GENERATING CERTAIN QUINTIC IRREDUCIBLE POLYNOMIALS OVER FINITE FIELDS. Youngwoo Ahn and Kitae Kim

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1 Korean J. Mah. 19 (2011), No. 3, pp 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, Revsed Augus 29, Acceped Augus 31, 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 ( ). Correspondng auhor.

2 264 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 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 =

3 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 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 x m (m+1) x3 + 2n (m+1) x m x + n. (m+1) 2 (m+1) 2 For a gven G(x) = x 5 +3a 2 x 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 α = (1 + β α) q5 1. γ 1

4 266 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 σ 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.

5 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 σ 2 + σ 4 2, n = 3σ 1 σ 3 5σ σ 2 + σ 4 2. Applyng Eq (4) no he equaons (5) and (6), we ge So m = 10 2σ σ 4 2 = 5 σ σ 4 2, n = 16σ σ 4 2 = 8σ σ 4 2. σ 4 2 = 5 3m and σ 5 2 = n m + 1 m + 1. Noe ha m 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 = m (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 b2 x 3 2ax 2 + bx a.

6 268 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 σ 5 2 = 2( 3σ 5 2 ) 2(2σ 5 2 ) + 10σ 5 2 = 0, σ 1 + σ 3 + σ 5 2 = 3σ σ 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 σ 5 2 )α = 0

7 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 σ 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 σ 2 + σ 4 2 and n = 3σ 1 σ 3 5σ 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 x m (m+1) x3 + 2n (m+1) x m 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 = m 3 5 3m 2 (m+1) (m+1) m + 5 3m = m, (m+1) (m+1) 2 2 n = 3 3n 2n 5 n 2 m+1 (m+1) (m+1) m + 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 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

8 270 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 ( b ) = a, 3+b 2 2 b = 5 3m (m + 1) = b2 3+b ( b ) = 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 = (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,

9 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 = = = 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 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.

10 272 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, [2] A.K. Lensra and E. Verheul, The XTR publc key sysem, Advances n Crypology (CRYPTO 2000), LNCS 1880, [3] A.K. Lensra and E. Verheul, Fas Irreducbly and subgroup membershp esng n XTR, Advances n Crypology (PKC 2001), LNCS 1992, Deparmen of Mahemacs Inha Unversy 253, Yonghyun-dong, Nam-gu Incheon, , Korea E-mal: ywahn@nha.edu Deparmen of Mahemacs Inha Unversy 253, Yonghyun-dong, Nam-gu Incheon, , Korea E-mal: kkm@nha.ac.kr