Resultants. summary and questions. December 7, 2011

Size: px

Start display at page:

Download "Resultants. summary and questions. December 7, 2011"

Cody Strickland
5 years ago
Views:

1 Resultants summary and questions December 7, An exercise about free modules Let A be a unitary commutative integral ring. Let K be the fraction field of A. Let n 1 be an integer. Set V = A n and V = K n. We have a natural inclusion V V. The canonical basis of the free A-module V is also the canonical basis of the K-vector space V. Let M n n (A) be the set of n n matrices with entries in A. Let M M n n (A) be such an n n matrix with entries in A. We can associate to M an endomorphism Φ of the A-module V, namely the endomorphism whose matrix in the canonical basis is M. We can associate to M an endomorphism φ of the K-vector space V, namely the endomorphism whose matrix in the canonical basis is M. Of course Φ is the restriction of φ to V. Question 1 Deduce that φ is injective if and only if Φ is injective, if and only if the determinant of M is non-zero. What about surjectivity? 2 The comatrix Let A be a communtative ring with unit and let m, n be two positive integers. Let M m n (A) be the set of m n matrices with entries in A. We note C m (A) = M m 1 (A) the set of column vectors of length m. This is a rank m free A- module. Let M M m m (A). We associate to M an endomorphism of A-modules (an A-linear map) Φ M : C m (A) C m (A) defined by Φ M (c) = Mc. We note N the transposed of the comatrix of M and Φ N : C m (A) C m (A) the associated A-linear map. We know that MN is det(m) times the identity matrix. So Φ M Φ N = det(m) Id. In particular, the image of Φ M contains det(m)c m (A) : if c has the form c = det(m)d with d C m (A) then set e = Φ N (d). We have Φ M (e) = Φ M (Φ N (d)) = det(m)d = c. 1

2 3 Definition of the resultant Let again A be a commutative integral ring with unit. Let K be its fraction field. Let f and g be two polynomials in A[x]. We set l = deg(f) and m = deg(g) and we assume that m 1 and l 1. For every integer k 0 we call A[x] k the A-module consisting of all polynomials with coefficients in A having degree k. The A-module V = A[x] m 1 A[x] l 1 is free of rank m + l. We choose a natural basis. The polynomials x m 1, x m 2,..., x, 1 form a basis of A-module A[x] m 1. The polynomials x l 1, x l 2,..., x, 1 form a basis of the A-module A[x] l 1. The pairs (x m 1, 0), (x m 2, 0),..., (1, 0), (0, x l 1 ), (0, x l 2 ),..., (0, 1) form a basis Θ for the A-module V = A[x] m 1 A[x] l 1. If P = 0 k m 1 p kx k and if Q = 0 k l 1 q kx k, then the pair (P, Q) lies in V and its coordinates in the basis Θ form the column vector t (p m 1, p m 2,..., p 0, q l 1, q l 2,..., q 0 ). So we have an isomorphism between V and C m+l. We are interested in the A-linear map defined by µ : V = A[x] m 1 A[x] l 1 A[x] m+l 1 µ(c, D) = Cf + Dg. We set W = A[x] m+l 1. We note that W is a rank m + l free A-module. The polynomials x m+l 1, x m+l 2,..., x, 1, form a basis Ω of this A-module. The matrix of the linear map µ in the bases Θ (of V) and Ω (of W) is called the Sylvester matrix of the polynomials f and g. We denote it Syl(f, g). This is a square matrix of size (m + l) (m + l). Its determinant is called the resultant of f and g and denoted Res(f, g). Remark : one sometimes writes Syl l,m (f, g) and Res l,m (f, g) to stress the dependency in l and m. Otherwise, l is assumed to be the degree of f and m the degree of g. We note that Res(f, g) lies in A and is zero if and only if the linear map µ is not injective. See section 1. So Res(f, g) = 0 if and only if the exist two polynomials C and D in A[x] such that deg(c) m 1 and deg(d) l 1 and Cf + Dg = 0 and (C, D) (0, 0). This amounts to saying that the gcd of f and g in K[x] is not a constant : there exist h, F, G K[x] such that f = F h and g = Gh and deg(h) 1. Question 2 Why? Another important remark : considerations in section 2 show that the image of µ contains Res(f, g)w. In particular, it contains the constant polynomial 2

3 Res(f, g). So there exist two polynomials C and D in A[x] such that Cf +Dg = Res(f, g). We write the Sylvester matrix in a simple case. We assume l = 2 and and m = 3 and f(x) = f 2 x 2 + f 1 x + f 0 g(x) = g 3 x 3 + g 2 x 2 + g 1 x + g 0. Here V = A[x] 2 A[x] 1. The image of (x 2, 0) by µ is x 2 f(x) + 0 g(x) = x 2 f(x). The image of (x, 0) is xf(x). The image of (1, 0) is f(x). The image of (0, x) by µ is 0 f(x) + x g(x) = xg(x). The image of (0, 1) by µ is g(x). The matrix of µ in the bases Θ (of V) and Ω (of W) is Syl(f, g) = f g 3 0 f 1 f 2 0 g 2 g 3 f 0 f 1 f 2 g 1 g 2 0 f 0 f 1 g 0 g f 0 0 g 0 If f = f 0 is a non-zero constant, it is natural to take for Syl(f, g) the scalar matrix of order deg(g) = m Syl(f 0, g) = Diag(f 0,..., f 0 ). So Res(f 0, g) = f deg(g) 0 and Res(f, g 0 ) = g deg(f) 0 if f 0 and g 0 are non-zero constants and f and g are non-zero polynomials. Finaly we set Res(0, g) = Res(f, 0) = 0. 4 First properties Some simple properties can be proved by operating on columns. If m l then Res(f, g) = ( 1) lm Res(g, f). Res(f, g) = f m d l Res(f, g%f) where d is the degree of g%f. If m l then where d is the degree of f%g. Res(f, g) = g l d m ( 1) m(l d) Res(f%g, g) 3

4 5 The K-algebra K[x]/f(x) Let f K[x] be a polynomial with degree l 1. We write Q for the K-algebra K[x]/f(x). For every polynomial g(x) K[x] with degree m 0 we denote by φ g the map from Q to Q that sends h mod f onto g h mod f. The family (x l 1 mod f, x l 2 mod f,..., x 2 mod f, x mod f, 1 mod f) is a K-basis of Q. We write the matrix of φ g in this basis. We compare it to the Sylvester matrix Syl(f, g). We check that Res(f, g) = f m l det(φ g ). We deduce that Similarly Res(f, g 1 g 2 ) = Res(f, g 1 ) Res(f, g 2 ). Res(f 1 f 2, g) = Res(f 1, g) Res(f 2, g). If K is an algebraic closure of K, then there exist α1,..., α l and β 1,..., β m in K such that f = f l (x α 1 )... (x α l ) and g = g m (x β 1 )... (x β m ). Applying repeatedly the above multiplicativity rules and noting that we prove that Equivalently Res(f, g) = f m l Res(x α, x β) = α β, g l m Res(f, g) = ( 1) lm g l m 6 First applications 1 i l 1 j m and xs 1 j m f(β j ) = f m l (α i β j ). 1 i l g(α i ). Question 3 Let K be a field such that P (x) = x 3 1 and Q(x) = x 2 + 3x + 1 have a common root. Which is the characteristic of K? We compute the resultant of these two polynomials. Since they are unitary, the Sylvester matrix does not depend on K. >> polylib::resultant(x^3-1,x^2+3*x+1,x); 20 Since 20 = the resultant is zero if and only if the characteristic p of K is 2 or 5. We assume that p = 2 and compute the gcd of P (x) and Q(x) in F 2 [x]. R := Dom :: IntegerMod(2); P := poly(x^3-1,r); Q:= poly(x^2+3*x+1,r); >> gcd(p,q); 2 poly(x + x + 1, [x], Dom::IntegerMod(2)) 4

5 We assume now that p = 5 and compute the gcd of P (x) and Q(x) in F 5 [x]. R := Dom :: IntegerMod(5); P := poly(x^3-1,r); Q:= poly(x^2+3*x+1,r); >> gcd(p,q); poly(x + 4, [x], Dom::IntegerMod(5)) Remark : Let A and B be two commutative integral rings with unit. Let φ : A B be a ring homomorphism. There exists a ring homomorphism Φ : A[x] B[x] such that Φ(x) = x and the restriction of Φ to A is φ. Let P (x) and Q(x) be two non-zero polynomials in A[x]. Let p 0 and q 0 be their leading coefficients. Assume φ(p 0 ) 0 and φ(q 0 ) 0. The the resultant of Φ(P ) and Φ(Q) is φ(res(p, Q)). Question 4 Let C 1 C 2 be the plane curve with equation x 2 + 2y 3 = 3. Let C 2 C 2 be the plane curve with equation x 2 + xy + y 2 = 3. Let P = (x P, y P ) C 2 be a point in the intersection of C 1 and C 2. What can we say about y P? We note f(x, y) = x 2 + 2y 3 3 and g(x, y) = x 2 + xy + y 2 3 the two polynomials involved in the question. Let V (f, g) be the set of points P = (x P, y P ) such that f(x P, y P ) = 0 and g(x P, y P ) = 0. We say that V (f, g) is an algebraic set. Let I be the ideal in C[x, y] generated by f and g. Let h be a polynomial in this ideal. Then for every point P = (x P, y P ) in V we have h(x P, y P ) = 0. Set A = C[y]. We identify C[x, y] and A[x]. We denote by Res x (f, g) the resultant of f and g seen as polynomials in the indeterminate x having coefficients in A = C[y]. We have seen that the resultant belongs to A and also to the ideal generated by f and g : Res x (f, g) C[y] I. So Res x (f, g) is an equations resulting from f and g and depending only on one parameter y. We say that we have eliminated x. We set r(y) = Res x (f, g) and we compute f:=x^2+2*y^3-3; g := x^2+x*y+y^2-3; >> r:=polylib::resultant(f,g,x); y - 2 y + y - 3 y >> factor(r); y (y - 1) (3 y + 2 y + 4 y + 3) So every point P = (x P, y P ) V satisfies y 2 P (y P 1)(4y 3 P + 2y 2 P + 3y P + 3) = 0. 5

6 If y P = 0 we have f(x P, 0) = 0 = x 2 P 3 and g(x P, 0) = 0 = x 2 P 3. So x P = ± 3. We check that ( 3, 0) and ( 3, 0) lie in the intersection of C 1 and C 2. If y P = 1 we have f(x P, 1) = 0 = x 2 P 1 and g(x P, 0) = 0 = x 2 P + x P 2. We compute the gcd if x 2 1 and x 2 + x 2 : >> gcd(x^2-1,x^2+x-2); x - 1 So we find the point with coordinates (1, 1). If y P is a root of 4y 3 + 2y 2 + 3y + 3 = 0 then we check that x P = 2y 2 P y P. The three roots of 4y 3 + 2y 2 + 3y + 3 = 0 are distinct : >> polylib::discrim(4*y^3+2*y^2+3*y+3,y); We have found 3 more intersection points. So have found a total of 6 intersection points. Question 5 Let α 1, α 2, α 3 be the three roots of A(x) = x 3 + 5x + 7. Compute B(x) = (x α 2 1 )(x α2 2 )(x α2 3 ) without computing the α i. Let C(x) be a polynomial with degree 1. Give a formula for (x C(α 1 ))(x C(α 2 ))(x C(α 3 )). A:=x^3+5*x+7; B:=x^2-y; polylib::resultant(a,b,x); 3 2 y + 10 y + 25 y - 49 Question 6 Let α 1, α 2, α 3 be the three roots of A(x) = x 3 + 5x + 7. Let β 1, β 2, β 3, β 4 be the roots of B(x) = x x Give a formula for (X α i β j ). 1 i 3 1 j 3 A:=x^3+5*x+7; B:=y^4+11*y+13; C:=collect(subs(B,y=z/x)*x^4,x); x + 11 x z + z C:=C/coeff(C,x,4); x z z 6

7 x >> r:=polylib::resultant(a,c,x); z 231 z 50 z z 1925 z 1605 z z z >> numer(r); z z z z z z z z >> 13^3*7^4; Question 7 Let A be an integral ring and K its fraction field. Let K be an algebraic closure of K. An element in K is said to be integral over A if it is canceled by a unitary polynomial with coefficients in A. Prove that the set of integral elements is a subring of K containing A. Question 8 We consider the plane curve C R 2 parametrized by the equations and Give an equation for this curve. A:=(1+t^2)^3*x-4*t*(1-t^2)^2; B:=(1+t^2)^3*y-8*t^2*(1-t^2); r:=polylib::resultant(a,b,t); x(t) = 4t(1 t2 ) 2 (1 + t 2 ) 3, y(t) = 8t2 (1 t 2 ) (1 + t 2 ) x x y x y x y y 7

8 >> factor(r); (- 4 x y + 3 x y + 3 x y + x + y ) 7 The discriminant Let K be a field. If P (x) K[x] is a degree d unitary polynomial we define the discriminant of P to be Disc(P ) = ( 1) d(d 1) 2 Res(P, P ) = ( 1) d(d 1) 2 P (α i ) (1) = ( 1) d(d 1) 2 1 i d 1 i d (α i α j ) = 1 j d j i 1 i<j d (α i α j ) 2 (2) where d is the degree of P, and α 1, α 2,..., α d are the roots of P in some algebraic closure of K. Question 9 Which is the degree of P? so In general, if P d is the leading coefficient of P, we set Disc(P ) = P 2d 2 d (α i α j ) 2 1 i<j d P d Disc(P ) = ( 1) d(d 1) 2 Res d,d 1 (P, P ). 8

1. Algebra 1.5. Polynomial Rings

1. ALGEBRA 19 1. Algebra 1.5. Polynomial Rings Lemma 1.5.1 Let R and S be rings with identity element. If R > 1 and S > 1, then R S contains zero divisors. Proof. The two elements (1, 0) and (0, 1) are