Chùng minh Bê · 1

Một phần của tài liệu Xây dựng Singer và bài toán hit (Trang 39 - 45)

Trong ph¦n n y ta tr¼nh b y mët chùng minh cho Bê · 1.4. Bê · ÷ñc chùng minh b¬ng quy n¤p. Tr÷íng hñp khði ¦u l  k¸t qu£ d÷îi ¥y, mët mð rëng cõa Bê · 5.1 trong [9].

Bê · 2.1. Gi£ sû R l  mët R3F(k)-ìn thùc vîi σ(R) = 0, v  u 6= 1 l  mët ph¦n tû tòy þ trong P3. Khi â

Ru2 ∈ Sq1(P3⊗F(k)) + Sq2(P3⊗F(k)).

Chùng minh. Ta chia th nh hai tr÷íng hñp. 1 i1(R) ≡0 (mod 2). Ta vi¸t R d÷îi d¤ng R =Qi00Qi11Qi22 X sym W12j1 · · ·Wk2jk. Suy ra Ru2 =Qi00Qi11Qi22u2 X sym W12j1 · · ·Wk2jk.

V¼ Sq (Q0Q1Q2u2) = 0 n¶n theo [8, Bê · 2.5] ta câ Qi00Qi11Qi22u2 = Sq1(A) vîi A ∈P3. Nh÷ vªy Ru2 = Sq1(A) X sym W12j1 · · ·Wk2jk= Sq1A X sym W12j1 · · ·Wk2jk. 2 i1(R) ≡ 1 (mod 2). °t S = R/Q1Q2i2, vîi i2 = i2(R). V¼ σ(R) = 0 n¶n i2(R) l  ch®n. Do â ta câ Ru2 =SQi22u2Q1 =SQi22u2Sq2(Q2) = Sq2(SQi22+1u2) + Sq2(SQi22u2)Q2 = Sq2(SQi22+1u2) + Sq2(Su2)Qi22+1. Ta câ Sq2(Su2) = Sq2(S)u2+S(Sq1u)2.

Chó þ r¬ng i1(S) =i1(Sq2S)≡ 0 (mod 2). N¶n theo 1 ta câ

Sq2(Su2) = Sq1v vîi v ∈ P3⊗F(k).

Nh÷ vªy

Sq2(Su2)Qi22+1 = Sq1(v)Qi22+1 = Sq1(vQi22+1).

Bê · ÷ñc chùng minh.

B¥y gií, ta tr¼nh b y chùng minh cho Bê · 1.4. Chùng minh ÷ñc chia l m ba b÷îc.

1 N¸u Bê · 1.4 (a) v  Bê · 1.4 (b) óng vîi måi n ≤ N th¼ Bê · 1.4 (c) công vªy.

Gi£ sû u= Sq1v1+ Sq2v2 vîi v1, v2 ∈ P3. Ta câ

Ru2n =R(Sq1v1+ Sq2v2)2n =R(Sq1v1)2n +R(Sq2v2)2n

=Sq2n(Rv21n) + Sq2n(R)v21n+Sq2n+1(Rv22n) + Sq2n(R)(Sq1v2)2n + Sq2n+1(R)v22n

Chó þ r¬ng

Sq2n(R)(Sq1v2)2n = Sq2nR(Sq1v2)2n+R(Sq1Sq1v2)2n = Sq2nR(Sq1v2)2n.

Do â,

Ru2n + Sq2n(R)v21n + Sq2n+1(R)v22n ∈A(P3⊗F(k)).

°t R := R/Q22n−1. Hiºn nhi¶n R l  mët R3F(k)-ìn thùc khæng chia h¸t cho Q2 vîi h(R) = h(R)−(2n −1)≡ 0 (mod 2n) v  i1(R) =i1(R) ≤2n−1.

p döng [Bê · 2.3, Ch÷ìng II] ta thu ÷ñc

Sq2n(R) = Sq2n(RQ22n−1) =XS1+XT1,

Sq2n+1(R) = Sq2n+1(RQ22n−1) =XS2+XT2,

trong â méi h¤ng tû S1 hay S2 l  mët R3F(k)-ìn thùc thäa m¢n σ(S1) < n

v  σ(S2)< n, cán méi h¤ng tû T1 hay T2 l  mët R3F(k)-ìn thùc vîi i2(T1) ≡

i2(T2)≡2n −1 (mod 2n) v  h(T1) 2n−1 = h2(nT2−1) = 0. Do â Ru2n +XSv12n +XS2v22n +XT1v12n +XT2v22n ∈A(P3⊗F(k)).

Theo giû thi¸t, Bê · 1.4 (a) óng vîi c¡c bë ba (S1, v1, n) v  (S2, v2, n); nh÷ th¸ S1v12n v  S2v22n ·u thuëc A(P3⊗F(k)).

T÷ìng tü, Bê · 1.4 (b) óng cho c¡c bë ba (T1, v1, n) v  (T2, v2, n) cho n¶n

T1v12n v  T2v22n ·u thuëc A(P3⊗F(k)). Nâi tâm l¤i,

Ru2n ∈ A(P3⊗F(k)).

B÷îc 1 ÷ñc chùng minh.

2 N¸u Bê · 1.4 (a) óng vîi måi n ≤N th¼ Bê · 1.4 (b) công vªy. p döng [Bê · 2.2, Ch÷ìng II] ta câ

R = Sq2n+1 RQ2i2(R)−2n−1+XS,

trong âR :=R/Qi22(R) v  méi S trong têng l  mët R3F(k)-ìn thùc thäa m¢n

σ(S)< n. Do â

Ru2n = Sq2n+1 RQ2i2(R)−2n−1u2n +XSu2n.

V¼ σ(S) < n n¶n ¡p döng Bê · 1.4 (a) cho bë ba (S, u, n) ta câ Su2n ∈

°t R1 :=RQi22(R)−2 . Theo cæng thùc Cartan ta câ Sq2n+1(R1)u2n = Sq2n+1(R1u2n) + Sq2n(R1)(Sq1u)2n +R1(Sq2u)2n = Sq2n+1(R1u2n) + Sq2nR1(Sq1u)2n+R1(Sq1Sq1u)2n +R1(Sq2u)2n = Sq2n+1(R1u2n) + Sq2nR1(Sq1u)2n+R1(Sq2u)2n. V¼ vªy Sq2n+1(R1u2n) +R1(Sq2u)2n ∈A(P3⊗F(k)). Ta câ σ(R1) =σ(Qi2(R)−2n −1

2 ) = n−1< n. Do â ¡p döng Bê · 1.4 (a) ta câ

R1(Sq2u)2n ∈A(P3⊗F(k)).

Tâm l¤i

Ru2n = Sq2n+1(R1)u2n +XSu2n ∈A(P3⊗F(k)).

B÷îc 2 ÷ñc chùng minh xong. 3 Bê · 1.4(a) óng vîi måi n.

Kh¯ng ành n y ÷ñc chùng minh b¬ng quy n¤p theo n.

Vîi n = 1, tø gi£ thi¸t σ(R)<1 ta câ σ(R) = 0. p döng Bê · 2.1 ta câ

Ru2 ∈ Sq1(P3⊗F(k)) + Sq2(P3⊗F(k)).

Do â, Bê · 1.4 (a) óng vîi n = 1.

B¥y gií x²t n > 1, gi£ sû Bê · 1.4 (a) óng vîi t§t c£ c¡c gi¡ trà nhä hìn

n. Theo B÷îc 1 v  B÷îc 2, Bê · 1.4 (b) v  Bê · 1.4 (c) công óng vîi t§t c£ c¡c gi¡ trà b² hìn n. Ta x²t ba tr÷íng hñp sau.

• Tr÷íng hñp 1. σ(R) = 0. Theo Bê · 2.1, ta câ

Ru2n =R(u2n−1)2 ∈ A(P3⊗F(k)).

• Tr÷íng hñp 2. Tçn t¤i mët sè nguy¶n m vîi 0≤m < σ(R) v  h(R) 2m

= 0. K¸t hñp vîi sü ki»n m < σ(R)< n v  i2(R) ≡ 2σ(R) −1 (mod 2σ(R)+1), ta câ m+ 1< n v  i2(R) ≡2m+1−1 (mod 2m+1).

V¼ m+ 1 < n v  theo gi£ thi¸t quy n¤p, ¡p döng Bê · 1.4 (b) cho bë ba

(R, u2n−m−1, m+ 1) ta câ

• Tr÷íng hñp 3. σ(R)> 0 v  h(R) 2m

= 1 vîi måi m thäa m¢n 0≤ m < σ(R). Khi â h(R)≡2σ(R)−1 (mod 2σ(R)).

Kþ hi»u p:=σ(R). Ta vi¸t R duy nh§t d÷îi d¤ng

R =RS2p,

trong âR l  mëtR3F(k)-ìn thùc vîi i0(R), i1(R) ≤2p−1, i2(R) = 2p−1, cán S l  mët ìn thùc theo c¡c bi¸n Q0, Q1, Q2 thäa m¢n σ(S) = 0.

p döng Bê · 2.1 cho S v  v :=u2n−p−1 6= 1, ta thu ÷ñc

Sv2 ∈A(P3⊗F(k)).

M°t kh¡c,

h(R) =h(R)−2ph(S) ≡2p−1 (mod 2p), i2(R) = 2p−1≥i1(R).

Sû döng gi£ thi¸t quy n¤p còng vîi p = σ(R) < n, ta ¡p döng Bê · 1.4 (c) cho bë ba (R, Sv2, p) ta thu ÷ñc

Ru2n = R(Sv2)2p ∈ A(P3⊗F(k)).

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Một phần của tài liệu Xây dựng Singer và bài toán hit (Trang 39 - 45)