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.

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Graduate student

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