The following observation will be crucial when developing the theory of the resul- tant:
Lemma D.1.1 Letp(X1, . . . , Xn)andq(X1, . . . , Xn)be non-constant polynomi- als over a fieldK. The following conditions are equivalent:
1. p(X1, . . . , Xn)andq(X1, . . . , Xn)have a common irreducible factor;
F. Borceux, An Algebraic Approach to Geometry, DOI10.1007/978-3-319-01733-4,
© Springer International Publishing Switzerland 2014
379
380 D Resultants 2. there exist non-zero polynomialsφ (X1, . . . , Xn)andψ (X1, . . . , Xn)such that
ψ (X1, . . . , Xn)p(X1, . . . , Xn)=φ (X1, . . . , Xn)q(X1, . . . , Xn), degreeφ <degreep, degreeψ <degreeq.
Proof If p(X1, . . . , Xn) and q(X1, . . . , Xn) have a common irreducible factor d(X1, . . . , Xn), there exist polynomialsφandψsuch that
p(X1, . . . , Xn)=d(X1, . . . , Xn)φ (X1, . . . , Xn), q(X1, . . . , Xn)=d(X1, . . . , Xn)ψ (X1, . . . , Xn).
This immediately implies
degreeφ <degreep, degreeψ <degreeq becaused(X1, . . . , Xn)is not constant. Moreover
ψ (X1, . . . , Xn)p(X1, . . . , Xn)=φ (X1, . . . , Xn)ψ (X1, . . . , Xn)d(X1, . . . , Xn)
=φ (X1, . . . , Xn)q(X1, . . . , Xn).
Conversely, given the situation of assertion 2 in the statement, every irre- ducible factor of p(X1, . . . , Xn) is an irreducible factor of q(X1, . . . , Xn) or φ (X1, . . . , Xn)(see TheoremB.4.9). But not all irreducible factors of the polyno- mialp(X1, . . . , Xn)can appear—with their multiplicity—in the decomposition of φ (X1, . . . , Xn)in irreducible factors, because the degree ofφis strictly less than the degree ofp. Thus at least one irreducible factor ofp(X1, . . . , Xn)is an irreducible
factor ofq(X1, . . . , Xn).
Rephrasing this result in terms of a polynomial domainDover a fieldK (that is, a ring of the formD=K[X1, . . . , Xn]withKa field; see DefinitionB.2.2), we obtain:
Corollary D.1.2 LetDbe a polynomial domain over a fieldKandp(X),q(X)be two polynomials inD[X]which are not constant overD. The following conditions are equivalent:
1. p(X) and q(X) admit a common irreducible factor which is non-constant overD;
2. there exist non-zero polynomialsφ (X), ψ (X)∈D[X]such that p(X)ψ (X)=q(X)φ (X)
and as far as the degrees overDare concerned
degreeφ (X) <degreep(X), degreeψ (X) <degreeq(X).
Proof The proof of(1⇒2)in LemmaD.1.1also proves the additional condition on the degrees overD, since nowd(X)is assumed to have degree at least 1 overD.
Analogously, the proof of the converse implication allows us to restrict our attention to those irreducible factors which are not constant overD.
Definition D.1.3 Let D be a polynomial domain over a fieldK. Consider two polynomialsp(X)andq(X)of respective degreesn >0 andm >0 overD:
p(X)=anXn+ ã ã ã +a0, q(X)=bmXm+ ã ã ã +b0. The resultant of these two polynomials is the value of the determinant
R(p, q)=det
⎛
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎝
a0 a1 ã ã ã an 0 ã ã ã 0 0 a0 a1 ã ã ã an 0 ã ã ã 0
ã ã ã ã 0 ã ã ã 0 a0 a1 ã ã ã an b0 b1 ã ã ã bm 0 ã ã ã 0
0 b0 b1 ã ã ã bm 0 ã ã ã 0
ã ã ã ã 0 ã ã ã 0 b0 b1 ã ã ã bm
⎞
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎠
∈D
which containsmrows ofai’s andnrows ofbj’s.
The reader should be aware that by TheoremB.3.1, the definition and the prop- erties of a determinant in the polynomial domainDare at once inherited from the corresponding considerations in the quotient fieldLof D, which containsDas a subring.
The key property of the resultant is:
Theorem D.1.4 LetDbe a polynomial domain over a fieldK. Consider two poly- nomialp(X)andq(X)of respective degreesn >0 andm >0 overD. The following conditions are equivalent:
1. p(X) and q(X) have a common irreducible factor which is not a constant overD.
2. The resultant ofp(X)andq(X)is equal to 0.
Proof Assume first thatp(X)andq(X)have a common irreducible factord(X)of degree at least 1 overD. We can then write
φ (X)=αn−1Xn−1+ ã ã ã +α0, ψ (X)=βm−1Xm−1+ ã ã ã +β0
for the two polynomials of CorollaryD.1.2, where at least one coefficientαi and at least one coefficientβj are non-zero. The equalityp(X)ψ (X)=q(X)φ (X)means
382 D Resultants a0β0=b0α0
a1β0+a0β1=b1α0+b0α1 ...
anβm−2+an−1βm−1=bmαn−2+bm−1αn−1 anβm−1=bmαn−1.
Consider nowL, the quotient field of D (see TheoremB.3.1), and the system of homogeneous equations inYi,Zj and coefficients inL
a0Z0=b0Y0 a1Z0+a0Z1=b1Y0+b0Y1
...
anZm−2+an−1Zm−1=bmYn−2+bm−1Yn−1 anZm−1=bmYn−1.
This system admits a non-zero solution inL given by the αi,βj. Therefore the determinant of the system is equal to zero. But this determinant is precisely (possibly up to its sign) the resultant ofp(X)andq(X).
Conversely if the resultant ofp(X)andq(X)is zero, the homogeneous system of equations indicated above, viewed as a system with coefficients in the fieldL, has a non-zero solutionαi,βj inL. Since the system is homogeneous, multiplying the solution by the product of all the denominators of all the fractionsαi,βj yields an- other non-zero solutionαi,βj, which now lies inD. Defineφandψby the formulổ above in terms ofαiandβj. The fact that the coefficientsαi,βj are solutions of the system of homogeneous equations means precisely thatp(X)ψ (X)=q(X)φ (X).
The result follows by CorollaryD.1.2.
Corollary D.1.5 Letp(X)be a polynomial of degree at least 2 over a polynomial domainD. The following conditions are equivalent:
1. p(X)admits a multiple irreducible factor which is not constant overD;
2. the resultant ofp(X)andp(X)is zero.
Proof As observed in Sect.B.5, Proposition A.8.5and its proof remain valid in the case of a polynomial domainD over a field K. The result follows by Theo-
remD.1.4.
The reader will immediately notice that when a polynomialp(X1, . . . , Xn)over a fieldKadmits a multiple irreducible factor, it trivially contains terms in which at least one of the variablesXi appears with exponent at least 2. So the assumption of degree 2 in CorollaryD.1.5is again not a real restriction.
Let us also observe that in TheoremD.1.4 and its Corollary, the fact that the common irreducible factor has degree at least 1 overDis essential:
Counterexample D.1.6 PutD=R[X]. InD[Y], the two polynomials p(Y )=X2Y2+X2, p(Y )=2X2Y
admitX as common irreducible factor of multiplicity 2, but their resultant is not equal to zero.
Proof Indeed the resultant is
det
⎛
⎝ X2 0 X2
2X2 0 0
0 2X2 0
⎞
⎠=4X6.
Let us conclude this section with a result which is reminiscent of the properties of the greatest common divisor, in the case of polynomials in one variable (see TheoremA.5.1).
Proposition D.1.7 LetD be a polynomial domain over a fieldK. Consider two polynomialp(X)and q(X)of respective degreesn >0 andm >0, overD. The resultantR∈Dof these two polynomials can be written in the form
R=α(X)p(X)+β(X)q(X)
withα(X),β(X) two polynomials of respective degrees at mostm−1 andn−1 overD.
Proof Consider the following equalities
p(X)=a0+a1X+ ã ã ã +anXn p(X)X=a0X+a1X2+ ã ã ã +anXn+1
...
p(X)Xm−1= ã ã ã + ã ã ã +anXn+m−1 q(X)=b0+b1X+ ã ã ã + ã ã ã q(X)X=b0X+b1X2+ ã ã ã + ã ã ã
...
q(X)Xn−1= ã ã ã + ã ã ã +bnXm+n−1.
Writeρ1, . . . , ρn+mfor the cofactors of the elements of the first column of the de- terminant defining the resultantR(see DefinitionD.1.3). Multiply thei-th equality
384 D Resultants above byγi and add the results. The sum on the left hand side takes the expected form
ρ0+ρ1X+ ã ã ã +ρmXm−1
p(X)+
ρm+1+ρm+2X+ ã ã ã +ρm+nXn−1 q(X) with indeed
α(X)=ρ0+ρ1X+ã ã ã+ρmXm−1, β(X)=ρm+1+ρm+2X+ã ã ã+ρm+nXn−1 polynomials inD[X]of degrees at mostm−1 andn−1.
It remains to prove that the sum on the right hand side is equal toR. By definition of a determinant
R=a0ρ0+b0ρm+1
(the sum of the elements of the first column, multiplied by their cofactors). This takes care of the elements in the “first column” of the equalities above. For the
“second column” of these equalities we have, again by the well-known properties of a determinant
a1ρ0+a0ρ1+b1ρm+1+b0ρm+2=0
(the sum of the elements of the second column, multiplied by the cofactors of the elements of the first column). Thus the sum of the elements of the second “column”
becomes
(a1ρ0+a0ρ1+b1ρm+1+b0ρm+2)X=0,
and so on for the other “columns”.