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If R is a commutative then we say that an ideal I in R is principal if there exists a R such that I = ar r R˙.Wecalla the generator of the and∈ we denote the ideal{ by| (∈a).If} R is an integral such that every ideal is principal then R is called a principal ideal domain (in Herstein p. 144 the definition of assumes the existance of an identity element Lemma 1 says that this is redundant). We will use the abreviation P.I.D. We have proved that a Euclidean ring is a principal ideal domain (see Theorem 3.7.1 p.144). Lemma 1. A principal ideal domain contains an identity element. Proof. R is an ideal in R so R =(u) with u R.Sinceu R we see that u = uc with c R.Ifa R then a = ub =(uc)b =(∈cu)b = c(ub)=∈ca.Thisc an identity ∈ ∈ (hence the identity). ¤ Assume that R is an with identity. If a, b R are non-zero and generate the same ideal then a b and b a thus b = va and a =∈ wb with v,w R.. Hence a = wva. This implies| that vw| =1. On the other hand of v ∈R is invertible then (a)=(va) for all a R. We therefore see ∈ ∈ Lemma 2. If R is an integral domain with identity and if a, b are non-zero elements of R then (a)=(b) if and only if there exists an invertible element v R with b = va. ∈ If R is an integral domain and if a, b R then we write a b if b = ua for some u R.Ifa, b R then a greatest common∈ divisor of a, b is an| element c R such that∈ c a and c ∈b and if d R and d a and d b then d c. ∈ | | ∈ | | | Theorem 1. Let R be a P.I.D. and let a and b be non-zero elements of R. Then there exists a of a and b in R whichisuniqueuptomul- tiplication by an invertible element. Furthermore, there exist s, t R such that sa + tb is a greatest common divisor of a and b. ∈ Proof. Suppose that c and d are greatest common divisors of a and b.Thend c and c d. Thus there exists an invertible element of R such that d = uc. Thisprovesthe| uniqueness| assertion. We will now prove the existance. Let I = ap+bq p, q R . We note that I is an ideal in R.SinceR is a P.I.D, I =(c) with{ c |R.Since∈ } R has an identity element, a, b I so c a and c b.Nowsupposethat∈d a and d b then a, b (d) so (c) (d) hence∈ d c. Finally,| since| c I there exist s, t| R such| ∈ ⊂ | ∈ ∈ that c = sa + tb. ¤ Let R be a P.I.D. then we will say that a, b are relatively prime if every greatest common divisor of a and b is invertible (see Exercise 1 below). Theorem 2. Let R be a P.I.D. and let a, b be non-zero relatively prime elements of R.Thenifa bc with c R then a c. | ∈ | Proof. Since a, b are relatively prime we may write an invertible element, u R, in the form u = as + bt.Writingp = s/u and q = t/u then p, q R and we∈ have 1=ap + bq. This implies that ∈ c =(ap + bq)c = apc + q(bc). 1 2

By assumption bc = va with v R.Thus ∈ c = apc + qav = a(pc + qv). Thus a c. | ¤ If R is an integral domain with identity then an element a R is said to be irreducible if it is not zero, not invertible and is such that if a = ∈bc R then b or c is invertible. ∈ Corollary 1. Let R be a P.I.D. and let a R be irreducible. If b, c R are such that a bc then a b or a c. ∈ ∈ | | | Proof. If a b there us nothing to prove. Suppose that a doesn’t divide b.Then if u a then| either u is invertible or u is an invertible element times a since a is assumed| to be irreducible. Thus if u divides both a, b then u must be invertible. Hence if a doesn’t divide b then a, b are relatively prime. The result now follows from Theorem 2. ¤ The next result is critical to our discussion since it implies that the procedures in the proofs of the main results in this addendum terminate after a finite number of steps.

Theorem 3. Let R be a principal ideal domain and assume that I1,I2, ..., In,In+1, ... are ideals in R.If I1 I2 ... In In+1 ... ⊂ ⊂ ⊂ ⊂ ⊂ then there exists no such that In = Ino for all n no. (That is increasing sequences of ideals stabilize.) ≥

Proof. Let I = In.Thenifa, b I then a Im and b In for some m, n thus if ∪ ∈ ∈ ∈ p is the larger of m and n we have a, b Ip so a b Ip I.ThusI is a subgroup ∈ ± ∈ ⊂ under addition. If a I then a In for some n thus since In is an ideal we have ∈ ∈ ar In for all r R.Thusar I for all r R.SinceR is a P.I.D., I =(ao). ∈ ∈ ∈ ∈ Now ao I so there exists no such that ao In .ThusI =(ao) In . Hence if ∈ ∈ o ⊂ o n no then I In In I. Hence In = In . ≥ ⊂ o ⊂ ⊂ o ¤

A such that increasing sequences of ideals stabilize is called a . These rings are the basic objects of study in algebraic geometry. Lemma 3. Let R be a P.I.D. and let a R be non-zero and not invertible. Then there exists an irreducible element b R∈such that b a. ∈ | Proof. If a is irreducible then the result is clearly true. If not then a = a1a10 with neither a1 or a0 invertible. If a1 is irreducible then a1 a and we are done. 1 | Otherwise, a1 = a2a20 with neither a2 nor a20 invertible. If a2 is irreducible then since a2 a1 and a1 a we see that a2 a and we are done. Otherwise we have a2 = a3a0 | | | 3 with neither a3 nor a30 invertible. Continuing this process then either we have proved the assertion after a finite number of steps or we have a = a1a0,a1 = a2a20 , ..., an = an+1an0 +1,... with neither ai nor ai0 invertible. Thus if Ii =(ai) we have I1 I2 ... In In+1 ... and Lemma 2 implies that Ii = Ii+1 for all i. This⊂ contradicts⊂ ⊂ the⊂ previous⊂ theorem. We conclude that the alternative6 to proving the assertion after a finitenumberofstepsisfalse. Thiscompletesthe proof. ¤ 3

Theorem 4. Let R be a P.I.D. and let a R be non-zero and not invertible. Then ∈ there exist irreducible elements p1, ..., pn R such that a = p1 pn furthermore, ∈ ··· if q1, ..., qm R are irreducible and such that a = q1 qm then m = n and we can ∈ ··· rearrange q1, ..., qn so that qi = uipi with ui invertible. Proof. We will divide the proof into two parts. First we will prove the product decomposition and second the uniqueness. We now begin the proof of the decom- position. If a is irreducible then take p1 = a and we have the desired product with n =1.Ifa is not irreducible then it is divisible by an irreducible element p1 so a = p1a1 and a1 is not invertible since a is not irreducible. The previous lemma now implies that a1 = p2a2.Ifa2 were invertible then a1 is irreducible and we can take p2 = a2 and we have a = p1p2.Otherwise,a2 = p3a3 with p3 irreducible. If a3 were a unit then we would be done. Continuing in this way we either find that after a finite number of steps that a = p1 pn or we have a ··· sequence a = p1a1,a1 = p2a2, ..., am = pmam, ... and pi is irreducible and ai is not invertible. We will now show that the latter is impossible. Indeed, it implies that if Ii =(ai) for i =1, 2, ... then Ii Ii+1 and since pi is irreducible for all i it follows ⊂ that Ii = Ii+1. ThiscontradictsTheorem3. Thisprovesthefirstpartofthe theorm.6 We now prove by induction on n that if a = p1 pn and a = q1 qm then ··· ··· n = m and after rearranging the qi we have qi = uipi.Ifn =1then a is irreducible. Thus if m>1 then we would have a = q1(q2 qm).Thiswould ··· imply that q2 qm is invertible. This is only possible if m =1and we have proved the result··· if n =1.Nowassumeforn and we prove for n +1.Thus we have a = p1 pnpn+1 = q1 qm.Nowp1 a so p1 q1(q2 qm).Sop1 q1 or ··· ··· | | ··· | p1 q2 qm.Ifp1 doesn’t divide q1 then we can argue the same way to see that | ··· p1 q2 or p1 q3 qm.Continuinginthiswaywefind that p1 qr for some r.Relable | | ··· | so that r =1.Sinceq1 isirreducibleweseethatq1 = u1p1 with u1 invertible. Now replace a by a/p1 = p2 pn+1 = u1q2 qm. Replace q2 with u2q2.The inductive hypothesis now applies··· and we see that··· m 1=n and after reordering we − have qi = vipi with vi invertible for i =2, ..., n +1. This completes the proof. ¤ Exercise 1. Let R be a P.I.D. If one greatest common divisor of a and b is invertible then a, b are relatively prime. Exercise 2. Let R be a Noetherian ring and let I be an ideal in I show that R/I is Noetherian. n Exercise 3. Let R = 2m n Z and m =0, 1, 2,... . Show that R is a of the field of rational numbers{ | that∈ is a P.I.D. Show} that the maximal proper ideals are the ideals (p)=pR where p is an odd prime in Z.