# Contradiction in proof that in an integral domain, every prime is irreducible.

Let $\pi$ be a prime element in an integral domain. So, $\pi$ is a non-unit and if $\pi \mid ab \$ then $\pi \mid a$ or $\pi \mid b$.

An irreducible element $z$ is an element such that if $z=ab$, then $a$ or $b$ is a unit, but not both.

Let $\pi =xy$. So, $\pi \mid xy \implies \pi \mid x$ or $\pi \mid y$.

Assume $\pi \mid x$. We also have $x \mid \pi$. So, $\pi$ and $x$ are associates, which implies $\pi = xu$ where $u$ is a unit.

Thus, $\pi=xy=xu$. So, $y=u$ and therefore $y$ is a unit.

However, what if $\pi$ divides $x$ and divides $y$? Then we would have that both $x$ and $y$ are units which contradicts the definition of irreducible, that $\pi$ is not the product of two units.

• Why is it not possible? – user46372819 Apr 29 '15 at 5:28
• If $\pi \mid x$, then $y$ is a unit; if $\pi \mid y$, then $x$ is a unit. So if $\pi \mid x$ and $\pi \mid y$, then $x$ and $y$ are both units, hence $\pi = xy$ is a unit, contradicting the original assertion that it's a prime element. – mjqxxxx Apr 29 '15 at 5:29
• Makes sense. But what about Euclid's lemma: If a prime $p$ divides $xy$ then $p \mid x$ or $p \mid y$ or both. – user46372819 Apr 29 '15 at 5:36
• Yes, but you're also assuming that $p$ is equal to $xy$... this rules out the "both" case. – mjqxxxx Apr 29 '15 at 6:15

## 1 Answer

But $\pi$ divides $x$ or $y$, not both. What you have shown is that a prime element is not the product of two elements that are divided by the prime.

• Why can't it divide both? – user46372819 Apr 29 '15 at 5:27
• Euclid's lemma says that if a prime $p$ divides $xy$ then $p \mid x$ or $p \mid y$ or both. – user46372819 Apr 29 '15 at 5:29
• Precisely, because it would be an unit. As you have shown. – ajotatxe Apr 29 '15 at 5:31
• I don't understand what you wrote. – user46372819 Apr 29 '15 at 5:33