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Fearless Symmetry by Robert Gross, Avner Ash

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222 CHAPTER 19
Here,
|
W
|
means as usual the absolute value of W (e.g.,
|
4
|
= 4
and
|
7
|
= 7). If you are patient enough, you can untangle what this
theorem and all of the preceding definitions are describing. You will
find out that it is quite close to Theorem 7.2, which we used to prove
quadratic reciprocity. Theorem 19.5 gives us a different approach.
To prove quadratic reciprocity starting from Theorem 19.5, we take
clever choices of W, and use the existence of the eigenelement α that
Theorem 19.3 gives us, together with the knowledge of N given by
Theorem 19.5.
In this section, we will be varying W, so we will put a subscript on
the corresponding α given to us by Theorems 19.3 and 19.5, writing
it as α
W
. That is, α
W
is the simultaneous eigenelement depending
on W described in Theorem 19.4. Because N = 4|W|, we have the
important fact that if a b (mod 4|W|), then α
W
(a) = α
W
(b). We will
use this fact several times in our derivation of quadratic reciprocity.
A Derivation of Quadratic Reciprocity
Suppose first that W =−1, so that N = 4. Because α
W
(p) =
1
p
can be thought of as a morphism defined on (Z/4Z)
×
,wecan
conclude that
1
p
is determined just by the value of p (mod 4).
Because
1
3
=−1, and
1
5
= 1, we can deduce the usual
formula for
1
p
given on page 79. For example, if p 3 (mod 4),
1
p
=
1
3
=−1. See chapter 7, if you have not already
glanced back to refresh your memory.
Next, take W = 2, and we know that α is a morphism defined on
(Z/8Z)
×
. Computation of
2
p
for p = 3, 5, 7, and 17 gives the usual
formula for
2
p
,whichwegaveonpage79.
Next, suppose that p and q are the odd primes that we want to
compare in quadratic reciprocity. First, assume that p q (mod 4),
QUADRATIC RECIPROCITY REVISITED 223
with p > q, and let W = (p q)/4. Then p q (mod 4W), which
means (because of Theorem 19.5) that α
W
(p) = α
W
(q), or
W
p
=
W
q
.Wehave
p
q
=
4W + q
q
=
4W
q
=
W
q
=
W
p
=
4W
p
=
p q
p
=
q
p
=
1
p

q
p
,
which implies quadratic reciprocity for p q 1 (mod 4) and p
q 3 (mod 4).
To derive the remaining cases, observe that for any W the
congruence
x
2
W 0 (mod 4W 1)
always has the solution x 2W. This tells us that if W is positive
and p is any prime dividing 4W 1, then α
W
(p) = 1. Factoring
4W 1 into a product of primes, and using the fact that α
W
is a
morphism, we see that α
W
(4W 1) = 1. In other words, we have
α
W
(1) = 1.
Now, suppose that p + q 0 (mod 4), and let W = (p + q)/4.
Then α
W
(p) = α
W
(4W q) = α
W
(q) = α
W
(1)α
W
(q) = α
W
(q), which
implies that
W
p
=
W
q
, and, reasoning as before, we can
conclude that
p
q
=
q
p
.
Note that Theorem 19.4 actually identified the mysterious simul-
taneous eigenelement α. It is just the Legendre symbol:
α
W
(p) =
W
p
.
In the case of more general reciprocity laws, we usually do not
understand the black box this well.
224 CHAPTER 19
The fact that the Galois representation χ was also closely
related to the Legendre symbol is a sort of coincidence occurring
because quadratic reciprocity is the simplest of all reciprocity laws.
Something similar happens with other one-dimensional Galois
representations of G. This relationship is called class field theory.
But as soon as you go to higher-dimensional Galois representations,
the black boxes diverge radically from the Galois representations
in their definitions and become more complicated too. This starts to
happen in the study of representations arising from elliptic curves.
EXAMPLE: Let W =−1 and p be an odd prime. Then Frob
p
either fixes or exchanges the roots i, i of x
2
+ 1. The fact
that χ(Frob
p
) =
1
p
means that Frob
p
(i) =
1
p
i. This
explains the connection between equation (7.4) in chapter 7
and the last exercise in chapter 16.

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