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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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Hint: By definition of associate, your $d$ must be a unit. So, you do not want to show that $d$ is prime. What you want to show is that if $x|ab$ then $x|a$ or $x|b$, and $x$ is not a unit.

An alternative way would be to observe that an element $p$ is prime if and only if $(p)$ is a prime ideal. Hence, show that $(x)$ is a prime ideal (this should follow very easily from the fact that $x$ is an associate of $p$).

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    You don't have to do the other approach if you know that if $(p)$ is a prime ideal then $(x)$ is a prime ideal, hence $x$ is prime.2012-11-29
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An element of your ring is prime if and only if it generates a prime ideal. Now you know that (p) is a prime ideal. What do you know is true about the ideal (d)?

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    Right, I meant (x). So (x) = (p), which is a prime ideal. Thus x generates a prime ideal, and x is prime.2012-11-28
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Hint $\, $ unit $\rm\,u,\, $ prime $\rm\, p,\,$ $\rm\ up\mid xy\:\Rightarrow\:p\mid xy\:\Rightarrow\:p\mid x\ \ or\ \ p\mid y\:\Rightarrow\:up\mid x\ \ or\ \ up\mid y\:\Rightarrow\: up\:$ prime.

Alternatively: $\ $ Note $\rm\: (up) = (p)\:$ hence $\rm\:D/(p) = D/(up),\ $ therefore

$\rm\begin{eqnarray} domain\ D/(p)&\iff&\rm \ domain\ D/(up)\\ \rm hence\ \ prime\,\ p&\iff&\rm\ prime\,\ up\end{eqnarray} $

Remark $\ $ Irreducibilty follows similarly to the hint since

$\rm irreducible\,\ q\, \iff\, [\, q = xy\:\Rightarrow\: q\mid x\ \ or\ \ q\mid y\,]$