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Let $R$ be an integral domain, and let $r \in R$ be a non-zero non-unit. Prove that $r$ is irreducible if and only if every divisor of $r$ is either a unit or an associate of $r$.

Proof. ($\leftarrow$) Suppose $r$ is reducible then $r$ can be expressed as $r = ab$ where $a$, $b$ are not units. This contradicts the fact that every divisor of $r$ is either a unit or an associate of $r$.

($\rightarrow$) Suppose every divisor of $r$ is neither a unit nor an associate of $r$. Then $r$ is reducible.

Can someone verify if my proof is correct. However, I prefer a direct proof. This proof doesn't look nice. Can someone show me how I can do this directly?

Bill Dubuque
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2 Answers2

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To obtain easy "direct" proofs it is convenient to work directly with divisibility relations, $ $ in particular to rewrite "$a$ is a unit" as $\,a\mid 1.\,$ Then we can use standard divisibility properties such as this property: $ $ if $\,a\mid r\,$ then $\,r\mid a\color{#c00}{\iff} r/a\mid 1,\,$ by scaling or canceling $\,a.\,$ Combining this with $\,a\mid r,\ r\mid a\,\Rightarrow\, a,r\,$ are $\,\rm\color{#0a0}{associate},\,$ then it is mechanical to derive all of the common equivalent characterizations of "irreducible", namely:

$\begin{align} r\ \text{is irred}\!\!\overset{\rm def\!\!}\iff&\ \ r = ab\,\Rightarrow\, \quad\!a\mid 1\ \ {\rm or}\ \ \ b\mid 1\quad\text{i.e. iff $\,a\,$ or $\,b\,$ is a unit}\\ \iff&\ \ r = ab\,\Rightarrow\,\quad\! a\mid 1\ \ {\rm or}\ \, {\small \frac{r}a}\mid 1\\ \color{#c00}\iff&\ \ r = ab\,\Rightarrow\,\quad\! a\mid 1\ \ {\rm or}\,\ \ r\mid a\quad\text{i.e. iff every divisor $a$ is a unit or }\color{#0a0}{\rm associate}\\ \iff&\ \ r = ab\,\Rightarrow\, r/b\mid 1\ \ {\rm or}\,\ \ r\mid a\\ \color{#c00}\iff&\ \ \color{#0af}{r = ab}\,\Rightarrow\, \quad r\mid b\ \ {\rm or}\,\ \ r\mid a\quad \text{i.e. iff $\,r\,$ is }\,\color{#0a0}{\rm associate\,} \text{ to $\,a\,$ or $\,b\,$}\\[.2em] \text{compare to } &\ \ \color{#90f}{r\,\mid\, ab}\:\Rightarrow\, \quad r\mid b\ \ {\rm or}\,\ \ r\mid a\quad\text{i.e. definition of $\,r\,$ is $\rm\color{#90f}{prime}$} \end{align}$

Note the $\rm\color{#0af}{final}$ form of "irred" makes trivial the deduction that $\rm\color{#90f}{primes}$ are irreducible, i.e.

$\qquad\ \ \color{#0af}{r = ab}\,\Rightarrow\, \color{#90f}{r\mid ab}\overset{r\ \text{is prime}}\Rightarrow\ r\mid b\ \ {\rm or}\ \ r\mid a\,\Rightarrow\, r\,$ irreducible by the final form of irred

Bill Dubuque
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$\leftarrow$: You say that $a,b$ are not units. You should also mention that they are not associates of $r$.

$\rightarrow$: You should not start with "Suppose every divisor of $r$ is neither a unit nor an associate of $r$" This condition is never fulfilled! Please review how to negate qunators.

Actually, proper techniques of negating quantors would almost immediately show the ful claim.

Hagen von Eitzen
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