Về dạng chuẩn Edwards và một vài ứng dụng

Q
x
2
+ y
2
= c
2
(1 + x
2
y
2
) k
ax
2
+ y
2
= 1 + dx
2
y
2
a = d, a, d ∈ k \ {0, 1}
Q
Q
P
1
× P
1
d Q

K
E K
E : y
2
+ a
1
xy + a
3
y = x
3
+ a
2
x
2
+ a
4
x + a
6
,
a
1
, a
2
, a
3
, a
4
, a
6
∈ K ∆ = 0 ∆ E




















∆ = −d
2
2
d
8
− 8d
3
4
− 27d
2
6

+ 9d
2
d
4
d
6
d
2
= a
2
1
+ 4a
2
d
4
= 2a
4
+ a
1
a
3
d
6
= a
2
3
+ 4a
6
d
8

= a
2
1
a
6
+ 4a
2
a
6
− a
1
a
3
a
4
+ a
2
a
2
3
− a
2
4
.
L K L − E
E(L) = {(x, y) ∈ L × L : y
2
+ a
1
xy + a

3
y −x
3
− a
2
x
2
− a
4
x −a
6
= 0} ∪ {∞}
∞
y
2
= x
3
− x y
2
= x
3
+ x
E K
E : y
2
+ a
1
xy + a
3
y = x

3
+ a
2
x
2
+ a
4
x + a
6
.
E
¯
E : y
2
z + a
1
xyz + a
3
yz
2
= x
3
+ a
2
x
2
z + a
4
xz
2

+ a
6
z
3
,
P E (x : y : z).
P (x, y)
(x : y : 1) P (x : y : z)
z = 0 (x/z, y/z)
z = 0 P ∞,
P = (0 : y : 0) = (0 : 1 : 0)
E K a
1
, a
2
, a
3
, a
4
, a
6
E K
E K E
K
∆ = 0 E
E
∞
L − E (x, y)
x, y L
L − L K

E
1
E
2
K
E
1
: y
2
+ a
1
xy + a
3
y = x
3
+ a
2
x
2
+ a
4
x + a
6
E
2
: y
2
+ ¯a
1
xy + ¯a

3
y = x
3
+ ¯a
2
x
2
+ ¯a
4
x + ¯a
6
K u, r, s, t ∈ K, u = 0
(x, y) → (u
2
x + r, u
3
y + u
2
sx + t)
E
1
E
2
E : y
2
+ a
1
xy + a
3
y = x

3
+ a
2
x
2
+ a
4
x + a
6
K (K) = 2, 3

y +
a
1
x
2
+
a
3
2

2
= x
3
+

a
2
+
a

2
1
4

x
2
+

a
4
+
a
1
a
3
2

x +

a
2
3
4
+ a
6

.














y
1
= y +
a
1
2
x +
a
3
2
,
a

2
= a
2
+
a
2
1

4
,
a

4
= a
4
+
a
1
a
3
2
,
a

6
=
a
2
3
4
+ a
6
,
y
2
1
= x
3

+ a

2
x
2
+ a

4
x + a

6
.
y
2
1
=

x +
a

2
3

3
+

a

4
−

a
2
2
3

x +

a

6
−
a
3
2
27

.







x
1
= x +
a

2

3
,
a = a

4
−
a
2
3
,
b = a

6
−
a
3
2
27
,
y
2
1
= x
3
1
+ ax + b.
∆ = −16(4a
3
+ 27b
2

)
E K E(K)
P = (x
1
, y
1
) Q = (x
2
, y
2
)
E P Q
P Q E
R

R

x
R R P Q R = P + Q
2P = P + P E
P + Q = R P + P = R
P E R

R

x R R
P R = P + P = 2P
E
y
2

+a
1
xy+a
3
y = x
3
+a
2
x
2
+a
4
x+a
6
a
1
, a
2
, a
3
, a
4
, a
6
∈
K P
1
= (x
1
, y

1
) P
2
= (x
2
, y
2
) E P
1
, P
2
= ∞.
P
1
+ P
2
= P
3
= (x
3
, y
3
)
x
1
= x
2
,
x
3

= −x
1
−x
2
−a
2
+ m(m + a
1
), y
3
= −y
1
−a
3
−a
1
x
3
+ m(x
1
−x
3
),
m =
y
2
−y
1
x
2

−x
1
x
1
= x
2
y
2
= −y
1
− a
1
x
1
− a
3
P
1
+ P
2
= ∞ P
2
P
1
−P
1
x
1
= x
2

P
2
= −P
1
x
3
= −x
1
−x
2
−a
2
+ m(m + a
1
), y
3
= −y
1
−a
3
−a
1
x
3
+ m(x
1
−x
3
),
m =

3x
2
1
+2a
2
x
1
+a
4
−a
1
y
2y
1
+a
1
x
1
+a
3
P + ∞ = P
P E.
E
K
∞.
P E
n ≥ 1 nP = ∞ P n −
E n ≥ 1 nP = ∞ P
E F
q

,
#E(F
q
) = #{P ∈ E(F
q
)}.
#E(F
q
)
E
F
q
. E(F
q
)
|q + 1 − #E(F
q
)| ≤ 2
√
q.
O(log
8
q) q
K
E
M,A,B
: Bv
2
= u
3

+ Au
2
+ u,
A ∈ K \ {−2, 2} B ∈ K \ {0}
B ∈ K \{0} B
3
(
v
B
)
2
= (
u
B
)
3
+
A
B
(
u
B
)
2
+
1
B
2
u
B

.
X = u/B, Y = v/B
Y
2
= X
3
+
A
B
X
2
+
1
B
2
X.
(u, v) → (u/B, v/B)
(1.1)
P
1
= (u
1
, v
1
) P
2
= (u
2
, v
2

)
E
M,A,B
• P
1
= ±P
2
P
3
= (u
3
, v
3
) = P
1
+ P
2
u
3
= Bλ
2
− A − u
2
− u
1
v
3
= λ(u
1
− u

3
) −v
1
,
λ = (v
2
− v
1
)/(u
2
− u
1
)
• P
1
= P
2
P
1
= −P
2
P
3
= (u
3
, v
3
) = 2P
1
u

3
= Bλ
2
− A − 2u
1
v
3
= λ(u
1
− u
3
) −v
1
,
λ = (3u
2
1
+ 2Au
1
+ 1)/(2Bv
1
)
• P
2
= −P
1
P
1
+ P
2

= ∞ −P
1
P
1
(u
1
, −v
1
)
K (K) = 2, 3
E E : y
2
= x
3
+ ax + b
x
3
+ ax + b = 0 K
3α
2
+ a K α
x
3
+ ax + b = 0 K
E
s (3α
2
+ a)
−1
K

B = s, A = 3αs (x, y) →
(u, v) = (s(x − α), sy) E E
M,A,B
E
M,A,B
Bv
2
= u
3
+ Au
2
+ u
E
E
M,A,B
: Bv
2
= u
3
+ Au
2
+ u (0, 0) ∈ E
M,A,B
(k)
2 E
x
3
+ ax + b = 0 K
(1)
E

E
M,A,B
(x, y) → (s(x−α

), t(y−β

)) s, t, α

, β

∈
K, s, t = 0 (α, 0) 2
E (0, 0)
α

= α, β

= 0 (x, y)
(s(x −α), ty) E
M,A,B
Bt
2
y
2
= s
3
(x −α)
3
+ As
2

(x −α)
2
+ s(x − α).
(x, y) E y
2
= x
3
+ ax + b
Bt
2
(x
3
+ ax + b) = s
3
(x −α)
3
+ As(x − α)
2
+ (x − α).
Bt
2
= s
3
s
s
2
(x
3
+ ax + b) = s
2

(x −α)
3
+ As(x − α)
2
+ (x − α).
x x = α
s
2
(3α
2
+ a) = 1,
3α
2
+ a K

k d ∈ k \ {0, 1}
E
E,d
E
E,d
: x
2
+ y
2
= 1 + dx
2
y
2
.
E

E,a,d
:
ax
2
+ y
2
= 1 + dx
2
y
2
a, d ∈ k \ {0, 1}, a = d
E k
E E K/k
[K : k] = 2
E
E,a,d
: ax
2
+ y
2
= 1 + dx
2
y
2
E
E,d/a
: X
2
+ Y
2

= 1 + (d/a)X
2
Y
2
(x, y) → (x
√
a, y) E
E,a,d
E
E,d/a
k(
√
a) a k E
E,a,d
E
E,d/a
k
x
2
+ y
2
= 1 − 200x
2
y
2
−4x
2
+ y
2
= 1 −100x

2
y
2
E
E,a,d
E
M,A,B
: Bv
2
= u
3
+ Au
2
+ u
A = 2(a + d)/(a − d) B = 4/(a − d)
A, B a = d B ∈ k \{0}
A ∈ k \ {−2, 2} A = 2 a −d = a + d d = 0
E
E,a,d
A = −2 −d −a = a −d a = 0
E
E,a,d
E
E,a,d
(k) E
M,A,B
(k)
k E
E,a,d
E

M,A,B
ϕ : E
E,a,d
(k) → E
M,A,B
(k)
(x, y) → (u, v)
(u, v) =

(1 + y)/(1 − y), (1 + y)/(1 − y)x

ϕ
E
E,a,d
(k) E
M,A,B
(u, v) → (x, y) =

(u/v), (u − 1)/(u + 1)

(x, y) ∈ E
E,a,d
(k) (u, v)
Bv
2
= u
3
+ Au
2
+ u A = 2(a + d)/(a − d)

B = 4/(a − d)
x
2
+ y
2
= 1 + dx
2
y
2
(u, v)
E
M,A,B
y = 1 x = 0 ϕ
(x, y) E
E,a,d
v = 0 u = −1
(u, v) E
M,A,B
ϕ E
E,a,d
(k) E
M,A,B
(k)
E
E,a,d
E
M,A,B

k 2
E k E(k)

d ∈ k \ {0, 1} x
2
+ y
2
= 1 + dx
2
y
2
k E
E(k)
d ∈ k x
2
+y
2
= 1 + dx
2
y
2
k E
k E(k)
d ∈ k x
2
+y
2
= 1+ dx
2
y
2
k E
E

s
2
+ a
1
rs + a
3
s = r
3
+ a
2
r
2
+ a
4
r + a
6
.
(k) = 2 ¯s = s + (a
1
r + a
3
)/2
E ¯s
2
= r
3
+ (a
2
− a
2

1
/4)r
2
+ (a
4
− a
1
a
3
)r + (a
6
− a
2
3
/4)
a
1
= 0 a
3
= 0 E
s
2
= r
3
+ a
2
r
2
+ a
4

r + a
6
P = (r
1
, s
1
) E 2P
2P = (r
2
, 0) ¯r = r − r
2
2P (0, 0)
2P = (0, 0) a
6
= 0 E
s
2
= r
3
+ a
2
r
2
+ a
4
r a
2
a
4
r

1
, s
1
P s
1
= 0 s
1
= 0 P
E r
1
= 0 2P = (0, 0)
E P (0, 0)
P s
1
−0 = (r
1
−0)λ λ
λ = (3r
2
1
+2a
2
r
1
+a
4
)/2s
1
3r
3

1
+2a
2
r
2
1
+a
4
r
1
= 2s
2
1
P E 2s
2
1
= 2s
3
1
+ 2a
2
r
2
1
+ 2a
4
r
1
r
3

1
= a
4
r
1
a
4
= r
2
1
a
2
= (s
2
1
− r
3
1
− a
4
r
1
)/r
2
1
a
4
= r
2
1

a
2
= s
2
1
/r
2
1
− 2r
1
d = 1 − 4r
3
1
/s
2
1
a
2
= 2((1 + d)/(1 − d))r
1
r
1
= 0 d = 1 d = 0 d = 0
a
2
= 2r
1
, a
4
= r

2
1
E
r
3
+ a
2
r
2
+ a
4
r = r
3
+ 2r
1
r
2
+ r
2
1
r = r(r + r
1
)
2
E d

r
1
(
√

d + 1)/
√
d − 1), 0

E
E E

E

(r
1
/(1 − d))s
2
= r
3
+ a
2
r
2
+ a
4
r
(dr
1
/(1 −d))s
2
= r
3
+ a
2

r
2
+ a
4
r
u = r/r
1
v = s/r
1
E

(1/(1 −d))v
2
= u
3
+ a
2
/r
1
u
2
+ a
4
/r
2
1
u = u
3
+ 2((1 + d)/(1 −d))u
2

+ u
a
2
= 2((1 + d)/(1 − d))r
1
a
4
= r
2
1
E

d/(1 −d)v
2
= u
3
+ 2((1 + d)/(1 −d))u
2
+ u.
x
2
+ y
2
= 1 + dx
2
y
2
E
A,B
: Bv

2
= u
3
+ Au
2
+ u

A, B
v = 2v

u = u

E
A,B
(1/(1 −d))v
2
= u
3
+ 2((1 + d)/(1 −d))u
2
+ u
x
2
+ y
2
= 1 + dx
2
y
2
E


1/d d x
2
+ y
2
= 1 + (1/d)x
2
y
2
(1/(1 − 1/d))v
2
= u
3
+ 2((1 + 1/d)/(1 − 1/d))u
2
+ u

v = v

u = −u

E

: (d/(1 − d))v
2
=
u
3
+ 2((1 + d)/(1 −d))u
2

+ u
k d
r
1
/(1 − d) dr
1
/(1 − d) k
¯s =

r
1
/(1 −d)s
¯s =

dr
1
/(1 −d)s E E

E

E
x
2
+ y
2
= 1 + dx
2
y
2
x

2
+ y
2
= 1 + (1/d)x
2
y
2
x
2
+ y
2
= 1 + dx
2
y
2
E

E
E d
x
2
+ y
2
= 1 + dx
2
y
2
E

E

k E d
E
E

E

E x
2
+y
2
= 1+dx
2
y
2
x
2
+ y
2
= 1 + (1/d)x
2
y
2

k
E k x
2
+ y
2
= 1 + dx
2

y
2
d ∈ k \ {0, 1} (x
1
, y
1
), (x
2
, y
2
)
E
(x
1
, y
1
) + (x
2
, y
2
) =

x
1
y
2
+ y
1
x
2

1 + dx
1
x
2
y
1
y
2
,
y
1
y
2
− x
1
x
2
1 −dx
1
x
2
y
1
y
2

.
(0, 1) (x
1
, y

1
)
E (−x
1
, y
1
)
E
dx
1
x
2
y
1
y
2
/∈ {−1, 1}
(x
1
, y
1
) + (x
2
, y
2
) =
(x
3
, y
3

) (x
3
, y
3
) E x
2
3
+y
2
3
=
1 + dx
2
3
y
2
3
dx
1
x
2
y
1
y
2
/∈ {1, −1}
N = x
2
3
+ y

2
3
− (1 + dx
2
3
y
2
3
)
N = x
2
3
+ y
2
3
− (1 + dx
2
3
y
2
3
)
=

x
1
y
2
+ y
1

x
2
1 + dx
1
x
2
y
1
y
2

2
+

y
1
y
2
− x
1
x
2
1 −dx
1
x
2
y
1
y
2


2
−
−

1 + d

x
1
y
2
+ y
1
x
2
1 + dx
1
x
2
y
1
y
2

2

y
1
y
2

− x
1
x
2
1 −dx
1
x
2
y
1
y
2

2

.
N T = (x
1
y
2
+ y
1
x
2
)
2
(1 − dx
1
x
2

y
1
y
2
)
2
+ (y
1
y
2
− x
1
x
2
)
2
(1 +
dx
1
x
2
y
1
y
2
)
2
−((1+dx
1
x

2
y
1
y
2
)
2
(1−dx
1
x
2
y
1
y
2
)
2
+d(x
1
y
2
+y
1
x
2
)
2
(y
1
y

2
−x
1
x
2
)
2
)
T = (x
2
1
+ y
2
1
−(x
2
2
+ y
2
2
)dx
2
1
y
2
1
)(x
2
2
+

y
2
2
−(x
2
1
+ y
2
1
)dx
2
2
y
2
2
) −(1 −d
2
x
2
1
x
2
2
y
2
1
y
2
1
)

2
(x
1
, y
1
), (x
2
, y
2
)
E x
2
1
+ y
2
1
= 1 + dx
2
1
y
2
1
x
2
2
+ y
2
2
= 1 + dx
2

2
y
2
2
T
T = (1 + dx
2
1
y
2
1
− (1 + dx
2
2
y
2
2
)dx
2
1
y
2
1
)(1 + dx
2
2
y
2
2
− (1 + dx

2
1
y
2
1
)dx
2
2
y
2
2
) −
−(1 −d
2
x
2
1
x
2
2
y
2
1
y
2
2
)
2
= (1 −d
2

x
2
1
x
2
2
y
2
1
y
2
2
)
2
− (1 − d
2
x
2
1
x
2
2
y
2
1
y
2
2
)
2

= 0,
x
2
3
+ y
2
3
= 1 +dx
2
3
y
2
3
E E dx
1
x
2
y
1
y
2
/∈ {1, −1} 
E E
E
E
e = 1 −d
E (1/e)v
2
= u
3

+ (4/e −
2)u
2
+ u i ∈ {1, 2, 3} P
i
E P
i
= ∞
(x
i
, y
i
) = (0, 1) P
i
= (0, 0) (x
i
, y
i
) = (0, −1) P
i
= (u
i
, v
i
) x
i
= 0
u
i
= (1 + y

i
)/(1 −y
i
) v
i
= 2(1 + y
i
)/(1 −y
i
)x
i
P
i
∈ E(k)
P
1
+P
2
= P
3
E(k) = {(u, v) ∈ k×k : (1/e)v
2
= u
3
+(4/e−2)+u}∪{∞}
P
1
+ P
2
P

1
P
2
E x
i
= 0 y
i
= 1
E
E
(1/e)
3
e
2
v
2
= e
3
u
3
+ (4 −2e)e
2
u
2
+ e
3
u
V = ev, U = eu
V
2

= U
3
+ (4 − 2e)U
2
+ e
2
U
P
i
(x
i
, y
i
)
x
i
= 0 U
i
= e(1 + y
i
)/(1 −y
i
) V
i
= 2e(1 + y
i
)/(1 −y
i
)x
i

P
i
∈ E(k) i = 1, 2, 3
(x
i
, y
i
) = (0, 1) P
i
= ∞ ∈ E(k) (x
i
, y
i
) = (0, −1) P
i
= (0, 0) ∈ E(k)
P
i
= (u
i
, v
i
) ∈ E(k)
P
1
+ P
2
= P
3
(x

1
, y
1
) = (0, 1)
(x
3
, y
3
) = (x
2
, y
2
) P
1
P
2
= P
3
P
1
+ P
2
= ∞ + P
2
= P
2
= P
3
(x
2

, y
2
) = (0, 1) (x
1
, y
1
) = (0, 1) (x
2
, y
2
) = (0, 1)
(x
3
, y
3
) = (0, 1) (x
2
, y
2
) = (−x
1
, y
1
) (x
1
, y
1
) = (0, −1)
(x
2

, y
2
) = (0, −1) P
1
= (0, 0) = P
2
x
1
, x
2
u
1
= (1+y
1
)/(1−y
1
) = u
2
v
1
= 2u
1
/x
1
= −2u
2
/x
2
P
1

= −P
2
P
1
+ P
2
= ∞ = P
3
(x
3
, y
3
) = (0, 1)
(x
1
, y
1
) = (0, −1) (x
3
, y
3
) = (−x
2
, −y
2
) (x
2
, y
2
) = (0, −1)

(x
3
, y
3
) = (0, 1) (x
2
, y
2
) = (0, 1) x
2
= 0
P
1
= (0, 0) P
2
= (u
2
, v
2
) u
2
= (1 + y
2
)/(1 − y
2
) v
2
= 2u
2
/x

2
(0, 0) + (u
2
, v
2
) = (r
3
, s
3
)
r
3
= (1/e)(v
2
/u
2
)
2
−(4/e − 2) −u
2
= 1/u
2
s
3
= (v
2
/u
2
)(−r
3

) = −v
2
/u
2
2
P
3
= (u
3
, v
3
) u
3
= (1+y
3
)/(1−y
3
) = (1−y
2
)/(1+y
2
) = 1/u
2
= r
3
v
3
= 2u
3
/x

3
= −2/u
2
x
2
= −v
2
/u
2
2
= s
3
. P
1
+ P
2
= P
3
(x
2
, y
2
) = (0, −1)
x
1
= 0 x
2
= 0 P = (u
1
, v

1
) u
1
= (1 +
y
1
)/(1 − y
1
) v
1
= 2u
1
/x
1
P
2
= (u
2
, v
2
) u
2
= (1 + y
2
)/(1 − y
2
)
v
2
= 2u

2
/x
2
(x
3
, y
3
) = (0, −1) (x
1
, y
1
) = (x
2
, −y
2
) u
1
= (1 + y
1
)/(1 −
y
1
) = (1 − y
2
)/(1 + y
2
) = 1/u
2
v
1

= 2u
1
/x
1
= 2/x
2
u
2
= v
2
/u
2
2
P
3
= (0, 0)
−P
3
+ P
2
= (0, 0) + P
2
= (1/u
2
, −v
2
/u
2
2
) = (u

1
, −v
1
) = −P
1
P
1
+ P
2
= P
3
x
3
= 0 P
3
= (u
3
, v
3
) u
3
= (1 +
y
3
)/(1 −y
3
) v
3
= 2u
3

/x
3
P
2
= −P
1
u
2
= u
1
v
2
= −v
1
x
2
= −x
1
y
2
=
(u
2
−1)/(u
2
+ 1) = (u
1
−1)/(u
1
+ 1) = y

1
(x
3
, y
3
) = (0, 1)
P
2
= −P
1
u
2
= u
1
v
2
= −v
1
(u
1
, v
1
) + (u
2
, v
2
) = (r
3
, s
3

) r
3
= (1/e)λ
2
− (4/e − 2) − 2u
1
s
3
= λ(u
1
−r
3
) −v
1
λ = (3u
2
1
+ 2(4/e −2)u
1
+ 1)/((2/e)v
1
)
(r
3
, s
3
) = (u
3
, v
3

)
u
2
= u
1
(u
1
, v
1
) + (u
2
, v
2
) = (r
3
, s
3
) r
3
= (1/e)/λ
2
−(4/e −2) −u
1
−u
2
s
3
= λ(u
1
− r

3
) − v
1
λ = (v
2
− v
1
)/(u
2
− u
1
)
(r
3
, s
3
) = (u
3
, v
3
)
P
3
= P
1
+ P
2

d
k

k 2
d, e 0 k e = 1 − d d
k x
1
, y
1
, x
2
, y
2
k
x
2
1
+ y
2
1
= 1 + dx
2
1
y
2
1
x
2
2
+ y
2
2
= 1 + dx

2
2
y
2
2
dx
1
x
2
y
1
y
2
= ±1
 = dx
1
x
2
y
1
y
2
 ∈ {1, −1} x
1
, x
1
, y
1
, y
2

= 0
dx
2
1
y
2
1
(x
2
2
+y
2
2
) = dx
2
1
y
2
1
+d
2
x
2
1
x
2
2
y
2
1

y
2
2
= dx
2
1
y
2
1
+
2
= dx
2
1
y
2
1
+1 = x
2
1
+y
2
1
(x
1
+ y
1
)
2
= x

2
1
+ y
2
1
+ 2x
1
y
1
= dx
2
1
y
2
1
(x
2
2
+ y
2
2
) + 2x
1
y
1
dx
1
x
2
y

1
y
2
= dx
2
1
y
2
1
(x
2
2
+ 2x
2
y
2
+ y
2
2
) = dx
2
1
y
2
1
(x
2
+ y
2
)

2
.
(x
2
+y
2
) = 0 d = ((x
1
+y
1
)/x
1
y
1
(x
2
+y
2
))
2
d k x
2
+ y
2
= 0
(x
1
−y
1
)

2
= dx
2
1
y
2
1
(x
2
−y
2
)
2
x
2
−y
2
= 0