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In Integral Domain, D, every associate of an irreducible [resp. prime] element of D is irreducible [resp. prime].

  • I am done with irreducible part.

  • For prime, I am stuch with this idea. So if p is prime, let say x is an associate of p then p=xd for some d in D. Since p is prime, then p|x or p|d. We need to show that d is prime. How?

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    Remember that if two ring elements $a,b \in R$ are associates, then $a = u b$ for some unit $u \in R$. So, use the definition of associate elements which makes your element $x$ necessarily a unit. Now how will the prime $p$ divide a unit? Then how must $p$ divide $d$?2012-11-28

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