5. The Problem and the Main Results
The family of prime ideals of the quantum plane has a simple structure as we shall presently review. Recall that an ideal P is prime if and if for any two elements a and b of the quantum plane from it follows that or
We denote by B the set of prime ideals in any ring B. consists of the following prime ideals: where .
denotes the two-sided ideal generated by set
Due to the commutation relation
the above set of ideals can be rewritten as
since, for example,
so
and
The fact that the ring structure of the quantum plane is so trivial prevents us from considering ”curves”. That is a motivation to attempt to introduce new rings that play the role of the jet spaces [
5] of the quantum plane and possess interesting nontrivial prime ideals.
Let us consider the noncommutative ring where are new indeterminates.
Consider the unique -derivation (a -linear map satisfying the usual Leibniz rule: such that and Assuming let us define the following elements of
, , , .
In similar fasion let us define the following elements of
for integers
We can consider a noncommutative ring as well as a ring of the usual (commutative) polynomials .
Definition 2. A polynomial F is called bi-homogeneous of bi-degree (p,q), if Fis homogeneous of degree p (resp.q) when considered as a polynomial in (resp. in the ). Any monomial has the bi-degree () where the total degree in is and the total degree in is .
Let us consider a
-linear bijective map
sending the class of any monomial into the same monomial viewed as an element of
Via this bijection we have a multiplication law
on
such that for any two bi-homogeneous polynomials of bi-degrees
and
respectively,
The bijection is not an isomorphism of rings. From now on we shall identify and as sets via above bijection. Note that is bi-graded in the usual way. In the following let q be not a root of unity. Our main results about can be presented as the following Theorems 1–4.
Theorem 1. If is a prime ideal then P contains a non-zero bi-homogeneous polynomial which is an irreducible element of .
Theorem 2. If bi-homogeneous such that its image is irreducible then is prime.
Theorem 3. Any prime ideal not containing any of the ideals or is of the form where T is the family of all bi-homogeneous polynomials in
Theorem 4a. Any prime idealsuch thatis of the formwhereforgenerates a prime ideal of
Theorem 4b. Any prime idealsuch thatis of the formwhereforgenerates a prime ideal of
7. Proofs of the Main Results
For the proofs of Theorems 1–4 we need the following definition of the lexicographical ordering in
.
Let us consider a polynomial Write , such that is bi-homogeneous of bi-degree Let’s consider the set . The sizeof a polynomial g in will be defined as , a number of points in If has a bi-degree then has a bi-degree and has a bi-degree The size of and will stay the same as the size of
Lemma 1. If and , then the size of will be strictly less than the size of
It follows that all points of with the first coordinate equal to will disappear in and the size of will be strictly less than the size of
Similarly, if and then the size of will be strictly less than the size of
7.1. Proof of Theorem 1
We start by showing the following claim: there exists a nonzero bi-homogeneous polynomial in Indeed take of smallest possible size. We claim that the size of g equal to 1 which means g is bi-homogeneous. Assume that the size of g is greater or equal than 2.
Case 1. g is not homogeneous in
Let’s consider such that there is at least one term with total degree in equal to Since g is not homogeneous in On the other hand by the Lemma 1 we have which contradicts the minimality of
Case 2. g is homogeneous in but not in
Let us consider such that there is at least one term in g with the total degree in equal to Since g is not homogeneous in , On the other hand by the Lemma 1 we have which contradicts with minimality of the size of g.
This proves our claim. To conclude the proof of Theorem 1, using our claim one can pick a nonzero bi-homogeneous polynomial of smallest bi-degree with respect to lexicographical order among the nonzero bi-homogeneous polynomials in P.
We claim that f is irreducible in
If we assume it is not irreducible, then .
Note the following properties of bi-degrees:
bideg
If and then
Let be the highest element of with respect to lexicographical order be the highest element of with respect to lexicographical order and let be the lowest element of be the lowest element of .
Then the highest element of will be and the lowest element of will be Since we have
Since and it follows that because if then has to be less then which contradicts with the choice of It immediately follows that Similarly, and , so g and h are both bi-homogeneous of degrees less than
Since P is a prime ideal, at least one of them belongs to This contradicts the choice of f.
7.2. Proof of Theorem 2
Assume f is irreducible in and bi-homogeneous of bi-degree
We prove by induction on the total degree N in that if f has a total degree N then from it follows that g or .
If the theorem is clear. Assume the theorem is true for total degree less or equal to
Let N be the total degree of We have that from it follows that g· where and belong to We may assume that and are bi-homogeneous.
Since
f is bi-homogeneous,
for some
Let
be the highest element of
with respect to lexicographical order
be the highest element of
and
be the highest element of
Then
for some
t and
Since f is irreducible in the commutative ring it follows that (f is bi-homogeneous and the bi-degree of is ) or (bi-degree of is .
Assume, for example, the former is the case. From we get where Obviously,
Since the total degree in of is less or equal to , by the induction hypothesis either and or and the proof is complete.
7.3. Proof of Theorem 3
It is obvious that
To prove assume on the contrary that P does not belong to Let be of minimal size. Since by this assumption f cannot be bi-homogeneous the size f more than 1. There are two cases.
Case 1. f is not homogeneous in .
For an arbitrary there exists a pair such that otherwise f ought to be homogeneous in
Let
Then by Lemma 1, the size of
h is less than the size of
f. It follows that
so
h can be written as
where
Let us pick out the bi-homogeneous components of bi-degree Then where are bi-homogeneous. Since because we have So
Similarly let . As above we get for all s. Since is not contained in P it follows that at least one of Because P is prime, However, is obviously bi-homogeneous so
Since the pair
is arbitrary it follows that
which is a contradiction.
Case 2.
f is homogeneous in
but not in
Write
For an arbitrary there exists a pair such that otherwise f ought to be homogeneous in
Let
Then by the Lemma 1 size of
is less than size of
f. It follows that
so
can be written as
Then we also have , where Let us pick out the bi-homogeneous components of bi-degree Then where are bi-homogeneous.
Since because of we have , so .
Similarly let
. As above we get
for all
s. Since
is not contained in
P it follows that at least one of
Since
P is prime,
but
is obviously bi-homogeneous so
The pair
is arbitrary so it follows that
which is a contradiction.
7.4. Proof of Theorem 4a
Let us consider the factor ideal
Then
Due to the structure of prime ideals of we have
Theorem 4b can be proved similarly.