Clearly, we have
. Since
, there is a non-zero solution vector
of the homogeneous linear equations
. For any
, we have
which implies that
or equivalently
. Thus,
.
For any
, we have
Hence,
which is equivalent to
We claim that
. Indeed, if any one of
is non-zero, then there are at least two non-zero linear independent vectors among
which are all solutions of
. This contradicts with
. Hence,
. In
, we have
(1) If
, we claim that
. Indeed, if there exist
such that
, then
which implies that
and hence
This contradicts with the assumption that
. Then, we obtain that
if
. On the other hand, we have
since
. Therefore,
and
In order to show
is a subalgebra of
, we need to show
for any
. If this not the case, we have
where
and
u are all linear combinations of monomials with non-negative even exponents. When
is even, we have
Considering the parity of exponents of the monomials that appear on both sides, the equation above implies that
and
Therefore,
is in the graded ideal
of
. By Lemma 10,
is a graded prime ideal of
. So,
. Hence, there exist
and
in
k such that
However, this contradicts with the fact that
, which we have proved above. Thus,
when
n is even.
When
is odd, we have
This implies that
Since
, there is at least one non-zero
. Then, we obtain
, which contradicts with the proved fact that
when
n is even. Therefore,
when
n is odd.