Let K be a cubic number field, and 
the image by
of K. We will first show that 
. As in
Proposition 4.1, write 
with 
a fundamental
discriminant.
Let p be a prime. If 
, then 
, so
. Hence we now assume that 
.
By Proposition 4.1 (1), it follows that p is totally ramified. By
Proposition 2.3, this means that 
splits as the cube of a linear form,
i.e that 
. We consider three cases.
p>3. In this case, it follows from Proposition 4.1 (2) and (3)
that 
, hence that 
by Proposition 3.3 (3).
p=2. Since 
, we have 
by Proposition 4.1 (2),
hence by Proposition 4.1 (3) we deduce that 
hence
as before.
p=3. This is the only difficult case. Since 3 cannot be an
inessential discriminantal divisor, there exists 
such that 
. Now since 3 is totally ramified, write
for a prime ideal 
of degree 1. Since 
is
isomorphic to 
, we have 
, hence either 
,
or 
belongs to 
. Replacing if necessary 
by 
or 
(which does not change the property 
, we
may assume that 
.
We clearly have
Assume first that 
. Thus 
has non-negative valuation
for every prime ideal except perhaps those above 3, but the only prime ideal
above 3 is 
and the 
-adic valuation of 
is at least equal to
1, hence 
. By the above formula we have
, and since
, it follows that
. Since 
is the only prime ideal above 3, this
means that the 
-adic valuation of 
is at least equal to 3,
hence that 
, in other words that
. Now since 
, 3 splits in K as
the characteristic polynomial 
of 
splits modulo 3. It follows that
for a certain 
. Write
We thus have 
and 
. Since 
we have 
.
But this is absurd since we would then have
Thus 
. Since we still have 
and 
we obtain
and since 
we have 
so
. Furthermore by our choice of 
we have
hence
Let 
be the integral basis used to define the form 
. Thus
for some u, v and w in 
, and hence by
definition we have 
. We have the following lemma:
Lemma 4.3 Assume that F is a primitive cubic form and p a
prime such that 
. Then 
if and only if there exists
such that 
with 
.
Assuming this lemma for a moment, we see that since 
represents
an integer congruent to 
modulo 9, we have 
, thus
finishing the proof that 
.
Let us prove the lemma. By lifting the condition 
to 
, we
can write
with 
, 
, 
in 
and G an integral cubic form.
Then 
if and only if 
. Thus if 
we have 
with 
.
Conversely, assume that there exists 
such that 
with 
. Then 
, hence
. Since F is primitive,
hence 
. Again F being
primitive, 
and 
cannot both be divisible by p, from which
it follows that there exists 
such that 
and 
, and since 
we have 
.
But then 
hence 
and so 
, so
, thus proving the lemma.
To finish the proof of the Davenport-Heilbronn Theorem 4.2, we must now
prove that if 
, there exists a cubic field K such that F is
equivalent to 
. For this, we introduce a definition.
Definition 4.4 We will say that two cubic forms 
and 
are rationally equivalent if there exists 
such that
for 
.
We first show the following lemma.
Lemma 4.5 Let F be any form in 
( i.e primitive and
irreducible). Then there exists a number field K such that F is rationally
equivalent to 
.
Proof In some algebraic closure of 
, write
. Since F is irreducible,
is a cubic irrationality, and we will take 
.
Write 
so that 
(we have seen
above that this is always possible). Since K is a 
-vector space of
dimension 3, there exist 4 integers k, l, m and n not all zero such
that 
. Taking conjugates, we obtain the
same equality with 
and 
. Using 
, we
obtain
with 
. Furthermore the determinant kn-lm is
nonzero since otherwise either 
or 
would be in 
. Thus
is rationally equivalent to F.
Finally, we have the following lemma.
Lemma 4.6 Let 
and 
be two forms belonging to U. Then
if 
and 
are rationally equivalent, they are equivalent.
Since 
belongs to U, it follows from Lemma 4.5 and this lemma that
any form in U is equivalent to 
for a certain K, and this finishes
the proof of Davenport-Heilbronn's Theorem 4.2.
Proof Assume that 
for some 
and
. Since we want to show that 
and 
are equivalent,
we can replace them by equivalent forms. In other words, without changing
the equivalence classes of 
and 
we may replace M by any matrix
of the form UMV with U and V in 
. The elementary divisor
theorem ( i.e the Smith normal form) tells us that we can choose U and V
so that
Replacing 
by 
, we may assume 
.
To summarize, by replacing 
and 
by equivalent forms
and modifying 
and M, we may assume that
with 
.
If m=1, then 
and, since 
and 
are in U, they
are primitive. Hence 
so 
and 
are equivalent.
Otherwise, there exists a prime p such that 
. Write 
and 
with 
and 
prime to p ( i.e of
zero p-adic valuation). Since 
we have 
. Now, writing for
i=1 or 2, 
, the equality
is equivalent to
where the 
are rational numbers prime to p.
Assume l-k>0, hence 
. Then 
and 
.
If on the other hand 
then 
hence 
and
. Replacing 
by 
we obtain again 
and 
.
The formula for the discriminant implies 
. If p>2
this immediately implies 
. If p=2, the formula for the
discriminant shows that
and this is congruent to 0 or
1 modulo 4, so 
.
Since 
we must have
, and 
with
. Since 
, 
means that 
, hence
. But then, since 
,
, contradiction. This finishes the proof
of Lemma 4.6, and hence of Davenport-Heilbronn's Theorem 4.2.
