Cryptography Reference
In-Depth Information
y
2
=
x
3
y
2
=
x
3
+
x
(
a
)
− x
(
b
)
Figure 2.1
Therefore, the roots of the cubic must be distinct. However, the case where the
roots are not distinct is still interesting and will be discussed in Section 2.10.
In order to have a little more flexibility, we also allow somewhat more
general equations of the form
y
2
+
a
1
xy
+
a
3
y
=
x
3
+
a
2
x
2
+
a
4
x
+
a
6
,
(2.1)
where
a
1
,...,a
6
are constants. This more general form (we'll call it the
gen-
eralized Weierstrass equation
) is useful when working with fields of char-
acteristic 2 and characteristic 3. If the characteristic of the field is not 2, then
we can divide by 2 and complete the square:
=
x
3
+
a
2
+
a
1
x
2
+
a
4
+
a
1
a
3
x
+
a
3
4
+
a
6
,
y
+
a
1
x
2
2
2
+
a
3
4
2
which can be written as
y
1
=
x
3
+
a
2
x
2
+
a
4
x
+
a
6
,
with
y
1
=
y
+
a
1
x/
2+
a
3
/
2 and with some constants
a
2
,a
4
,a
6
. If the charac-
teristic is also not 3, then we can let
x
1
=
x
+
a
2
/
3andobtain
y
1
=
x
1
+
Ax
1
+
B,
for some constants
A, B
.
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