UFDs (Pt. II)
Point of Post: This is a continuation of this post.
What we’d now like to do is substantiate our claim that . The properness of any such containment is clear since as we have already mentioned is a UFD (still to be proven) which is not a PID. Thus, it clearly suffices to prove the containment. We do this in two steps corresponding to the two defining properties of UFDs. First we prove that irreducibility and primality are really the same thing in PIDs. Indeed:
Theorem: Let be a PID and be irreducible, then is prime.
Proof: Consider the ideal . We claim that from where it will follows that and so is prime. To see this suppose that , since is a PID we know that for some and so for some . Since is irreducible this implies that either is associate to , in which case , or is a unit in which case . Thus, no proper ideal properly contains and since we know that and so as claimed.
What we now show is that every element of a PID can be factored into a product of irreducibles. To do this we shall first need a lemma which says that every ascending chain of ideals in a PID eventually terminates (this is actually equivalent to the more general notion of a Noetherian ring (one in which every ideal is finitely generated)). Indeed:
Lemma: Let be a PID, then if is a ascending sequence of ideals in then is eventually constant (i.e. there exists such that implies ).
Proof: Since is a PID we know that for some . Let , we know that, being the union over a chain of ideals, is an ideal and so by assumption that is a PID we have that for some . Now, since is the union over the chain we have that for some and so from where it readily follows that . Now, if we know that and since this implies that . Thus, the chain is eventually constant.
From this we can prove that every element of a PID can be factored into irreducibles:
Theorem: Let be a PID and be non-zero and non-unit. Then, there exists irreducibles such that .
Proof: If is irreducible, we’re done, so assume not. Then, where . If and are irreducible, then we’re done. So, assume not, without loss of generality say that is not irreducible. Write . We see then that . If are all irreducible we’re done, so assume not, say . Continuing in this way we arrive at a chain of ideals . By the lemma we have to have that this chain is eventually constant, say this contradicts that is not irreducible. Thus, is irreducible. Continuing this process factors into irreducibles as desired.
Combining these two gives:
Theorem: Let be a PID, then is a UFD, but not conversely.
and a corollary:
Corollary: Let be a Euclidean domain, then is a UFD, but not conversely.
With this we can finish a proof from our last post on PIDs
Theorem: Let be a Euclidean domain with satisfying the -inequality and let . Then, is a greatest common divisor of if and only if and if then .
Proof: As remarked earlier it’s clear that if is a greatest common divisor then these properties hold. So, conversely suppose that and has maximal degree among all such common divisors. If is not a greatest common then there exists a such that . Now, since is a UFD we can factor and as and where and is irreducible. Now, since we know that and where and . Since we must have that for some and so is also a divisor of and since it’s equal to multiplied by an irreducible (and so non-unit) we know (since satisfies the -inequality) that which is a contradiction. Thus, . Since was arbitrary the conclusion follows.
 Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.
 Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.
 Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.
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