多項式作輾轉相除法所需次數的估計

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(1)THE NUMBER OF STEPS IN THE POLYNOMIAL EUCLIDEAN ALGORITHM JIA-WEI GUO. Abstract. Let M be a monic polynomial over Fq of deg M ≥ 2. For polynomials a with deg a < deg M and (a, M ) = 1, the Euclidean Algorithm requires an average value of deg M + O ln4 deg M ζA (2) divisions.. 1. Introduction Let a and N be positive integers, 1 ≤ a < N, (a, N ) = 1, and let a = N. 1 c1 +. ,. 1 c2 + .. .. +. 1 cl(a,N ). where c1 , c2 , · · · ,cl(a,N ) are positive integers, cl(a.N ) > 1. Put X L(N ) = l(a, N ). 1≤a<N (a,N )=1. Denote by r(N ) the number of solutions of the equation N = xx0 + yy 0 in positive integers x, x0 , y, y 0 , for which x > y, x0 > y 0 , (x, y) = (x0 , y 0 ) = 1. H. Heilbronn prove that if N > 2, then 3 L(N ) = φ(N ) + 2r(N ) 2 and  r(N ) =. ln 2 φ(N ) ln N + O N  ζ(2) 1. 3 .  X1 d|N. d.  ,.

(2) 2. JIA-WEI GUO. where φ(N ) is the Euler phi-function, which yields the following asymptotic ) formula for L(N φ(N ) : L(N ) 2 ln 2 = ln N + O ln4 ln N . φ(N ) ζ(2) In this paper, let Fq denote the finite field with q elements. Let A = Fq [T ] be the polynomial ring with coefficients over Fq , A+ be the set of monic polynomials in A and p denote the irreducible polynomial in A. Given any 0 6= f ∈ A, the degree of f is denoted by deg f and the absolute value of f is denoted by |f | = q deg f . The notation X0 Y0 and denote respectively the sum and product over all monic polynomials in A and the notation µ(f ) denotes the polynomial m¨obius function of A+  1 if f = 1,    (−1)r if f = p p · · · p where p , p , · · · , p are the 1 2 r 1 2 r µ(f ) =  distinct monic irreducible polynomials in A+ ,    0 otherwise . Furthermore, let Q ∈ A+ , the polynomial Euler phi-function φ(Q) denote the order of (A/QA)× and we have Y0 1 . φ(Q) = |Q| 1− |p| p|Q. The polynomial zeta-function is given by X0 1 Y0 1 −1 1 ζA (s) = = 1− s = . s |f | |p| 1 − q 1−s p f. Hence, we immediatly have ζA (2) =. 1 . 1−q −1. Suppose A1 , . . . , An ∈ A with deg Ai ≥ 1 (1 ≤ i ≤ n). Define the continued fraction expansion of A1 , . . . , An by 1 A1 +. = [A1 , A2 , · · · , An ] .. 1 A2 + .. .. +. 1 An. Furthermore, define polynomials Pi , Qi by (1). P0 = 0, P1 = 1,. Pi = Ai Pi−1 + Pi−2 (1 < i ≤ n),. Q0 = 1, Q1 = A1 , Qi = Ai Qi−1 + Qi−2 (1 < i ≤ n),.

(3) THE NUMBER OF STEPS IN THE POLYNOMIAL EUCLIDEAN ALGORITHM. 3. where Pi = Pi (A1 , A2 , · · · , Ai ) and Qi = Qi (A1 , A2 , · · · , Ai ). It is clear that (2). Pi (A1 , · · · , Ai ) = Qi−1 (A2 , · · · , Ai ), deg Pi−1 < deg Pi , 0 ≤ deg Qi−1 < deg Qi , (Qi−1 , Qi ) = 1 for 1 ≤ i ≤ n.. By [2] corollaries 5.55 and 5.58 we know that Pi and Qi are the numerator and the denominator, respectively, of the ration function in reduced form represented by [A1 , A2 , · · · , Ai ]. Let a and M be two polynomials over Fq with 0 ≤ deg a < deg M and (a, M ) = 1. The usual polynomial Eculidean algorithm for computing the greatest common divisor will give a series of equations (3). r0 = M, r1 = a and ri−1 = Ai ri + ri+1 (1 ≤ i ≤ s),. where 0 ≤ deg ri+1 < deg ri , i = 1, 2, . . . , s − 1 and rs+1 = 0. Furthermore, A1 , A2 , . . . , As ∈ A with each deg Ai having positive degree and a = [A1 , A2 , · · · , As ] . M Suppose M ∈ A+ and define l(a, M ) = s as above. Put X L(M ) = l(a, M ). 0≤deg a<deg M (a,M )=1 ) The purpose of this paper is to estimate the value L(M φ(M ) , where deg M ≥ 2 and φ(M ) is the polynomial Euler phi-function. In theorem 3.3, we obtain. L(M ) deg M = + O ln4 deg M , φ(M ) ζA (2) where the implied constant does not depend on q and M . 2. Continued Fractions Lemma 2.1. Suppose [A1 , A2 , · · · , An ] and [B1 , B2 , · · · , Bm ] are two continued fractions, deg Ai ≥ 1 for 1 ≤ i ≤ n, deg Bi ≥ 1 for 1 ≤ i ≤ m. If [A1 , A2 , · · · , An ] = [B1 , B2 , · · · , Bm ], then n = m, Ai = Bi for 1 ≤ i ≤ n. Proof. Suppose m ≥ n, we proceed by induction on n. If n = 1 and m ≥ 2, then A11 = B1 +[B21,··· ,Bm ] . It follows A1 − B1 = [B2 , · · · , Bm ]. Since deg Bi ≥ 1 for 2 ≤ i ≤ m, we have deg[B2 , · · · , Bm ] < 0. It is impossible to the above equality. Suppose that the result is true for n = k. When n = k + 1, we have 1 1 = . A1 + [A2 , · · · , Ak+1 ] B1 + [B2 , · · · , Bm ].

(4) 4. JIA-WEI GUO. It follows A1 − B1 = [B2 , · · · , Bm ] − [A2 , · · · , Ak+1 ]. Since deg Ai ≥ 1 for 2 ≤ i ≤ k + 1 and deg Bi ≥ 1 for 2 ≤ i ≤ m, we have deg[A2 , · · · Ak+1 ] < 0 and deg[B2 , · · · , Bm ] < 0. Therefore, we obtain A1 = B1 and [A2 , · · · Ak+1 ] = [B2 , · · · , Bm ]. By induction hypothesis, we have m = k + 1 and Ai = Bi for 2 ≤ i ≤ k + 1. This completes the induction and so proves the lemma. Lemma 2.2. Suppose n ≥ 2 and A1 , · · · , An ∈ A. For 1 ≤ i ≤ n − 1, we have Qn (A1 , · · · , An ) =Qi (A1 , · · · , Ai )Qn−i (Ai+1 , · · · , An ) + Qi−1 (A1 , · · · , Ai−1 )Qn−i−1 (Ai+2 , · · · , An ). Proof. By the definition of Qn (A1 , · · · , An ) in (1), the conclution is true for all n ≥ 2 and i = n − 1. Suppose the conclution is true for all n ≥ k + 1 and i = n − k. By induction hypothesis, we have Qk+1 (An−k , · · · , An ) = An−k Qk (An−k+1 , · · · , An ) + Qk−1 (An−k+2 , · · · , An ) and Qn (A1 , · · · , An ) =Qn−k (A1 , · · · , An−k )Qk (An−k+1 , · · · , An ) + Qn−k−1 (A1 , · · · , An−k−1 )Qk−1 (An−k+2 , · · · , An ) for any n ≥ k + 2. By above equality and (1) again, we have Qn (A1 , · · · , An ) ={An−k Qn−k−1 (A1 , · · · , An−k−1 ) + Qn−k−2 (A1 , · · · , An−k−2 )}Qk (An−k+1 , · · · , An ) + Qn−k−1 (A1 , · · · , An−k−1 )Qk−1 (An−k+2 , · · · , An ) =Qn−k−1 (A1 , · · · , An−k−1 ){An−k Qk (An−k+1 , · · · , An ) + Qk−1 (An−k+2 , · · · , An )} + Qn−k−2 (A1 , · · · , An−k−2 )Qk (An−k+1 , · · · , An ) =Qn−k−1 (A1 , · · · , An−k−1 )Qk+1 (An−k , · · · , An ) + Qn−k−2 (A1 , · · · , An−k−2 )Qk (An−k+1 , · · · , An ) =Qi (A1 , · · · , Ai )Qn−i (Ai+1 , · · · , An ) + Qi−1 (A1 , · · · , Ai−1 )Qn−i−1 (Ai+2 , · · · , An ) for any n ≥ k + 2 and i = n − (k + 1). This completes the induction and so proves the lemma.. . Lemma 2.3. Suppose n ≥ 0 and A1 , . . . , An ∈ A. Then we have Qn (A1 , · · · , An ) = Qn (An , · · · , A1 ). Proof. The conclution is immediate by lemma 2.2 with i = 1 and induction on n. .

(5) THE NUMBER OF STEPS IN THE POLYNOMIAL EUCLIDEAN ALGORITHM. 5. 3. Main Theorem Suppose M ∈ A+ and define r(M ) as the number of solutions of the equation M = f f 0 + gg 0 in polynomials f, f 0 , g, g 0 ∈ A, for which deg f > deg g ≥ 0, deg f 0 > deg g 0 ≥ 0, (f, g) = (f 0 , g 0 ) = 1 and we further define R(M ) as the number of solutions of the equation subject to deg f > deg g ≥ 0, deg f 0 > deg g 0 ≥ 0. Theorem 3.1 below shows that r(M ) L(M ) = + φ(M ). q−1 By the definition of R(M ), we have X0 r(M (hh0 )−1 ) (4) R(M ) = hh0 |M. and comparing deg f with deg f 0 in R(M), we have (5). R(M ) = 2R1 (M ) + R2 (M ),. where R1 (M ) is the number of solutions of the equation M = f f 0 + gg 0 in polynomials f, f 0 , g, g 0 ∈ A, for which deg f > deg g ≥ 0, deg f 0 > deg g 0 ≥ 0, deg f < deg f 0 and R2 (M ) is the number of solutions of the equation subject to deg f > deg g ≥ 0, deg f 0 > deg g 0 ≥ 0, deg f = deg f 0 . By a repeated application of the m¨obius inversion formula to (4), we get X0 (6) r(M ) = µ(h)µ(h0 )R(M (hh0 )−1 ). hh0 |M. For integer n ≥ 0, define %0 (M, n) as the number of solutions of the equation M = f f 0 + gg 0 in polynomials f, f 0 , g, g 0 ∈ A, for which deg f > deg g ≥ 0, deg f 0 > deg g 0 ≥ 0, (f, g) = 1, n + deg f = deg f 0 and we further define %(M, n) as the number of solutions of the equation subject to deg f > deg g ≥ 0, deg f 0 > deg g 0 ≥ 0, (f, g) = 1, n + deg f < deg f 0 . By (5) and the definition of R1 (M ) and R2 (M ), we obtain X0 X0 R(M ) = 2 %(M u−1 , deg u) + %0 (M u−1 , deg u). u|M. u|M. Hence, by (6) and above equality, we have (7) X0 X0 r(M ) = 2 µ(h)µ(h0 )%(M (uhh0 )−1 , deg u)+ µ(h)µ(h0 )%0 (M (uhh0 )−1 , deg u). uhh0 |M. uhh0 |M.

(6) 6. JIA-WEI GUO. Theorem 3.1. Suppose M ∈ A+ . Then we have L(M ) =. r(M ) + φ(M ). q−1. Proof. If deg M ≤ 1, then r(M ) = 0. Hence the equality is trivial by direct calculation. Now we can suppose deg M ≥ 2 and given a ∈ A with a 1 ≤ deg a < deg M , (a, M ) = 1. Develop M as a continued fraction a = [A1 , · · · , As ], deg Ai ≥ 1, i = 1, . . . s. M We have M = αQs (A1 , · · · , As ) × for some α ∈ Fq and s = l(a, M ) > 1. Choose i in the interval 1 ≤ i ≤ s − 1 in s − 1 different ways. For any β ∈ Fq× , put (8). f = βQi (A1 , · · · , Ai ), g = βQi−1 (A1 , · · · , Ai−1 ), f 0 = β −1 αQs−i (Ai+1 , · · · , As ), g 0 = β −1 αQs−i−1 (Ai+2 , · · · , As ).. These polynomials, by virture of lemma 2.2, satisfy the quation M = f f 0 + gg 0 and by lemma 2.3 we have f 0 = β −1 αQs−i (As , · · · , Ai+1 ), g 0 = β −1 αQs−i−1 (As , · · · , Ai+2 ). By (2) we have deg f > deg g ≥ 0, deg f 0 > deg g 0 ≥ 0, (f, g) = (f 0 , g 0 ) = 1. Moreover, by lemma 2.3, (2) and [2] corollary 5.55, we have (9) Qi−1 (Ai−1 , · · · , A1 ) Pi (Ai , · · · , A1 ) g = = = [Ai , · · · , A1 ] f Qi (Ai , · · · , A1 ) Qi (Ai , · · · , A1 ) and g0 Qs−i−1 (Ai+2 , · · · , As ) Ps−i (Ai+1 , · · · , As ) = = = [Ai+1 , · · · , As ]. f0 Qs−i (Ai+1 , · · · , As ) Qs−i (Ai+1 , · · · , As ) 0. a For any a0 ∈ A, 1 ≤ deg a0 < deg M , (a0 , M ) = 1 and M = [B1 , · · · , Br ], we have M = α0 Qr (B1 , · · · , Br ) for some α0 ∈ Fq× and r > 1. As (8), there are (q − 1)(r − 1) corresponding (f1 , f10 , g1 , g10 ). If one of them is equal to some (f, f,0 g, g 0 ), by (9) and lemma 2.1, we have a = a0 . Hence we know that when a is different, no corresponding (f, f 0 , g, g 0 ) are equal. Conversely, given a representation of M , deg M ≥ 2, by M = f f 0 + gg 0 with. deg f > deg g ≥ 0, deg f 0 > deg g 0 ≥ 0, (f, g) = (f 0 , g 0 ) = 1. Suppose. g g0 = [Bj , · · · , B1 ] and 0 = [Bj+1 , · · · , Br ] f f.

(7) THE NUMBER OF STEPS IN THE POLYNOMIAL EUCLIDEAN ALGORITHM. 7. with deg Bi ≥ 1 (1 ≤ i ≤ r). By [2] corollary 5.55, lemma 2.3 and (2), we have f = γ1 Qj (Bj , · · · , B1 ) = γ1 Qj (B1 , · · · , Bj ), g = γ1 Pj (Bj , · · · , B1 ) = γ1 Qj−1 (Bj−1 , · · · , B1 ) = γ1 Qj−1 (B1 , · · · , Bj−1 ), f 0 = γ2 Qr−j (Bj+1 , · · · , Br ), g 0 = γ2 Pr−j (Bj+1 , · · · , Br ) = γ2 Qr−j−1 (Bj+2 , · · · , Br ) for some γ1 , γ2 ∈ Fq× . By lemma 2.2 we have M = f f 0 + gg 0 = γ1 γ2 Qr (B1 , · · · , Br ). Putting a = γ1 γ2 Pr (B1 , . . . , Br ). Since r ≥ 2, by (1), [2] corollaries 5.55 and 5.58, we have deg a ≥ 1 and a = [B1 , · · · , Br ], M where (a, M ) = 1 and deg a < deg M . To sum up, we have a one-to-one correspondence between all suitably restricted representation of M by the blinear form in the definition of r(M ), and all of the sequences βQi (A1 , . . . , Ai ), i = 1, . . . , s − 1, a = [A1 , · · · , As ], (a, M ) = 1 and 1 ≤ where β is any element in Fq× , M deg a < deg M . Hence X r(M ) = (q − 1) (l(a, M ) − 1) 1≤deg a<deg M (a,M )=1.  = (q − 1)  . X 1≤deg a<deg M (a,M )=1.   (q − 1) + − (q − 1) l(a, M ) + (q − 1)  . by l(a, M ) = 1 for all a ∈ Fq× X = (q − 1) l(a, M ) − (q − 1) 0≤deg a<deg M (a,M )=1. . . . . X. X 1≤deg a<deg M (a,M )=1. 1. 0≤deg a<deg M (a,M )=1. = (q − 1)L(M ) − (q − 1)φ(M ). We have L(M ) =. r(M ) + φ(M ). q−1. Therefore we prove the theorem.. . Theorem 3.2. Suppose M ∈ A+ and deg M ≥ 1. Then we have   3  0 X (q − 1)φ(M ) deg M 1   r(M ) = + O q |M |  , ζ(2) |h| h|M. where the implied constant does not depend on q and M ..  1 .

(8) 8. JIA-WEI GUO. Proof. If deg M = 1, then r(M ) = 0. Hence the equality is trivial by direct calculation. Now we can suppose deg M ≥ 2. By (7), lemma 4.4 (b) and lemma 4.4 (a), we have (10) r(M ) =2. X0. µ(h)µ(h0 )%0 (M (uhh0 )−1 , deg u). uhh0 |M. uhh0 |M. X0. X0. µ(h)µ(h0 )%(M (uhh0 )−1 , deg u) +.

(9) q − 1

(10)

(11) M (uhh0 )−1

(12) deg(M (uhh0 )−1 ) − deg u 2ζ(2) uhh0 |M     0 X X0

(13)

(14)

(15)

(16)

(17) M (uhh0 )−1

(18) 

(19) M (uhh0 )−1

(20)  + O q + O 2q. =2. µ(h)µ(h0 ). uhh0 |M. uhh0 |M.

(21)

(22) −1 (q − 1) |M | X0 = µ(h)µ(h0 )

(23) uhh0

(24) deg M − deg(u2 hh0 ) ζ(2) uhh0 |M   3  X0 1  . + O q|M |  |h| h|M. Since X0 X0 X0

(25) −1

(26) µ(h) |h|−1 |m|−1 µ(h0 ) µ(h)µ(h0 )

(27) uhh0

(28) =. X0 uhh0 |M. m|M h−1. h|M. =. (11). X0. h0 |m. µ(h) |h|−1. h|M. = and (12) X0. φ(M ) |M |.

(29)

(30) −1 µ(h)µ(h0 )

(31) uhh0

(32) deg(u2 hh0 ). uhh0 |M. =. X0. X0

(33)

(34) −1

(35)

(36) −1 µ(h)µ(h0 )

(37) uhh0

(38) deg(uh0 ) µ(h)µ(h0 )

(39) uhh0

(40) deg(uh) + uhh0 |M. uhh0 |M. X. =2. 0.

(41)

(42) −1 µ(h)µ(h0 )

(43) uhh0

(44) deg(uh0 ). uhh0 |M. =2. X0 h|M. µ(h) |h|−1. X0 m|M h−1. |m|−1 deg m. X0. µ(h0 ). h0 |m. = 0, combining (10) with (11) and (12), the proof is complete.. .

(45) THE NUMBER OF STEPS IN THE POLYNOMIAL EUCLIDEAN ALGORITHM. 9. Theorem 3.3. Suppose M ∈ A+ and deg M ≥ 2. Then we have L(M ) deg M = + O ln4 deg M , φ(M ) ζ(2) where the implied constant does not depend on q and M . Proof. By theorem 3.1 and theorem 3.2, we have. (13). L(M ) r(M ) = +1 φ(M ) (q − 1)φ(M )  3   0 X 1  deg M |M |  + 1 . = +O ζ(2) φ(M ) |h| h|M. By lemma 4.2, lemma 4.3 and deg M ≥ 2, we have  3 0 X 1  |M |  + 1 (ln deg M )(ln deg M )3 + 1 φ(M ) |h| (14) h|M. ln4 deg M. Combining (13) with (14), the proof is complete.. . Given M, a ∈ A with deg M ≥ deg a. As (3), we can suppose r0 = M, r1 = a and ri−1 = Ai ri + ri+1 (1 ≤ i ≤ s), where 0 ≤ deg ri+1 < deg ri , i = 1, 2, . . . s − 1 and rs+1 = 0. Suppose deg ri = ti , i = 0, 1, . . . , s. We define sequence P (M, a) = {ti }si=0 and N (t0 , t1 , . . . , ts ) = #{(M, a)| deg M = t0 , deg a = t1 , P (M, a) = {ti }si=0 }. By division alogorithm and induction, we have N (t0 , t1 , . . . , ts ) = q t0 (q − 1)s+1 . This is obvious if s = 1. We now suppose s ≥ 2. For any fixed r1 , r2 ∈ A with deg r1 = t1 and deg r2 = t2 , there are exactly q t0 −t1 (q − 1) number of r0 with deg r0 = t0 satisfies r0 = A1 r1 + r2 , where A1 depends on r0 . Hence, N (t0 , t1 , . . . , ts ) = q t0 −t1 (q − 1) × N (t1 , t2 , . . . , ts ) = q t0 −t1 (q − 1) × q t1 (q − 1)s = q t0 (q − 1)s+1 . Thus, we have an easy method to estimate an average value of the Euclidean algorithm in another way by the following theorem: Theorem 3.4. We have q −(m+n) (q − 1)−2. X deg M =m deg a=n m≥n. l(a, M ) =. n + 1. ζA (2).

(46) 10. JIA-WEI GUO. Proof. Suppose t0 = m and t1 = n. We have X. l(a, M ) =. tX 1 +1. X. N (t0 , t1 , . . . , ts )s. s=1 t1 >t2 >···>ts ≥0. deg M =m deg a=n m≥n. =. tX 1 +1. X. q t0 (q − 1)s+1 s. s=1 t1 >t2 >···>ts ≥0 tX 1 +1 . t1 q t0 (q − 1)s+1 s = s−1 s=1 t1 X t1 t0 2 (q − 1)s (s + 1) = q (q − 1) s s=0. = q (q − 1) {(1 + q − 1)t1 + t1 (q − 1)(1 + q − 1)t1 −1 } t0. 2. = (t1 + 1)q t0 +t1 (q − 1)2 − t1 q t0 +t1 −1 (q − 1)2 = (n + 1)q m+n (q − 1)2 − nq m+n−1 (q − 1)2 . By dividing q m+n (q − 1)2 in both sides, we complete the proof.. . 4. Auxiliary Lemmas Lemma 4.1. We have X. φ(M ) = q 2m−1 (q − 1)2 .. deg M =m. Proof. See [3] Proposition 2.7. Lemma 4.2. Suppose M ∈ A+ and deg M ≥ 2. Then we have |M | ln deg M, φ(M ) where the implied constant does not depend on q and M . Proof. See [4] Lemma 2.1.. . Lemma 4.3. Suppose M ∈ A+ and deg M ≥ 2. Then we have X0 1 ln deg M, |h| h|M. where the implied constant does not depend on q and M ..

(47) THE NUMBER OF STEPS IN THE POLYNOMIAL EUCLIDEAN ALGORITHM. 11. Proof. Suppose M = pe11 pe22 · · · pet t , where each pi is monic irreducible, pi 6= pj for i 6= j and each ei is a positive integer. We have t X0 1 Y 1 1 ) = + ··· + (1 + |h| |pi | |pi |ei i=1. h|M. <. t Y ( i=1. =. 1 ) 1 − |p1i |. |M | φ(M ). by lemma 4.2 ln deg M.. . Lemma 4.4. Suppose M ∈ A+ and deg M ≥ 2. Then we have (a) (b). %0 (M, n) q |M | , q−1 %(M, n) = |M | (deg M − n) + O (q |M |) , 2ζA (2). where the implied constants do not depend on q and M . Proof. Suppose deg M = m and fix a pair of f , g∈ A with m ≥ deg f > deg g ≥ 0 and (f, g) = 1. 2 We have to count the number of pairs of f 0 , g 0 ∈ A which satisfy the equation M = f f 0 + gg 0 , deg f 0 > deg g 0 ≥ 0. 0. f This restrictions imply deg f 0 = m − deg f and g 0 = M −f . g 0 0 It is easy to check that the number of pairs of f , g ∈ A is equivalent to the number of polynomials f 0 which satify the equation. ff0 ≡ M (15). (mod g), deg f 0 = m − deg f,. deg f 0 > deg. M − ff0 ≥ 0. g. Since (f, g) = 1, the congruence ff0 ≡ M. (mod g). has a unique solution f0 with deg f0 < deg g and we can suppose f 0 = f0 +gh with h ∈ A. Since deg f 0 = m−deg f ≥ deg f and deg f > deg g > deg f0 , we.

(48) 12. JIA-WEI GUO. have deg f 0 > deg f0 and deg f 0 = deg(gh). Thus deg h = m − deg f − deg g. By (15), we have M − f f0 m − deg f > deg( − f h) ≥ 0. g f0 = deg(f h) = Since m − deg f = (m − deg g) − (deg f − deg g) and deg M −f g m − deg g, the first deg f − deg g + 1 coefficients of f h are fixed. Thus the first deg f −deg g +1 coefficients of h are fixed. By the last inequality above, we have ( f0 q m−2 deg f − 1 if f | M −f , g m−2 deg f q otherwise ,. solutions of h, which is exactly the number of solutios of f 0 in (15). By the definition of %0 (M, n), we get X (q m−2 deg f + O (1)) %0 (M, n) = 0≤deg g<deg f = m−n 2 (f,g)=1. X. . φ(f )q n. deg f = m−n 2. by lemma 4.1 = q m−n−1 (q − 1)2 q n q |M | . This proves part (a). Similarly, we have X. %(M, n) =. (q m−2 deg f + O (1)). 0≤deg g<deg f < m−n 2 (f,g)=1. =. [ m−n−1 ] X 2 X k=1. φ(f )(q m−2k + O (1)). deg f =k. by lemma 4.1 =. [ m−n−1 ] 2 X. q 2k−1 (q − 1)2 (q m−2k + O (1)). k=1. =.   [ m−n−1 ] 2 m−n  X  + O (1) q m−1 (q − 1)2 + O  q 2k+1  2 k=1. = This proves part (b).. q−1 |M | (deg M − n) + O (q |M |) . 2ζA (2) .

(49) THE NUMBER OF STEPS IN THE POLYNOMIAL EUCLIDEAN ALGORITHM. 13. References [1] H. Heilbronn, On the average length of a class of finite continued fractions, Abhandlungen aus Zahlentheorie und Analysis, VEB Deutsher Verlag, Berlin 1968. [2] R. Lidl and H. Niederreiter, Finite Fields, Cambridge University Press (1997). [3] M. Rosen, Number Theory in Function Fields, Springer-Verlag, GTM 210 (2002). [4] Chih-Nung, Hsu, A Polynomial Additive Divisor Problem. [5] J. D. Dixon, The number of steps in the Euclidean algorithm, J. Number Theory 2 (1970), 414–422. [6] T. Tonkov, On the average length of a class of finite continued fractions, Acta Arith. 26 (1974), 47–57. [7] J. W. Porter, On a theorem of Heilbronn, Mathematika. 22 (1975), 20–28. [8] H. Davenport, Multiplicative Number Theory, Springer-Verlag, GTM 74 (1980). [9] G. H. Norton, On the asymptotic analysis of the Euclidean algorithm, J. Symbolic Computation 10 (1990) 53–58. [10] G. W. Effinger and D. R. Hayes, Additive Number Theory of Polynomials Over a Finite Field, Clarendon Press Oxford (1991).. Department of Mathematics National Taiwan Normal University 88 Sec. 4 Ting-Chou Road Taipei, Taiwan E-mail address: swordsad@yahoo.com.tw.

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