Using the
-linearity of the trace map we have an exact sequence
where
is the kernel of the trace map. Note the trace map defines an
-linear subspace of
C. Let
K be the subspace of
that is the inverse image of evaluation at
. That is
We can obtain the dimension of
by determining
. In practice, this is difficult. Consider the space of functions
Using Bombieri’s estimate and a degree argument, we will determine a sufficient condition (condition
2 of Theorem 1) when
and
E is isomorphic to
. But to make this useful we will first show Theorem 1 is sufficient to determine the dimension of
E.
3.1. Dimension of E
For , by definition . Counting with multiplicity, each pole in h corresponds to q poles in f. For , we have .
Consider the map where . By definition, the kernel is . Note that for a general G the map φ is not surjective.
Lemma 3. When the map φ is surjective.
Proof. Recall that the divisor only changes the positive coefficients of G and does not change . When , there is no restriction on zeros in . Therefore, in this case φ is onto.
If , then for some point and negative integer . Every function in must have a zero at P. In the factorization , this zero must occur in at least one factor . Though h may not be in , there will always exist some such that . Observe . In this case, φ is onto. ☐
If , then the kernel of φ is . If , then φ is injective. Therefore the δ defined in Theorem 1 is merely . Using Lemma 3, we have the following proposition.
Proposition 4. If , then the sequenceis exact. Therefore we have a dimension formula for E: In general,
φ may not be surjective. There is still a dimension bound:
Note in [
7], a similar result is obtained with the use of group cohomology and other auxillary constructions.
3.2. Bombieri’s Estimate
A key step in determining when
is a bound developed by Bombieri [
8].
Theorem 5 (Bombieri’s estimate).
Let X be a complete, geometrically irreducible, nonsingular curve of genus , defined over . Let for , with pole divisor on X. Thenwhere is any primitive p-th root of unity and t is the number of distinct poles of f on X. Let . On this subspace of K the conditions of Bombieri’s Estimate are not met.
Proof. Recall that
. Therefore, for
we have
Let . From this we see clearly that . ☐
Lemma 7. For each , there exists an and such that . Therefore, Proof. Suppose there is an
and an
such that
. Consider
, the coefficient-wise
-Frobenius endomorphism on
. Observe
Furthermore,
. Rearranging by exponents of
p we see
By considering the order of poles of , we determine that must be a constant . There is a b in such that . Then and . Therefore, . Let . Observe . Also, , so . Therefore, . ☐
Consider
, and
as defined in our definition of the trace code. By the definition of
K, we have
. Observe that if
then
f satisfies the conditions of Bombieri’s Estimate. Hence,
for each
P. For such
f, each term of the sum in the left-hand-side in Theorem 5 contributes 1. This is a total contribution of
. Hence for
, we have
Observe
and
. Using these two inequalities, we obtain a more general bound:
The condition presented in Proposition 8 is exactly condition (
2) from Theorem 1.
3.3. E and
Recall the definitions of
E and
:
In the case presented in [
7], Van der Vlugt had
. In the current more general case, Proposition 8 provides conditions forcing all elements of
K to be of the form
, for
. However, it may be that elements of this form that are not of the form
, with
. We will show that this is not the case and that condition (
2) of Theorem 1 is sufficient to force
. It will be useful to develop our understanding of the interplay of
K,
and
E, and the nature of the degree of functions therein.
As is the case in Lemma 6, elements of the form , for , can also be written in the form , for . Also notice that for any and , the function is an element of K. Furthermore, for any and , is in E. Consider the following:
Definition 9. For , let be the elements such that , for some .
We see that for , when , there is a y such that .
Proposition 10. For , is an -subspace of .
Proposition 11. If and then for each .
Lemma 12. Let for some and . Then Proof. Let
,
. Then
for some
. Hence
Rearranging terms we see that
Therefore
is in
. Hence, for every
,
and
are in
. Suppose
but
. Then
The only elements of equal to their own power are elements of . Hence for some . From this we see that can be identified with a subgroup of . Hence ☐
Definition 13. For , define the p-linear degree of f, denoted , to be the largest possible integer such that , where , .
The following properties of are straightforward:
Proposition 14. - 1.
For , we have . This is a restatement of Lemma 7.
- 2.
For , we have .
- 3.
For , we have .
Proposition 15. Suppose such that either or , for some . Then we have the inequality: Proof. We proceed by induction on . Suppose . By Lemma 7, we have , for any , any , and . Therefore and .
Now consider a positive integer
k and
such that
. Without loss of generality, we may assume that
. There is a nonzero
such that
. Therefore,
. From this we obtain
. Since
we have
. By the inductive hypothesis,
. Combining this with Lemma 12 we have
☐
Corollary 16. If , then .
Proposition 17. Suppose condition (2) from Theorem 1 holds. That is, suppose If , then is nonempty.
Proof. Suppose
and
for each
. Such an
cannot be constant. Choose
with the least number of poles. In other words,
is minimal and positive. Applying Corollary 16, there is some
,
and
such that
This may be rewritten as
where
Observe
and
. Hence
. By Proposition 11,
. But
and
This contradicts the choice of an f with minimal poles. Hence, when , we can choose an not of the form . ☐