every PID is a UFD
The first step of the proof shows that any PID is a Noetherian ring in which every irreducible is prime. The second step is to show that any Noetherian ring in which every irreducible is prime is a UFD.
We will need the following
The second part of the lemma follows since if are two such gcd’s, then , so and so that are associates. ∎
Let be a chain of (principal) ideals in . Then is also an ideal. Since is a PID, there is such that , and thus for some . Then for each , . So satisfies the ascending chain condition and thus is Noetherian.
To show that each irreducible in is prime, choose some irreducible , and suppose . Let . Now, , but is irreducible. Thus either is a unit, or is an associate of . If is an associate of , then so that and is a unit. If is itself a unit, then we can assume by the lemma that . Then so that there are such that . Multiplying through by , we see that . But and . Thus so that is a unit. In either case, is prime. ∎
If is Noetherian, and if every irreducible element of is prime, then is a UFD.
We show that any nonzero nonunit is is expressible as a product of irreducibles (and hence as a product of primes), and then show that the factorization is unique.
Let be the set of ideals generated by each element of that cannot be written as a product of irreducible elements of . If , then has a maximal element since is Noetherian. is not irreducible by construction and thus not prime, so is not prime and thus not maximal. So there is a proper maximal ideal with , and .
Since is maximal in , it follows that and thus that is a product of irreducibles. Choose some irreducible ; then and
for some . If (note that this includes the case where is a unit), then and hence is a product of irreducibles, a contradiction. If then (since ). since is not a unit, and thus . This contradicts the presumed maximality of in . Thus and each element of can be written as a product of irreducibles (primes).
The proof of uniqueness is identical to the standard proof for the integers. Suppose
where the and are primes. Then ; since is prime, it must divide some . Reordering if necessary, assume . Then where is a unit. Factoring out these terms since is a domain, we get
We may continue the process, matching prime factors from the two sides. ∎
|Title||every PID is a UFD|
|Date of creation||2013-03-22 16:55:51|
|Last modified on||2013-03-22 16:55:51|
|Last modified by||rm50 (10146)|