Irreducible implies prime (PID)
Let be a principal ideal domain and an irreducible element in . Then, if , then or .
Suppose and . We need to prove that .
Then, the ideal generated by and is principal, and is generated by a factor of both and . Since is irreducible, the only possibility for this is the whole ring. Thus:
for suitable choices of .
Multiplying both sides by , we get:
Since , divides the entire right-hand-side, and hence .