# 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?

-
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 $d$? – Charles Boyd Nov 28 '12 at 17:20

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 $p$).

-
So p is prime if and only if (p) is a prime ideal. And p and x are associates if and only if (p)=(x) So (x) is a prime ideal (????) that implies x is also prime. Does it make sense? – ghet Nov 28 '12 at 17:31
Depending on how you define a prime element, this should be straightforward using the definition of prime ideal. – Rankeya Nov 28 '12 at 17:32
I used a theorem that says p and x are associates if and only if (p)=(x). Does it follow that (x) is a prime ideal since (p) is prime ideal? Then does it imply that x is prime? – ghet Nov 28 '12 at 17:38
Dear @ghet: If the two ideals are equal, then doesn't one of them having property $y$ imply the other also has property $y$, because they are equal? – Rankeya Nov 28 '12 at 17:43
Ok thanks so much. – ghet Nov 28 '12 at 17:45

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)?

-
ideal (d) or ideal (x). Since p and x are associates then (p)=(x). – ghet Nov 28 '12 at 18:12
Right, I meant (x). So (x) = (p), which is a prime ideal. Thus x generates a prime ideal, and x is prime. – andybenji Nov 28 '12 at 23:04

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

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

-