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I would like to prove that in an integral domain $R$, every prime element $p$ is irreducible. I understand the case where $p = ab$ but the textbooks I have read do not address the case where $p \neq ab$, i.e.,$px = ab$.

I was wondering why they do not discuss the case where $px = ab$.

Navies
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3 Answers3

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You don't need to examine the case $p\neq ab$.

In order to prove every prime element $p$ is irreducible you have to show IF $p=ab$, then $a$ or $b$ is an unit (see the definition of irreducible element).

However, if we have $p\neq ab$? It doesn't matter, we don't care, what matters to us is just the case whenever $p=ab$.

user42912
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  • By $p \neq ab$, I am referring to the case where $px = ab$. – Navies Apr 16 '15 at 03:27
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    @EdwardPoon Why do you like to analyse this case? you want to prove that $p$ is irreducible. If you prove the case for $px=ab$, you are proving $px$ is irreducible, instead of $p$. – user42912 Apr 16 '15 at 03:31
  • If i recall correctly, the definition of a non-zero and non-unit element $p$ is prime if $p | ab \implies p | a$ or $p | b$ and $p | ab \implies px = ab$ for some $x \in R$. – Navies Apr 16 '15 at 03:33
  • @EdwardPoon yes, you are right. However, you want to prove that $p$ is irreducible. You already know that $p$ is prime. – user42912 Apr 16 '15 at 03:36
  • Right, you are correct. – Navies Apr 16 '15 at 03:37
  • @EdwardPoon you have to use the fact $p$ is prime in the proof, but what you want to prove is $p$ is irreducible and to prove this you have to prove if $p=ab$, then $a$ or $b$ is unit. – user42912 Apr 16 '15 at 03:38
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Let $R$ be an integral domain, if $p \in R$ is prime, then $p$ is irreducible.

Put

$$p=ab.$$

It suffices to show $a$ or $b$ is a unit. As $ab \in (p)$ and $(p)$ is prime, either $a \in (p)$ or $b \in (p)$. So WLOG suppose $a \in (p)$ then $a=pr$ for some $r \in R$ and thus

\begin{align} p&=ab\\ &=rpb \end{align}

And since were in an integral domain this imples $rb=1$ thus $b$ is a unit and $p$ is irreducible.

MyMathYourMath
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Proof: suppose P is a prime element of an integral domain R, and p=ab we know that, P|P => P|ab => P|a or P|b P|a and p=ab or a|P => a is unit similarly P|b and p=ab ir b|p => b is unit, so either a is a unit or b is a unit, hence p must be irreducible element,

Mohammad Essa Nazari
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