18 MATH 101A: ALGEBRA I PART B: RINGS AND MODULES5. Principal rings(Lecture by Alex Charis, notes by Andrew Gainer)Definition 5.1. Given a ring R, a norm is a function N : R → N∪{0}with N(0) = 0.Definition 5.2. A domain R is a Euclidean domain if there is a normon R such that, for all a, b ∈ R with b $= 0, there exists q ∈ R so thata = bq + r with r = 0 or N(r) < N(b).Example 5.3. (1) R = Z with N = | · |.(2) R = Z[i] with N = | · |.(3) R = F [x] with F a field and N = deg.Definition 5.4. Let A be a domain. An element a ∈ A with a $= 0 isirreducible if a = bc only if b or c is a unit in A.Proposition 5.5. Let a ∈ A be such that a $= 0 and (a) is prime.Then a is irreducible.Proof. Suppose a = bc. Then bc ∈ (a). So, one is in the ideal. Supposethat b ∈ (a). Then b is not a unit because (a) $= A. But b = ar. So,b = bcr. Since A is a domain, cr = 1 and c is a unit as required. !Definition 5.6. Let A be a domain and a ∈ A, a $= 0. Then a is saidto have a unique factorization into irreducibles if there exist a unit uand irreducibles piso that a = up1p2· · · prand, if a = vq1· · · qsfor v aunit and qiirreducible, then s = r and qi= uipiup to reordering.Remark 5.7. • If p is irreducible and u ∈ U(A), then up is irreducible.• By convention, for u ∈ U(A), u = u is a factorization into irre-ducibles (with r = 0).Definition 5.8. A domain A is a unique factorization domain (UFD)if every a ∈ A with a $= 0 has a unique factorization into irreducibles.Lang calls this a “factorial entire ring.”Definition 5.9. a divides b (denoted a|b) if there exists c ∈ A so thatac = b.Definition 5.10. If d ∈ A such that d $= 0 then d is a greatest commondivisor (gcd) of a and b if d|a, d|b and, if, for e $= 0, e|a, e|b, then e|d.Proposition 5.11. If A is a PID and a, b ∈ A with a, b $= 0 then c isthe gcd of a and b if (a, b) = (a) + (b) = (c).Proof. Since a, b ∈ (c) it is clear that c|a and c|b. Since c ∈ (a, b), wecan write c = va + sb. So, if d|a and d|b then c = vdx + sdy for somex, y ∈ R. So, d|c as required. !MATH 101A: ALGEBRA I PART B: RINGS AND MODULES 19[I reversed the order in the notes and put uniqueness first:]Lemma 5.12. If A is a PID, p is irreducible in A and a, b ∈ A withp|ab then p|a or p|b.Proof. Suppose p|/b. Then 1 = gcd(p, b). So, 1 = px + qb. So, a =pax + qab. So, p|a. !This lemma implies the uniqueness of factorization in a PID. If a =p1· · · pr= q1· · · qsthen p1|a implies p1|qifor some i. So, qi= up1.Then up2· · · pr= q1· · · !qi· · · qs. By induction on r it follows that r = sand the factorization is unique.Theorem 5.13. Let A be a PID. Then A is a UFD.Proof. Let S b e the set of nonzero principal ideals whose generatorsdo not have unique factorizations into irreducibles. Suppose S $= ∅.Consider a chain (a1) ! (a2) ! · · · which is as long as possible (orinfinite) and take its union. This union must be an ideal and is thereforeprincipal. So, we call its generator a. Then a ∈ (an). So, (a) = (an)and the chain is finite. So, the generator of any ideal strictly containing(a) has a factorization.Note that a cannot be irreducible (because then a = a is a factoriza-tion). So, we can write a = bc where b, c are not units. Then (b) " (a)and (c) " (a). So, b, c have factorizations. Then the product of thesefactorizes a and the factorization is unique by the lemma. So, (a) /∈ S,a contradiction. So, S = ∅ which means that every nonzero element ofA has a unique factorization.