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.

Published by ayushkhaitan3437

Hello! My name is Ayush Khaitan, and I'm a graduate student in Mathematics.

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