2 JOEL V. BRAWLE, SHUHONG GAO, AND DONALD MILLS for all ; 2 G. Then for monc polynomals f and g whose coecents are n F q and whose roots le n G, the c

Transcription

1 Computng Composed Products of Polynomals Joel V. Brawley, Shuhong Gao, and Donald Mlls Abstract. If f(x) and g(x) are polynomals n F q[x] of degrees m and n respectvely, then the composed sum of f and g, denoted f g, s the degree mn polynomal whose roots are all sums of roots of f wth roots of g. Lkewse, the composed multplcaton of f and g, denoted f g, s the degree mn polynomal whose roots are all products of roots of f wth roots of g. In 1987, Brawley and Carltz dened a more general noton of polynomal composton, denoted by f g, for whch f g and f g are specal cases. They prove that when f and g are rreducble wth degrees m and n coprme, then f g s rreducble of degree mn. Ths gves us a way to obtan rreducbles of relatvely large degree usng rreducbles of smaller degrees. In ths paper, we descrbe several methods of computng polynomal compostons of the above form and compare ther tme complextes. 1. Introducton Let F q denote the nte eld of order q = p m, p a prme, and let f(x) and g(x) be monc polynomals n F q [x]. The composed sum of f and g s the polynomal dened by (1.1) f g = (x? ( + )) whle the composed multplcaton of f and g s the polynomal dened by (1.2) f g = (x? ): In both cases the product runs through all roots of f and of g, ncludng multplctes. In [1] these compostons are generalzed as follows. Let G be a nonempty subset of the algebrac closure? q of F q wth the property that G s nvarant under the Frobenus automorphsm () = q, and suppose there s dened on G a bnary operaton satsfyng (1.3) ( ) = () () 1991 Mathematcs Subject Classcaton. Prmary 1205; Secondary 68Q25. 1

2 2 JOEL V. BRAWLE, SHUHONG GAO, AND DONALD MILLS for all ; 2 G. Then for monc polynomals f and g whose coecents are n F q and whose roots le n G, the composed product, denoted f g, s the polynomal dened by (1.4) f g = (x? ( )); where agan the products are over all roots of f and of g. It s clear that deg f g = (deg f)(deg g), and t s also clear that when G =? q and s the usual addton (respectvely, the usual multplcaton) on? q, then (1.4) becomes (1.1) (respectvely (1.2)). Whle the roots of f and g are n G and not necessarly n F q, t s easy to prove that (1.3) mples that the composed product (1.4) has ts coecents n F q [1]. Further, under the addtonal assumpton that G s a group under, the composton (1.4) has the followng property whch allows for the constructon of rreducbles over F q of a relatvely large degree from rreducbles over F q of relatvely small degrees. Theorem 1.1 ([1]). Let (G; ) be a -nvarant group satsfyng (1.3) and let f; g be monc polynomals n F q [x] wth roots n G. If deg f = m and deg g = n, then the composed product f g s rreducble n F q [x] f and only f f and g are rreducble n F q [x] and gcd(m; n) = 1. The proof of Theorem 1.1 uses among other thngs the followng property: If f and g factor as f = f 1 f 2 f r and g = g 1 g 2 g s where the f and g j are rreducble n F q [x], then (1.5) ( r s f ) ( j=1 g j ) = r s (f g j ): j=1 Our man goal s to show how to compute f g ecently. In leu of ths, (1.5) shows that gven the factorzaton of f and g, to nd f g one needs only to compute the compostons of the ndvdual rreducble factors. Consequently, we shall focus on methods of computng compostons of rreducble polynomals. In Secton 2, we dscuss the representaton of the operaton, as well as show a connecton between ellptc curve groups and the composed product. In Secton 3, we show how to ecently compute the general composed product. In Secton 4, we specalze to f g and f g. We rst determne how to ecently compute them usng (1.1) and (1.2) drectly, and then gve formulas for f g and f g whch are dened as determnants of matrces whose entres are polynomals. For these, we demonstrate how to nd f g and f g ecently usng fast nterpolaton. We also show how to compute f g and f g usng lnear recurrng sequences. We conclude wth a bref summary. Recall that f h(x); k(x) 2 F q [x] wth degrees at most n, then the number of F q -operatons needed to compute h + k or h? k s O(n), whle the tme needed to multply h by k s O(M(n)). Here M(n) = n 2 for classcal multplcaton and nl(n) for fast multplcaton (Schonhage & Strassen 1971, Schonhage 1977, and Cantor & Kaltofen 1991), where we use L(n) n place of log n log log n, as used by Shoup [13]. If we operate n the rng F q [x]= < f(x) > (and so reduce the product

3 COMPUTING COMPOSED PRODUCTS OF POLNOMIALS 3 of h and k modulo f, where deg f = n), then the number of F q -operatons needed s O(M r (n)). For classcal multplcaton, M r (n) = n 2 ; for fast multplcaton, M r (n) = nl(n) log n. The addtonal log n term s needed when usng fast multplcaton because the tme needed to do long dvson of f nto hk s O(nL(n) log n). In ths paper, f we are gven two polynomals f and g of degrees m and n respectvely, we wll always assume (for ease of exposton) that m n when dscussng the runnng tmes. 2. Propertes of the Damond Bnary Operaton 2.1. Representatons of the Bnary Operaton on G. We consder the form of the operaton dened on G. The followng s stated n [1] but not proved. Theorem 2.1. Suppose G s a nte -nvarant subset of? q on whch there s dened a bnary operaton satsfyng (1.3). Then can be represented by a polynomal h(x; y) 2 F q [x; y] such that h(; ) = for all ; 2 G. Proof. Suppose G = f 1 ; 2 ; :::; t g where t = jgj. By Lagrange nterpolaton, dene h(x; y) = where S and W are dened as S (x) = 1;jt ( j ) S (x) 2G;6= (x? ); W = W S j (y) W j 2G;6= (? ): It s clear that h(; ) = for all ; 2 G. We need to prove that h(x; y) 2 F q [x; y]. Snce G s -nvarant, nduces a permutaton on the elements n G. Hence [h(x; y)] q = ( j ) S q (x) S j (y) W W j (;j) = ( j ) q (S (x)) q (S j (y)) q W q (;j) W q j = ( () (j) ) S ()(x q ) S (j) (y q ) W (;j) () W (j) where q = () and q j = (j). But snce nduces a permutaton on the elements n G, and snce the sum s taken over all (; j), we have [h(x; y)] q = ( j ) S (x q ) S j (y q ) = h(x q ; y q ): W W j (;j) So each coecent of h(x; y) belongs to F q. Note that G does not have to be nte n order for us to represent composed products as polynomals n F q [x; y], as seen n Secton 1 for f g and f g. Also note that f G s not nte, the operaton may not have a polynomal representaton [3].

4 4 JOEL V. BRAWLE, SHUHONG GAO, AND DONALD MILLS In ths paper we are concerned only wth polynomal representatons of composed products. We can also go the other way; that s, f we are gven h(x; y) 2 F q [x; y], we can use h(x; y) to dene a operaton on a sutable subset G of the closure, where by \sutable" we mean that G s closed under. Theorem 2.2. For every h(x; y) 2 F q [x; y] and every nonempty subset S of? q, there exsts a smallest subset G contanng S such that G s -nvarant and The operaton dened by = h(; ) s a bnary operaton on G satsfyng (1.3). Moreover, f S s nte then so s G. Proof. Put S 0 = f q : 2 S; 0g. By denton, we see that S 0 s -nvarant. Now for = 0; 1; 2; ::: dene S +1 by S +1 = fh(; ) : ; 2 S g [ S : Ths produces a nested sequence of sets S 0 S 1 S. Note that each S s -nvarant, and ( ) = () () for each S. Moreover, for each 1, 2 S for all ; 2 S?1. Now let G = [ 0 S. Then G s closed under. If S s nte, then there s an extenson of F q, say F q m for some m such that S F q m. Obvously, S F q m for all, so that G F q m Ellptc Curve Groups: A Specal Case. We show that almost any ellptc curve group s somorphc to a nte -nvarant subset G of? q, wth an approprate bnary operaton satsfyng (1.3). For a bref ntroducton to ellptc curves over nte elds, see [8]. Theorem 2.3. If (E; +) s an ellptc curve group over the eld F q (consstng of F q -ratonal ponts), q > 3 odd, then there exsts a nte -nvarant subset G of? q and a group operaton on G such that (G; ) s somorphc to (E; +) and s an automorphsm of (G; ). S Proof. Let E = f(x; y)g f[0]g denote the ponts on an ellptc curve over F q, q > 3 odd, and let a 2 F q be a quadratc nonresdue. Then (x 2? a) 2 F q [x] s rreducble and f 2 F q 2 s a root of x 2? a, then satses q =?. Let e 2 F q have the property that the pont (e; 0) s not on the curve. Such a pont must exst snce at most three ponts on E have the form (x; 0). Consder the mappng : E F q 2 dened as follows: e f P = [0] (P ) = x + y f P = (x; y). Clearly s a one-to-one correspondence between E and a subset G of? q, and further, the ellptc curve addton nduces an operaton on G whch makes G somorphc to E; namely, (2.1) (P ) (Q) = (P + Q):

5 COMPUTING COMPOSED PRODUCTS OF POLNOMIALS 5 It remans to show that G s -nvarant and that s an automorphsm of (G; ). Now f P = (x; y) s on E, then so s ts negatve,?p = (x;?y). Thus, we may wrte ((P )) q = (x + y) q = x + y q = x + y(?) = x? y = (?P ). Also, by choce of e, the relaton (2.2) ((P )) q = (?P ) holds when P = [0] so (2.2) s vald for all P on E, mplyng that G s -nvarant. Usng both (2.1) and (2.2), we have ((P )) ((Q)) = (?P ) (?Q) = (?(P + Q)) whch by (2.2) s ((P + Q)) q. But then (2.1) gves us ((P + Q)) q = ((P ) (Q)) q = ((P ) (Q)), whch shows that s an automorphsm of the group G. 3. Computng the General Composed Product Suppose that f g s dened as n (1.4), and suppose the damond product s represented by h(x; y) 2 F q [x; y],.e. = h(; ) for all ; 2 G. Our goal s to compute f g ecently. We assume by (1.5) that f and g are rreducble of degrees m and n respectvely. By the denton (1.4), one needs to construct an extenson eld of F q that contans the roots of f and g. When gcd(m; n) = 1, the smallest such eld s F q mn, and t can be constructed as F q m[y]= < g(y) > where F q m = F q [x]= < f(x) >. In other words, F q mn ' R = F q [x; y]=i where I =< f(x); g(y) > s the deal n F q [x; y] generated by f(x) and g(y). Let x represent the class of x and y the class of y n R. Then the roots of f(z) are x q, 0 m? 1, and the roots of g(z) are y qj, 0 j n? 1. When computng n R, one always reduces powers of x modulo f(x) and powers of y modulo g(y). The composton f g can be computed by Algorthm 3.1 below. When gcd(m; n) > 1, the smallest eld that contans the roots of f and g s F q mn=d where d = gcd(m; n). One can stll construct F q m = F q [x]= < f(x) >, but F q m[y]= < g(y) > s no longer a eld. In fact, g(y) factors over F q m as a product of d rreducbles of degree n=d. We wll prove below that factorng s not necessary, and the same algorthm that s used for the case where gcd(m; n) = 1 works for all m and n. Algorthm 3.1. Computng f g. INPUT: f; g 2 F q [x] of degrees m and n respectvely; h(x; y) = P (;j) c jx y j 2 F q [x; y]. OUTPUT: f g. Step 0: Form the rng R = F q [x; y]=i, where I =< f(x); g(y) > and each class s represented by a unque polynomal of degree m? 1 n x and degree n? 1 n y. Step 1: Compute the polynomals u x q mod f(x), 0 m? 1

6 6 JOEL V. BRAWLE, SHUHONG GAO, AND DONALD MILLS and v j y qj mod g(y), 0 j n? 1. Step 2: In R, compute h j = h(u ; v j ), 0 m? 1 and 0 j n? 1. Step 3: END. m?1 n?1 In R[z], compute and output the polynomal (z? h j ). j=0 Theorem 3.2. Algorthm 3.1 computes f g correctly, and when m n, t uses O(nM r (n) log q + n 2 M r (n)t log E + n 4 M 2 r (n)) F q -operatons, where E s the largest degree of x or y n h(x; y) and T s the number of nonzero terms n h(x; y). Proof. The correctness follows from Lemma 3.4 below. We gve the tme needed to compute f g. In Step 1, u and v j can be computed teratvely by rasng to the q th power usng O(nM r (n) log q) F q -operatons. In Step 2, the h j are computed usng O(n 2 M r (n)t log E) F q -operatons. In Step 3, the product polynomal can be computed teratvely. At the k th teraton, we multply (z? h j ) by a degree k polynomal n R[z] wth O(k) multplcatons n R. Hence P the total number mn?1 of R-operatons used to compute the product polynomal s O( k=1 k) = O(n 4 ). Now each multplcaton n R can be done wth O(M r (m)m r (n)) = O(Mr 2 (n)) operatons n F q by vewng R as F q m[y]= < g(y) > where F q m = F q [x]= < f(x) >. So the total tme for Step 3 s O(n 4 Mr 2 (n)) operatons n F q. Hence the total tme for Algorthm 3.1 s as stated n the theorem. Usng classcal multplcaton, the tme for Algorthm 3.1 s O(n 3 log q + n 4 T log E + n 8 ); usng fast multplcaton, the tme s O(n 2 L(n) log n log q + T n 3 L(n) log n log E + n 6 (L(n)) 2 (log n) 2 ), where L(n) = log n log log n. Corollary 3.3. Algorthm 3.1 computes f g and f g correctly usng O(nM r (n) log q + n 4 M 2 r (n)) F q -operatons. Lemma 3.4. Let f(x) 2 F q [x], g(y) 2 F q [y] be rreducble of degrees m and n respectvely. Dene the algebra R = F q [x; y]= < f(x); g(y) > wth x = x+ < f(x); g(y) > and y = y+ < f(x); g(y) >. Then the polynomal (3.1) s(z) = m?1 n?1 z? x qk y ql 2 R[z] k=0 l=0 equals (f g)(z), whch s vewed as a polynomal over R by the natural embeddng of F q n R. Proof. Let K represent any extenson eld of F q over whch f and g splt completely, say

7 COMPUTING COMPOSED PRODUCTS OF POLNOMIALS 7 f(x) = m (x? ); g(y) = n j=1 (y? j ). Dene the algebra W = K[x; y]= < f(x); g(y) >. Note that K W and F q R W by means of natural embeddngs. The polynomals s(z) and (f g)(z) are vewed as polynomals over W ; we prove that they are equal over W and thus equal over R. Dene n W = Q Q l6=(x? l) l6= (? l ) ; j = Q Q l6=j(y? l) l6=j ( j? l ), P m where 1 m and 1 j n. Note that = 1, j = 0 for 6= j and P 2 = for 1 P m; smlar statements can be made for the j. Also, m x = n and y = j=1 j j. The elements f j g, 1 m and 1 j n, are prmtve dempotents n W and form an orthogonal bass for W over K. Denote j = j for each and j. We have P (;j) j = 1, x = P (;j) j and y = P (;j) j j. Also W = K 11 K 12 K mn. Snce the j are orthogonal dempotents, we have x y = P (;j) j( j ) and z = P (;j) z j. Hence s(z) = (z? x qk y ql ) = (k;l) (k;l) z? (;j) (;j) j ( qk = (k;l) = j (z? qk (;j) (k;l) j (z? qk = (;j) j (f g)(z) 1 ql j ) A 1 ql j ) A ql j ) Ths proves the lemma. = (f g)(z): 4. Computng f g and f g In ths secton, we explore alternate methods of computng (1.1) and (1.2), and mprove the runnng tmes gven n Corollary Computng f g and f g va the Dentons. Note that and f g = (x? ( + )) = ; f g = (x? ) = ; ((x? )? ) = ((x=)? ) = g(x? ) n g(x=).

8 8 JOEL V. BRAWLE, SHUHONG GAO, AND DONALD MILLS Smlarly, f g = f(x? ) and f g = m f(x=). The products are taken over all roots of f and of g. We dscuss the number of F q -operatons needed to nd f g. Let F q n = F q [y]= < g(y) >. Then the roots of g are j y qj mod g(y) for 0 j n? 1. To compute f g, we rst compute f(x? j ), 0 j n? 1, and then multply them as follows: par the factors f(x? j ) together and calculate the product of each par; then par these results and repeat the process untl all factors are combned. Note that j, 0 j n? 1, can be computed by repeatedly rasng to the q th power usng O(nM r (n) log q) F q -operatons. For each j, f(x? j ) can be computed va ts Taylor seres at x = 0: f(x? j ) = f(? j ) + f 0 (? j )x + f 00 (? j ) 2 x f (m) (? j ) x m. m Note that each coecent f (k) (? j ) s just a lnear combnaton of the elements k 1; j ; j 2; :::; m j wth scalars n F q. We compute 1; j ; j 2; :::; m j usng O(mM r (n)) F q -operatons. Each coecent of f(x? j ) can be computed n O(n 2 ) F q -operatons, and so each f(x? j ) can be computed n tme O(mn 2 ) = O(n 3 ). Hence all the f(x? j ) can be computed n tme O(n 4 ). Note that the f(x? j ) can also be computed by Horner's Rule; however, the cost s greater, even f one uses fast multplcaton. To multply the f(x? j ), 0 j n? 1, we requre log n teratons. At the k th teraton (0 k d(log n)e? 1), we multply (n=2 k+1 ) pars of polynomals n F q n[x], each of degree at most 2 k m. The k th teraton needs O((n=2 k+1 )M r (2 k m)) = O(M r (mn)) = O(M r (n 2 )) F q n-operatons. All the teratons need O(M r (n 2 ) log n) F q n-operatons, or O(M r (n 2 )M r (n) log n) F q -operatons. The total tme needed to nd f g s stated n Theorem 4.1. We can compute f g smlarly by parng factors, but each factor m f(x=) can be found drectly, wthout usng Taylor seres. The number of F q -operatons needed to nd all the j mf(x= j) s O(mnM r (n)) = O(n 2 M r (n)). Theorem 4.1. The composed sum f g can be computed wth the formula f g = f(x? ) usng O(n 4 + M r (n)m r (n 2 ) log n + nm r (n) log q) F q -operatons. Lkewse, the composed multplcaton f g can be computed wth the formula f g = m f(x=) usng F q -operatons. O(M r (n)m r (n 2 ) log n + nm r (n) log q) Usng classcal multplcaton, the tme gven n Theorem 4.1 s O(n 6 log n + n 3 log q) for both f g and f g; usng fast multplcaton, the tme s O(n 4 + n 3 L(n)L(n 2 )(log n) 3 + n 2 L(n) log n log q) for f g, and O(n 3 L(n)L(n 2 )(log n) 3 + n 2 L(n) log n log q) for f g. These tmes mprove upon the tme gven n Corollary 3.3.

9 COMPUTING COMPOSED PRODUCTS OF POLNOMIALS Matrx Methods. The methods below are matrx-orented, and stem from the use of the tensor, or Kronecker, product operaton to calculate f g [1]. We wll specalze to f g and f g followng a presentaton of the general approach. Recall that f A and B are square matrces over F q of szes m and n respectvely, then the Kronecker product of A and B s the square matrx over F q of sze mn gven by AB = (a j B) [1]. The egenvalues of AB are of the form [6] where and range over the egenvalues for A and B, respectvely, so that det(xi? A B) = (x? ). From ths we conclude that f f and g are monc nonconstant polynomals over F q wth companon matrces A and B, respectvely, then f g = det(xi? A B). More generally, we consder the case where G s a -nvarant subset of? q, and we dene a bnary operaton on G where xy = h(x; y) = P (;j) c jx y j 2 F q [x; y] [1]. For square matrces A and B over F q, dene a bnary operaton H(A; B) where H(A; B) = (;j) c j (A B j ), and the exponents ; j for A and B denote ordnary, as opposed to tensor, matrx products. These exponents correspond to the ones gven n the above equaton for xy. The egenvalues of H(A; B) are the numbers h(; ) [6], where and range over the egenvalues for A and B, respectvely. Hence det (xi? H(A; B)) = (x? h(; )), so that f f and g are any two polynomals over F q wth degrees m and n respectvely and wth companon matrces A and B, respectvely, then from [1] we have f g = det (xi? H(A; B)). For the case where H(; ) = +, we have f g = det (xi mn? (A I n + I m B)), where I n denotes the n-square dentty matrx. Whle we now have the advantage of workng n F q, the szes of the tensor product matrces can quckly become huge, so we prefer to use a method that employs matrces of smaller sze. The formulas presented below accomplsh ths by buldng on the formulas gven above, usng matrx theory and symmetrc functon theory to obtan better formulas for f g and f g. m Theorem 4.2 ([8]). Let f(x) = a x and g(x) = b j x j be two monc j=0 polynomals over F q of degrees m and n, respectvely, each havng dstnct roots. Let A and B be ther respectve companon matrces. Then (4.1) and f g = det n 0 n m b j x j A n?j A = det a x B m? j=0

10 10 JOEL V. BRAWLE, SHUHONG GAO, AND DONALD MILLS (4.2) f g = det 0 n m b j (xi m? A) j A = det a (xi n? B) j=0 are both polynomals over F q of degree mn. Snce f and g are each devod of repeated roots, ther companon matrces are each smlar to a dagonal matrx whose nonzero entres contan the polynomal's roots. A proof for (4.1) s gven below. The proof that follows s smlar to the proof of (4.2) gven n [8], and s gven for sake of completeness. Proof. Use the facts that, for matrces A, B, C, and D of approprate szes, (A B)(C D) = (AC) (BD) and (A B)?1 = A?1 B?1 (provded A?1 and B?1 exst). Let the egenvalues of A, whch are the roots of f, be represented by 1 ; :::; m and let D be the dagonal matrx whose nonzero entres are the. Then there s an nvertble matrx P (whose elements le n some extenson eld of F q ) such that A = P DP?1. Hence f g = det(xi mn? A B) = det(xi mn? (P DP?1 ) B) = det(x(p I n )(P?1 I n )? (P DP?1 ) B) = det((p I n )(xi mn? D B)(P?1 I n )) = det(xi mn? D B) = det(dag(xi n? B)) = m = det = det det(xi n? B) m m (xi n? B) a x B m? : In order to compute f g and f g ecently usng (4.1) and (4.2), we use Newton's nterpolaton method (see [4] for detals). Ths nvolves choosng mn + 1 dstnct ponts c, = 0; 1; :::; mn, from an extenson eld of F q of sucent sze, and evaluatng f g and f g at these ponts. These values are then used to recover the polynomals, whch have degree mn. Note that an extenson eld of sucent sze s F q e where e = O(log q mn). Also, we choose the ponts c so that fast nterpolaton can be done (as n FFT). Now to the mplementaton, where for convenence we let m J(x) = det a x B m? and K(x) = det m a (xi n? B).

11 COMPUTING COMPOSED PRODUCTS OF POLNOMIALS 11 Note that the eld F q e can be constructed n neglgble tme for our applcaton (see [12] and [13] for detals). The number of F q e-operatons requred to nd J(c ) (smlarly, K(c )) for any s O(n 3 ); here we use Horner's Rule and the sparseness of the companon matrx B. Hence all the values det[j(c )] (smlarly, det[k(c )]) are found n tme O(n 5 ). Usng fast nterpolaton, the number of F q e-operatons needed to nd the coecents of f g (smlarly, f g) s O(n 2 log n 2 ) (agan, we refer the reader to Schonhage & Strassen 1971, Schonhage 1977, and Cantor & Kaltofen 1991). The total number of F q e-operatons needed to nd f g (smlarly, f g) s O(n 5 ), or O(n 5 M r (e)) F q -operatons. Whle ths method costs more than the one gven n Secton 4.1 for fast multplcaton, t s not dependent upon the sze of q. The polynomals f g and f g can also be wrtten as varants of the resultant of f(x) and g(x). To be specc, we use the fact [4] that the resultant of f and g can be wrtten as Q m Q n j=1 (? j ) (assumng that f and g are monc). If we ntroduce a new varable z, then (4.3) f g = res z (f(x? z); g(z)) = res z (g(x? z); f(z)); where res denotes the resultant. Lkewse we have (4.4) f g = (?1) mn res z z n g x ; f(z) z Use Schwartz's algorthm [11] for fast resultant calculaton to compute (f g)(c ) = res z (f(c? z); g(z)) n O(n log 2 n) F q e-operatons, where c has the same meanng as above. Then the number of F q e-operatons needed to nd the values (f g)(c ) (smlarly, (f g)(c )) s O(mn(n log 2 n)) = O(n 3 log 2 n). The tme needed to recover the polynomal s absorbed nto the above runnng tme, so that the tme needed to nd f g and f g usng F q -operatons s O((n 3 log 2 n )M r (e)), the best of the tmes gven yet. As wth the method whch uses (4.1) and (4.2), ts tme s ndependent of the sze of q. We summarze these results n the next theorem. Theorem 4.3. The composed products f g and f g can be computed wth formulas (4.1) and (4.2), respectvely, usng O(n 5 M r (e)) F q -operatons. If (4.3) and (4.4) are used nstead, the number of F q -operatons requred to compute f g and f g s O((n 3 log 2 n)m r (e)). In both cases, e = O(log q n). Note n closng that the method used to nd (4.1) and (4.2) can be extended to the case where ether or s lnear n the polynomal representaton of. Let 1 ; :::; m represent the roots of f, where f s dened as n Theorem 4.2. We have " m f g = det h(b)[h?1 (B)(xI n? t(b))? k I n ] k=1 = det(h m (B)f( )) = det m k=0 a k (xi n? t(b)) k h m?k (B) where = h?1 (B)(xI n? t(b)), and h(b) and t(b) are polynomals n B. In partcular, h(b) s assocated wth those terms n the polynomal representaton of : #

12 12 JOEL V. BRAWLE, SHUHONG GAO, AND DONALD MILLS n whch the exponent of s one, and t(b) corresponds to the remander of the terms Lnear Recursve Sequences and Composed Products. A method of computng f g and f g whch allows us to work entrely wthn F q s to use lnear recursve sequences (LRS) and the Berlekamp-Massey Algorthm. The method for ndng f g usng LRS's was descrbed to us by John Brllhart [2], who attrbuted the method to D.H. Lehmer. We derve a formula for the k th term of the sequence whose mnmal polynomal s f g, and use ths to compute f g. Recall that a lnear recursve sequence fa j g s generated by a polynomal f(x) = m?1 m x m + f k x k 2 F [x], F a eld, f a j =? f m?k a j?k for j = m; m + 1; :::: Such k=0 k=1 a polynomal of smallest degree s called the mnmal polynomal of the sequence. We rst show how to nd f g va LRS's. Suppose (nonzero) LRS's fa k g and fb k g have mnmal polynomals f(x); g(x) 2 F q [x], respectvely. Suppose further that f(x) and g(x) have dstnct roots 1 ; :::; m and 1 ; :::; n, respectvely. Then there exst elements A and B j n some extenson of F q so that for all k 0, m a k = A k and b k = B j j k. j=1 Then the k th element of the product sequence fa k b k g s gven by m n a k b k = A B j ( j ) k. j=1 Hence the products j, 1 m and 1 j n, are all the canddates for the roots of the mnmal polynomal of fa k b k g. If all the j are dstnct, then the mnmal polynomal s ndeed f g. For our purposes, f f and g are rreducble and of coprme degrees (so that the products of ther roots are dstnct), then f g s the mnmal polynomal of the correspondng product sequence. It s well known that the mnmum number of elements of fc k g = fa k b k g needed to determne f g s 2mn. Gven a 0 through a 2mn?1 and b 0 through b 2mn?1 we compute c 0 through c 2mn?1 ; ths new sequence provdes the nput for the Berlekamp-Massey Algorthm, whch outputs f g (see [7] and [9] for detals). The steps taken to nd f g are as follows. Frst we select nonzero startng states for fa k g and fb k g; the smplest such states are ones wth zeros for the rst m? 1 terms (n? 1, respectvely), and wth a one for the m th term of the sequence (n th term, respectvely). We use these startng states to generate a m through a 2mn?1 and b n through b 2mn?1 (va f and g respectvely) usng O(mn(m + n)) = O(n 3 ) F q -operatons. The number of F q -operatons needed to nd the subsequence c 0 ; :::; c 2mn?1 usng componentwse multplcaton s O(n 2 ). The number of F q - operatons requred for the Berlekamp-Massey Algorthm s O((mn) 2 ) = O(n 4 ) [9]. So the total tme needed to nd f g s O(n 4 ) F q -operatons. Ths tme, whle slower than the runnng tme gven n Secton 4.1 (usng fast multplcaton), does not depend upon the value of q. It s also slower than the resultant-based nterpolaton method. The advantage n usng ths method, though, s that we work entrely n the ground eld F q. We turn our attenton to ndng a lnear recursve sequence whose mnmal polynomal s f g, where f; g 2 F q [x] are rreducble and of coprme degree. n

13 COMPUTING COMPOSED PRODUCTS OF POLNOMIALS 13 Theorem 4.4. Suppose fa k g and fb k g are (nonzero) lnear recursve sequences havng rreducble mnmal polynomals f(x) and g(x) n F q [x], respectvely. Suppose further that the degrees of f and g are relatvely prme. Then f g s the mnmal polynomal of the sequence fc k g whose k th term s gven by k k a b k? : (4.5) c k = To prove ths, we show that the elements of any nonzero LRS of a gven polynomal can be wrtten n terms of the trace functon (ths was stated n [10] but not proven), and then use the fact that the degrees of f and g are coprme to obtan (4.5). Lemma 4.5 ([5]). Let A = f 0 ; 1 ; :::; m?1 g and B = f 0 ; 1 ; :::; m?1 g be a par of dual bases of F q m=f q, and let be any element of F q m. Then the coordnate x of n = x x x m?1 m?1 equals T r( ), and lkewse the coordnate a of n = a a a m?1 m?1 equals T r( ). m?1 Lemma 4.6. Let f = x m + f k x k be rreducble, and let represent a root k=0 of f. Then a nonzero sequence fa k g s generated by f(x) there exsts a 2 F q m such that a = T r( ) for all 0. Proof. Suppose a = T r( ) for all 0. Snce f g s generated by f(x) (because s a root of f(x)) and snce the trace functon s lnear, fa g s also generated by f(x). Now let f 0 ; 1 ; :::; m?1 g represent the dual bass for f1; ; :::; m?1 g. Take = m?1 a 2 F q m. Then a = T r( ) for = 0; 1; :::; m? 1. Suppose a = T r( ) for 0 k? 1 where k m. Then T r( k ) = T r? m The lemma follows by nducton on k. m f m? k? =? f m? T r( k? ) m f m? a k? =? = a k : So under the condtons set forward for f, we can generate any nonzero LRS by usng the trace functon on an approprate multple of the elements of the bass correspondng to f. Smlar statements can be made for the rreducble g of degree n over F q mentoned above. For ths polynomal, we use the notaton for ts roots, fb k g for ts assocated nonzero LRS, and let 2 F q n play the same role as above. We requre one more result before provng Theorem 4.4. Lemma 4.7 ([5]). Let A = f 0 ; 1 ; :::; m?1 g and B = f 0 ; 1 ; :::; n?1 g be bases for K = F q m and L = F q n over F = F q, respectvely, and assume that m and n are coprme. Then one has the followng results, where we wrte E = F q mn.

14 14 JOEL V. BRAWLE, SHUHONG GAO, AND DONALD MILLS C = f j : = 0; :::; m? 1 ; j = 0; :::; n? 1g s a bass for E=F. T r E=F () = T r K=F ()T r L=F () for all 2 K and all 2 L. Proof of Theorem 4.4. Apply Lemma 4.6 to f and g. Thus there exst elements 2 F q m and 2 F q n such that a k = T r( k ) and b k = T r( k ) for each k. Snce f and g are rreducble and of coprme degree, f g s rreducble. Let c k = T r( ( + ) k ). Then by Lemma 4.6, fc k g s generated by f g. Snce f g s rreducble, t s the mnmal polynomal of fc k g. Note that c k = T r[ ( + ) k ] = T r = = = k k k " k k k k k k? # T r( k? ) T r( )T r( k? ) a b k? where we use Lemma 4.7 n the thrd lne. We have proved the theorem.? = k?1? + k?1??1, one can compute all the k and thus? k Usng the formula the c k, k 2n 2, n tme O((mn) 2 ) = O(n 4 ). So the total tme needed to nd f g s agan O(n 4 ) F q -operatons. The results of ths porton of the paper are summarzed n the followng theorem. Theorem 4.8. Suppose that fa k g, fb k g are nonzero lnear recurrng sequences whose elements come from F q, wth rreducble mnmal polynomals f(x); g(x) 2 F q [x] respectvely. Suppose further that the degrees of f and g are relatvely prme. Then the composed products f g and f g can be computed wth the sequences fa k g and fb k g usng O(n 4 ) F q -operatons. 5. Summary We have shown that a general composton f g can be computed n O(nM r (n) log q + n 2 M r (n)t log E + n 4 M 2 r (n)) F q -operatons. In the specal cases of f g and f g, we presented several fast methods for computng them. All of the methods gven n Secton 4 were shown to be faster than the general method n Secton 3. The fastest method s the nterpolaton method as appled to (4.3) and (4.4), wth the runnng tme beng O((n 3 log 2 n)m r (e)) F q -operatons, where e = O(log q mn) = O(log q n). Whle the LRS method costs more (t requres O(n 4 ) F q -operatons), t s better than the resultant method n that we operate entrely wthn the ground eld F q. The resultant method requres us to work n an approprate extenson of F q so that fast nterpolaton can be done.

15 COMPUTING COMPOSED PRODUCTS OF POLNOMIALS 15 Recall that the runnng tme for the method of Secton 4.1 (as appled to f g) s O(n 3 L(n)L(n 2 )(log n) 3 + n 2 L(n) log n log q) when fast multplcaton s used. Ths method s necent for large q (n comparson wth the LRS method, the Secton 4.1 method s necent for values of q such that log q n 2 ). We conclude that the most ecent methods for computng f g and f g are the resultant-based nterpolaton method and the LRS method. References [1] Brawley, J.V. and Carltz, L. Irreducbles and the Composed Product for Polynomals Over a Fnte Feld. North-Holland, Amsterdam; Dscrete Mathematcs 65 (1987), [2] Brllhart, John. Prvate Communcaton. [3] Brown, Davd D. Iterated Presentatons and Module Polynomals Over Fnte Felds. Ph.D. Thess, Clemson Unversty, [4] Geddes, Keth O.; Czapor, Stephen R.; Labahn, George. Algorthms for Computer Algebra. Kluwer, Boston, [5] Jungnckel, Deter. Fnte Felds: Structure and Arthmetcs. Wssenschaftsverlag, Mannhem, [6] MacDuee, C.C. The Theory of Matrces. Chelsea, New ork, [7] Massey, J.L. Shft-Regster Synthess and BCH Decodng. IEEE Transactons on Informaton Theory, vol. IT-15, January 1969, [8] Menezes, Alfred J. (edtor); Blake, Ian F.; Gao, uhong; Mulln, Ronald C.; Vanstone, Scott A.; aghooban, Tomk. Applcatons of Fnte Felds. Kluwer, Boston, [9] Menezes, Alfred J.; van Oorschot, Paul C.; Vanstone, Scott A. Handbook of Appled Cryptography. CRC Press, Boca Raton, FL, [10] Rueppel, Raner A. and Staelbach, Othmar J. Products of Lnear Recurrng Sequences wth Maxmum Complexty. IEEE Transactons on Informaton Theory, vol. IT-33, no. 1, January, 1987, [11] Schwartz, J.T. Fast Probablstc Algorthms for Vercaton of Polynomal Identtes. J. Assoc. Comput. Mach. 27 (1980), no. 4, [12] Shoup, Vctor. New Algorthms for Fndng Irreducble Polynomals Over Fnte Felds. Math. Comp. 54 (1990), no. 189, [13] Shoup, Vctor. Fast Constructon of Irreducble Polynomals Over Fnte Felds. J. Symbolc Comput. 17 (1994), no. 5, Department of Mathematcal Scences, Clemson Unversty, Clemson, South Carolna E-mal address: jvbrw@clemson.edu, sgao@math.clemson.edu, ddmlls@math.clemson.edu