8430 HANDOUT 6: PROOF OF THE MAIN THEOREM
PETE L. CLARK
1.
Proof of the main theorem for maximal orders
We are now going to take a decisive step forward by proving the Main Theorem
on which primes
p
are represented by the principal quadratic form
q
D
(
x, y
) of
discriminant
D <
0, when
D
is a
fundamental
discriminant, i.e., the discriminant
of the full ring of integers of
Q
(
√
D
).
Of course, to solve our original problem
of representation of primes by
x
2
+
ny
2
for
arbitrary
n
, we need to look also at
quadratic orders of not necessarily maximal discriminant. But notice that all our
efforts so far have yielded solutions for only finitely many values of
n
, whereas here
we can take
n
to be any squarefree negative number which is not 3 (mod 4). Here
is the result:
Theorem 1.
Let
D <
0
be a fundamental quadratic discriminant, the discriminant
of (the maximal order of) the imaginary quadratic field
K
=
Q
(
√
D
)
. Let
h
(
K
) =
# Pic(
O
(
D
))
be the class number of
K
.
Then there exists a monic polynomial
P
D
(
t
)
∈
Z
[
t
]
such that:
(a)
P
D
(
t
)
is irreducible of degree
h
(
K
)
, and has at least one real root.
(b) A prime number
p
which is prime to
D
and to
disc(
P
D
)
is of the form
q
D
(
x, y
)
iff
(
D
p
) = 1
and the mod
p
reduction
P
D
(
t
)
∈
F
p
[
t
]
has at least one rational root.
The proof occurs in two steps: first we bring our heavy artillery to bear and almost
immediately deduce the following result:
Theorem 2.
With notation as above, a prime
p
prime to
D
is of the form
q
D
(
x, y
)
iff
p
splits completely in the Hilbert class field
K
1
of
K
=
Q
(
√
D
)
.
Then we need to use a bit of comparatively elementary (but slightly tricky) reason
ing to get from Theorem 2 to Theorem 1.
Well, second things first:
Proof of Theorem 2: Suppose
p
is prime to
D
. We already know that a necessary
condition for
p
to be represented by
q
D
is the fundamental congruence (
D
p
) = 1.
(Note that this is to be interpreted as the Kronecker symbol when
p
= 2; or just
assume
p
is odd; you’re not missing out on much.) Further, we saw that the fun
damental congruence buys us precisely a Galois conjugate pair
p
,
p
of prime ideals
of
O
(
D
) =
O
K
lying over
p
. And we also saw that
p
is of the form
q
D
(
x, y
) iff one
of these ideals (equivalently, both) are
principal
.
But now we know more:
there is an abelian field extension
K
1
/K
, called the
Hilbert class field, which has the following three remarkable properties: (i) its Ga
lois group is isomorphic to Pic(
O
K
), the ideal class group of
K
; (ii) it is unramified
1
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PETE L. CLARK
at every prime ideal
p
of
K
; and (iii) a prime
p
is principal iff its Frobenius ele
ment
τ
p
∈
Pic(
O
K
) is trivial. As we saw, it immediately follows from Chebotarev
Density that the set of principal primes
p
has density
1
h
(
K
)
. Compiling these two
conditions, we see that indeed what we want is for the rational prime
p
to split
completely from
Q
to
K
1
, the Hilbert class field of
K
. We’re done.
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 Fall '11
 Clark
 Geometry, Algebraic number theory, pete l. clark

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