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Lemma 19
Let
be a field.
Assume
is injective
(which is always the case if
).
Then
.
Let be the maximum of total degrees of elements
.
Then we see that
holds for any .
Thus for any , we have
This together with Lemma 18 gives the result.
Lemma 20
Let
be a subalgebra of a polynomial algebra
over a field
.
If there exists an positive integer
such that
holds for all integers
, then we have
Suppose on the contrary that
.
Take elements
which are linearly independent over .
By multiplying a ``common denominator'', we may assume that are
elements of .
Then it follows that the sum
is direct in .
Let be the maximum of total degrees of
.
Then the directness above implies that an inequality
holds for any integer .
Since the left hand side is not greater than
,
we obtain the following inequation
which leads to a contradiction when is large enough.
The first statement is an easy consequence of
Lemma 19.
This in turn implies (2) (except for finite primes).
In precise, let
be elements in
such that
multi degrees of
are linearly
independent.
Then for almost all primes,
their reductions
are defined as elements of
and
their multi degrees stays invariant under the reduction.
We apply Lemma 18 to see that
there exists a positive real number which is independent of
such that an inequality
holds for any large integer .
(1) is already proved in subsection 3.1.
To prove (4), we denote by
the quotient field with characteristic
.
Since is a free -module of rank with generators
, we have
Then Corollary 7 gives us an relationship of total degrees
of both hands sides. Namely, there exists a positive number such that
holds for any positive integer .
Combining these, we obtain
Then we use Lemma20 to see that (4)
is true.
(One should be very careful about grading of here.
degrees of generators
of are all . not .)
(3) follows from (4) and a consideration on transcendence degrees.
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2003/3/3