Korean J. Mah. 19 (2011), No. 3, pp. 263 272 GENERATING CERTAIN QUINTIC IRREDUCIBLE POLYNOMIALS OVER FINITE FIELDS Youngwoo Ahn and Kae Km Absrac. In he paper [1], an explc correspondence beween ceran cubc rreducble polynomals over F q and cubc rreducble polynomals of specal ype over F q 2 was esablshed. In hs paper, we show ha we can mmc such a correspondence for qunc polynomals. Our ransformaons are raher consrucve so ha can be used o generae rreducble polynomals n one of he fne felds, by usng ceran rreducble polynomals gven n he oher feld. 1. Inroducon Generang rreducble polynomals and deermnng her rreducbly have played an mporan role n he heory of fne felds and s applcaons, especally codng heory and crypography. For desgnng crypographc proocols, A. K. Lensra and E. Verheul used a cubc rreducble polynomal f(x) = x 3 cx 2 + c p x 1 over he fne feld F p 2 where p s an odd prme ([2], [3]). In order o oban compac represenaon of he elemens n F p 6, hey made use of he absolue race map. Along wh he work, Km e al. suded n [1] cubc rreducble polynomals of he same form defned over F q 2 where q s a power of prme p, and esablshed a correspondence beween he se of such rreducble polynomals and he se of rreducble polynomals of he form g(x) = x 3 ax 2 + bx + a n F q [x] where s a quadrac non-resdue Receved May 31, 2011. Revsed Augus 29, 2011. Acceped Augus 31, 2011. 2000 Mahemacs Subjec Classfcaon: 11T71, 12E10, 12E20. Key words and phrases: fne felds, rreducble polynomal, correspondence. Ths research was suppored by Basc Scence Research Program hrough he Naonal Research Foundaon of Korea(NRF) funded by he Mnsry of Educaon, Scence and Technology (2011-0011654). Correspondng auhor.
264 Youngwoo Ahn and Kae Km n F q. Ther correspondence s so explc ha one can use o generae a cubc rreducble polynomal over F q 2 from a cubc rreducble polynomal over F q, or vce versa. In hs paper, we esablsh a one o one correspondence beween a famly of rreducble polynomals of he form x 5 cx 4 + c q x 1 over F q 2 and a famly of rreducble polynomals of he form x 5 + 3a 2 x 4 + 5+b 2 x 3 2ax 2 + bx a over F q. Lke [1], our approach s somewha heorecal bu he ransformaons are consrucve. Thus provdes an effcen mehod o generae qunc rreducble polynomals over F q 2, sarng wh ceran rreducble polynomals of over F q. 2. Ceran qunc rreducble polynomals In hs secon, we gve a one o one correspondence beween he se of ceran qunc rreducble polynomals over F q 2 and over F q. Le p be an odd prme. We assume ha every feld of characersc p ha we consder s a subfeld of a fxed algebrac closure of he prme feld F p. Then he Galos feld F q s unquely deermned by he number q = p k of elemens ha conans. Snce p s odd, half of he elemens of F q are non-squares n F q. Le be a non-square elemen n F q. Then becomes a square n he quadrac exenson F q 2, say = α 2 for some α F q 2. From now on, α wll sand for an elemen of F q 2 \ F q such ha α 2 = F q. Then (α q ) 2 = (α 2 ) q = q = and hence α q = α. Moreover, we have F q 2d = F q d(α) for any posve odd neger d. Suppose ha F (x) = x 5 cx 4 +c q x 1 s an rreducble polynomal n F q 2 whose all roos are h where 1 5. Snce x 5q F (x q ) = F (x) q, h q h q5 s represen fve dfferen roos of F (x), for f h q = h q j = h q5 j = h 1 j. The complee facorzaon of F over s splng feld F q 10 wren as F (x) = We clam ha h = h q5 (x h ) = (x h q ). hen h 1 = can be for each = 1,..., 5. Frs noe ha h h q = 1 for some and so h q2 1 = 1. Tha s, for each. If no, hen h q+1 h F q 2 whch conradcs o he rreducbly of F (x). If h 1 = h q 2, h 2 =
Generang ceran qunc rreducble polynomals over fne felds 265 h q 1 hen h 1 = h q2 1 so h 1 F q 2 whch also conradcs o he rreducbly of he polynomal. Moreover, f h 1 = h q 2, h 2 = h q 3, h 3 = h q 1 hen h 4 mus be h q2 4 whch s a conradcon. Hence he clam s proved. Snce h = h q5 s 1: N q 10 /q 5(h ) = h h q5 for each. for each, h h q5 = 1 and so he norm of h over F q 5 = 1. I follows from Hlber 90 ha h = g q5 1 Now we wll dscuss our one o one correspondence beween qunc rreducble polynomals of he form x 5 cx 4 +c q x 1 over F q 2 and ceran qunc rreducble polynomals over F q. Theorem 1. Le F q be a fne feld of characersc p and a quadrac non-resdue n F q wh = α 2 for some α F q 2. There s a one o one correspondence beween he se of rreducble polynomals n F q 2 of he form (1) x 5 cx 4 + c q x 1 and he se of rreducble polynomals n F q of he form (2) x 5 + 3a 2 x 4 + 5+b2 x 3 2ax 2 + bx a The correspondence s gven by: For a gven F (x) = x 5 cx 4 + c q x 1 wh c = m + nα, we assocae G(x) = x 5 3n m+1 x4 + 10+2m (m+1) x3 + 2n (m+1) x2 + 5 3m x + n. (m+1) 2 (m+1) 2 For a gven G(x) = x 5 +3a 2 x 4 + 5+b2 x 3 2ax 2 +bx a, we assocae F (x) = x 5 cx 4 + c q x 1 wh c = 5 b2 + 8a2 α. 3+b 2 3+b 2 Proof. Le h 1, h 2,, h 5 be all he roos of F (x) = x 5 cx 4 + c q x 1. Then for each, h q = h 1 and h = g q5 1 for some g F q 10. Recall ha α q = (q 1)/2 α = α and F q 10 = F q 5(α) s he quadrac exenson of F q 5. Snce g F q 10, for each, g can be represened as g = γ 1 + γ 2 α for some γ 1, γ 2 F q 5. Noce ha γ,1 canno be 0. If no, h = (γ 2 α) q5 1 = α q5 α 1 = αα 1 = 1, a conradcon. Thus we can rewre h as, by leng β = γ 2 /γ 1 F q 5, ( h = g q5 1 = (γ 1 + γ 2 α) q5 1 = 1 + γ ) q 5 1 2 α = (1 + β α) q5 1. γ 1
266 Youngwoo Ahn and Kae Km Now he polynomal F (x) defned over F q 2 can be expressed by ( F (x) = x 1 β ) α (3). 1 + β α Here we used he fac ha (1 + β α) q5 1 = (1 + β α) q5 1 + β α = 1 + βq5 α q5 1 + β α = 1 β α 1 + β α. Now we assocae he rreducble polynomal F (x) over F q 2 o an rreducble polynomal F (x) defned n he feld F q whose roos are β 1, β 2,..., β 5 : F (x) = (x β ). Le us denoe σ he h elemenary symmerc polynomal of β 1,..., β 5. We hen calculae he consan erm of F (x): 1 β α 1 + β α = (1 + σ 2 + σ 4 2 ) (σ 1 + σ 3 + σ 5 2 )α (1 + σ 2 + σ 4 2 ) + (σ 1 + σ 3 + σ 5 2 )α. Snce he consan erm of F (x) s 1 and p s an odd prme, we oban σ 1 + σ 3 + σ 5 2 = 0. Noe ha 1 + σ 2 + σ 4 2 0, for oherwse 5 h = 1 hs leads o a conradcon. Smlarly, by sraghforward calculaon and comparng he coeffcens, we have c = (5 + σ 2 3σ 4 2 ) + (3σ 1 σ 3 5σ 5 2 )α 1 + σ 2 + σ 4 2, 0 = (10 2σ 2 + 2σ 4 2 ) + (2σ 1 2σ 3 + 10σ 5 2 )α 1 + σ 2 + σ 4 2. Thus we ge he followng equaons So (4) σ 1 + σ 3 + σ 5 2 = 0, σ 1 σ 3 + 5σ 5 2 = 0, 5 σ 2 + σ 4 2 = 0. σ 1 = 3σ 5 2, σ 2 = 5 + σ 4 2 and σ 3 = 2σ 5 2.
Generang ceran qunc rreducble polynomals over fne felds 267 Furhermore, by leng c = m + nα for m, n F q, we have (5) (6) m = 5 + σ 2 3σ 4 2 1 + σ 2 + σ 4 2, n = 3σ 1 σ 3 5σ 5 2 1 + σ 2 + σ 4 2. Applyng Eq (4) no he equaons (5) and (6), we ge So m = 10 2σ 4 2 6 + 2σ 4 2 = 5 σ 4 2 3 + σ 4 2, n = 16σ 5 2 6 + 2σ 4 2 = 8σ 5 2 3 + σ 4 2. σ 4 2 = 5 3m and σ 5 2 = n m + 1 m + 1. Noe ha m + 1 0 because f m = 1 hen c = 1 + nα and so 1 s a roo of F (x) whch conradcs o he rreducbly of F (x). Thus, we have he coeffcens of F (x) = x 5 σ 1 x 4 + σ 2 x 3 σ 3 x 2 + σ 4 x σ 5 as follows: σ 1 = 3σ 5 2 = 3n m + 1, σ 2 = 5 + σ 4 2 = 10 + 2m (m + 1), σ 3 = 2σ 5 = 2n (m + 1), σ 4 = 5 3m (m + 1) 2, σ 5 = n (m + 1) 2. Recall ha β F q 5 and h = (1 + β α) q5 1. So f β F q hen h F q 2 whch leads o a conradcon. Hence β s are no conaned n F q and hence we conclude ha he polynomal F (x) s rreducble over F q of he requred form. Conversely, suppose ha G(x) s an rreducble polynomal defned n F q [x] of he form (7) x 5 + 3a 2 x 4 + 5+b2 x 3 2ax 2 + bx a.
268 Youngwoo Ahn and Kae Km Le β 1, β 2,..., β 5 be all he roos of F (x) n F q 5. Then G(x) = (x β 1 )(x β 2 ) (x β 5 ) and by rearrangng, we may assume ha β q = β +1 mod 5. Defne a polynomal F (x) over F q 2 o be G (x) = ( x (1 + β α) q5 1 ). As above, le h = (1 + β α) q5 1 and σ be he h elemenary symmerc polynomal of β j s. Then h = 1 β α 1+β. Before compung coeffcens of α F, we noe ha he roos of he polynomal F are conjugae o each oher over F q 2. Ths means ha he polynomal s rreducble over F q 2. From he defnon of F (x), we have he followng equales 10 2σ 2 + 2σ 4 2 = 10 2(5 + σ 4 2 ) + 2σ 4 2 = 0, 2σ 1 2σ 3 + 10σ 5 2 = 2( 3σ 5 2 ) 2(2σ 5 2 ) + 10σ 5 2 = 0, σ 1 + σ 3 + σ 5 2 = 3σ 5 2 + 2σ 5 2 + σ 5 2 = 0. In order o descrbe coeffcens of F (x) n erms of values n F q 2, we frs noe ha 5 (1 + β α) = (1 + σ 2 + σ 4 2 ) + (σ 1 + σ 3 + σ 5 2 )α = 1 + σ 2 + σ 4 2, whch s no zero because f no hen 1 + β α = 0 for some and so α F q, a conradcon. A sraghforward compuaon shows ha (8) (9) (10) (1 + β α)h 1 h 2 h 3 h 4 h 5 = (1 + σ 2 + σ 4 2 ) (σ 1 + σ 3 + σ 5 2 )α = 1 + σ 2 + σ 4 2 (1 + β α) h 1 h 2 h 3 h 4 1 < 2 < 3 < 4 = (5 + σ 2 3σ 4 2 ) (3σ 1 σ 3 5σ 5 2 )α (1 + β α) h 1 h 2 h 3 1 < 2 < 3 = (10 2σ 2 + 2σ 4 2 ) (2σ 1 2σ 3 + 10σ 5 2 )α = 0
Generang ceran qunc rreducble polynomals over fne felds 269 (11) (12) (1 + β α) 1 < 2 h 1 h 2 = (10 2σ 2 + 2σ 4 2 ) + (2σ 1 2σ 3 + 10σ 5 2 )α = 0 5 (1 + β α) h = (5 + σ 2 3σ 4 2 ) + (3σ 1 σ 3 5σ 5 2 )α. The equaons (11) and (12) say ha he coeffcens of x 2 and x 3 are zero, respecvely. The equaon (9) ells us he consan erm s 1. Now we le c = m + nα where m = 5 + σ 2 3σ 4 2 1 + σ 2 + σ 4 2 and n = 3σ 1 σ 3 5σ 5 2 1 + σ 2 + σ 4 2. Then he coeffcen of x 4 s c and he coeffcen of x s c q by he facs α q = α and he equaons (10) and (13). The one o one correspondence s mmedae from our ransformaons. To be precse, suppose ha F (x) = x 5 cx 4 + c q x 1 s an rreducble polynomal over F q 2 where c = m + nα. Then, by he ransformaon above, we ge F (x) = x 5 3n m+1 x4 + 10+2m (m+1) x3 + 2n (m+1) x2 + 5 3m x + n. Then he lfed rreducble polynomal (F (m+1) 2 (m+1) 2 ) (x) = x 5 c x 4 +(c ) q x 1 o F q 2 s F (x) agan, because, f we le c = m +n α, m = 5 + 10+2m 3 5 3m 2 (m+1) (m+1) 2 1 + 10+2m + 5 3m = m, (m+1) (m+1) 2 2 n = 3 3n 2n 5 n 2 m+1 (m+1) (m+1) 2 1 + 10+2m + 5 3m = n. (m+1) (m+1) 2 2 Thus (F ) and F are he same. Conversely, suppose ha G(x) = x 5 + 3a 2 x 4 + 5+b2 x 3 2ax 2 + bx a s an rreducble polynomal defned n F q [x] and G s he lfed polynomal by our ransformaon, say G (x) = x 5 cx 4 + c q x 1 wh c = m + nα, where m = 5 b2 8a2 and n = 3 + b2 3 + b. 2
270 Youngwoo Ahn and Kae Km Then he qunc polynomal (G ) obaned from G s agan G: f a and b are consan erm and coeffcen of x, respecvely, hen a = n 8a2 (m + 1) = 3+b ( 2 2 5 b 2 + 1 ) = a, 3+b 2 2 b = 5 3m (m + 1) = 5 3 5 b2 3+b ( 2 2 5 b 2 + 1 ) = b. 3+b 2 2 Snce he remanng erms are compleely deermned by he consan erm and he degree 1 erm we conclude he correspondence s one o one as requred. 3. Examples In hs secon, we shall gve wo examples o explan our ransformaon for fne felds F 5 and F 17. In order o oban an rreducble polynomal over F p 2, we should frs fnd an rreducble polynomal G(x) of he desred form n F p [x] where p s an odd prme. Noe ha = 2 s a quadrac non-resdue of p = 5. If we se a = 1 and b = 0, hen G(x) = x 5 + 2x 4 + x 2 1 sasfes he condon as n he heorem. To be precse, we show ha G(x) s rreducble over F 5. Suppose ha G has a roo γ n F 5. Then G(γ) = γ 5 + 2γ 4 + γ 2 1 = γ 2 + γ 1 + 1 = (γ 2) 2 3. Snce 3 s no a square n F 5, G(γ) canno be zero. Hence, G(x) has no roos n F 5. Now consder he followng rreducble polynomals over F 5 : x 2 ± 2, (x ± 1) 2 ± 2, (x ± 2) 2 ± 2. Snce ±2 are quadrac non-resdues mod 5 and he number of quadrac rreducble polynomals over F 5 s 10, hose are all of he rreducble polynomals of degree 2 over F 5. If x 2 ± 2 dvdes G(x) hen x 5 + 2x 4 + x 2 1 = (x 2 ± 2)(x 3 + d 2 x 2 + d 1 x + d 0 ). Comparng he coeffcens, we have d 1 = 0 and d 1 ± 2 = 0. So, x 2 ± 2 canno dvde G(x). Smlarly, no quadrac rreducble polynomals dvde G(x), and hence we can conclude ha G(x) s rreducble over F 5. Now, by applyng our ransformaon, we have G (x) = x 5 cx 4 + c 5 x 1,
Generang ceran qunc rreducble polynomals over fne felds 271 where c = m + nα. Snce m = 5 b 3+b 2 c = α and c 5 = 2 α = α. Hence = 5 3 G (x) = x 5 αx 4 αx 1. = 0 and n = 8a2 3+b 2 = 32 3 = 1, For a second example, we frs noe ha, for a prme p wh p 2, 3 (mod 5), 5 s a quadrac non-resdue mod p. When p = 17, 5 s also a quadrac non-resdue mod 17. As n he above example, we se a = 1, b = 0 and compue he lfed rreducble polynomals of he followng polynomals: G(x) = x 5 + 7x 4 + x 3 + 7x 2 1, = 5 G(x) = x 5 + 7x 4 x 3 7x 2 1, = 5. From he values a, b and, we ge m = 5 3 = 5 6 = 4, n = 8 52 3 = 4 3 1 = 7 and c = 4 + 7α. Then G(x) and G(x) are ransformed no he same rreducble polynomal x 5 + (4 7α)x 4 (4 + 7α)x 1. In concludng remarks, we nvesgae some properes of such polynomals as n our heorem. Frs, le us denoe he polynomal x 5 + 3a 2 x 4 + 5+b2 x 3 2ax 2 + bx a by G(x, a, b), or smply G(x). Then we have G(x) = G( x) = G(x, a, b). Tha s, f G s rreducble hen so s G, and vce versa. Second, consder he lfed rreducble polynomals G (x) and G (x) of G(x) and G(x), respecvely, where he polynomals sasfy he above propery. Then G and G have he form G (x) = x 5 cx 4 + c q x 1 and G (x) = x 5 cx 4 + c q x 1 where c = m + nα and c = m + ñα, respecvely. Snce m = m and ñ = 8( a)2 3+b 2 = 8a2 3+b 2 = n, we have c q = (m + nα) q = m q + n q α q = m + n( α) = m nα = c. Smlarly, c q = m ñα = c. x 5 G (x 1 ) = x 5 (x 5 cx 4 + c q x 1 1) = x 5 c q x 4 + cx 1 = x 5 cx 4 + c q x 1 = G (x) Thus G s he recprocal of G.
272 Youngwoo Ahn and Kae Km Fnally, n he paper [1] he auhors gave anoher one-o-one correspondence beween cubc rreducble polynomals of ceran ypes. Namely, here s a one-o-one correspondence beween he se of rreducble polynomals n F q 2[x] of he form f(x) = x 3 cx 2 + c q x 1 and he se of rreducble polynomals n F q [x] of he form x 3 + ux 2 x + v. In fac, hs correspondence s obvous n he sense ha one can easly ge such a correspondence by assocang f(x) o he recprocal g (x) = 1 a x3 g(x 1 ) of g(x) nsead of g self, where g(x) = x 3 ax 2 + bx + a as menoned n he nroducon. In he same argumens, one can have anoher correspondence beween ceran rreducble polynomals n F q 2[x] and n F q [x], of degree 5. References [1] H. Km, J. Km, and I. Ye, Ceran Cubc Polynomals over Fne Felds, J. Korean Mah. Soc. 46 (2009), no. 1, 1 12. [2] A.K. Lensra and E. Verheul, The XTR publc key sysem, Advances n Crypology (CRYPTO 2000), LNCS 1880, 1 19. [3] A.K. Lensra and E. Verheul, Fas Irreducbly and subgroup membershp esng n XTR, Advances n Crypology (PKC 2001), LNCS 1992, 73 86. Deparmen of Mahemacs Inha Unversy 253, Yonghyun-dong, Nam-gu Incheon, 402-751, Korea E-mal: ywahn@nha.edu Deparmen of Mahemacs Inha Unversy 253, Yonghyun-dong, Nam-gu Incheon, 402-751, Korea E-mal: kkm@nha.ac.kr