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5 Factorisation in Euclidean Rings

5.1 Ideals are principal

We recall a trivial fact and a definition:

Lemma 5.1.1. LetR be a ring and a R; then the set aR := {ar | r R} is an ideal of R.

Definition 1. Let R be a ring. The ideal I R is called principal if it is of the form aR for some a R.

Definition 2. A ring R is called a principal ideal ring if every ideal is principal; it iscalled a principal ideal domain if it is an integral domain and a principal ideal ring.

Here is the crucial property of Euclidean rings.

Theorem 5.1.2 (E.R. P.I.D.). Let R be a Euclidean ring; then R is a principal idealdomain.

Proof. Let IR. If I = {0} then I = 0R, so suppose I = {0}. Choose some a I with d(a)as small as possible.

As IR we clearly have that aR I.Suppose that x I; then x = qa + r for some q, r with r = 0 or d(r) < d(a). Now

r = x qa I + aR = I, so d(r) < d(a) is impossible. Hence x = qa aR.

5.2 Highest Common Factors

Theorem 5.2.1 (EDs have HCFs). Let R be a Euclidean domain, and let a, b R, with

b = 0. Then there exists a highest common factor d of a, b, expressible as d = ar + bs forr, s R.

Proof. We mimic in a suitably abstract way the Euclid algorithm proof for Z.So consider the ideals aR and bR of R. We can then form the ideal aR + bR. As every

ideal in R is principal (EDPID) we have that aR + bR = dR for some d. We claim that dis a highest common factor; clearly d = ar + bs.

For condition (i) note that aR aR + bR = dR, so that a = a1 dR, yielding a = da

for some a. Hence d|a; similarly d|b.For condition (ii) suppose that e|a and e|b. Then a = ex so that a eR. Similarly b eR.

Then for all r, s we will get ar + bs eR since eR is an ideal. That is dR = aR + bR eR.

Henced

=d

1 eR

, andd

=ed

for somed. That is,

e|d

.

As in the proof we used only the fact that R is a PID we have this more general result:

Corollary 5.2.2 (PIDs have HCFs). Let R be a Principal Ideal Domain, and let a, b R,with b = 0. Then there exists a highest common factor of a, b.

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5.3 Factorisation

Proposition 5.3.1 (Irreducible Prime in a PID). Let R be a Principal Ideal Domain,and let x R be irreducible. Then x is prime.

We first prove an important lemma.

Lemma 5.3.2 (Eulers Lemma). Suppose that h|ab and h, a have 1 as highest common factor. Thenh|b

Proof. We have that 1 = hr + as, and that ab = hk for some r,s,k. Then

b = b(hr + as) = bhr + bas = bhr + hks = (br + ks)h

and we are done.

Now we prove the proposition.

Proof. Now suppose that x is irreducible and that x|ab. Suppose that x |a. Let d be ahighest common factor ofx and a. Then d|x by definition, and so x = de for some e. As x isirreducible we have that d or e is a unit. Suppose that e is a unit, with ef = 1. Then d = xf,and so xf|a; then at once x|a, a contradiction. So we have that d is a unit, an associate of1. hence 1 is also a highest common factor, and then we are done by the lemma.

The following is an abstract version of the concrete fact that in Z we cannot go onfactorising a number into smaller and smaller pieces.

Proposition 5.3.3 (PID ascending ideal chains terminate). Let R be a Principal IdealDomain, and Ik R, and Ik Ik+1 for k = 0, 1, . . . . Then for some N, Ik = IN for all

k N.

Proof. Consider the subset I :=k

Ik; we prove that it is an ideal. Clearly 0 I0 I.Suppose that x, y I and a R; then x Ir and y Is for some r, s 0. Hence withn := max(r, s) we have that x, y In R. Then x, x + y, r x In I and we are done.

Now R is a PID, so we have that I = aRfor some a R. Then a = a1 aR = I =k

Ik,so that a IN for some N, and indeed a Ik R for all k N. Now we have thataR Ik I = aR and are done.

5.4 Unique Factorisation

This is an area where our intuition developed in Z and K[X] notoriously leads us astray inmore general domains.

Definition 3. We will say that the integral domain R is aUnique Factorisation Domain(UFD), or that it has unique factorisation, if

(i) every element 0 = a R is either a unit or can be written a = f1f2 . . . f M as a productof (a finite number of) irreducible elements f1, f2, . . . , f M R;

(ii) iff1f2 . . . f M = g1g2 . . . gN for irreducibles f1, f2, . . . , f M andg1, g2, . . . , gN thenM = N,and for each k = 1, . . . , M there exists a unit uk such that after relabelling the gk wehave gk = ukfk.

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As a trivial example, every field is a UFD. From our experience we know that Z is aUFD.

Theorem 5.4.1 (PID UFD). Let R be a Principal Ideal Domain. Then R is a UniqueFactorisation Domain.

Proof. There are two different things to be proved: the existence of factorisation into irre-ducibles and the uniqueness.

For the existence, consider 0 = x R, and suppose that we cannot express x in therequired form. Clearly x cant be a unit, so x = yz where neither y nor z is a unit. If bothy and z are expressible as products of irreducibles so is x; so one of these, without loss ycant be so expressed. Now put I0 := xR and I1 := yR. As x = yz we have that I0 I1.Moreover, if I1 = I0 we would have that y = y1 yR = xR, so that y = xa for some a.Then x = yz = xaz, and as x = 0 we have 1 = az and z is a unitwhich it is not. Sowe have found unequal ideals I0 I1. But we can repeat this process, starting now with y,and so construct eventually a properly ascending chain of idealswhich we have seen to be

impossible.As for uniqueness, we prove something slightly stronger. We prove that if uf1f2 . . . f M =

vg1g2 . . . gN for irreducibles f1, f2, . . . , f M and g1, g2, . . . , gN and units u, v then M = N andafter relabelling fk and gk are associates. We do this by induction on M + N, the resultbeing vacuously true when M + N = 0.

As f1 is irreducible it is prime, and so f1 divides some gk; relabel so that f1|g1 and g1 =f1u1 say. But g1 is irreducible, and so u1 is a unit. We therefore have that uu1f1f2 . . . f M =vu1g1g2 . . . gN = vf1g2 . . . gM, and so (as R is an integral domain) (uu1)f2 . . . f M = vg2 . . . gM.We can now apply the inductive hypothesis and get M 1 = N 1 and after relabelling theremaining fk and gk are associates.

Remark 5.4.2. We have shown the inclusions

{Euclidean rings} {Principal ideal domains} {Unique factorisation domains}

All these are proper inclusions. We will see in the exercises thatZ[X] is a UFD but not a

PID. It can be shown thatZ[1+192

] is a PID but it is not a Euclidean ring.

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Which of the following are true?

1. Z[X] is a UFD.

2. Let f(X), g(X) Q(X). Then there is some h(x) Q(X) such that< f(X) > + < g(X) >=< h(X) >.

3. Let f(X), g(X) Q(X). Then there is some h(x) Z(X) such that

< f(X) > + < g(X) >=< h(X) >.

4. Let a be a prime element in a PID. Then the ideal < a > is maximal.

5. IfF is a field and f : F[X] C is a homomorphism then ker f is a maximal ideal ofF[X].

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