Rapid Review Of Factorization In Integral Domains Euclidean Domains Pid Ufd Noetherian Domains

Rapid Review Of Factorization In Integral Domains Euclidean D It is important to notice that every euclidean domain is a pid and every pid is a ufd. however, as demonstrated by our examples, the converse of each of these statements fails. there are principal ideal domains that are not euclidean domains, and there are unique factorization domains that are not principal ideal domains (\({\mathbb z}[x]\)). Let r be a ufd in which every nonzero prime ideal is maximal. then r is a pid. lemma 1. if p is a nonzero irreducible element of r, then (p) is a prime ideal of r. proof. let a and b be nonzero elements of r such that ab ∈ (p), i.e. ab = pf for some nonzero element f ∈ r.

Domains Classes Ed Pid Ufd 6 Euclidean Domains Pids And Ufds We Now Ris a principal ideal domain (pid) if every ideal iof ris principal, i.e. for every ideal iof r, there exists r2rsuch that i= (r). example 1.10. the rings z and f[x], where fis a eld, are pid’s. we shall prove later: a principal ideal domain is a unique factorization domain. however, there are many examples of ufd’s which are not pid’s. Since z is a euclidean domain, it is automatically a pid and ufd by theorem 3.4.5, so that z [x] is a ufd is immediate from theorem 3.4.10. similarly, we know that q [x] is a euclidean domain, hence a ufd, so that q [x, y] = q [x] [y] is a ufd. it then follows from the exercise below that z [x] and q [x, y] are not pids, providing our examples. Ed implies pid implies ufd. theorem: every euclidean domain is a principal ideal domain. proof: for any ideal i, take a nonzero element of minimal norm b. then i must be generated by b, because for any a ∈ i we have a = b q r for some q, r with n (r) <n (b), and we must have r = 0 otherwise r would be a nonzero element of smaller norm than. Euclidean domains proposition if r is euclidean, then r is a pid. proof let i 6= 0 be an ideal and pick some b 2i with d(b) minimal. pick a 2i, and write a = bq r with either r = 0, or d(r) <d(b). this latter case is impossible: r = a bq 2i, and by minimality, d(b) d(r). therefore, r = 0, which means a = bq 2(b). since a was arbitrary, i = (b).