Prime and Irreducible 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? 
 A: 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 $p$).
A: 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)?
A: 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
$$\rm irreducible\,\ q\, \iff\,  [\, q = xy\:\Rightarrow\: q\mid x\ \ or\ \ q\mid y\,]$$
