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Fearless Symmetry by Robert Gross, Avner Ash

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ONE- AND TWO-DIMENSIONAL REPRESENTATIONS 209
The 2-Torsion
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
q
) explicitly (which is much more difficult if
p > 2) and test the theorem: For q any prime except for divisors of
2(4A
3
+ 27B
2
), we must have the number of points on E(F
q
) odd or
even depending on whether χ
ψ
(Frob
q
) is odd or even.
Pick an elliptic curve E defined by the equation y
2
= x
3
+ 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.
7
It
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
3
+ Ax + B = 0.
(This is a cubic equation, so it will have three solutions. One of the
consequences of the restriction that 4A
3
+ 27B
2
= 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.
An Example
We take a particularly easy example to get started understanding
the theorem. Let E be the elliptic curve defined by the equation
y
2
= x
3
x (in other words, we are taking A =−1 and B = 0). To
find the points in E[2], we need to solve the equation x
3
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
7
Remember that this means P + P = O, where the plus sign means adding points
according to the recipe given in chapter 9.
210 CHAPTER 18
complicated group law on E, you can check that P + P = O, Q + Q =
O, and R + R = O. In fact, you can also check that P + Q = R.
Now, let q be any odd prime. We want to understand how Frob
q
permutes these three points. In this case, we can easily understand
what happens: Frob
q
is an element of the absolute Galois group G,
and the absolute Galois group leaves fixed the rational numbers
0, 1, and 1. In other words, Frob
q
(P) = P, Frob
q
(Q) = Q, and
Frob
q
(R) = R.
8
In this case, the matrix ψ(Frob
q
) is well-defined, and using the
prescription on page 206, we see that
ψ(Frob
q
) =
10
01
.
The trace of this matrix is 2. Therefore, χ
ψ
(Frob
q
) = 2, and Theo-
rem 18.5 is the assertion that
2 q + 1 #E(F
q
) (mod 2).
Because 2 0 (mod 2), this is a fancy way of saying that the
number q + 1 #E(F
q
) is even. Because q is an odd prime, q + 1
is even, so in the end we have to understand why the number of
points in E(F
q
) is even.
How can we arrange the points on E(F
q
) to make them easy to
count? Sort them into three classes:
1. O.
2. P, Q, and R.
3. The rest of the solutions of y
2
x
3
x (mod q).
We can see that there is one element in the first class and three
elements in the second class. What about the third class? The other
solutions of y
2
x
3
x (mod q) will have y ≡ 0 (mod q). If a typical
point in this class is (j, k), then (ptj, k) will also be in this class and
will be different from (j, k), because k ≡ 0 (mod q).
Thus, because we can pair them up, there are an even number
of points in the third class. There are four other points on the
8
The third equation in fact is implied by the other two, and the fact that P + Q = R.
Why? If σ is any element of the absolute Galois group G, then σ (P + Q) = σ (P) + σ (Q)if
P and Q are any points on the elliptic curve E = (Q
alg
).

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