Cryptography Reference
In-Depth Information
2 (
For example, consider the hyperbola xy
=
1in
A
R
). Projective geometry allows an
interpretation of the behaviour of the curve at x
=
0or y
=
0; see Example 5.2.7 .
k
Definition 5.2.1 Projective space over
of dimension n is
n (
n
+
1 (
P
k
)
={
lines through (0 ,..., 0) in
A
k
)
}
.
n (
A convenient way to represent points of
P
k
)isusing homogeneous coordinates .Let
a 0 ,a 1 ,...,a n ∈ k
with not all a j =
0 and define ( a 0 : a 1 :
···
: a n ) to be the equivalence
class of ( n
+
1)-tuples under the equivalence relation
( a 0 ,a 1 ,
···
,a n )
( λa 0 ,λa 1 ,
···
,λa n )
∈ k .
n (
for
any λ
Thus
P
k
)
={
( a 0 :
···
: a n ): a i ∈ k
for 0
i
n and a i =
n
n (
0forsome0
i
n
}
. Write
P
= P
k
).
···
: a n ) is the set of points on the line between
(0 ,..., 0) and ( a 0 ,...,a n ) with the point (0 ,..., 0) removed.
There is a map ϕ :
In other words, the equivalence class ( a 0 :
A
n
→ P
n given by ϕ ( x 1 ,...,x n )
=
( x 1 :
···
: x n : 1). Hence,
A
n is
n .
identified with a subset of
P
1 (
1 (
Example 5.2.2 The projective line
P
k
) is in one-to-one correspondence with
A
k
)
{∞}
P
1 (
k
={
( a 0 :1): a 0 ∈ k}∪{
}
P
2 (
k
since
)
(1 : 0)
.The projective plane
) is in one-to-
A
2 (
k
∪ P
1 (
k
one correspondence with
)
).
n (
Defi n ition 5.2.3 A point P
=
( P 0 : P 1 :
···
: P n )
∈ P
k
)is defined ov er
k
if there is some
such that λP j ∈ k
n and σ
λ
∈ k
for all 0
j
n .If P
∈ P
Gal(
k
/
k
) then σ ( P )
=
( σ ( P 0 ):
···
: σ ( P n )).
Exercise 5.2.4 Show that P is defined over
k
if and only if there is some 0
i
n such
n (
that P i =
0 and P j /P i ∈ k
for all 0
j
n . Show that
P
k
) is equal to the set of points
n (
P
∈ P
k
) that are defined over
k
. Show that σ ( P ) in Definition 5.2.3 is well-defined (i.e.,
P =
( P 0 ,...,P n ) then σ ( P )
σ ( P )).
if P
=
( P 0 ,...,P n )
n (
Lem m a 5.2.5 A point P
∈ P
k
) is defined over
k
if and only if σ ( P )
=
P for all σ
Gal(
k
/
k
) .
n (
Proof Let P
=
( P 0 :
···
: P n )
∈ P
k
) and suppose σ ( P )
P for all σ
Gal(
k
/
k
). Then
such that σ ( P i )
there is some ξ :Gal(
k
/
k
)
→ k
=
ξ ( σ ) P i for all 0
i
n . One can
. It follows by Theorem A.7.2 (Hilbert 90) that ξ ( σ )
verify 2
that ξ is a 1-c oc ycle in
k
=
. Hence, σ ( P i )
σ ( γ ) for some γ
∈ k
=
P i for all 0
i
n and all σ
Gal(
k
/
k
).
Hence, P i
∈ k
for all 0
i
n and the proof is complete.
Recall that if f is a homogeneous polynomial of degree d then f ( λx 0 ,...,λx n )
=
λ d f ( x 0 ,...,x n ) for all λ
n
+
1 (
∈ k
and all ( x 0 ,...,x n )
∈ A
k
).
2
At least, one can verify the formula ξ ( στ ) = σ ( ξ ( τ )) ξ ( σ ). The topological condition also holds, but we do not discuss this.
 
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