ONE- AND TWO-DIMENSIONAL REPRESENTATIONS 209
We conclude this chapter with a relatively simple case where we
can actually prove Theorem 18.5, using various facts stated earlier
in the book. Namely, we will look at the Galois representation
ψ coming from an elliptic curve where we set p = 2. We will be
able to compute ψ(Frob
) explicitly (which is much more difﬁcult if
p > 2) and test the theorem: For q any prime except for divisors of
), we must have the number of points on E(F
) odd or
even depending on whether χ
) is odd or even.
Pick an elliptic curve E deﬁned by the equation y
+ Ax + B.
We are letting p = 2 in the above theorem, meaning that we want
to identify the points on E that satisfy the equation 2P = O.
happens that these points are particularly easy to describe. In
addition to O itself, they are the points P = (a,0), Q = (b, 0), and
R = (c, 0) where a, b, and c all solve the equation x
+ Ax + B = 0.
(This is a cubic equation, so it will have three solutions. One of the
consequences of the restriction that 4A
= 0 is that the three
solutions will be unequal, so there really are three different points
in our list.) It will always happen that P + Q = R.
Be forewarned that some of the analysis gets quite complicated,
and you can skip it if you just want to go on to the next chapter. But
if you plow through it, you should get an idea of how many of the
ideas in this chapter and chapter 16 ﬁt together beautifully.
We take a particularly easy example to get started understanding
the theorem. Let E be the elliptic curve deﬁned by the equation
− x (in other words, we are taking A =−1 and B = 0). To
ﬁnd the points in E, we need to solve the equation x
− x = 0.
This equation has three solutions: x = 0, x = 1, and x =−1. Cor-
responding to those solutions there are three points on the el-
liptic curve: P = (0, 0), Q = (1, 0), and R = (−1, 0). If you use the
Remember that this means P + P = O, where the plus sign means adding points
according to the recipe given in chapter 9.