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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 $px = ab$.

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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 $p=ab$.

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  • $\begingroup$ By $p \neq ab$, I am referring to the case where $px = ab$. $\endgroup$
    – Navies
    Apr 16 '15 at 3:27
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    $\begingroup$ @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 $p$. $\endgroup$
    – user42912
    Apr 16 '15 at 3:31
  • $\begingroup$ 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 $x \in R$. $\endgroup$
    – Navies
    Apr 16 '15 at 3:33
  • $\begingroup$ @EdwardPoon yes, you are right. However, you want to prove that $p$ is irreducible. You already know that $p$ is prime. $\endgroup$
    – user42912
    Apr 16 '15 at 3:36
  • $\begingroup$ Right, you are correct. $\endgroup$
    – Navies
    Apr 16 '15 at 3:37
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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,

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    $\begingroup$ It would help to typeset your solution using MathJax. See this introduction. $\endgroup$ Apr 17 '17 at 18:47

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