Page 37 - RMGO 3

Lema lui Hensel. Aplicat , ii 37

Conform Lemei lui Hensel (prima form˘a), rezult˘a c˘a f = X p−1 − 1 are p − 1
r˘ad˘acini distincte ˆın O p .

s
∗
2
2. Determinarea p˘atratelor perfecte din O p . Fie α = p · ε ∈ O s , i α = β ,
p
t
β = p · ε 1 ∈ O p .
Atunci
2
s
2
2
2t
α = p · ε = β = p · (ε 1 ) ⇒ s = 2t s , i ε = ε .
1
Fie ε = a 0 + a 1 p + . . . s , i 1 ≤ a 0 ≤ p − 1.
∗ 2
Cazul 1: Fie p ≥ 3. Atunci ε ∈ O p ⇔ a 0 este ,,rest p˘atratic modulo p”.
Demonstrat¸ie. ,,⇒”: Implicat , ia este evident˘a.
,,⇐”: Fie a 0 un rest p˘atratic modulo p i.e. ∃b 0 ∈ Z, (b 0 , p) = 1 astfel ˆıncˆat
2
a 0 ≡ b (mod p).
0
0
2
Fie g = X −ε ∈ O p [X|. Atunci g(b 0 ) ≡ 0 (mod p) s , i g (b 0 ) = 2b 0 6≡ 0 (mod p);
conform Lemei lui Hensel rezult˘a c˘a g are o r˘ad˘acin˘a ˆın O p .
∗ 2
Cazul 2: p = 2. Atunci ε = a 0 + a 1 · 2 + . . . ∈ (O ) ⇔ ε ≡ 1 (mod 8), adic˘a
2
4
3
ε = 1 + a 3 · 2 + a 4 · 2 + . . ..
Demonstrat¸ie. ,,⇒”: Rezult˘a din observat , ia simpl˘a c˘a a ∈ Z s , i (a, 2) = 1 ⇔
2
2
2
a = 2k + 1, deci a = (2k + 1) = 4k + 4k + 1 = 4k(k + 1) + 1 ≡ 1 (mod 8), ce
trebuia demonstrat.
3 4
,,⇐”: Dac˘a ε ≡ 1 (mod 8) ˆın O 2 ⇒ ε = 1 + a 3 · 2 + a 4 · 2 + . . . ⇒ ∃b ∈ Z \ 2Z
2
2
cu ε ≡ b (mod 8) (pentru c˘a ∀b ∈ Z \ 2Z, b ≡ 1 (mod 8)).
2
Fie g = X − ε ∈ Z 2 [X]. Atunci
2
g(b) = b − ε ≡ 0 ( mod 8),
0
g (b) = 2b ≡ 0 ( mod 2)
s , i
0
g (b) = 2b ≡ 2 ( mod 4).
0
Deci v 2 (g(b)) ≥ 3 si v 2 (g (b)) = 1.
Atunci din Lema lui Hensel, forma a treia, rezult˘a c˘a ∃β ∈ O 2 astfel ˆıncˆat

2
g(β) = β − ε = 0 iar β ≡ b ( mod 4).