • $\begingroup$ thanks so much I had not even thought about how units would come into play here, you've been very helpful! $\endgroup$ – Nicole Mar 14 '15 at 22:13

Finiteness is out of the question. You have to prove two statements:

  1. If $p$ is prime, then $p$ is irreducible

  2. If $p$ is irreducible, then $p$ is prime

The first statement is true on every domain. Hint: if $p=ab$, then obviously $p\mid ab$. Apply the hypothesis that $p$ is prime and conclude.

The second statement, instead, requires the hypothesis that $D$ is a principal ideal domain (although such a hypothesis could be relaxed).

Suppose $p$ is irreducible and that $p\mid ab$. We want to prove that either $p\mid a$ or $p\mid b$, so we can as well assume that $p\nmid a$ and prove that $p\mid b$.

Consider the ideal $pD+aD$; since $D$ is a principal ideal domain, we can write it as $pD+aD=cD$ for some $c\in D$. Since $p\in cD$, we have $c\mid p$ or $cd=p$ for some $d\in D$. Since $p$ is irreducible, either $c$ or $d$ is a unit.

Suppose $d$ is a unit; then $c=pd^{-1}\in pD$, so $cD\subseteq pD$ and therefore $aD\subseteq pD$, against the assumption that $p\nmid a$.

Thus $c$ is a unit. Now you should be able to go on.

Since $c$ is a unit, $pD+aD=D$, which means that $1=px+ay$ for some $x,y\in D$. Hence $b=1b=(px+ay)b=p(xb)+(ab)y$ and so $p\mid b$.


Hint $\ $ Primes are always irreducible since $\ p = ab\ \Rightarrow\ p\mid a\, (\Rightarrow b\mid 1)$ or $\ p\mid b\,(\Rightarrow\,a\mid 1)$
For the converse you can employ one of the proofs below.

Euclid's Lemma in Bezout form, gcd form and ideal form

$ \smash[t]{\begin{align}\\ \\ px\!+\!ay=&\,\color{#c00}1,\,\ p\ \mid\ ab\ \ \ \Rightarrow\, p\ \mid\ b.\ \ \ {\bf Proof}\!:\,\ p\ \mid\ pb,ab\, \Rightarrow\, p\,\mid pbx\!\!+\!aby\! = (\overbrace{px\!+\!ay}^{\large\color{#c00} 1}\!) b = b\\ (p,\ \ \ a)=&\,\color{#c00}1,\,\ p\ \mid\ ab\ \ \ \Rightarrow\, p\ \mid\ b.\ \ \ {\bf Proof}\!:\,\ p\ \mid\ pb,ab\, \Rightarrow\, p\,\mid (pb,\ \ ab) = (p,\ \ \ a)\ \ b =\, b\\ P\! +\!A\ =&\,\color{#c00}1,\, P\supseteq AB\, \Rightarrow P \supseteq B.\,\ {\bf Proof}\!:\! P \supseteq\! PB,\!AB\!\Rightarrow\!\! P\supseteq\! PB\!+\!\!AB =(P\!+\!A)B = B \end{align}}$


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