We extend the Euclidean algorithm to the polynomial ringD[X] whereDis a unique factorization do- main. The success of this enterprise depends on the theory of subresultants. Subresultant sequences are special remainder sequences which have many applications including Diophantine equations, Sturm theory, elimination theory, discriminants, and algebraic cell decompositions. Our approach to subresultants follows Ho and Yap [84, 83], who introduced the pseudo-subresultants to carry out Loos’ program [119] of studying subresultants via specialization from the case of indeterminate coefficients. This approach goes back to Habicht [76].
One of the most well-studied problems in the early days of computer algebra (circa 1970) is the problem of computing the GCD in the polynomial ringD[X] whereDis a UFD. See the surveys of Loos [33] and Collins [45]. This led to a development of efficient algorithms whose approach is quite distinct from the HGCD approach of the previous lecture. The reader may be surprised that any new ideas are needed: why not use the previous techniques to compute the GCD inQD[X] (whereQDis
the quotient field ofD) and then “clear denominators”? One problem is that computing remainders in QD[X] can be quite non-trivial for some D (say,D =F[X1, . . . , Xd]). Another problem is that
clearing denominators is really a multiple GCD computation (in its dual form of a multiple LCM computation). Multiple GCD is expensive in practical terms, even when it is polynomial-time as in the case D = Z. Worst, in case D = F[X1, . . . , Xd], we are involved in a recursive situation
of exponential complexity. Hence the challenge is to develop a direct method avoiding the above problems.
In this lecture,Drefers to a unique factorization domain with quotient fieldQD.
The reader may safely take D=Z and soQD=Q.
§1. Primitive Factorization
The goal of this section is to extend the arithmetic structure of a unique factorization domainD to its quotient fieldQD and toQD[X]. It becomes meaningful to speak of irreducible factorizations in
QD andQD[X].
Content of a polynomial. Let q ∈ D be an irreducible element. For any non-zero element a/b∈QD wherea, b∈D are relatively prime, define the q-orderofa/bto be
ordq(a/b) :=
n if qn|abut not qn+1|a,
−n if qn|bbut not qn+1|b. (1) Exactly one of the two conditions in this equation hold unlessn= 0. In this case,ordq(a/b) = 0 or
equivalently, q does not divideab. For example, inD =Z we haveord2(4/3) = 2, ord2(3/7) = 0 andord2(7/2) =−1.
We extend this definition to polynomials. IfP ∈QD[X]\ {0}, define theq-order ofP to be
ordq(P) := min
§1. Primitive Factorization Lecture III Page 78
where ci ranges over all the non-zero coefficients ofP. For example,ord2(X3−54X+ 2) = −2 in
Z[X]. Any associate ofqordq(P)is called aq-contentofP. Finally, we define acontentofP to be
u
q
qordq(P)
whereqranges over all distinguished irreducible elements ofD,uis any unit. This product is well- defined since all but finitely manyordq(P) are zero. For the zero element, we defineordq(0) =−∞
and theq-content and content of 0 are both 0.
Primitive polynomials. A polynomial ofQD[X] isprimitiveif it has content 1; such polynomials
are elements ofD[X]. Thus every non-zero polynomialP has a factorization of the form P =cQ
wherecis a content ofP andQis primitive. We may always chooseQso that its leading coefficient is distinguished. In this case,cis called thecontent ofP, andQtheprimitive partofP. These are denotedcont(P) andprim(P), respectively. We call the product expression “cont(P)prim(P)” the
primitive factorization ofP.
Ifprim(P) =prim(Q), we say thatP, Qaresimilarand denote this by P ∼Q.
HenceP ∼Qiff there existα, β∈D such thatαP =βQ. In particular, ifP, Q are associates then they are similar.
For instance, the following are primitive factorizations:
−4X3−2X+ 6 = (−2)·(2X3+X−3) (15/2)X2−(10/3)X+ 5 = (5/6)·(9X2−4X+ 6). Also,−4X3−2X+ 6∼6X3+ 3X−9.
The following is one form of a famous little lemma1:
Lemma 1 (Gauss’ Lemma) If D is a UFD and P, Q ∈ D[X] are primitive, then so is their product P Q.
Proof. We must show that for all irreducible q ∈ D, ordq(P Q) = 0. We can uniquely write any
polynomialP ∈D[X] as
P =qP0+P1, (P0, P1∈D[X])
where deg(P0) is less than the tail degree of P1 and the tail coefficienttail(P1) is not divisible by q. [Iftail(P) is not divisible byq, thenP0= 0 andP1=P.] Moreover,
ordq(P) = 0 iff P1= 0.
ThusP1= 0. LetQ=qQ0+Q1be the similar expression forQand againQ1= 0. Multiplying the expressions forP and Q, we get an expression of the form
P Q=qR0+R1, R1=P1Q1= 0. 1We refer to Edwards [63] for a deeper investigation of this innocuous lemma.
§1. Primitive Factorization Lecture III Page 79
By the uniqueness of such expressions, we conclude thatordq(P Q) = 0. Q.E.D.
If Pi = aiQi (i = 1,2) is a primitive factorization, we have cont(P1P2) = cont(a1Q1a2Q2) = a1a2cont(Q1Q2) and prim(P1P2) = prim(Q1Q2). By Gauss’ lemma, cont(Q1Q1) = and prim(Q1Q2) =Q1Q2, where , are units. Hence we have shown
Corollary 2 ForP1, P2∈QD[X],
cont(P1P2) =·cont(P1)cont(P2), prim(P1P2) =·prim(P1)prim(P2).
Another corollary to Gauss’ lemma is this:
Corollary 3 IfP(X)∈D[X]is primitive andP(X)is reducible inQD[X]then P(X)is reducible inD[X].
To see this, suppose P = QR with Q, R ∈ QD[X]. By the above corollary, cont(P) =
·cont(Q)cont(R) = for some unit. ThenP =·prim(P). By the same corollary again, P =··prim(Q)prim(R).
Sinceprim(Q),prim(R) belongs to D[X], this shows P is reducible.
We are ready to prove the non-trivial direction of the theorem in§II.1: ifD is a UFD, thenD[X]
is a UFD.
Proof. Suppose P ∈ D[X] and without loss of generality, assume P is not an element of D. Let its primitive factorization be P = aP. Clearly P is a non-unit. We proceed to give a unique factorization ofP (as usual, unique up to reordering and associates). In the last lecture, we proved that a ring of the formQD[X] (being Euclidean) is a UFD. So if we viewPas an element ofQD[X], we get a unique factorization, P =P1P2· · ·P where eachPi is an irreducible element ofQD[X]. Letting the primitive factorization of eachPi be ciPi, we get
P=c1· · ·cP1· · ·P.
Butc1· · ·c= (some unit). Thus
P = (·P1)P2· · ·P
is a factorization ofPinto irreducible elements ofD[X]. The uniqueness of this factorization follows from the fact thatQD[X] is a UFD. Ifais a unit, then
P = (a··P1)P2· · ·P
gives a unique factorization of P. Otherwise, since D is a UFD, a has a unique factorization, say a=a1· · ·ak. Then
P=a1· · ·ak(·P1)P2· · ·P
§2. Pseudo-remainders Lecture III Page 80
The divide relation in a quotient field. We may extend the relation ‘bdividesc’ to the quotient field of a UFD. Forb, c∈QD, we saybdividesc, written
b|c,
if for all irreducible q, either 0≤ordq(b)≤ordq(c) orordq(c)≤ordq(b)≤0. ClearlyQD is also
a “unique factorization domain” whose irreducible elements are q, q−1 where q∈ D is irreducible. Hence the concept of GCD is again applicable and we extend our previous definition to QD in a
natural way. We callb apartial contentofP ifbdivides cont(P).
Exercises
Exercise 1.1: Assume that elements in QD are represented as a pair (a, b) of relatively prime
elements ofD. Reduce the problem of computing GCD inQD to the problem of GCD inD.
2 Exercise 1.2: (Eisenstein’s criterion) Let D be a UFD and f(X) = ni=0aiXi be a primitive
polynomial inD[X].
(i) If there exists an irreducible element p∈D such that an≡0(modp),
ai≡0(modp) (i= 0, . . . , n−1),
a0≡0(modp2), thenf(X) is irreducible inD[X].
(ii) Under the same conditions as (i), conclude that the polynomial g(x) =ni=0aiXn−i is
irreducible. 2
Exercise 1.3: (i)Xn−pis irreducible overQ[X] for all primep∈Z.
(ii)f(X) =Xp−1+Xp−2+· · ·+X+ 1 (= Xp−1
X−1) is irreducible inQ[X] for all primep∈Z.
HINT: apply Eisenstein’s criterion tof(X+ 1).
(iii) Letζ be a primitive 5-th root of unity. Then√5∈Q(ζ).
(iv) The polynomialg(X) =X10−5 is irreducible overQ[X] but factors as (X5−√5)(X5+√5)
overQ(ζ)[X]. 2