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Let $p$ be a nonzero prime element of an integral domain D. This means that whenever $p$ divides a product $ab$ with $a, b \in D$, it must divide $a$ or $b$. Show that $p$ is irreducible.

I tried to solve this question by assuming p is reducible, then there exists $a,b \in D$ such that $ab = p$.

$\Rightarrow p\vert ab$

Since p is a prime element

$\Rightarrow p\vert a$ or $p\vert b$

Assume $p\vert a $ WLOG then $a= pc $ for some $c \in D$

$\therefore p=ab =pcb$

$\Rightarrow p(1-cb) = 0 $

$\Rightarrow p = 0 $ or $bc =1$

Since p is nonzero, so bc=1. Then b and c are units.

After this I don't know how to prove it. Could you please help me to solve it?

Nhay
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    You began your proof with $p=ab$. What does it mean if $b$ is a unit? :) – Arkady Mar 24 '16 at 21:38
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    It's proved: the only possible decomposition of $p$ as a product of two elements is trivial (one of the factors must be a unit). – Bernard Mar 24 '16 at 21:40
  • To me, a prime element is by definition an irreducible element... – Surb Mar 24 '16 at 21:40
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    @Surb Consider 3 in $\mathbb{Z}_6$. 3 is prime in this ring, yet $3 = 3 \times 3 \times 3 \times \ldots$ Of course $\mathbb{Z}_6$ is not an integral domain. I don't remember where I got this example from. – Robert Soupe Mar 31 '16 at 04:17

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Your proof started out much like the proof of Theorem 1.21 in Alaca & Williams Introductory Algebraic Number Theory, but then you overcomplicated things with $a = pc$.

Let $p \in D$ be a prime and suppose that $p = ab$, where $a, b \in D$. As $ab = p \cdot 1$, we have $p \mid ab$, and so, as $p$ is prime, we deduce $p \mid a$ or $p \mid b$ ... Since $1 = a / p \cdot b$ or $1 = a / b \cdot p$, either $b$ is a unit or $a$ is a unit of $D$.

Robert Soupe
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