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

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 $b$.

• 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
• @utkarshk5, I subscribe to the persuasion that integral domains are defined to have a unity. I know some authors prefer otherwise - should have mentioned my preference explicitly. – Andreas Caranti Jan 11 at 15:34

Prime $$\Rightarrow$$ irreducible is obvious if we employ 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.$$

Beware that factorization theory is more complicated in non-domains. Basic notions such as associate and irreducible bifurcate into a few inequivalent notions. See here for more on that.

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.

• The link for the book is not working, can i ask exactly what book this was and the page number of the proof? – user531499 Feb 17 '19 at 14:13