### Ah. Primes, irreducible elements, and Number Theory

Proving that primes are irreducible in any integral domain is simple.

Assume that a prime $p$ is not irreducible. Then $p=ab$, where $a$ and $b$ are not units. Now we know that $p|p$. Therefore, $p|ab$. Let $p|a$. Then $a=pc$, where $c\in R$. Now we have $p=pcb$. This implies $cb=1$, which is a contradiction as $b$ cannot be a unit.

But proving that every irreducible is a prime too requires more specialized conditions. Say $a$ is an irreducible such that $a|bc$. Also assume that $a\not|b$. If euclidean division is valid in the ring, then $(a,b)=R$. If every ideal has to be a PID, then $(a,b)=R$.

This might not be an exhaustive list of conditions for which $(a,b)=R$ if $a$ and $b$ are co-prime, but this is all we have right now.

Anyway, if $(a,b)=R$ and $a|bc$, then $ax+by=1$. Multiplying by $c$ on both sides, we ultimately get $(c)\subset (a)$, which proves $a|c$. This is valid for both PIDs and Euclidean domains. Note that the commutativity of the domain makes all this possible.