Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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 $p$).

share|cite|improve this answer
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)?

share|cite|improve this answer
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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.