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§1. Homogeneous Spaces over Elliptic Curves 159

Explicitly in terms of coordinates, a fixed point (x, y) in E(k t ) is a point satis-
2
fying the relation (x, y) = (b/x, −by/x ). This fixed point relation is equivalent
to
by
(FP) xx = b, xy = =−x y.
x
Now we analyze these fixed-point relations further with the notations k t = k(c),

where c is a square root of u(t),and x = x + cx , y = y + cy ,and xy = cz

2
with z ∈ k since xy =−x y. Setting d = c, we observe that xx = b if and only if
2
2
b = x − dx . Next, we calculate using the equation of the curve E = E[a, b]

3
2
2
2
2
dz = (cz) = (xy) = x 2 x + ax + bx
2 2 2 2

= b x + b a + b x = b 2x + 2a .
This yields the first relation between x and z
d 2 a

(1) x = z − .
2b 2 2
Further for x we have

2
1 d
2 2 2
dx = x − b = z − a − b.
4 b 2

From xy = (x − cx )(y + cy ) = (x y − dx y ) + (x y − x y )c, we deduce

the relations

(2) x y = dx y and z = x y − x y .

1
(1.6) Assertion. Let t ∈ H (k, T ) be given by the quadratic extension k t and let P t
be the corresponding curve as above. The curve P t over k can be described as the
locus of the quartic equation
2

1 d 2
2
dx = z − a − b.
4 b 2

In terms of (x, y) we recover x from the relation (1) in terms of z, and we recover

y and y from the two linear relations (2) in terms of the other variables.

2

(1.7) Remark. In (1.6) substitute M = z/b, N = 2x , d = d,and d = a − 4b.
Then the quartic equation for P t becomes
2 4 2
N = d M − 2aM + d
2

with d d = a − 4b.
(1.8) Remark. Returning to (1.5), we point out that the statement: the element P t in
1
H (k, E(k s )) is zeroifand onlyif P t (k) is nonempty is related to the transformation
2
2
in 4(3.1) of a quartic y = f 4 (x) to a cubic y = f 3 (x) given a simple root of f 4 (x).