Cryptography Reference
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P and such that the line between P and Q is not
contained in U (this is generically the case for an irreducible quadratic hypersurface). Then
the line between P and Q does not lie in T and so is given by an equation of the form 7
Let Q
U (
k
) be such that Q
=
( x,y,z )
=
P
+
t (1 ,a,b )
(6.3)
for some a,b
∈ k
(in other words, the equations x
=
x P +
t,y
=
y P +
at , etc). Such a
line hits U at precisely one point Q
U (
k
) with Q
=
P . Writing U
=
V ( F ( x,y,z ))
it follows that F ( x P +
t,y P +
at,z P +
bt )
=
0 has the form t ( h ( a,b ) t
g ( a,b ))
=
0for
some quadratic polynomial h ( a,b )
∈ k
[ a,b ] and some linear polynomial g ( a,b )
∈ k
[ a,b ].
2
Hence, we have a rational map
A
U given by
g ( a,b )
h ( a,b ) (1 ,a,b ) .
( a,b )
P
+
The inverse is the rational map
=
(( y Q
y P ) / ( x Q
x P ) , ( z Q
z P ) / ( x Q
p U ( x Q ,y Q ,z Q )
x P ))
2 .
such that p U : U
→ A
1 (
1 (
Recall the map comp 2 : G q 3 , 2 → A
F q 3 ) from the study of
T 2 . We identify
A
F q 3 )
3 (
2 . This motivates
with
A
F q ). The image of comp 2 is U , which is birational via p U to
A
the following definition.
2
Definition 6.4.3 The
T 6 compression map is comp 6 : G q, 6 → A
is given by comp 6 =
decomp 2 p U .
p U comp 2 . The inverse of comp 6 is the
T 6 decompression map decomp 6 =
Example 6.4.4 Let q
2 , 5 (mod 9) be an odd prime power so that
F q 6
= F q ( ζ 9 ) where
ζ 1
9
ζ 9
ζ 9 is a primitive 9th root of unity (see Exercise 6.4.5 ). Let θ
=
and α
=
ζ 9 +
. Then
= F q ( α ). Note that α 3
3 (
1 (
F q 2
= F q ( θ ) and
F q 3
3 α
+
1
=
0. Identify
A
F q ) with
A
F q 3 )
z ( α 2
by f :( x,y,z )
x
+
+
2). As in the proof of Lemma 6.4.1 one can verify that
the equation
N F q 6 / F q 2 (( f ( x,y,z )
+
θ ) / ( f ( x,y,z )
+
θ ))
=
1
is equivalent to
x 2
y 2
z 2
F ( x,y,z )
=
x
+
yz
=
0 .
Denote by U the hyperplane V ( F ( x,y,z )) in
A
3 .Let P
=
(0 , 0 , 0). The tangent plane to
U at P is given by the equation x
=
0. Note that, since
3 is not a square in
F q , the o n ly
solution to F (0 ,y,z )
=
0 over
F q is ( y,z )
=
(0 , 0) (but this statement is not true over
F q ;
U contains, for example, the line (0 ,
ζ 3 t,t )). Given a,b
∈ F q the line ( t,at,bt ) hits U at
t
=
0 and
a 2
b 2 ) .
t
=
1 / (1
+
ab
7
Here, and below, P + Q denotes the usual coordinate-wise addition of 3-tuples over a field.
 
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