# Let $R$ be an integral domain. If $x \in R$ is prime, then $x$ is irreducible.

I am trying to understand the proof for the following theorem:

Let $$R$$ be an integral domain. If $$x \in R$$ is prime, then $$x$$ is irreducible.

Here is the proof:

I typed this a while ago and I don't understand the part where if $$x | bc$$, then $$x=bc$$? Is something wrong at this step?

Also, is the definition of prime elements where $$p$$ is prime if whenever $$p|ab$$, then either $$p|a$$ or $$p|b$$? Now I don't feel so sure. This could be the reason why I am not understanding the proof...

You have it reversed. What you seek is the inference $$(1)\Rightarrow(2)$$ below.

Theorem $$\,\ (1)\,\Rightarrow\,(2)\!\iff\! (3)\$$ below,  for a nonunit $$p\neq 0$$

$$(1)\ \ \ \color{#c00}{p\ \mid\ ab}\ \Rightarrow\ p\:|\:a\ \ {\rm or}\ \ p\:|\:b\quad$$ [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.$$

I think you were trying to make the argument that if $$x =bc$$, then one of them must be a unit (and you meant $$x=bc$$ in your first line.)

Then, it definitly follows that $$x \mid bc$$, and since this is a prime, then $$x \mid b$$ or $$x \mid c$$, and the rest works.

An element is irreducible if it is not the product of two non units.

The proof must be suppose that $$x$$ is the product of two non units, $$x=bc$$, since $$x$$ is irreduclible, $$x|b$$ or $$x|c$$, suppose that $$b=rx$$, $$x=rxc$$ implies that $$(1-rc)x=0$$ since $$A$$ is integral, $$1=rc$$ and $$c$$ is a unit. Contradiction.

• One should exclude units (invertibles) in the definition of an irreducible to be consistent with standard conventions. – Bill Dubuque Nov 4 '18 at 16:47