Unirational Fields of Transcendence Degree One and Functional Decomposition

Transcription

1 Unirational Fiels of Transcenence Degree One an Functional Decomposition Jaime Gutierrez, Rosario Rubio an Davi Sevilla Dept of Mathematics, Statistics an Computation University of Cantabria Santaner, Spain ABSTRACT In this paper we present an algorithm to compute all unirational fiels of transcenence egree one containing a given finite set of multivariate rational functions In particular, we provie an algorithm to ecompose a multivariate rational function f of the form f = g(h), where g is a univariate rational function an h a multivariate one 1 INTRODUCTION Let K be an arbitrary fiel an K(x) = K(x 1,, x n) be the rational function fiel in the variables x = (x 1,, x n) A unirational fiel over K is an intermeiate fiel F between K an K(x) We know that any unirational fiel is finitely generate over K (see [6]) In the following whenever we talk about computing an intermeiate fiel we mean that such finite set of generators is to be calculate The problem of fining unirational fiels is a classical one In this paper we are looking for unirational fiels F over K of transcenence egree one over K, treg(f/k) = 1 In the univariate case, the problem can be state as follows: given univariate rational functions f 1,, f m K(y), we wish to know if there exists a proper intermeiate fiel F such that K(f 1,, f m) F K(y); an in the affirmative case, to compute it By the classical Lüroth theorem (see [14]) the problem is ivie in two parts: first to compute f such that K(f 1,, f m) = K(f), an secon to ecompose the rational function f, ie, to fin g, h K(y) such that, F = K(h) with f = g(h) Constructive proofs of Lüroth s theorem can be foun in [7], [12] an [1] Algorithms for ecomposition of univariate rational functions can be foun in [17] an [1] In the multivariate case, the problem is: given f 1,, f m in K(x) we wish to know if there exists a proper inter- This work is partially supporte by Spanish DGES Grant Project PB Permission to make igital or har copies of all or part of this work for personal or classroom use is grante without fee provie that copies are not mae or istribute for profit or commercial avantage, an that copies bear this notice an the full citation on the first page To copy otherwise, to republish, to post on servers or to reistribute to lists, requires prior specific permission an/or a fee ISSAC 2001, UWO, Canaa c 2001 ACM / 00/ 0008 $500 meiate fiel F such that K(f 1,, f m) F K(x) with treg(f/k) = 1; an in the affirmative case, to compute it A central result is the following generalization of Lüroth s theorem: Theorem 1 (Extene Lüroth s Theorem) Let F be a fiel such that K F K(x) If treg(f/k) = 1, then there exists f K(x) such that F = K(f) Such an f is calle a Lüroth s generator of the fiel F This theorem was first prove in [2] for characteristic zero an in [4] in general, see also [11] Theorem 3 Using Gröbner basis computation, the paper [5] provies an algorithm to compute a Lüroth s generator, if it exits See also [9] for another algorithmic proof of this result In this paper we present a new algorithm, which only requires to compute gc s, to etect if a unirational fiel has transcenence egree 1 an, in the affirmative case, to compute a Lüroth s generator We also present a constructive proof of the above theorem for polynomials (see [8]): if the unirational fiel contains a nonconstant polynomial, then it is generate by a polynomial By the Extene Lüroth s theorem, to fin an intermeiate fiel of transcenence egree one is equivalent to the following: first to fin a Lüroth s generator f, ie, K(f 1,, f m) = K(f), if it exists, an secon to ecompose the multivariate rational function f, ie, to fin g K(y) an h K(x) such that f = g(h) in a nontrivial way The pair (g, h) is calle a uni multivariate ecomposition of f We present two algorithms to compute a nontrivial uni multivariate ecomposition of a multivariate rational function, if it exits This paper is ivie in four sections In section 2 we state the proof of the Extene Lüroth s theorem an its polynomial version In section 3 we present an analyze two algorithms to compute a uni multivariate ecomposition of a rational function, if it exists In section 4 we iscuss the performance of these algorithms 2 THE EXTENDED LÜROTH THEOREM In this section we present an algorithm to the following computational problem: given f 1,, f m K(x), to compute a Lüroth generator f for F = K(f 1,, f m) if it exists, moreover we etect if F contains a non-constant polynomial, an in the affirmative case we fin a generating polynomial We start with the following efinition: 1

2 Definition 1 Let p K[x 1,, x n, y 1,, y n] = K[x, y] be a non constant polynomial We say that p is near separate if there exist non constant polynomials r 1, s 1 K[x] = K[x 1,, x n] an r 2, s 2 K[y] = K[y 1,, y n], such that neither r 1, s 1 are associate, nor r 2, s 2 are associate an p = r 1s 2 r 2s 1 In the particular case when p = r(x 1,, x n)s(y 1,, y n) s(x 1,, x n)r(y 1,, y n), we say that p is a symmetric near separate polynomial We say that (r, s) is a symmetric near separate representation of p In this paper, eg x1,,x n will enote the total egree with respect to the variables x 1,, x n an eg will enote the total egree with respect to all the variables Also, given a rational function f we will also e note as f n, f the numerator an enominator of f, respectively In the following theorem we give some basic properties of near separate polynomials, for later use Theorem 2 Let p K[x, y] be a near separate polynomial an r 1, s 1, r 2, s 2 as in the above efinition Then (i) If gc(r 1, s 1) = 1 an gc(r 2, s 2) = 1, p has no factors in K[x] or K[y] (ii) eg x1,,x n p = max{eg r 1, eg s 1} an eg y1,,y n p = max{eg r 2, eg s 2} (iii) If p is symmetric an (α 1,, α n) K n verifies p(x 1,, x n, α 1,, α n) 0, then there exists (r, s), a symmetric near separate representation of p, such that r(α 1,, α n) = 0 an s(α 1,, α n) = 1 (iv) If p is symmetric, the coefficient of x i 0 k y j 0 k near separate polynomial a i0 b j0 b i0 a j0, in p is the where a i is the coefficient of x i k in r an b i is the coefficient of x i k in s Proof (i) Suppose v K[x 1,, x n] is a non constant factor of p Then there exists i such that eg xi v 1 Without loss of generality we will suppose that i = 1 Let α be a root of v, consiering p as a univariate polynomial in the variable x 1, in a suitable extension of K[x 2,, x n] If α is a root of any of the polynomials r 1 or s 1, then it is also a root of the other This is a contraiction, because gc(r 1, s 1) = 1 Therefore α is neither a root of r 1 nor of s 1 Then, r 1(α, x 2,, x n) r2(y1,, yn) = s 1(α, x 2,, x n) s 2(y 1,, y K n) A contraiction again, since r 2, s 2 are not associate in K (ii) If eg r 1 eg s 1, the equality is trivial Otherwise, if eg r 1 = eg s 1 > eg x1,,x n p, the terms with greatest egree with respect to x 1,, x n vanish This is a contraiction, because r 2, s 2 are not associate The proof is the same for r 2, s 2 (iii) Let (r, s) be a representation of p If r(α 1,, α n) = 0, since p(x 1,, x n, α 1,, α n) 0, we have s(α 1,, α n) 0 Then we have a new near separate representation: r s(α 1,, α n), s s(α 1,, α n) If s(α 1,, α n) = 0, then the representation ( s, r) we are in the previous case If r(α 1,, α n), s(α 1,, α n) 0, then we consier the representation s r s(α 1,, α n) s r(α 1,, α n), s(α 1,, α n) (iv) This is a simple routine confirmation Now, we state an important theorem that relates uni multivariate ecompositions to near separate polynomials, that is prove in [10]: Theorem 3 Let A = K(x) an B = K(y) be rational function fiels over K Let f, h A an f, h B be non constant rational functions Then the following statements are equivalent: A) There exists a rational function g K(t) satisfying f = g(h) an f = g(h ) B) h h ivies f f in A K B As a consequence, a rational function f K(x) verifies f = g(h) for some g, h if an only if h n(x) h (y) h (x) h n(y) ivies f n(x) f (y) f (x) f n(y) Given an amissible monomial orering > in a polynomial ring an a nonzero polynomial G in that ring, we enote by lm G the leaing monomial of G with respect to > an lc G its leaing coefficient Algorithm 1 Input: f 1,, f m K(x) Output: f K(x) such that K(f) = F = K(f 1,, f m), if it exists Otherwise, returns null A Let > be a grae lexicographical orering for y = (y 1,, y n) Let i = m B Let F k = f k n(y) f k (x)f k (y) for k = 1,, i C Compute H i = gc({f k, k = 1,, i}) with lc H i = 1 D If H i = 1, RETURN NULL (F oes not have transcenence egree 1 over K) If there exists j {1,, i} such that lm H i = lm F j, then RETURN f j Otherwise, let f i+1 be a coefficient of H i in F \ K Increase i an go to B 2

3 Correctness proof If F has transcenence egree 1 over K, we can write F = K(f) By Theorem 3, f n(y) f(x)f (y) ivies H i for any i Therefore, H i is non constant if a Lüroth generator exists If there are i, j such that lm H i = lm F j, then F j is a greatest common ivisor of {F k, k = 1,, i} Therefore, F j ivies F k for every k Fix such a k Let q = f k n (y){f j }, s = f k (y){f j } the normal form with respect to the monomial orering >; then there exist p, q, r, s F[y] such that f j n (y) = p(y)f i q(y) f j (y) = r(y)f i s(y) where lm F j oes not ivie any monomial of q or s By theorem 2(i), q, s 0 By the efinition in step B, F k = F j(p r f k (x)) (q s f k (x)) Hence F j ivies q s f k (x 1,, x n) an we conclue that q s f k (x) = 0, since otherwise we woul get that lm F j ivies lm (q s f k (x 1,, x n)), which contraicts the choice of the polynomials q, s Thus f k (x 1,, x n) = q s F = K(f j) Now we suppose that lm H i < lm F k for all k Again, fix a value for k Then there exists a C F[y] \ F such that F k = H ic Let, α be the lowest common multiples of the enominators of the coefficients of H i an C, respectively Then D = H i, C = αc K[x, y] Since H i is monic, the polynomial D is primitive Then, f k n(y) f k (x) f k n(x) f k (y) = D an by theorem 2, C α f k f k n(y) f k (x) f k n(x) f k (y) = D b C, bc K[x, y] On one han, D K[y], thus D (an H i) have a non constant coefficient On the other han, C b K[y], then the non constant coefficients of D in the ring K(x)[y] have smaller egree than that of f k (x) The choice of assures that the coefficients of H have smaller egree than f k Summarizing, there exists a coefficient a F of H i that can be ae to the list of generators an has smaller egree than them If treg (F/K) = 1, H i is non constant for all i, an the generator has smaller egree than the others Therefore, the algorithm ens in a finite number of steps Finally, we note that complexity is ominate in the step C by computing gc s of multivariate polynomials, so the algorithm is polynomial in the egree of the rational functions an in n (see [16]) From the fact that the Lüroth generator can be foun with only some gc computations, we obtain that if f is a Lüroth generator of K(f 1,, f n) then it is also a Lüroth generator of K (f 1,, f n) for any fiel extension K of K, K K Example 1 Let Q(f 1, f 2) Q(x, y, z) where Let f 1 = y2 x 4 2y 2 x 2 z + y 2 z 2 + x 2 2xz + z 2 yx 3 yxz yzx 2 + z 2 y f 2 = Compute y 2 x 4 2y 2 x 2 z + y 2 z 2 x 2 2xz + yx 3 yxz + z 2 yzx 2 + z 2 y F i = f i n(s, t, u) f i(x, y, z)f i (s, t, u), i = 1, 2 H 2 = gc(f 1, F 2) = tu + s 2 t + x2 y zy u + x2 y + zy s Since lm H 2 < lm F i with respect to the lexicographical orering s > t > u, we take a non constant coefficient of H 2: f 3 = x2 y zy Now H 3 = tu + s 2 t + x2 y zy u + x2 y + zy s an H 3 = F 3, since H 3 = H 2 The algorithm returns f 3, a Lüroth generator of Q(f 1, f 2) It is important to highlight that when the fiel F contains a non constant polynomial you can compute a polynomial as a generator, an this generator neither epens on the groun fiel K This result was prove in [8], for zero characteristic A general proof can be foun in [11] Algorithm 2 Input: f 1,, f m K(x) Output: f K[x] such that K(f) = F = K(f 1,, f m), if it exists Otherwise, returns null A Compute a Lüroth generator f of K(f 1,, f m) using Algorithm 1 B Let s be the egree of f If s > eg f n an f n is not constant, return null Otherwise, let f = 1/f If s > eg f an f is not constant, return null Otherwise, let f = f Let f n (s) (s), f be the homogeneous components of egree s of f n, f, respectively Let a = f n (s) If f (s) a or f n af are not constant, return null Otherwise, let f = 1 y a f Correctness proof Once a Lüroth s generator has been compute, take a generator f with egree m such that if f = fn an f f n f = f n (s) + + f n (0), = f (s) + + f (0), 3

4 the sum in homogeneous polynomials, then either f (s) = 0 or f n (s) K f (s) If p K(f) is a polynomial, then there exists g K(y) with egree r such that p = g(f) If g = aryr + + a 0 b ry r + + b 0, p = arf n r + + a 0f r b rfn r + + b 0f r (s) ar(f n + + f n (0) ) r + + a 0(f (s) + + f (0) = )r b r(f n (s) + + f n (0) ) r + + b 0(f (s) + + f (0) )r Since p is a polynomial, the egree of the previous enominator is smaller than the egree of the numerator Therefore b rf n (s) r + + b0f (s) r = 0 If f (s) = 0 then b r = 0 an p = a rf r n + + a 0f r f (b r 1f r 1 n + + b 0f r 1 ) Hence f ivies the numerator of p, an therefore ivies f n This proves that f is a polynomial! If, on the contrary, f (s) f n (s) 0, g = 0 Contraiction, since f n (s) K f (s) f (s) 3 TWO UNI MULTIVARIATE DECOMPO- SITION ALGORITHMS We efine the egree of a rational function f = f n/f K(x) as eg f = max {eg f n, eg f } if gc(f n, f ) = 1 The following efinition was introuce in [15] for polynomials Definition 2 Let f, h K(x) an g K(y) such that f = g(h) Then we say that (g, h) is a uni-multivariate ecomposition of f It is non-trivial if 1 < eg h < eg f The rational function is uni-multivariate ecomposable if there exits a non-trivial ecomposition If f is a polynomial having a nontrivial uni-multivariate ecomposition, then by Algorithm 2 we get that there exits a uni-multivariate ecomposition (g, h) with g an h polynomials The paper [15] provies an algorithm to compute a nontrivial uni-multivariate ecomposition of a polynomial f of egree m that only requires O(nm(m + 1) n log m) arithmetic operations in the groun fiel K The known techniques for ecomposition all ivie the problem into two parts Given f, in orer to fin a ecomposition f = g(h), 1 one first computes caniates h, 2 then computes g given h Determining g from f an h is a subfiel membership problem The paper [13] gives a solution to this part We also present another faster metho, that only requires solving a linear system of equations Usually, the harer step is to fin caniates for h One goal in ecomposition is to have components of smaller egree than the compose polynomial This will be the case here 31 Preliminary results First, we state some results that will be use in the algorithms presente later On the properties to highlight out of uni-multivariate ecomposition is the goo behaviour of the egree with respect to this composition Theorem 4 Let g K(y) an h K(x), an f = g(h) Then eg f = eg g eg h Proof Let g = gn g an h = hn h with gc(g n, g ) = 1 an gc(h n, h ) = 1 Then there exist polynomials A, B K[y] such that g n(y) A(y) + g (y) B(y) = 1 Homogenizing the polynomials g n, g, A, B we obtain, respectively, the bivariate polynomials eg n(y 1, y 2), eg (y 1, y 2), ea(y 1, y 2), e B(y 1, y 2) verifying eg n(y 1, y 2) e A(y 1, y 2) y u 2 + eg (y 1, y 2) e B(y 1, y 2) y v 2 = y w 2 with either u = 0 or v = 0 an w =max{u, v} Therefore, eg n(h n, h ) e A(h n, h ) h u + eg (h n, h ) e B(h n, h ) h v = h w If is an irreucible factor of gc(eg n(h n, h ), eg (h n, h )), then ivies h On the other han, ivies eg n(h n, h ) an eg (h n, h ); this implies that ivies h n As a consequence, gc(eg n(h n, h ), eg (h n, h )) = 1 So, f = egn(hn, h ) eg (h n, h ) ha, a = eg h n eg h is in reuce form Without loss of generality, we can take eg g n = r n r =eg g with g n(y) = a rn y rn + + a 0 g (y) = b r y r + + b 0 Then, eg f = max {eg eg n(h n, h ), eg eg (h n, h ) h rn r } an eg n(h n, h ) = a rn h rn n + + a 0h rn eg (h n, h ) = b r h r n + + b 0h r If eg eg n(h n, h ) = r n eg h, we immeiately obtain that eg f = eg g eg h If eg eg n(h n, h ) < r neg h, then s = eg h n = eg h Write h n = h (s) n h = h (s) + h n (s 1) + + h (0) n + h (s 1) + + h (0) where h n (j), h (j) are the homogeneous components of h n an h with egree j, respectively Since the! egree ecreases, eg eg n(h n (s), h (s) h n (s) h (s) ) = 0 an eg h (s) egn n h (s) = 0 Therefore, K In this case, you can take h K(x) a rational function with eg h = egh, eg h n eg h an such that f = g (h ) for some g K(y) with eg g = eg g Uner these hypothesis, we prove before that eg f = eg g eg h = eg g eg h 4

5 Corollary 1 Let g = g n/g with g n = a uy u + + a 0, g = b vy v + +b 0 an h = h n/h verifying gc(g n, g ) = gc(h n, h ) = 1 If f = f n/f = g(h) with f n = (a u h u n + + a 0 h u ) h max{v u,0} f = (b v h v n + + b 0 h v ) h max{u v,0} then gc(f n, f ) = 1 Proof It is easy to prove that eg f n, eg f max{u, v} max{eg h n, eg h } If gc(f n, f ) 1, then eg f < eg g eg h, contraicting theorem 4 Corollary 2 Given f an h, if there exists g such that f = g(h), then g is unique Furthermore, it can be compute from f an h by solving a linear system of equations Proof If f = g 1(h) = g 2(h), then (g 1 g 2)(h) = 0, an by theorem 4, eg (g 1 g 2) = 0, thus g 1 g 2 is constant It is then clear that it must be 0, that is, g 1 = g 2 Again by theorem 4, the egree of g is etermine by those of f an h We can write g as a function with the corresponing egree an unetermine coefficients Equating to zero the coefficients of the numerator of f g(h), we obtain a linear homogeneous system of equations in the coefficients of g If we compute a non trivial solution to this system, we fin g Definition 3 Let f K(x) be a rational function Two uni multivariate ecompositions (g, h) an (g, h ) of f are equivalent if there exists a composition unit l K(y), ie, eg l = 1, such that h = l(h ) Corollary 3 Let f K(x) be a non constant rational function Then the equivalence classes of uni multivariate ecompositions of f correspon bijectively to intermeiate fiels F, K(f) F K(x), with transcenence egree 1 over K Proof The bijection is {[(g, h)], f = g(h)} {K(f) F, (F/K) = 1} [(g, h)] F = K(h) Suppose we have a uni multivariate ecomposition (g, h) of f Since f = g(h), F = K(h) is an intermeiate fiel of K(f) K(x) with transcenence egree 1 over K Also, if (g, h ) is equivalent to (g, h), h = l(h ) for some composition unit l K(y) Consequently, h = l 1 (h) an K(h) = K(h ) Let (g, h) an (g, h ) be two uni multivariate ecompositions of f such that K(h) = K(h ) Then there exists rational functions l, l K(y) such that h = l(h ) an h = l (h) By theorem 4, eg l(l ) = 1 an eg l = eg l = 1 By the uniqueness (see Corollary 2) of the left component, y = l(l ) So, l K(y) is a composition unit an (g, h), (g, h ) are equivalent By Theorem 1, there exist h K(x) an g K(y) such that F = K(h) an f = g(h) 32 First algorithm We now procee with the first algorithm for computing caniates h = h n/h This algorithm is base on Theorem 3 Since h n(x) h (y) h (x) h n(y) ivies f n(x) f (y) f (x)f n(y), one can compute caniates for h from f merely looking at the near-separate ivisors H = r(x) s(y) r(y)s(x) Next, the problem is: given a multivariate polynomial H = (x, y), how can one etermine if it is a symmetric near-separate polynomial? This is a consequence of theorem 2: Corollary 4 Given a polynomial p K[x, y], it is possible to fin a near separate representation (r, s) K[x] 2 of p, if it exists, by solving a linear system of equations with coefficients in K Moreover, any other solution (r, s ) of this linear system of equations gives an equivalent ecomposition Algorithm 3 Input: f K(x) Output: A uni multivariate ecomposition (g, h) of f, if it exists A Factor the symmetric polynomial p = f n(x) f (y) f (x) f n(y) Let D = {H 1,, H m} the set of factors of p (up to prouct by constants) Let i = 1 B Check if H i is a symmetric near separate polynomial If H = r(x) s(y) r(y) s(x), then h = r is a right s component for f; compute the left component g by solving a linear system (see Corollary 2) an RETURN (g, h) C If i < m, then increase i an go to B Otherwise, RE- TURN NULL (f is uni multivariate inecomposable) Example 2 Let f = 4z 4 y 2 8z 3 y 3 + 8z 2 yx + 4z 2 y 4 8zy 2 x +4x 2 2z 2 y + 2zy 2 2x + 10 The factorization of the polynomial f(x, y, z) f(s, t, u) is 2 2x 1 + 2s 2ut 2 + 2u 2 t 2zy 2 + 2z 2 y x s + z 2 y zy 2 u 2 t + ut 2 The first factor f 1 = 2x 1+2s 2ut 2 +2u 2 t 2zy 2 +2z 2 y is not symmetric near separate because f 1(x, y, z, x, y, z) 0 On the other han, the secon factor f 2 = x s+z 2 y zy 2 u 2 t + ut 2 oes satisfy f 2(x, y, z, x, y, z) = 0 Note that by a previous remark, the components of the ecomposition can be consiere as polynomials Then f 2 can be written as f 2 = h(x, y, z) h(s, t, u) Taking h(x, y, z) = f 2(x, y, z, 0, 0, 0) = x+z 2 y zy 2, we check that it satisfies the previous equation (see theorem 2) The left component g is also a polynomial, an by theorem 4, has egree 2 Solving the equation f = g(h) we have the multi univariate ecomposition: Example 3 Let f = fn f (4t 2 2t + 10, x + z 2 y zy 2 ) with f n = y 2 x 2 + 2x 2 yz 2 2y 6 x + z 4 x 2 2z 2 xy 5 + y 10 81x 2 450xyz 625y 2 z 2, f = y 2 x 2 + 2x 2 yz 2 2y 6 x + z 4 x 2 2z 2 xy 5 + y x 2 900xyz 1250y 2 z 2 We look for all the intermeiate fiels of Q(f) Q(x, y, z) with transcenence egree 1 over Q First, we will factor the polynomial f n(x, y, z) f (s, t, u) f n(s, t, u) f (x, y, z) = 625f 1 f 2, 5

6 where f 1= xtz 2 u xt5 zsty zu 2 sy + zt 5 y 9 25 xz2 s 9 25 xu2 s 9 9 xys xyut xts sy5 + uty 5, f 2= xtz 2 u 9 25 xt5 + zsty + zu 2 sy zt 5 y 9 25 xz2 s xu2 s 9 9 xys xyut + xts sy5 + uty 5 We have f 1(x, y, z, x, y, z) 0, thus f 1 is not symmetric near separate On the other han, f 2(x, y, z, x, y, z) = 0 Moreover, f 2 = zt 5 y + uty t5 tz 2 u yut x + zty y5 + zu 2 y s z t U y sx We check that f 2 is symmetric near separate, by solving a linear system of equations Define f 2(x, y, z, 1, 0, 0) = r(x, y, z) = 9 25 xz xy y5 = 925 z2 925 y x y5 Next, we compute s 0(y, z) such that 9 25 y5 s 0(t, u) 9 25 t5 s 0(y, z) = zt 5 y + uty 5 Let s 0(y, z) = a 5(z) y a 0(z) Then a 1 = 25 9 z an a 0 = a 2 = a 3 = a 4 = a 5 = 0 Hence, s 0 = 25 zy an 9 = 1 Thus s = x zy, r1(y, z) s0(t, u) c10 s 1(y, z) = r 0(t, u) s(1, 0, 0) = 1 an (r, s) is a symmetric near separate representation of p: r = 9 25 xz xy y5 s = x zy Now we compute g, which is a univariate function with egree 2 Solving the corresponing linear system of equations we obtain 33 Secon algorithm g = 625t t For this algorithm, we suppose that K has sufficiently many elements If it is not the case, then we can ecompose f in an algebraic extension K[ω] of K, an then check if it is equivalent to a ecomposition with coefficients in K; this last step can be one by solving a system of linear equations in the same fashion as the computation of g The algorithm is base on Corollary 1; we will nee several technical results too Lemma 1 Let f K(x) Then for any amissible monomial orering > there are units u K(y), v i K(x i), i = 1,, n such that, if f = f n / f = u f(v 1,, v n), then lm f n > lm f, f n (0,, 0) = 0 an f (0,, 0) 0 Proof Let > be any amissible monomial orering Let u 1 K(y) be a unit such that f 1 = f 1n / f 1 = u 1(f) verifies lm f 1n > lm f 1 Such a unit always exists: If lm f n < lm f, let u 1 = 1/y If lm f n = lm f, let u 1 = (1/y) (y If lm f n > lm f, let u 1 = y lc fn lc f ) Let α = (α 1,, α n) K n such that f 1 (α) 0 (such a α exists if K has sufficiently many elements) Let v i = x i + α i, i = 1,, n an f 2 = f 2n/f 2 = f 1(v 1,, v n) Then f 2 (0,, 0) 0, an we can take u = y f2n(0,, 0) f 2 (0,, 0) so that f = u f(v 1,, v n) verifies all the conitions Lemma 2 Let i, j, k N with i < j k, P, Q K[x] an > an amissible monomial orering such that lm P > lm Q Then lm P j Q k j > lm P i Q k i Lemma 3 Let f = f n /f K(x) such that lm f n > lm f, f n (0,, 0) = 0 an f (0,, 0) 0 Then, for every uni multivariate ecomposition f = g(h) there exists an equivalent ecomposition f = g(h) with g = g n /g, eg g n > eg g an g n (0) = 0 (thus g (0) 0) Proof As in the proof of Lemma 1, there exists a unit u 1 such that if h 1 = u(h) = h 1n/h 1, then h 1n(0,, 0) = 0 Let g 1 = g(u 1 ) = (a uy u + + a 0)/(b vy v + + b 0) Then f = auhu 1n + + a 0h u 1 b vh v 1n + + h v u b0hv 1 1 an by Corollary 1, f = (b vh v 1n + + b 0h v 1)h max{u v,0} 1 As f (0,, 0) 0 an h 1n(0,, 0) = 0 we must have h 1 (0,, 0) 0 But f n (0,, 0) = 0 an f n = (a uh u 1n + + a 0h u 1)h max{v u,0} 1, thus a 0 = 0 Next, we will prove that there is an equivalent ecomposition verifying the conition on the egrees of the left component To that en, we will consier three cases Let > be any amissible monomial orering an w = eg g 1 = max{u, v} If lm h 1n < lm h 1 then using repeately Lemma 2, lm f n = lm h 1nh w 1 1 which contraicts our hypothesis < lm h w 1 = lm f If lm h 1n > lm h 1, then applying Lemma 2, lm f n = lm h u 1nh max{v u,0} 1 lm f = lmh v 1nh max{u v,0} 1 As lm f n > lm f by hypothesis, by Lemma 2 again we must have u > v, that is, eg g 1n > eg g 1 If lm h 1n = lm h 1 then, as in Lemma 1, we can cancel the leaing monomial of h 1 with a unit u 2 on the left, so that f = g 2(h 2) with lm h 2n > lm h 2 which is the previous case Let f = g(h) be a uni multivariate ecomposition of f with f = f n/f, g = (a uy u + + a 0)/(b vy v + + b 0) an 6

7 h = h n/h By the previous lemma, we can suppose u > v an g(0) = 0, ie a 0 = 0 Then, as f = auhu n + + a 1h nh u 1 (b vh v n + + b 0h v )hu v we have that h n f n an h f This is the key to the following algorithm Algorithm 4 Input: f K(x) Output: (g, h) a uni multivariate ecomposition of f, if it exists A Compute u, v 1,, v n as in Lemma 1 Let f = f n /f = u f(v 1,, v n) B Factor f n an f Let D = {(A 1, B 1),, (A m, B m)} be the set of pairs (A, B) such that A, B ivie f n, f respectively (up to prouct by constants) Let i = 1 C Check if there exists g K(y) with f = g(a i/b i); if such a g exists, RETURN u 1 (g), h(v 1 1,, v 1 n ) D If i < m, increase i an go to C, otherwise RETURN NULL (f is uni multivariate inecomposable) Example 4 Let f = 4z 4 y 2 8z 3 y 3 + 8z 2 yx + 4z 2 y 4 8zy 2 x +4x 2 2z 2 y + 2zy 2 2x + 10 as in Example 2 We take u = t 10 K(t) an v 1 = x, v 2 = y, v 3 = z Then f = 4z 4 y 2 8z 3 y 3 + 8z 2 yx + 4z 2 y 4 8zy 2 x +4x 2 2z 2 y + 2zy 2 2x We factor f = 2(x + z 2 y zy 2 )(2x 1 + 2z 2 y 2zy 2 ) We first take the caniate (x + z 2 y zy 2 ) We have to check if there are values of a 1, a 2 for which g = a 2t 2 + a 1t verifies f = g(x + z 2 y zy 2 ) We fin the solution a 2 = 4, a 1 = 2 Thus f = (4t 2 2t + 10)(x + z 2 y zy 2 ) Example 5 Let f = fn f with f n = y 2 x 2 + 2x 2 yz 2 2y 6 x + z 4 x 2 2z 2 xy 5 +y 10 81x 2 450xyz 625y 2 z 2, f = y 2 x 2 + 2x 2 yz 2 2y 6 x + z 4 x 2 2z 2 xy 5 +y x 2 900xyz 1250y 2 z 2, as in Example 3 Let > be the pure lexicographical orering with y > x > z Then lm f n = lm f = y 10 Following the proof of lemma 1, let u 1 = 1/(t 1), then u 1(f) = f 1n/f 1 with f 1n = y 2 x 2 + 2x 2 yz 2 2y 6 x + z 4 x 2 2z 2 xy 5 +y x 2 900xyz 1250y 2 z 2, f 1 = 81x xyz + 625y 2 z 2 Now, let α = (1, 0, 0), so that the enominator of the previous expression is non zero at the point α Then f 2n/f 2 = u(f(x + 1, y, z)) with f 2n = y 2 x 2 + 2y 2 x + y 2 + 2x 2 yz 2 + 4yz 2 x + 2yz 2 2y 6 x 2y 6 + z 4 x 2 + 2z 4 x + z 4 2z 2 xy 5 2z 2 y 5 + y x 2 324x xyz 900yz 1250y 2 z 2, f 2 = 81x x xyz + 450yz + 625y 2 z 2 Table 1: Average computing times (in secons) n Alg 3 Fact Alg 4 Fact As f 2n(0, 0, 0) = 162 an f 2 (0, 0, 0) = 81, if u 2 = t + 2, we have that u 2(u 1(f(x + 1, y, z))) = f = f n f verifies the conitions of Lemma 3 We factor f n an f : f n = z 2 + z 2 x + y + xy y 5 2, f = (9x yz) 2 As the egree is multiplicative an eg f = 10, an also lm h n > lm h, the possible values of h n, h are h n = z 2 + z 2 x + y + xy y 5, h {1, 9x yz, (9x yz) 2 } To check them, let g = a 2t 2 +a 1 t b 1 t+b 0 We substitute h in g an solve the homogeneous linear system obtaine by comparing the coefficients with those of f If h = 1, there is only the trivial solution, thus h is not a caniate for f If h = 9x yz, we get the non trivial solution a 2 = b 0 = 1, a 1 = b 1 = 0, thus f has a uni multivariate ecomposition (u 1 1 (u 1 2 (g)), h(x 1, y, z)) = ( 1+t2, z2 x+yx y 5 ) t 2 2 9x+25yz If h = (9x yz) 2, the only solution is the trivial one Therefore, any uni multivariate ecomposition of f is equivalent to the ecomposition (g, h) compute before 4 PERFORMANCE Both algorithms run in exponential time, since the number of caniates to be teste is, in the worst case, exponential in the egree of the input; the rest of the steps in both algorithms are in polynomial time However, in practical examples it seems that most of the time is spent in the factorization of the associate symmetric near separate polynomial, in Algorithm 3, or the numerator an enominator in Algorithm 4 We show in Table 1 the average times obtaine by running implementations of these algorithms in Maple VI (see [3]) on a collection of ranom functions The parameters are the number of variables n an the egree of the rational function We have also inclue the factorization times for each algorithm 7

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