Chapter 4 Factorization

Chapter 4

Factorization

01/22/19

Throughout this chapter, R stands for a commutative ring with unity, and D stands for an integral domain, unless indicated otherwise.

The most important theorem in this course, after the Fundamental The- orem of Galois Theory, can be brieﬂy stated as follows.

Theorem 4.0.1 Field ED PID UFD ID ) ) ) )

In other words, every field is an Euclidean Domain; every Euclidean Do- main is a Principal Ideal Domain; every Principal Ideal Domain is a Unique Factorization Domain; and every Unique Factorization Domain is an Inte- gral Domain. So far, we only have the definition for the first, third, and last of these expressions. In the following sections we will give the appropriate definitions of the other terms, prove each of the implications, and show, by counterexample, that all implications are strict. See Propositions 4.1.2, 4.2.1, 4.4.3, as well as Examples 4.1.1.2, 4.4.1.2, 4.4.2 and Exercise 4.6.1.

11 12 CHAPTER 4. FACTORIZATION 4.1 Euclidean Domain

Theorem 4.1.1 [Gallian 16.2] let F be a ﬁeld, and F [x] the ring of polynomials with coecients in F . For any f(x),g(x) F [x] with g(x) =0, there are unique q(x),r(x) F [x] such that 2 6 2 f(x)=q(x)g(x)+r(x)andr(x)=0or deg(r(x)) < deg(g(x).

Proof.

Deﬁnition 4.1.1. An Euclidean Domain consists of an integral domain D and a function : D 0 N,satisfyingthefollowingconditions: For any f,g D withg{=0,thereexist}! q, r D s.t. 2 6 2 f = gq + r and, either r =0, or (r) <(g). (4.1)

Notes 4.1.1.

The function is referred to as division function,ormeasure,ordegree.

Some authors require the function to satisfy (a) (ab)foralla, b D 0 . However, it can be shown that this extra requirement is superﬂuous.2 { } The q and r are not required to be unique. In fact, in some EDs they are not unique, as we will see below in Example ??.

Examples 4.1.1. 1. Z with (n)= n . | | 2. Theorem 4.1.1 above shows that for a ﬁeld F ,theringF [x]ofpolyno- mials over F with (f)=deg(f)formsanEuclideanDomain.

Proposition 4.1.2 If F is a ﬁeld, then it is an Euclidean Domain, with (a)=1for all a =0. 6

Note that the function is never used here, since the residue is always 0. In other words, a ﬁeld has exact division.

We get the following corollaries from Theorem 4.1.1. 4.2. PID 13

Corollary 4.1.3 Let F be a field, a F , and f(x) F [x]. Then f(a) is the multiplicity remainder in the division of f(x) by2x a. 2 Corollary 4.1.4 Let F be a field, a F , and f(x) F [x]. Then a is a root or a zero of f(x) i↵ x a is a factor2 of f(x). 2 Definition 4.1.2. Let F be a field, a F , f(x) F [x], and m N0.We say that a is a root of f(x)ofmultiplicity2m,ifthereis2 g(x) F [x]suchthat2 f(x)=(x a)mg(x), and (x a)isnotafactorofg(x). In2 other words, the m multiplicity of a as a root of f(x)isthelargestm N0 such that (x a) is afactoroff(x). To say that a is a root of multiplicity2 0, simply means that a is NOT aroot. Theorem 4.1.5 [Gallian 16.3] A polynomial of degree n over a field F has at most n zeros, counting multiplicity.

01/23/19

Proposition 4.1.6 The ring of Gaussian integers

Z[i]= m + ni C m, n Z { 2 | 2 } with (m + ni)=m2 + n2 = N(m + ni), is an Euclidean Domain.

Proof.

4.2 Principal Ideal Domain

Deﬁnition 4.2.1. A Principal Ideal Domain (PID) is an integral domain in which every ideal is principal, i.e. generated by a single element.

Example 4.2.1. Z is an PID.

We get more examples of PIDs from the following proposition. Proposition 4.2.1 [Gallian 18.4] If D is an Euclidean Domain, then it is a PID. Moreover, every non-zero ideal I is generated by a non-zero element of I with smallest value.

Version 2019.5.9 14 CHAPTER 4. FACTORIZATION

Proof. Let I E D.IfI = 0 then I = 0 . Assume I = 0 .Let0={a} I,suchthath i (a) is smallest among non-zero elements of6 I{. } 6 2 Claim: I = a . Clearly a I.Letb I.SinceD is an ED there exist q, r D suchh thati h i✓ 2 2 b = aq + r and, either r =0, or (r) <(a).

Since a, b I,wethatr = b aq I. Minimality of (a)forcer =0,so b = aq a2 . 2 2h i We get from this proposition and Examples 4.1.1 that in addition to Z, the ring Z[i]ofGaussianintegersisaPID,andforanyﬁeldF ,theringF [x] of polynomials over F is also a PID.

Corollary 4.2.2 R[x]/ x2 +1 C h i⇡

Proof. Consider the evaluation map

evi : R[x] C f(x) ! f(i) 7! It is an epimorphism, and its kernel is generated by a non-zero element of 2 lowest degree. Clearly evi(x + 1) = 0, and the kernel has no elements of degree 1. Now apply the ﬁrst isomorphism theorem.

Deﬁnition 4.2.2. Let a, b D.Wesaythata divides b if there is c D such that b = ac,i.e. b a2 .Wemayalsosaythata is a divisor of b2,or that b is a multiple of a.2h i

Lemma 4.2.3 Let a, b, b ,b ,c,r R. 1 2 2

1. a a | 2. a 0 | 3. 1 a | 4. a b and b c a c | | ) | 4.3. IRREDUCIBLE AND PRIME ELEMENTS 15

5. a b ,ab a (b + b ) | 1 | 2 ) | 1 2 6. a b a rb | ) | 7. b R a b is an ideal of R. In fact, it is non other than a . { 2 | | } h i

01/25/19

Deﬁnition 4.2.3. a, b are said to be associates in D if a b and b a.Wewrite a b. | | ⇠ Note 4.2.1. Note that the binary relation “is associate of”isanequivalence relation. We’ll use the expression “up to associates”torefertoelementsin the same equivalence class. For example, in Lemma 4.4.7 we will prove that greatest common divisors are unique, up to associates.

Lemma 4.2.4 Let a R. 2

1. a 0 i↵ a =0. ⇠ 2. a 1 i↵ a is a unit. ⇠ Proposition 4.2.5 Let a, b D. TFAE: 2

1. a, b are associates in D

2. there is a unit u D⇤ such that a = ub 2 3. a = b h i h i

Proof. (1) (2) (3) (1) ) ) )

4.3 Irreducible and Prime Elements

Recall that for p Z,notzero,not 1, TFAE: 2 ±

1. p = ab a = 1orb = 1. ) ± ± Version 2019.5.9 16 CHAPTER 4. FACTORIZATION irreducible 2. p ab p a or p b,i.e.ab p a p or b p . prime | ) | | 2h i) 2h i 2h i

We extend these two ideas to an arbitrary ID D.Ingeneraltheyarenot equivalent, like they are in Z. We have an example of that in Examples 4.3.1 below.

Deﬁnition 4.3.1. Let p D be a non-zero non-unit element (nznu for short). 2

1. We say p is irreducible if

p = ab a is a unit, or b is a unit. ) 2. We say p is prime if

p ab p a or p b, i.e. ab p a p or b p . | ) | | 2h i) 2h i 2h i Notes 4.3.1. 1. Note that the condition for irreducible is equivalent to

p = ab p a or p b. ) ⇠ ⇠ 2. The condition “p is prime”isequivalentto“p is a prime ideal”. h i 3. When checking that p is a prime, it suces to consider the case when a and b are nznu. Otherwise, the condition is trivially satisﬁed.

Exercise 4.3.1. Show that the properties of being irreducible or being prime are shared by associates. Given a, b D, such that a b,then 2 ⇠

1. a is irreducible i↵ b is irreducible.

2. a is prime i↵ b is prime.

Proposition 4.3.1 [Gallian 18.1, 18.2] Let p D be nznu. 2

1. If p is prime then p is irreducible.

2. Assume D is a PID. If p is irreducible then p is prime. 4.3. IRREDUCIBLE AND PRIME ELEMENTS 17

Proof. 1. Assume p is prime. WTS p is irreducible. Suppose p = ab. Then p ab and by primality p a or p b. WLOG p a,soa = pc for some c D,and| p = ab = pcb.Since| D is| an ID, and| p =0weget1=cb, so2b is a unit. 6

2. Assume now that D is a PID and p is irreducible. WTS p is prime. Suppose p ab,wherea, b D.Let | 2 I = xa + yp x, y D = a + p D { | 2 } h i h i E Since D is a PID, there is d D such that I = d . Now we have p I so there is z D such that2p = zd.Sincep ish irreducible,i either d2 is aunitorp 2d.Ifd is a unit then I = D,and1 D,sothereare x, y D such⇠ that 1 = xa + yp.Wegetb = xab + ypb2 ,andsincep ab, we get2 p b.Ontheotherhand,ifp d then p = d = I,andsince| a I we| get p a. ⇠ h i h i 2 | Examples 4.3.1. 1. The norm function, N(z), deﬁned as the square of the absolute value for any z C,isamultiplicativefunction,i.e. 2 N(zw)=N(z)N(w). In the domain of Gaussian integers, Z[i], the norm function takes non-negative integer values. It follows that the only units of Z[i]are 1, i,i.e.theelementswithnorm1. ± ± 01/28/19 Now, since N(1 + i)=2,itfollowsthat1+i is irreducible. Since Z[i] is PID, 1 + i is also prime.

2. Consider now the subring of C,

Z p 5 = a + bp 5 a, b Z . | 2 Clearly, it is an ID since⇥ it⇤ is a subring of an ID. The norm is given by N(a + bp 5) = a2 +5b2,andittakesnon-negativeintegervalues.The units are 1, the only elements with norm 1. The elements 1 p 5 ± ± 2 Z[p 5] have norm N(1 + p 5) = N(1 p 5) = 6. Since there are no elements in Z[p 5] with norm equal to 2 or 3, it follows that the elements 1 p 5 are irreducible. However, 1 + p 5isnotaprime since (1 + p± 5)(1 p 5) = 6 = 2 3butN(2) = 4, N(3) = 9, and therefore 1+p 5cannotdivideeither2or3.FromProposition4.3.1.2, · we conclude that Z[p 5] is not a PID. A similar argument shows that 2and3areirreduciblebutnotprimeinZ[p 5]. It also shows that Z[p 5] is not a UFD. See Deﬁnition 4.4.1 below. Version 2019.5.9 18 CHAPTER 4. FACTORIZATION

Unique Recall that for a ﬁeld F ,theringofpolynomialsF [x] is a PID. Hence, Factorization Domain irreducible and prime are equivalent. Deciding whether a polynomial is irreducible or not, is not always an easy task. The following is the ﬁrst of several irreducibility criteria we will use for that purpose.

Proposition 4.3.2 [Irreducibility Criterion in F [x],Gallian17.1]Let F be a ﬁeld.

1 Every linear polynomial in F [x] is irreducible.

2 For f F [x], with deg(f) > 1, if f has a root in F then it is reducible in F [x2].

3 For f F [x], with deg(f)=2or 3, f is reducible in F [x] i↵it has a root in2F .

Proof. This follows immediately from the deﬁnition of irreducible and Corol- lary 4.1.4.

The result of this proposition does not extend beyond degree 3.

Example 4.3.2. The polynomial x4 +5x2 +6 Q[x] is reducible, yet it has 2 no root in Q.Itfactorsasx4 +5x2 +6=(x2 +2)(x2 +3).

4.4 Unique Factorization Domain

Deﬁnition 4.4.1. An integral domain D is called a Unique Factorization Do- main (UFD), if every nznu a D can be factored as a product of irreducible elements, in a unique way up2 to order and associates. The uniqueness means that if a = p p = q q , 1 ··· r 1 ··· s with all the pisandqjsirreducible,thenr = s,and,aftersomepossible reordering, we get p q for i =1,...,r. i ⇠ i

By deﬁnition, every UFD is an integral domain, so we get for free the implication UFD ID in Theorem 4.0.1. ) 4.4. UFD 19

To show that an integral domain D is a UFD, we usually break the Ascending Chain Condition argument into two parts. For any nznu a D, 2 Noetherian

Existence: there exists a factorization of a into irreducible factors;

Uniqueness: two such factorizations have the same number of factors, and di↵er only on the order of the factors, and up to associates.

That is what we will do in the proof of Proposition 4.4.3 below.

Examples 4.4.1. 1. Since Z is a PID, primes and irreducible elements are the same. The Fundamental Theorem of Arithmetic tells us that Z is a UFD.

2. Example 4.3.1.2 above shows that in R = Z[p 5] 6=2 3=(1+p 5)(1 p 5) · The four factors have norms 4, 9, 6 and 6, respectively. Therefore, they are irreducible in Z[p 5]. Since the units of R have norm 1, associate elements in R must have the same norm. Therefore, the two factorizations of 6 are not the same up to associates. In conclusion, Z[p 5] is not a UFD. This shows that the implication UFD ID in Theorem 4.0.1 is strict. )

3. We will show in Proposition 4.4.3 below, that every PID is a UFD. It follows that for any ﬁeld F ,theringF [x] is a UFD.

Deﬁnition 4.4.2. We say that a ring R satisﬁes the Ascending Chain Condi- tion (ACC), if every ascending chain of ideals

I I 1  2 ··· has to become constant after a finite number of steps. That is, there is n 1, such that I = I = .AringthatsatisfiestheACCiscalledNoetherian . n n+1 ··· Proposition 4.4.1 Every PID satisfies the ACC, i.e. it is Noetherian.

Version 2019.5.9 20 CHAPTER 4. FACTORIZATION

Proof. Let D be a PID, and let I I be an ascending chain of ideals 1  2 ··· 1 of D.LetA = Ii. i=1 We ﬁrst claim[ that A is an ideal of D.Givena, b A,therearei, j N such that a I ,andb I . WLOG we may assume2j i.Sincetheideals2 2 i 2 j  satisfy Ij Ii,wehavea, b Ii,andIi being an ideal, yields a + b Ii,so a + b A.For r D,wehave2 ra I ,sora A. 2 2 2 2 i 2 Since D is a PID, there is a A such that A = a .Butthenthereisi N such that a I ,andtherefore2 h i 2 2 i a I A = a , h i i  h i which yields I = A.Foranyk i,wehave i I I A = I , i  k  i so Ik = Ii.

01/29/19

Lemma 4.4.2 Let D be an integral domain, a, b, c, u, v, p, q, a ,...,a D. 1 n 2

1. uv is a unit i↵ u and v are units.

2. If p is a prime, and p a a then p a for some i =1,...,n. | 1 ··· n | i 3. If p is irreducible, q is nznu, and q p then p q. | ⇠ 4. If a is nznu, a = bc and a b then c is a unit. ⇠ Proposition 4.4.3 Let D be an integral domain satisfying:

1. ACC, i.e D is Noetherian,

2. Every irreducible in prime.

Then D is a UFD. 4.4. UFD 21

Proof. Let a D be a nznu. We want to show that a has a factorization into irreducible2 factors, unique up to order and associates. Existence:Weﬁrstprovethefollowing Claim: Any nznu a has at least one irreducible factor. If a is irreducible, we are done. Otherwise, it can be factored as a = b1c1 with b1,c1 nznu. If b1 is irreducible, we are done. Otherwise, note that a b1 since c1 is not a h i h i unit, and b1 can be factored as b1 = b2c2 with b2,c2 nznu. If b2 is irreducible, we are done. Otherwise, note that b1 b2 since c2 is not a unit, and b2 h i h i can be factored as b2 = b3c3 with b3,c3 nznu. This process has to stop, for otherwise we would have an inﬁnite strictly ascending chain of ideals

a b1 b2 h i h i h i ··· contradicting the ACC assumption. Now we use the claim above to prove that any nznu a is a product of irreducible factors. If a is irreducible, we are done. Otherwise, write a = p1d1 where p1 is irreducible and d1 a nznu. Note that a d1 .Ifd1 is irre- h i h i ducible, we are done. Otherwise, write d1 = p2d2 where p2 is irreducible and d2 a nznu. Note that a = p1p2d2,and d1 d2 .Ifd2 is irreducible, we are h i h i done. Otherwise, write d2 = p3d3 where p3 is irreducible and d3 a nznu. Note that a = p1p2p3d3,and d2 d2 .Thisprocesshastostopforotherwise we would have an inﬁniteh strictlyi h ascendingi chain of ideals

a d1 d2 h i h i h i ··· Uniqueness:Supposea = p1p2 pr = q1q2 qs,withpi,qj irreducible, and prime by the second assumption.··· WLOG··· consider r s,andproceed by induction on r.  Base case (r =1)Wehavea = p1 = q1q2 qs.ByLemma4.4.2,p1 q1, and if we had s>1 the product q q would··· be a unit, contradicting⇠ the 2 ··· s assumption that all qj’s are irreducible. Hence s =1. Inductive step From a = p p p = q q q ,andLemma4.4.2.2we 1 2 ··· r 1 2 ··· s have p1 qj for some 1 j s.Switchqj with q1 to have p1 q1,andby Lemma| 4.4.2.3, we get p q .Thereisaunitu D such that |q = p u. 1 ⇠ 1 2 1 1 p p p p = p uq q q . 1 2 3 ··· r 1 2 3 ··· s Note that uq2 q2,so,replacingq2 with uq2,andusingthecancellation property of D we⇠ obtain p p p = q q q . 2 3 ··· r 2 3 ··· s Version 2019.5.9 22 CHAPTER 4. FACTORIZATION

By inductive hypothesis r = s and, after some possible reordering, pi qi for i =2,...,r. ⇠

From this proposition, together with Propositions 4.4.1 and 4.3.1.2, we get the following corollary.

Corollary 4.4.4 [Gallian 18.3] Every PID is a UFD.

This is the last implication needed to complete the proof of Theorem 4.0.1. The following example shows that this implication is strict.

Example 4.4.2. Z[x]isnotaPID.ItfollowsfromTheorem4.4.5below, that it is a UFD. Let

I = f(x) Z[x] constant term is even . { 2 | }

It is easy to see that I is an ideal of Z[x], but there is no polynomial f Z[x] that generates I.Anon-constantf would not have the constant 2 as a multiple in Z[x]. A constant f would have to be even to be in I,andwould not have x as a multiple in Z[x].

We have already seen that when D is an integral domain, the ring D[x]of polynomials over D is also an integral domain. One may ask what properties of D survive when we move to the larger ring D[x]. We have seen (see Example 4.1.1.2) that when F is a ﬁeld, F [x]eventhoughitisnotaﬁeld,it is an Euclidean Domain. Example 4.4.2 above shows that the properties ED and PID do not survive in general. We will prove that the property of UFD does survive.

Theorem 4.4.5 [Gallian 18.5] If D is a UFD, then the ring D[x] of polynomials over D is a UFD.

Based on this theorem, and the fact that Z is an Euclidean Domain, but Z[x] is not a PID, the best we can say for the polynomial ring of an integral domain is summarized in the following table. 4.4. UFD 23

D D[x] greatest common divisor ID ID UFD UFD PID UFD ED UFD Field ED

Corollary 4.4.6 If D is a UFD then the ring D[x1,...,xn] of polynomials in several variables over D is a UFD.

We will need several lemmas and propositions before we can prove Theorem 4.4.5. Deﬁnition 4.4.3. Let a, b, d D.Wesaythatd is a greatest common divisor of a and b provided: 2

d is a common divisor, i.e. • d a and d b, | |

if d0 is a common divisor of a and b then d0 d,i.e. • |

d0 a and d0 b d0 d. | | ) | Lemma 4.4.7 Greatest common divisor of a and b is unique, up to associates, when it exists.

We will denote by g.c.d.(a, b)“the”greatestcommondivisorofa and b when it exists, understanding that it is well-deﬁned up to associates.

01/30/19

Lemma 4.4.8 Let D be an ID, a, b, u D. 2

1. g.c.d.(a, b) g.c.d.(b, a), whenever one of them exists, ⇠ 2. g.c.d.(a, 0) a, ⇠ 3. if u is a unit g.c.d.(a, u) 1, ⇠ 4. if a b then g.c.d.(a, b) a. | ⇠ Version 2019.5.9 24 CHAPTER 4. FACTORIZATION

In general, g.c.d.(a, b) does not have to exist. We will show that in a UFD it does always exist. Because of Lemma 4.4.8 we only need to consider the case when a and b are nznu.

Example 4.4.3. Consider the ring R = Z p 5 from Example 4.3.1.2. Note that ⇥ ⇤ (1 + p 5)(1 p 5) = 6 = 2 3 · (1 + p 5)(1 + 2p 5) = 9+3p 5=3( 3+p 5) We claim that 6 and 9+3p 5donothaveag.c.d.inR.Supposeotherwise, and let d g.c.d.(6, 9+3p 5). Since 3 and 1 + p 5) are both common divisors of⇠ 6 and 9+3 p 5, we must have 3 d and 1 + p 5) d.Thenorm is a multiplicative functionR taking integer values.| Therefore, | we must have N(3) N(d)andN(1 + p 5) N(d)inZ.Thismeans9N(d)and6N(d), and 18 N|(d). On the other hand,| we have d 6andd 9+3| p 5, so| N(d) 36 and| N(d) 126. This yields N(d) 18, and since| N(d|)isanon-negativeinteger, | we must| have N(d)=18.Buttheequation| a2 +5b2 =18hasnointeger solutions.

Lemma 4.4.9 Let D be an ID, a, b, c D. We have g.c.d.(a, g.c.d.(b, c)) g.c.d.(g.c.d.(a, b)2,c) whenever one side exists. ⇠

Lemmas 4.4.9 and 4.4.8.1 tell us that g.c.d., as a binary operation when deﬁned, is associative and commutative, up to associates. Therefore, the expression g.c.d. a ,...,a is well-deﬁned, up to associates. { 1 n} Lemma 4.4.10.1 gives us a converse to Proposition 4.3.1.1 for UFDs. Proposition 4.3.1.1 already gave us the converse for PIDs. This converse is not true for IDs in general, as Example 4.3.1.2 shows.

Lemma 4.4.10 Let D be a UFD, a, p D, nznu. 2

1. If p is irreducible, then it is prime.

2. There are ﬁnitely many (up to associates) irreducible divisors of a.

Proof. (1) Assume p D is irreducible and let b, c D be nznu such that p bc.Thereisd D2such that pd = bc.Writeeach2 b, c, d as a product of | 2 4.4. UFD 25 irreducibles, b = p p 1 ··· j c = q1 qk d = r ···r 1 ··· l so we have pr r = p p q q .Byuniquenessofthefactorization, 1 ··· l 1 ··· j 1 ··· k we must have either p pi for some 1 i j or p qi for some 1 i k. It then follows that either⇠ p a or p b.So, pis prime.⇠   (2) Write a as a product of irreducibles| | a = p p .Ifp is an irreducible that 1 ··· j divides a,byPart(4.4.10.1),p pi for some 1 i j.Sincepi is irreducible and p is a nznu, by Lemma 4.4.2.3| we get p p .Therefore,uptoassociates, ⇠ i the only irreducible factors of a are p1,...,pj. Proposition 4.4.11 Let D be a UFD, a, b D nznu. 2 Let p1,...,pr be the list, without repetitions, of all irreducible factors of either a or b, up to associates. Write

a = p↵1 p↵r ,↵ 0, 1 ··· r i b = p1 pr , 0. 1 ··· r i Then a and b have a greatest common divisor, given by

g.c.d.(a, b)=p1 pr , where =min ↵ , 1 ··· r i { i i}

Proof. Let d = p1 pr .Since ↵ ,wehave↵ 0, so 1 ··· r i  i i i ↵1 1 ↵r r a = dp p 1 ··· r and d a.Similarly,d b,sod is a common divisor of a and b.Ifd0 is also a | | common divisor of a and b,theneveryirreduciblefactorofd0 divides a,and, up to associates, it has to be one of p1,...,pr.Therefore,wewrited0 as a 1 r product irreducibles, it must take the form d0 = p p with 0. Since 1 ··· r i d0 divides a,thereise D such that a = d0e. Now, e a,sothesameargument 2 | used with d0,showsthatthefactorizationofe into irreducible factors must have the form e = pµ1 pµr with µ 0. Now we have 1 ··· r i ↵1 ↵r 1+µ1 r+µr a = p p = d0e = p p 1 ··· r 1 ··· r Since p1,...,pr are pairwise non-associates, the uniqueness of factorization forces ↵i = i + µi,soi ↵i.Asimilarargumentshowsthati i,so  1 r 1 1 r r min ↵ , = .Therefore,d = p p = d0p p ,and i  { i i} i 1 ··· r 1 ··· r d0 d. | Version 2019.5.9 26 CHAPTER 4. FACTORIZATION least common Corollary 4.4.12 In any PID and in any ED the operation g.c.d. is well- multiple deﬁned, up to associates.

Note that Proposition 4.4.11 gives us a procedure to ﬁnd the g.c.d. of two elements in a UFD. The following exercise shows alternative ways to ﬁnd the g.c.d. of two elements in a PID, and in an ED.

Exercise 4.4.1. 1. Let D be a PID, a, b D.Letd be a generator of the ideal a + b .Showthatd g.c.d.(2a, b). h i h i ⇠ 2. Let D be an ED, a, b D,withb =0.Considerthesequence 2 6 r0,r1,r2,...,rn deﬁned recursively as follows: r0 = a, r1 = b,andusing Property 4.1 of an Euclidean Domain, until obtaining a residue 0,

r0 = q1r1 + r2 and (r2) <(r1), r1 = q2r2 + r3 and (r3) <(r2), . . rn 3 = qn 2rn 2 + rn 1 and (rn 1) <(rn 2), rn 2 = qn 1rn 1 + rn and rn =0.

Why does the sequence r1,r2,...,rn have to eventually attain the value rn =0?Provethatthelastnon-zeroentryintheresidueslist,i.e. rn 1 g.c.d.(a, b). ⇠

02/01/19

Corollary 4.4.13 Let a, b, c D. g.c.d.(ac, bc) c g.c.d.(a, b). 2 ⇠ · Deﬁnition 4.4.4. Let a, b, m D.Wesaythatm is a least common multiple of a and b provided: 2

m is a common multiple, i.e. • a m and b m, | |

if m0 is a common multiple of a and b then m m0,i.e. • |

a m0 and b m0 m m0. | | ) | 4.4. UFD 27

Note that the “least common multiple”ofa and b is unique, up to associates, when it exists. We denote it by l.c.m.(a, b), understanding that it is well-deﬁned up to associates.

Lemma 4.4.14 Let D be an ID, a, b, u D. 2

1. l.c.m.(a, b) l.c.m.(b, a), whenever one of them exists, ⇠ 2. l.c.m.(a, 0) 0, ⇠ 3. if u is a unit l.c.m.(a, u) a, ⇠ 4. if a b then l.c.m.(a, b) b. | ⇠

In general, l.c.m.(a, b) does not have to exist. We will show that in a UFD it does always exist. Because of Lemma 4.4.14 we only need to consider the case when a and b are nznu.

Lemma 4.4.15 Let D be an ID, a, b, c D. We have l.c.m.(a, l.c.m.(b, c)) l.c.m.(l.c.m.(a, b),c2 ) whenever one side exists. ⇠

Lemmas 4.4.15 and 4.4.14.1 tell us that l.c.m., as a binary operation when deﬁned, is associative and commutative, up to associates. Therefore, the expression l.c.m. a ,...,a is well-deﬁned, up to associates. { 1 n} Exercise 4.4.2. 1. Let D be an UFD, a, b D.Provethata and b have aleastcommonmultiple,and 2

g.c.d.(a, b) l.c.m.(a, b)=a b, · · so that when at least one of a and b is non-zero, then

a b l.c.m.(a, b)= · . g.c.d.(a, b)

2. Let D be an ID, a, b D.Provethatifa and b have a least common 2ab multiple l in D,then is a greatest common divisor of a and b. l

Version 2019.5.9 28 CHAPTER 4. FACTORIZATION content 4.5 Factorization in D[x] primitive

Recall that for any ring R, R is a subring of the ring of polynomials R[x], consisting of all constant polynomials. When D is an ID, the units of D[x] are all constant, and precisely the units of D.

Example 4.5.1. The previous statement about units, requires D to be an 2 ID. In Z4[x], the polynomial 1 + 2x is a non-constant unit, as (1 + 2x) =1. Deﬁnition 4.5.1. Let D be a UFD, 0 = f D[x]with 6 2 f(x)=a + a x + + a xn. 0 1 ··· n

The content of f(x), denoted c(f(x)), is deﬁned as g.c.d. a0,...,an .We’ll often write c(f)insteadofc(f(x)). Note that c(f)iswell-deﬁned,upto{ } associates. We say that f(x)isprimitive if c(f) 1. ⇠ Example 4.5.2. If we take f(x)=3x2 12x+9 Z[x]thenc(f) 3. Note that f(x)=3(x2 4x +3)andthesecondfactorisprimitive.Thisholdsin 2 ⇠ general.

Lemma 4.5.1 Let D be a UFD, 0 = f(x),g(x) D[x], 0 = a D. 6 2 6 2

1. c(af(x)) a c(f(x)). ⇠ · 2. There is a primitive f(x) D[x] such that f(x)=c(f) f(x). This 2 · primitive is unique, up to associates, and deg(f(x)) = deg(f(x)). b b 3. If f(x)=ag(x) with g(x) primitive, then c(f) a. ⇠b Proposition 4.5.2 [Gauss’ Lemma] Let D be a UFD, 0 = f(x),g(x) D[x].Iff(x) and g(x) are primitive, then f(x)g(x) is primitive.6 2

Proof. (By contradiction) Assume f(x)andg(x)areprimitive,andf(x)g(x) is not primitive. Write

n f(x)=a0 + a1x + + anx ··· m g(x)=b0 + b1x + + bmx f(x)g(x)=c + c x + ···+ c xn+m 0 1 ··· n+m 4.5. FACTORIZATION IN D[X]29

so that ck = aibk i.Sincef(x)g(x)isnotprimitive,d = c(f(x)g(x)) = i=0 g.c.d. c ,...,cX is not a unit, and there is p D irreducible such that { 0 n+m} 2 p d.Iffollowsthatp ck for every k =0,...,n+ m.Ontheotherhand,since f|(x)andg(x)areprimitive,thereare| a and b such that p a and p b . j l 6| j 6|l Choose j and l smallest, so that p a0,...,aj 1,b0,bl 1. Then, since | j+l j l j+l cj+l = aibj+l i = aibj+l i + ajbl + aibj+l i i=0 i=0 ! i=j+1 ! X X X

we get that p divides ajbl.Sincep is prime, this forces p aj or p bl,acontra- diction. | | Corollary 4.5.3 Let D be a UFD, 0 = f(x),g(x) D[x]. 6 2 c(fg) c(f) c(g). ⇠ ·

Proof. Let f(x), g(x) D be primitive such that f(x)=c(f)f(x)andg = 2 c(g)g(x). Then f(x)g(x)=c(f)c(g)f(x)g(x). By Gauss’s lemma f(x)g(x)is primitive, sob by Lemmab 4.5.1 we get c(fg) c(f) c(g), as desired.b ⇠ · b b b b b The converse of Gauss’ lemma follows from this corollary and Lemma 4.4.2.

02/04/19

Corollary 4.5.4 Let D be a UFD, 0 = f(x),g(x) D[x]. f(x)g(x) is primitive i↵ f(x) and g(x) are primitive.6 2 Lemma 4.5.5 Let D be a UFD, f(x) D[x] with deg(f(x)) 1.Iff(x) is irreducible then it is primitive. 2

Proof. Let f(x) D[x]beprimitivesuchthatf(x)=c(f) f(x). Since f(x) 2 · is irreducible, one of c(f)andf(ˆx)isaunit.Butdeg(f(x)) = deg(f(x)) 1, so f(x) is notb a unit, so c(f)isaunit,i.e. c(f) 1, and f(x)isprimitive.b ⇠ b bRecall that when R is a subring of S,thepolynomialringR[x]isa subring of S[x]. The next proposition relates three rings, a UFD D,itsring of polynomials D[x], and the ring of polynomials of its ﬁeld of quotients. It connects the concept of irreducible in these three rings.

Version 2019.5.9 30 CHAPTER 4. FACTORIZATION

Proposition 4.5.6 Let D be a UFD, and Q its ﬁeld of fractions. Let p D and f(x) D[x] with deg(f(x)) 1, and g(x) Q[x]. 2 2 2

1. p is irreducible in D i↵it is irreducible in D[x]. a 2. There are a, b D and a primitive g(x) D[x] such that g(x)= g(x). 2 2 b 3. If f(x) is irreducible in D[x] thenb it is irreducible in Q[x]. b

4. If f(x) is primitive and irreducible in Q[x], then it is irreducible in D[x].

Proof. (1) Since p D has degree 0, the only way to write p as a product of two polynomials is2 by both factors being constant. Since D and D[x]have the same units, the result follows at once. (2) Note ﬁrst that the coecients of any g(x) Q[x]arefractionsofele- ments of D.Eachofthosefractionscanbetakentobe“2 reduced”, meaning that the g.c.d. of numerator and denominator is a unit. If we let b be the l.c.m. of the denominators of the coecients of g(x), then bg(x) D[x]. Let a = c(bg(x)), and g(x) D[x] primitive, such that bg(x)=ag(x2). Thus, we a 2 have g(x)= g(x) b b b (3) By contradiction. Assume f(x)isirreducibleinD[x]butreducibleinQ[x] and write f(x)=b g(x)h(x)withg(x),h(x) Q[x] nznu, i.e. non-constant. 2 a Let a, b, c, d D and g(x), h(x) D[x]primitivesuchthatg(x)= g(x) 2 2 b c ac and h(x)= h(x). Web getbf(x)= g(x)h(x), i.e. bd f(x)=ac g(x)hb(x). d bd By Gauss’s Lemma g(x)h(x)isprimitive.ByLemma4.5.5,f(x)isprimitive. b b b b b ac Taking content of both sides, we get ac bd,andu = D is a unit. We b b ⇠ bd 2 have f(x)=u g(x)h(x), and g(x), h(x)arenon-constant.Thiscontradicts the irreducibility of f(x)inD[x]. (4) Suppose f(bx)=b g(x)h(xb)withb g(x),h(x) D[x], nznu. By Corol- lary 4.5.4, g(x)andh(x)areprimitive,butbeingnon-units,theymustbe2 non-constant, contradicting the irreducibility of f(x)inQ[x]. 4.5. FACTORIZATION IN D[X]31

Note 4.5.1. Note that the condition in Proposition 4.5.6.4 of f(x)being primitive is necessary. For example, the polynomial f(x)=2x +2isirre- ducible in Q[x], but it is not irreducible in Z[x]. Scholium 4.5.7 If f(x) is reducible in Q[x], then it factors as a product of non-constant polynomials of lower degree in D[x].

Using Proposition 4.5.6.3 we obtain another corollary to Gauss Lemma. Corollary 4.5.8 Let D be a UFD, and Q its ﬁeld of quotients. Let f(x) D[x] and g(x) Q[x].Iff(x) is primitive and f(x)g(x) D[x], then g(x) 2 D[x]. 2 2 2

a Proof. Let a, b D and g(x) D[x]primitive,suchthatg(x)= g(x). Let 2 2 b h(x)=f(x)g(x). Then bh(x)=af(x)g(x)inD[x]. By Corollary 4.5.3 and the fact that both f(x)andb g(x)areprimitive,weget b b b c(h(x)) a, b · ⇠ a so b a and D. | b 2 We can combine Lemma 4.5.5 with Proposition 4.5.6.3,4 and obtain the following proposition. Proposition 4.5.9 Let D be a UFD, Q its ﬁeld of fractions, and f(x) D[x] non-constant polynomial. f(x) is irreducible in D[x] i↵it is primitive2 and irreducible in Q[x].

We now have all the ingredients needed to prove Theorem 4.4.5. Theorem 4.5.5 f D is a UFD, then the ring D[x] of polynomials over D is aUFD.

Proof. Let f(x) D[x]benznu.Wewanttoshowthatf(x)canbefactored as a product of irreducibles2 in a unique way, up to order and associates. Existence: By induction on deg(f(x)). Base case.Ifdeg(f(x)) = 0, then f(x)isaconstantpolynomial,f(x)= a D.SinceD is a UFD, we can factor a as a product of irreducibles in D. 2 Version 2019.5.9 32 CHAPTER 4. FACTORIZATION

By Proposition 4.5.6.1, all these irreducibles in D are irreducible in D[x] Inductive step.Ifdeg(f(x) > 1, write f(x)=c(f) f(x) with f(x) primitive. Write c(f)asaproductofirreduciblesinD,whicharealso irreducible in D[x]. Now, if f(x) is irreducible, we are done.b Otherwise,b write f(x)=g(x)h(x)withg(x),h(x) D[x]nznu.ByCorollary4.5.4 g(x)andh(x) are primitive,b so neither2 one is a constant, and therefore deg(g(xb)), deg(h(x)) < deg(f(x)), and by induction hypothesis, each of g(x) and h(x)canbefactoredasaproductofirreducibles. Uniqueness: Assume f(x)=p (x) p (x)=q (x) q (x)witheach 02/05/19 1 ··· r 1 ··· s pi(x)andeachqj(x)irreducibleinD[x]. Rearrange, if needed, so that

p1(x),...,pk(x)arenon-constantand pk+1,...,pr are constant q1(x),...,ql(x)arenon-constantand ql+1,...,qs are constant

By Lemma 4.5.5 and 4.5.2, p1(x) pk(x)andq1(x) ql(x)areprimitive, and therefore by Lemma 4.5.1.3, we··· get c(f) p ···p q ...q .We ⇠ k+1 ··· r ⇠ l+1 s can replace some pi with an associate and some qj with an associate, such that p p = q ...q D,sop (x) p (x)=q (x) q (x) D[x]. By k+1 ··· r l+1 s 2 1 ··· k 1 ··· l 2 Proposition 4.5.6.3 we have p1(x),...,pk(x),q1(x),...,ql(x) are irreducible in Q[x], where Q is the ﬁeld of fractions of D.ButweknowthatQ[x] is a UFD, so we must have k = l and after some possible rearrangements pi(x) qi(x) (associates in Q[x]) for i =1,...k. Since the units in Q[x]areallnon-zero⇠ ai constants, there must be ai,bi D such that pi(x)= qi(x). This means 2 bi that in D[x], bipi(x)=aiqi(x), and taking the content, we get b c(b p (x)) c(a q (x)) a . i ⇠ i i ⇠ i i ⇠ i

There is a unit ui D such that bi = uiai,soqi(x)=uipi(x), and pi(x) q (x)(associatesin2 D[x]). Now consider k

4.6 Irreducibility Criteria

We now establish some additional irreducibility criteria. Recall that Proposi- tion 4.3.2 gives us some irreducibility criteria for polynomials in F [x], where 4.6. IRREDUCIBILITY CRITERIA 33

F is a field. These additional criteria consider polynomials over a UFD, and relates them to polynomials over the its field of fractions. Proposition 4.6.1 [Gallian 17.3] Let D be a UFD, and Q its field of fractions. Let S an ID, and ' : D S a ring homomorphism. Let ! ' : D[x] S[x] r ! '(r) for r R x 7! x 2 7! be the induced homomorphism on polynomials. Let f(x) D[x] be non- 2 constant and let f(x)='(f(x)). If the leading coecient of f(x) is not in

the kernel of ', and f(x) is irreducible in S[x], then f(x) is irreducible in Q[x]. Moreover, if f(x) is primitive, then it is also irreducible in D[x].

02/06/19

Proof. Assume f(x)isreducibleinQ[x]. By Scholium 4.5.7 there are non- constant g(x),h(x) D[x], such that f(x)=g(x)h(x). Since the leading coecient of f(x)isnotinker(2 '), neither are the leading coecients of g(x) and h(x). Thus, we have f(x)=g(x)h(x)withg(x), h(x)non-constant, showing that f(x) is reducible in S[x]. The moreover part follows from Proposition 4.5.6.4.

Example 4.6.1. Let f(x)=x4 +2x2 +2x x +1 Z[x]. Reducing 2 4 the coecients (mod 2), we get f(x)=x + x +1 Z2[x]. We claim this 2 polynomial is irreducible. First, it has no root in Z2,so,byCorollary4.1.4,it has no linear factors. To factor it as a product of two quadratic polynomials, it would take the form

x4 + x +1=(x2 + ax +1)(x2 + bx +1),

but looking at the coecients of x and x3 we get 1 = a + b = 0, a contradiction. Now, using Proposition 4.6.1, f(x)isirreducibleinQ[x]. Since f(x)is also primitive, it is irreducible in Z[x].

Version 2019.5.9 34 CHAPTER 4. FACTORIZATION

Proposition 4.6.2 [Eisenstein’s Criterion] Let D be an integral domain, P a prime ideal of D and f(x) D[x] primitive. Write 2 f(x)=a + a x + + a xn. 0 1 ··· n 2 If a0,...,an 1 P , an / P and a0 / P , then f(x) is irreducible in D[x]. 2 2 2

Proof. If we had f(x)reducibleinD[x], with the use of Proposition 4.5.6.4 and Scholium 4.5.7, it would factor as a product of two non-constant polynomials of lower degree, say f(x)=g(x)h(x)where

k g(x)=b0 + b1x + + bkx , ··· n k h(x)=c0 + c1x + + cn kx , ··· and 1 k

at = b0ct + b1ct 1 + btc0. ···

Since P is a prime ideal, the last summand on the right hand side, btc0 / P . By the choice of t all other summand in the right hand side are in P . And,2 since t k

Corollary 4.6.3 [Eisenstein’s Criterion for Z,Gallian17.4]Let

n f(x)=a0 + a1x + + anx Z[x] ··· 2 be primitive, and p Z a prime number such that p a0,...,an 1, p an and p2 a . Then f(x) is2 irreducible. | 6| 6| 0 Examples 4.6.2. 1. The polynomial x4 +10x +5 Z[x]isirreducible. Apply Eisenstein’s criterion with p =5. 2

2. If 1 = a Z is divisible by a prime number p,butnotbyp2,then 6 2 xn a is irreducible in Z[x]. 4.6. IRREDUCIBILITY CRITERIA 35

3. For a prime number p,thecyclotomicpolynomial

p 1 p 2 p(x)=x + x + + x +1 Z[x] ··· 2 is irreducible. To see this, observe that

p p p x 1 (x +1) 1 p i 1 (x)= , so (x +1)= = x , p x 1 p x i i=1 X ✓ ◆

so p(x +1)isirreducible,byEisenstein’scriterion. Itfollowsthat p(x)isalsoirreducible. 4. A similar argument to that of previous example, shows that x4 +1 2 Z[x]isirreducible.

The only remaining piece of Theorem 4.0.1 is an example of a PID that is not an ED. 1+p 19 Exercise 4.6.1. Let = and consider the subring of C given by: 2

R = a + b a, b Z { | 2 } Prove that R is a PID but not an ED. A detailed proof can be found in [3]. The readers is encourage to prove as much as possible before looking up the reference.

02/08/19 Problem Set 01 presentations

Version 2019.5.9 36 CHAPTER 4. FACTORIZATION Part III

Vector Spaces