# In a integral domain every prime element is irreducible

I'm trying to understand a proof of Hungerford's book which says that in a integral domain every prime element is irreducible:

I didn't understand why this implication $p=ab\implies p|a$ or $p|b$, is not the contrary $p=ab\implies a|p$ and $b|p$ ?

I'm a little confused

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If $p = ab$, then in particular $p$ divides $ab$, because $ab = p \cdot 1$. Since $p$ is prime, it has to divide either $a$ or $b$.

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Of course,thank you very much for your help –  user74141 May 29 '13 at 14:26
@user74141, you're welcome. –  Andreas Caranti May 29 '13 at 14:29
@user74141 One can make the inference prime $\,\Rightarrow\,$ irreducible more clear by using an alternative definition of irreducible - see my answer. –  Key Ideas May 29 '13 at 16:40
I don't understand the part, 1=xb hence b=1 (or b is a unit), how does it tie together, can you help? –  ciceksiz kakarot Nov 18 '13 at 19:10

Prime $\Rightarrow$ irreducible is clear if one employs a definition of irreducible in associate (vs. $\rm\color{#0a0}{unit}$) form. Then $\,\color{#c00}{ p=ab\,\Rightarrow\, p\mid ab}\,$ immediately yields the sought inference, as follows.

Theorem $\ \$ In the following, $\,\ (1)\,\Rightarrow\,(2)\!\iff\! (3)$

$(1)\ \ \ \color{#c00}{p\ \mid\ ab}\ \Rightarrow\ p\:|\:a\ \ {\rm or}\ \ p\:|\:b\quad$ [Definition of $\:p\:$ is prime]

$(2)\ \ \ \color{#c00}{p=ab}\ \Rightarrow\ p\:|\:a\ \ {\rm or}\ \ p\:|\:b\quad$ [Definition of $\:p\:$ is irreducible, in associate form]

$(3)\ \ \ p=ab\ \Rightarrow\ a\:|\:1\ \ {\rm or}\ \ b\:|\:1\quad$ [Definition of $\:p\:$ is irreducible, in $\rm\color{#0a0}{unit}$ form]

Proof $\ \ \ (1\Rightarrow 2)\,\ \ \ \color{#c00}{p = ab\, \Rightarrow\, p\mid ab}\,\stackrel{(1)}\Rightarrow\,p\mid a\:$ or $\:p\mid b.\$ Hence prime $\Rightarrow$ irreducible.

$(2\!\!\iff\!\! 3)\ \ \$ If $\:p = ab\:$ then $\:\dfrac{1}b = \dfrac{a}p\:$ so $\:p\:|\:a\iff b\:|\:1.\:$ Similarly $\:p\:|\:b\iff a\:|\:1.$

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I think the OP was asking why the first implication is there instead of the alternate implication he proposed. The answer is that the first implication is part of the definition of a prime. Look at the definition of prime that precedes the theorem in question in Hungerford: An element p of R is prime provided that: (i) p is a nonzero non unit; (ii) p|ab $\Rightarrow$ p|a or p|b.

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