(1) follows directly from the definitions, after noting that
is the determinant of the matrix
Note that if we take one of the expressions on the RHS of (1) for definition of
there is a sign ambiguity. However, this will not matter in the sequel since
, so that 
and 
are
-equivalent.
(2) The definition of 
shows that 
is the restriction to
of a cubic form with coefficients in 
.
Furthermore Galois theory shows that the coefficients of 
are
invariant under the Galois group of 
over 
, so that in
fact 
is a rational cubic form. Note that the rationality of
when x and y are rational also follows from the remark made
before the proposition.
(3) The roots of 
(with any choice for the square
root) are by assumption the 
for 
,
and this respects the convention for the choice of roots made at the
beginning. Hence one checks that
So that 
since, by the explicit formula for
, we have 
.
(4) This is the longest part of the proof. Let 
with a,
b, c, d in 
by (2). First, we note that if x and y are in 
then 
is in 
, hence by the remark made above,
where 
so 
. Applying
this to 
and 
we deduce that 
and 
.
Then applying this to 
and 
we deduce that
and 
. It follows that b and c belong to 
with 
. Now, by (3) we know that 
is
an integer. By the explicit formula for 
and using that b
and c belong to 
we obtain that
. Since 
, we cannot have
, hence 
, thus proving
that 
is integral.
Recall that by (1), if x and y are in 
, we have 
, where for 
, 
.
Now let 
be the GCD of the coefficients of 
. We thus have for
each x, y in 
, 
. But since
for any 
and since 
is an
integral basis, it follows that for any 
we have
.
This means by definition that 
is an inessential discriminant
divisor, and according to a theorem of Dedekind, for a cubic field this
implies that either 
, or else 
and 2 is totally split
in K. Assume the latter. We thus have 
for an integral cubic
form G, and 
by (3). However since
2 is totally split in K, it is in particular unramified, hence
, which is absurd since this implies that the integral cubic
form G has non-integral discriminant. Thus we have 
hence
is primitive.
Finally, let us show that 
is irreducible. Since we are in the cubic
case, this means that 
has no linear factor in 
. Assume
the contrary. By definition of 
, we may assume that such a linear
factor is proportional to 
. It follows that
there exist integers r and s not both zero such that
, so that 
.
Taking conjugates, we see that we have
, so that the conjugates of
are equal, hence by Galois theory 
, and
since r and s are not both zero, this is in contradiction with
being an integral basis. It follows that 
is
irreducible.
(5) Let 
be another integral basis. This means that there exists
integers A, B, C, D, E, F such that
with 
,
since 
and 
must generate the same lattice. It
follows that
where the second equality follows from 
for all 
and 
. Since 
, it follows from (1) that 
with 
, so 
and
are equivalent.
(6) This follows from a straightforward but tedious computation.
