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.