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Fearless Symmetry by Robert Gross, Avner Ash

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ONE- AND TWO-DIMENSIONAL REPRESENTATIONS 211
curve—O, P, Q, and R—so in all there must be an even number
of points on the curve.
Another Example
We now try a slightly more complicated example. Let E be the
elliptic curve y
2
= x
3
1, so that A = 0 and B =−1. We can apply
the theorem for any prime q other than 2 and 3. The three points in
E[2] (other than O)areP = (1, 0), Q = (ω, 0), and R = (ω
2
, 0), where
ω is the number
1+
3
2
that showed up above in our discussion of
the solutions of x
3
1 = 0.
EXERCISE: Show that P + Q = R.
Now choose your favorite prime q (other than 2 and 3, which are
ramified), and consider how Frob
q
shifts these points around. We
know that Frob
q
(P) = P, because the coordinates of P are in Z, and
elements of the absolute Galois group G always fix elements of Z.
What about Frob
q
(Q)? The two possibilities are Frob
q
(Q) = Q and
Frob
q
(Q) = R.
We take q = 5. We saw above in equation (18.3) that Frob
5
(ω) =
ω
5
= ω
2
, and so Frob
5
(Q) = R. Because R = P + Q, we see, using the
prescription on page 206 as usual, that
ψ(Frob
5
) =
11
01
.
Therefore, χ
ψ
(Frob
5
) = 1 + 1 0 (mod 2).
What about Frob
7
? Repeat the same computation, and you will
see that Frob
7
(ω) = ω. Therefore, Frob
7
(Q) = Q, and so
ψ(Frob
7
) =
10
01
.
This means that χ
ψ
(Frob
7
) = 1 + 1 0 (mod 2).
In fact, you can see that these two examples show the two
possibilities for ψ(Frob
q
), and in either case χ
ψ
(Frob
q
) 0 (mod 2).
Again, from Theorem 18.5, we conclude that #E(F
q
) is always even.

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