The level of pairs of polynomials
Alberto F. Boix Marc Paul Noordman Jaap Top


A. F. Boix]Alberto F. Boix

Department of Mathematics, Ben-Gurion University of the Negev, P.O.B. 653 Beer-Sheva 8410501, Israel.
fernanal@post.bgu.ac.il
A.F.B. was supported by Israel Science Foundation (grant No. 844/14) and Spanish Ministerio de Economía y Competitividad MTM2016-7881-P

M. P. Noordman]Marc Paul Noordman
Bernoulli Institute, University of Groningen, P.O. Box 407, 9700 AG Groningen, The Netherlands.
m.p.noordman@rug.nl








J. Top]Jaap Top

Bernoulli Institute, University of Groningen, P.O. Box 407, 9700 AG Groningen, The Netherlands.
j.top@rug.nl



The level of pairs of polynomials]The level of pairs of polynomials



[2010]Primary 13A35; Secondary 13N10, 14B05, 14F10, 34M15

[
    [
    December 30, 2023
=====================




Given a polynomial f with coefficients in a field of prime characteristic p, 
it is known that there exists a differential operator that raises 1/f to its pth power. We first discuss a relation
between the `level' of this differential operator and the
notion of `stratification' in
the case of hyperelliptic curves. 

Next we extend the notion of
level to that of a pair of
polynomials.
We prove some basic properties 
and we compute this level in certain special cases. 
In particular we present
examples of polynomials g and f such that there is no differential
operator raising g/f to its pth power.




§ INTRODUCTION


Let k be any perfect field and R=k[x_1,...,x_d] its polynomial ring in d variables. In this case it is known <cit.> that the ring  of k–linear differential operators on R is the R-algebra (which we take here as a definition)

    :=R ⟨ D_x_i,t| i=1,…,d  t≥ 1 ⟩⊆End_k(R),

generated by the operators D_x_i,t, defined as

    D_x_i,t(x_j^s)=stx_i^s-t, if i=j and s≥ t,
    
    0, otherwise .

For a non-zero f∈ R, let R_f be the localization of R at f; the natural action of  on R extends to R_f in such a way that R_f=1/f^m, for some m≥ 1. Whilst there are examples of m>1 in characteristic 0 (e.g. <cit.>), in positive characteristic one may always take m=1 (<cit.>). This is shown by proving the existence of a differential operator δ∈ such that δ(1/f)=1/f^p, i.e., δ acts as Frobenius on 1/f. We want to mention here that the existence of this differential operator was used as key ingredient in <cit.> to prove that local cohomology modules over smooth –algebras have finitely many associated primes. On the other hand, the fact that R_f is generated by 1/f as –module remains valid for more
general classes of rings R: the interested reader may consult <cit.>, <cit.>, <cit.> and <cit.> for details. We will suppose that k=𝔽_p and fix an algebraic closure k of k from now on.

For an integer e≥ 0, let R^p^e⊆ R be the subring of all the p^e powers of all the elements of R and
set ^(e):=_R^p^e (R), the ring of R^p^e-linear ring-endomorphism of R. Since R is a finitely generated R^p-module, by <cit.>, it is 

    =⋃_e≥ 0^(e).

Therefore, for δ∈, there exists e≥ 0 such that δ∈^(e) but δ∉^(e') for any e'<e. This number e is called the level of δ. For a polynomial f, the level is defined
as the lowest level of an operator δ such that δ(1/f)=1/f^p.

The level of a polynomial has been studied in <cit.> and <cit.>. In <cit.>, an algorithm is given to compute the level and a number of examples are exhibited. Moreover, if f is a cubic smooth homogeneous polynomial defining an elliptic curve 𝒞=V(f)={(x:y:z)∈ℙ^2_k: f(x,y,z)=0}, then the level of f characterizes the supersingularity of 𝒞 in the following way:

(<cit.>)
Let f∈ R be a cubic homogeneous polynomial such that 𝒞=V(f) is an elliptic curve over k. Denote by e the level of f. Then


  * 𝒞 is ordinary if and only if e=1.

  * 𝒞 is supersingular if and only if e=2.




This result was generalized for hyperelliptic curves of arbitrary genus
g≥ 2; indeed, let 𝒞:={(x:y:z)∈ℙ^2_k: f(x,y,z)=0}, where f is a homogeneous polynomial of degree 2g+1 defined over k. If
Jac(𝒞) denotes its Jacobian, then it is well known <cit.> that, for any integer n>0,

    Jac(𝒞)[n](k)=
    (ℤ/nℤ)^2g  if char(k)∤ n,
    
    (ℤ/p^mℤ)^i   if n=p^m, p=char(k)  and m>0,

where i can take every value in the range 0≤ i ≤ g, and is called the
p-rank of 𝒞. For the convenience of the reader, we recall here the following standard terminology:

The curve 𝒞 is said to be ordinary if its
p-rank is maximal, i.e., equal to the genus of 𝒞. The curve 𝒞 is said to be supersingular (resp.superspecial) if Jac(𝒞) is isogenous (resp. isomorphic)
over k to the product of g supersingular elliptic curves.
If 𝒞 is supersingular then the p-rank of 𝒞 equals 0,
however the converse of this statement does not hold.

The generalization of Theorem <ref> reads as follows
<cit.>:


Let f∈ R be a homogeneous polynomial in three variables and of degree 2g+1, such that 𝒞≅ V(f)⊂ℙ^2 defines a hyperelliptic curve over k of genus g. Denote by e the level of f. Assume p>2g^2-1. Then

        
  (i) e=2 if 𝒞 is ordinary,
        
  (ii) e>2 if 𝒞 is supersingular but not
          superspecial.
    

We also want to mention here that the level of a polynomial f 
is closely related to the so–called Hartshorne–Speiser–Lyubeznik–Gabber number of the pair (R,f), and that this number can be explicitly calculated using Macaulay2, see <cit.> for further information. On the other hand, one can also calculate the level of f in terms of F–jumping numbers <cit.>.

The goal of this paper is to introduce and study the level of a pair
of polynomials. Given f,g polynomials
defined over 𝔽_p, one may ask whether there is a differential operator
δ∈𝒟_R mapping g/f to (g/f)^p. Such an operator exists when g=1 by <cit.>, and
more generally, when f itself has level one, as pointed out in
<cit.>. Keeping in mind all of this, it seems natural to define the level
of g and f as

    (g,f):=inf{e≥ 0: ∃δ∈𝒟^(e) such that δ(g/f)=(g/f)^p}.

As we already mentioned, our goal in this paper is to study this notion, and to calculate it in several interesting examples. 

Part of our motivation for introducing it comes from <cit.>, where the author gave a conceptual proof of a polynomial identity obtained in <cit.> using hypergeometric series algorithms.
This polynomial identity, and the corresponding results obtained by Singh concerning associated primes of local cohomology modules <cit.> were the basis of <cit.>, where the authors proved, among other remarkable results, that local cohomology modules H_I_t (X)^ k ([X]) are rational vector spaces for any k>height (I_t(X)), where X is a matrix of indeterminates, and I_t (X) is the ideal of size t minors of this matrix <cit.>. 
The proof presented in <cit.> used as key ingredient certain differential operators defined over the integers that, modulo a prime p, act as the Frobenius endomorphism on quotients of polynomials <cit.>. 

Another motivation comes from <cit.>, where the authors use higher order differential operators to measure various kind of singularities in all characteristics.
These higher order operators also play a key role in recent developments in the study of symbolic powers of ideals (see <cit.> and <cit.> for details). We hope that the calculation of the level of a pair of polynomials might help in the understanding of these differential operators. 
The interplay between differential operators over the integers and their reduction modulo a prime p (which is a delicate issue, see <cit.> for details) was a key technical ingredient to prove in <cit.> that local cohomology modules over  can have p–torsion for at most finitely many primes p.

Now, we provide a more detailed overview of the contents of this
manuscript for the convenience of the reader; first of all, in
Section <ref> we give some
connection between being stratified for a nonlinear differential
equation and the level of a polynomial in the case of
hyperelliptic curves. Second, in Section
<ref>, we formally define
the level of a pair of polynomials, listing some of the properties
it satisfies. In Section <ref>, we focus on specific calculations when
f and g are both homogeneous polynomials; in particular, we will
show, among other things, that (g,f) is, in general, not
finite (see Proposition <ref>). We end this paper by raising some open questions to stimulate
further research on this subject.


§ ACKNOWLEDGEMENT

This research started when the first named author visited the university of Groningen in the
Fall of 2017. We thank Marius van der Put for valuable discussions and for his interest in this work.



§ STRATIFIED DIFFERENTIAL EQUATIONS AND HYPERELLIPTIC CURVES

The notion of stratification for nonlinear differential
equations was introduced in <cit.>; we briefly recall
it here. Let
C⊇𝔽_p be an algebraically closed field, let
C(z) be the one variable differential field extension
of C with derivation d/dz and let K be a finite separable extension of C(z). Consider
the differential equation f(y',y)=0, where f∈ K[S,T] is an
absolutely irreducible polynomial such that the image d of
df/dS in K[S,T]/(f) is nonzero; the differential algebra
A:=K[y',y,1/d] is given by the derivation D with D(z)=1 and
D(y)=y'. One says that f(y',y)=0 is stratified if and
only if D^p=0 <cit.>; it was also
proved in <cit.> that, if
p≥ 3 and f is the defining equation of an elliptic curve E, then
f(y',y)=0 is stratified if and only if E is supersingular, which is
equivalent to say, by Theorem <ref>, that the
homogeneous polynomial corresponding to f has level
two. Keeping in mind these characterizations, one may ask what is the
connection between being stratified and the level of a polynomial. For this we will use the next technical result,
involving among other notions the a-number of (the Jacobian variety of) a curve X of genus g.
This number equals the dimension of the kernel of the
Cartier-Manin matrix associated to X. Many properties
of it are discussed in the textbook <cit.>;
the p-rank f_X and the a-number a_X
satisfy f_X+a_X≤ g.
Here equality does not hold in general, but
a_X=0⇔ f_X=g⇔ X is ordinary, and a_X=g⇔ X is superspecial (see <cit.> and <cit.> for the latter).


Given an algebraically closed field k of prime characteristic p≥ 3, consider the hyperelliptic curve
ℋ of genus g≥ 1 defined by the equation
y^2=h(x), where h(x)∈ k[x] is squarefree
and has degree 2g+1. 
The following statements are equivalent.



  * ℋ is not ordinary.


  * There exist a_0,a_1,…, a_g-1∈ k with a_j≠ 0 for at least one j,
such that 
the differential equation

    (x')^2=h(x)/(a_g-1x^g-1+…+a_1 x+a_0)^2

is stratified.


  * The a-number of the Jacobian of ℋ 
is not zero.




Let 𝒞' be the modified Cartier operator defined
in <cit.>; by the argument pointed out
in <cit.>, our differential equation
is stratified if and only the differential
form ω:=((a_g-1x^g-1+…+a_1 x+a_0)/y)dx is exact, which
is equivalent to the condition 𝒞'(ω)=0. Our goal now is
to write down this condition in terms of the basis of differentials
ω_i:=(x^i-1/y)dx (1≤ i≤ g); it is easy to see that 𝒞'(ω)=0
if and only if

    ∑_i=1^g a_i-1^1/p𝒞'(ω_i)=0.

Now, if one writes h(x)^(p-1)/2=∑_j=0^N
c_j x^j, (where N=((p-1)/2)(2g+1)) then one has
<cit.> that

    𝒞'(ω_i)=∑_j=1^g c_jp-iω_j,

and therefore one ends up with the following equality:

    ∑_j=1^g (∑_i=1^g a_i-1^1/pc_jp-i)ω_j=0.

Equivalently, since the ω_j's are k–linearly
independent,  for any 1≤ j≤ g,

    ∑_i=1^g a_i-1^1/pc_jp-i=0.

Summing up, if one denotes by v the column vector
(a_0^1/p,…, a_g-1^1/p) and by C the Cartier–Manin
matrix of the hyperelliptic curve y^2=h(x)
<cit.>, one has that our differential
equation is stratified if and only if C· v=0, which, by <cit.>, 
is equivalent to the statement that
the hyperelliptic curve y^2=h(x) is not ordinary. This proves the equivalence between (i) and (ii); finally, the equivalence between (i) and (iii) follows immediately from the fact that the a-number of Jac(ℋ) 
equals the corank of the Cartier–Manin matrix of ℋ <cit.>.







































Combining Proposition <ref>
with Theorem <ref>, we obtain the following result.


Preserving the assumptions and notations of Proposition <ref>, let
g≥ 2, p>2g^2-1, and let f=y^2z^2g-1-z^2g+1h(x/z). If (f)≥ 3, then there are a_0,a_1,…, a_g-1∈ k with a_j≠ 0 for at least one j such that the equation

    (x')^2=h(x)/(a_g-1x^g-1+…+a_1 x+a_0)^2

is stratified.

The next examples  illustrate some of the results obtained above. 


Given 0≠ b∈𝔽_p, and p>7, consider the equation

    (x')^2=x^5+b/(a_1x+a_0)^2,

and assume that p≡ 3 5 (e.g. p=13). The hyperelliptic curve of genus two ℋ defined
by y^2=x^5+b has the following Cartier–Manin matrix:

    [ 0 0; c 0 ], where c:=(p-1)/2(2p-1)/5b^(p-3)/10.

In particular, ℋ is not ordinary. In this case, ℋ
is supersingular (but not superspecial) and therefore (y^2z^3-x^5-bz^5)≥ 3 by <cit.>. The equation (<ref>) is stratified, if and only if  a_1=0, as follows from the fact that the differential form dx/y is in the kernel of the Cartier operator, whereas  for a_1≠ 0 the form a_0dx/y+a_1xdx/y is not in the kernel.

Assume that p≡ 4 5 (e.g. p=19). In this case, by either <cit.> or <cit.>, ℋ is superspecial and therefore 
(<ref>) is stratified for any value of a_1,a_0. In this case, (y^2z^3-x^5-bz^5)≥ 3 by <cit.>. In contrast, where p≡ 1 5 (e.g. p=11), one can easily check that ℋ is ordinary (this also follows from <cit.>) and therefore 

(<ref>) is not stratified for any choice of a_1,a_0. In this case, by Theorem <ref>  (y^2z^3-x^5-bz^5)=2.



Given p>17, consider the equation

    (x')^2=(x-1)^8-x^8/(a_2x^2+a_1x+a_0)^2.

One can check that, under a Möbius transformation of the form

    (x,y)⟼(1/x+1,y/(x+1)^4),

the hyperelliptic curve ℋ defined by y^2=(x-1)^8-x^8 corresponds to y^2=x^8-1, and therefore both have the same p–rank. As shown in
<cit.>, ℋ is ordinary if and only if p≡ 1 8, and supersingular (that is, its p–rank is 0) if and only if p≡ 7 8. In the ordinary case, we know that the level is 2, and at least three in the supersingular (not superspecial) case. However, in the remaining cases (where p≡ 3,5 8) the curve has p–ranks 1 and 2 respectively, and in these two cases, while we can ensure that there are non–zero choices of a_2,a_1,a_0 such that (<ref>) can be either stratified or not, we can not predict in general what is the level.





















§ THE LEVEL OF A PAIR OF POLYNOMIALS

Hereafter, let k be a perfect field
of prime characteristic p, and let
R be the polynomial ring k[x_1,…, x_d]. The aim of this section is to study the following concept.


Given polynomials f,g with coefficients in k and f ≠ 0, one defines
the level of (g,f) as 

    (g,f):=inf{e≥ 0: ∃δ∈𝒟^(e) such that δ(g/f)=(g/f)^p}∈ℕ_0 ∪{∞}.

When g=1, one denotes (f) instead of
(1,f); this is the
notion of level of a polynomial introduced in
<cit.>. 


Note that (g,f) only depends on the quotient g/f, so one could also reasonably denote this notion by (g/f) instead. But this alternative notation is inconsistent with the one in <cit.> in the case f = 1, so we stick with the notation (g,f). In any case, one can usually assume that g and f are coprime, since common factors do not change the level of the pair. 

Note also that (g,f) = 0 if and only if g/f ∈ R. If g and f are coprime, this only happens if f is a constant. 

In Proposition <ref> we give an example of polynomials f and g such that (g,f) = ∞. 


Before going on studying this notion, we review
the so–called ideals of p^eth roots; the interested reader
can find a more detailed treatment in <cit.>, <cit.> and
<cit.>. For an ideal I ⊂ R we denote by I^[p^e] the ideal generated by the p^e-th powers of elements of I. 

 
Given g∈ R and an integer e ≥ 0, we define the ideal of p^eth roots I_e (g) to be the smallest ideal J⊆ R such that g∈ J^[p^e].


Under our assumptions, R is a free R^p^e-module with basis given by the monomials {𝐱^α|α≤ p^e-1}. A polynomial g ∈ R can therefore be written as 

    g=∑_0≤α≤ p^e-1g_α^p^e𝐱^α,

for unique g_α∈ R. Then I_e (g) is the ideal of R generated by elements g_α <cit.>. 



The main relation between these ideals and differential operators is the following equality, valid for any polynomial g ∈ R and any integer e ≥ 0 (see <cit.>): 

    𝒟^(e)· g = I_e(g)^[p^e].

Using this, one can relate the level of a pair of polynomials to ideals of p^eth roots as follows. 

 
Let f, g ∈ R and e ≥ 0 be given. Then the following are equivalent:

    
  * (g,f) ≤ e;
    
    
  * I_e (g^pf^p^e-p)⊆
I_e (gf^p^e-1);

    
  * I_e (g^pf^p^e-p)^[p^e]⊆ I_e (gf^p^e-1)^[p^e].

In particular, (g,f) =inf{e≥ 0: I_e (g^pf^p^e-p)⊆
I_e (gf^p^e-1)}.




The equivalence of (ii) and (iii) is proved in the last paragraph of the proof of <cit.>. We prove that (i) and (iii) are equivalent. Suppose that there is δ∈𝒟^(e) such that δ(g/f)=(g/f)^p. Since δ is linear over p^e-powers, this
implies that δ (gf^p^e-1)=g^p f^p^e-p. By (<ref>), this implies g^pf^p^e-p∈ I_e(g^pf^p^e-p)^[p^e], so that I_e (g^pf^p^e-p)^[p^e]⊆
I_e (gf^p^e-1)^[p^e].

Conversely, suppose now that I_e (g^pf^p^e-p)^[p^e]⊆
I_e (gf^p^e-1)^[p^e]. Again using
(<ref>), one has that
𝒟^(e) (g^p f^p^e-p)⊆𝒟^(e) (gf^p^e-1). In particular
g^p f^p^e-p∈𝒟^(e) (gf^p^e-1), hence
there is δ∈ D^(e) such that
δ (gf^p^e-1)=g^p f^p^e-p. Multiplying
this equality by 1/f^p^e and using that δ is linear over p^eth powers, we get δ(g/f)=(g/f)^p. 


Observe that the equality
𝒟^(e)· g = I_e(g)^[p^e]
is made explicit in, e.g., the proof of
<cit.>.
Using these techniques one can in case
e=(g,f)<∞, algorithmically
construct an explicit operator 
δ∈𝒟_R^(e)
with δ(g/f)=g^p/f^p. However we do
not know how to decide whether the level
of a given pair is finite.

Our next goal is to show that the level of a pair is invariant under coordinate transformations. Although <ref>, 
<ref> and
<ref> can also be found in
<cit.>, we review it here for the convenience
of the reader.

Denote G:=GL_d (k) and observe that R has a right action of G defined by (f|A)(x_1,...,x_d):=f(y_1,...,y_d), where

    [ y_1;   ⋮; y_d ]=A·[ x_1;   ⋮; x_d ],

for A∈ G.
Observe as well that a matrix A∈ G induces an isomorphism ϕ_A of k–algebras @1R[r]^-ϕ_A    R defined by ϕ_A(f)=f|A, the inverse being given by ϕ_A^-1. 

Given homogeneous f,g∈ R, we say that f and g are G–equivalent if there is A∈ G such that ϕ_A(f)=g.

We need the following easy fact.


Notations as before, let y_1,… ,y_d∈ R be homogeneous elements of degree 1 such that

    [ y_1;   ⋮; y_d ]=A·[ x_1;   ⋮; x_d ],

for some A∈ G. Then, for any e≥ 0 the set

    ℬ:={𝐲^α:=y_1^a_1⋯ y_d^a_d: α =(a_1,… ,a_d)∈^d, 0≤ a_i≤ p^e-1 for any 1≤ i≤ d}

is a basis of R as R^p^e–module.



We have 𝐲^α = ϕ_A(𝐱^α) for every multi-index α. Therefore, the set ℬ is the image of the R^p^e-basis {𝐱^α: α≤ p^e-1} under the k-algebra isomorphism ϕ_A. Since ϕ_A restricts to an isomorphism on R^p^e, the result follows.  


In this way, we are ready to prove the following:

For any f∈ R, A∈ G and e≥ 0, it holds that
ϕ_A (I_e (f))=I_e(ϕ_A(f)). In particular, for all f, g ∈ R we have (f,g) = (ϕ_A(f), ϕ_A(g)). 





Setting

    [ y_1;   ⋮; y_d ]=A·[ x_1;   ⋮; x_d ],

and applying Lemma <ref> we see that the set

    {𝐲^α:=y_1^a_1⋯ y_d^a_d: α =(a_1,… ,a_d)∈^d, 0≤ a_i≤ p^e-1 for any 1≤ i≤ d}

is a basis of R as R^p^e–module.
Write

    f=∑_0≤α≤ p^e-1f_α^p^e𝐱^α,

for f_α∈ R. Then

    ϕ_A(f)=∑_0≤α≤ p^e-1ϕ_A(f_α)^p^e𝐲^α,

which shows that ϕ_A (I_e (f))⊆ I_e (ϕ_A(f)). Equality holds because 
ϕ_A is an isomorphism. The second claim follows from the first together with Lemma <ref>.

In the next statement, our aim is to collect some properties that
the level of a pair of polynomials satisfy.


Let f, g ∈ R be non-zero polynomials such that g/f∉ R. Then the following statements hold.


  * (g,f)=1 if and only if g∈ I_1 (gf^p-1).

  * If (f)=1, then (g,f)=1.

  * If either I_e (g^pf^p^e-p)⊈I_e(f^p^e-1)
or I_e (g^pf^p^e-p)⊈I_e (g), then
(g,f) > e.





  * If f and g are homogeneous, and e≥ 1 is an integer such that p^e >  g -  f, then
I_e (gf^p^e-1) is generated by polynomials of degree at
most f.



The assumption that f does not divide g in R implies that (g,f) > 0. Then (i) follows from Lemma <ref> together with the easy observation that I_1(g^p) = (g). Part (ii) was already proved in <cit.>; we
repeat the proof for the sake of completeness. Let
δ'∈𝒟^(1) such that δ ' (1/f)=1/f^p. Then define 
δ:=δ '∘ (· g^p-1). We find that
δ (g/f)=δ ' (g^p/f)=g^p δ ' (1/f)=(g/f)^p, as
required.

Part (iii) follows immediately combining
Lemma <ref> with the fact that
I_e (g f^p^e-1)⊆ I_e(g) I_e (f^p^e-1)
<cit.>. Finally, to
prove part (iv) fix e≥ 1 an integer and write

    gf^p^e-1=∑_0≤α≤ p^e-1c_α^p^e𝐱^α,

for some c_α∈ R. Since both f and g are homogeneous
it follows that

    (g)+(p^e-1) (f)=p^e  (c_α)+ (𝐱^α),

which implies that

    (c_α)≤(p^e-1) (f)+ (g)/p^e=  (f)+  g -  f/p^e.

The second term on the right hand side is smaller than 1 by assumption, and since both sides are integers, we get c_α≤ f. The result follows. 




§ SOME EXAMPLES

The goal of this section is to calculate the level of a pair of polynomials (g,f) for several particular choices of g and f; we will quickly see that, even for low degrees, most of the calculations are highly non–trivial. In particular, we show that (g,f) is, in general, not always finite (see Example <ref>).

We want to start with the case considered by Singh, 
see for example <cit.>.


Let p be a prime number, X=[ u v w; x y z ] be a matrix of indeterminates
defined over R=k [u,v,w,x,y,z], and set Δ_1:=vz-wy, Δ_2:=wx-uz and
Δ_3:=uy-vx. Then, (g,f)=1 for each pair
(g,f)∈{(w,Δ_1Δ_2),  (v,Δ_1Δ_3), (u,Δ_2Δ_3) }.



By symmetry, it is enough to show that (g,f)=1 when (g,f)=(w,Δ_1Δ_2).








Set f:=Δ_1Δ_2, and notice that f=1· (xzvw)+ (-z^2)· (uv)+(-w^2)· (xy)+1· (yzuw). This shows that, if p=2, then I_1 (f)=R so (f)=1 and therefore (g,f)=1. Now, assume that p≥ 3, one can check that in the support of f^p-1 appears the monomial (xyuv)^(p-1)/2(zw)^p-1 with coefficient p-1(p-1)/2; this shows again that (f)=1 and therefore (g,f)=1.



Notice that, in the setting considered in Lemma <ref>, Singh shows in <cit.> that the differential operator
δ:=D_u,p-1D_y,p-1D_z,p-1 (which is clearly of level one) is such that δ(g/f)=(g/f)^p, for g/f any of the three fractions considered in Lemma <ref>.



Let k be a field of characteristic p, let f=x^d, assume that p≥ d, and let
g∈ R=k[x,y] be a homogeneous polynomial of degree
d which is not a multiple of f. Then, (g,f)=2 unless g∈ (x^d-1), in which
case (g,f)=1.



Write g=∑_i=0^d a_ix^i y^d-i; now, notice that

    gf^p-1=∑_i=0^d a_ix^i+d(p-1) y^d-i.

Given 0≤ i≤ d write i+d(p-1)=(d-1)p+(p+i-d), and notice that, unless
i=d, 0≤ p+i-d≤ p-1 (here, we are also using that
d≤ p). This shows that
I_1 (gf^p-1)= (a_d^1/px^d, a_i^1/px^d-1: 
1≤ i≤ d-1)= (x^d-1),
so (g,f)≠ 1 unless g∈ (x^d-1), in which case (g,f)=1. So, from
now on, assume that g∉ (x^d-1).

We have I_2(g^pf^p^2-p) = I_1(gf^p-1) = (x^d-1). 
Now, write

    gf^p^2-1=∑_i=0^d a_ix^i+d(p^2-1) y^d-i.

Again, the equality i+d(p^2-1)=(d-1)p^2+(p^2+i-d) and the fact
unless i=d, p^2+i-d≤ p^2-1, shows that 
I_2 (gf^p^2-1) = (a_d^1/p^2x^d, a_i^1/p^2x^d-1: 
1≤ i≤ d-1)= (x^d-1) = I_2(g^pf^p^2-p),
and therefore (g,f)=2, as claimed.


Lemma <ref> has the following interesting
consequence.


Let k be a field of prime characteristic p, and let f,g∈ k[x,y] be quadratic forms. If
√((f)) denotes the radical of (f), then

    (g,f)= 0,  if g is a multiple of f, 
    
    1, if either f is not the square of a linear form, or if g∈√((f)),
    
    2, otherwise.
 



First of all, if f is not the square of a linear form, then by
<cit.> (f)=1 and
therefore part (ii) of Proposition <ref>
implies that (g,f)=1. So, hereafter we assume that
f is the square of a linear form; thanks to
Theorem <ref> we
can assume, without loss of generality, that
f=x^2 and that g is again a quadratic form. Then, in this
case, Lemma <ref> says exactly
that (g,f)=2 unless g∈ (x), in which case
(g,f)=1; the proof is therefore completed.


As a more elaborate example we now consider
(g,f) with f=x^3+y^3+z^3 and g
any homogeneous cubic in 3 variables which is not a scalar multiple of f.
Since (f)=1 in case the characteristic
p≡ 1 3, Proposition <ref> (ii)
shows (g,f)=1 for p≡ 1 3
and any such g.

We expect that the same holds for all 
characteristics p≥ 5. The next two special
cases show that this is correct for
most g. By Example <ref>, the same
does not hold in characteristics p=2,3.



Let p≥ 5 with p≡ 2  3, let f=x^3+y^3+z^3, and let
g∈ R=k[x,y,z] be a homogeneous polynomial
of degree 3 such that, if one writes g=∑_a+b+c=3g_a,b,cx^a y^b z^c, and set B:=p-1(p-2)/3,(p-2)/3,(p+1)/3, 
C:=p-1(p-2)/3, D:=p-11,2(p-2)/3,(p-2)/3,
E:=p-11,(2p-1)/3,(p-5)/3, and
F:=p-1(p+4)/3,(p-2)/3,(p-5)/3, then the rank of

    A:=[                      Bg_1,1,1                      Cg_2,0,1                      Cg_2,1,0;                      Cg_0,2,1                      Bg_1,1,1                      Cg_1,2,0;                      Cg_0,1,2                      Cg_1,0,2                      Bg_1,1,1;                     B g_2,0,1                      Dg_0,2,1    Cg_3,0,0+Eg_0,3,0+Dg_0,0,3;                     B g_2,1,0    Cg_3,0,0+Eg_0,3,0+Dg_0,0,3                      Dg_0,1,2;   B g_3,0,0+Fg_0,3,0+Fg_0,0,3                      Dg_1,2,0                      Dg_1,0,2;                     D g_2,0,1                      Bg_0,2,1    Eg_3,0,0+Cg_0,3,0+Dg_0,0,3;    Eg_3,0,0+Cg_0,3,0+Dg_0,0,3                      Bg_1,2,0                      Dg_1,0,2;                     D g_2,1,0   F g_3,0,0+Bg_0,3,0+Fg_0,0,3                      Dg_0,1,2;                     D g_2,1,0    Eg_3,0,0+Dg_0,3,0+Cg_0,0,3                      Bg_0,1,2; E g_3,0,0+D g_0,3,0+C g_0,0,3                      Dg_1,2,0                      Bg_1,0,2;                     D g_2,0,1                      Dg_0,2,1 F g_3,0,0+F g_0,3,0+ Bg_0,0,3 ]

is three. Then (g,f)≤ 1, with equality exactly if g is not a multiple of f.



Write g=∑_a+b+c=3g_a,b,cx^a y^b z^c, and

    gf^p-1=∑_a+b+c=3∑_i+j+k=p-1
    g_a,b,cp-1i,j,kx^3i+ay^3j+b
    z^3k+c.

Then, if one picks i=j=(p-2)/3 and k=(p+1)/3, then
the corresponding term of gf^p-1 is

    ∑_a+b+c=3g_a,b,cp-1i,j,k
    z^p · (x^p-2+ay^p-2+bz^c+1).

Again, if i=k=(p-2)/3 and j=(p+1)/3, then
the corresponding term of gf^p-1 is

    ∑_a+b+c=3g_a,b,cp-1i,j,k
    y^p · (x^p-2+ay^b+1z^p-2+c).

By the same reason, if j=k=(p-2)/3 and i=(p+1)/3, then
the corresponding term of gf^p-1 is

    ∑_a+b+c=3g_a,b,cp-1i,j,k
    x^p · (x^a+1y^p-2+bz^p-2+c).

The above expansions show that the
basis elements x^p-1y^p-1z^2, x^p-1y^2z^p-1 and
x^2y^p-1z^p-1 contain respectively in their coefficient the below term, where B:=p-1(p-2)/3,(p-2)/3,(p+1)/3:

    g_1,1,1B z^p,
    g_1,1,1B y^p,
    g_1,1,1B x^p.

Hereafter, we only plan to prove that the coefficient of
x^p-1y^p-1z^2 is exactly Cg_0,1,2x^p+
Cg_1,0,2y^p+Bg_1,1,1 z^p
and one can show using the same arguments that the
coefficient of x^p-1y^2z^p-1 (resp. x^2y^p-1z^p-1)
is exactly Cg_0,2,1x^p+Bg_1,1,1y^p+Cg_1,2,0 z^p resp. 
Bg_1,1,1x^p+Cg_2,0,1y^p+Cg_2,1,0 z^p.

Indeed, we want to calculate the coefficient of x^p-1y^p-1z^2, so
suppose that there are non–negative integers λ,μ,γ
such that 3i+a=λ p+p-1, 3j+b=μ p+p-1, 3k+c=γ p+2.
Since (gf^p-1)=3p, it follows that 3p=3i+a+3j+b+3k+c=(λ+μ+γ +2)p,
which implies that λ+μ+γ=1, so we only have
three possibilities for these integers; namely, (1,0,0), (0,1,0) and
(0,0,1). For (1,0,0), we get i=(2p-1-a)/3, j=(p-1-b)/3, k=(2-c)/3.
Since p≡ 2 3, this forces a=0, b=1 and c=2.  By the
same argument, for (0,1,0) one gets a=1, b=0 and c=2, and
finally, for (0,0,1) one ends up with
a=b=c=1. This shows that the coefficient of
x^p-1y^p-1z^2 is exactly B(g_0,1,2x^p+
g_1,0,2y^p+g_1,1,1 z^p), as claimed.

One might ask from where the other rows of matrix A appearing in
our assumption comes from; following the same arguments, these rows
corresponds to the calculation of the coefficients of the below
basis elements:

    x^3y^p-2z^p-1, x^3y^p-1z^p-2, x^4y^p-2z^p-2,
        x^p-2y^3z^p-1, x^p-1y^3z^p-2, x^p-2y^4z^p-2,
        x^p-2y^p-1z^3, x^p-1y^p-2z^3, x^p-2y^p-2z^4.


Summing up, the foregoing implies, since by assumption
the rank of A is 3, that (x,y,z)=I_1 (gf^p-1), hence
g∈ I_1 (gf^p-1) and this shows that (g,f)=1 by using
part (i) of Proposition <ref>.




Let p≥ 5, let f=x^3+y^3+z^3, and let
g∈ R=k[x,y,z] be a non–zero monomial
of degree 3. Then, (g,f)=1.


If p≡ 1  3, then (f)=1 and therefore (g,f)=1
by part (ii) of Proposition <ref>, so hereafter we will assume that p≡ 2 3. By symmetry, it is enough to consider the monomials g = x^3, g = x^2y and g = xyz. In each of these cases, we will simply construct an explicit differential operator of level 1 that does what is needed. For g = x^3, consider first

    δ = D_x,p-1∘ D_y, p-2∘ D_z, 3

(see the Introduction for the notation D_x,n). Clearly δ is of level 1, since p > 3. 
We have that 

    gf^p-1 = ∑_i+j+k = p-1p-1i,j,k x^3i+3y^3jz^3k.

Applying δ gives us

    δ(gf^p-1) = ∑_i+j+k = p-1p-1i,j,k3i+3p-13jp-23k3 x^3i+4-py^3j+2-pz^3k-3,

where we use the convention that nk = 0 for k > n. We investigate for which indices i, j, k the coefficient in this term is zero. The first factor is never zero, since p-1, i, j and k are all between 0 and p-1. The second factor is zero unless 3i + 3 ≡ -1  p, as can be seen by writing out the product. Since i lies between 0 and p-1, and since p ≡ 2  3, the only integer value for i such that 3i + 3 ≡ -1  p is i = (2p-4)/3. This means that j is at most (p+1)/3. The third factor 3jp-2 is zero unless 3j is either -1 or -2 modulo p. In the allowed range for j, the only integer possibility is j = (p-2)/3. This leaves k = 1, and for this value of k we have 3k3 = 1 ≠ 0. So we see that the only non-zero term in δ(gf^p-1) is the one for indices (i,j,k) = (2p-4/3, p-2/3, 1). This gives

    δ(gf^p-1) = p-12p-4/3, p-2/3, 12p-1p-1p-2p-233x^p = p-12p-4/3, p-2/3, 1 x^p

Define now

    Δ = p-12p-4/3, p-2/3, 1^-1· x^2p·δ,

then Δ is also a differential operator of level 1, and by construction we have Δ(gf^p-1) = x^3p = g^p. Using that Δ is R^p-linear, we may divide both sides by f^p and get Δ(g/f) = g^p/f^p, as needed.

For the other cases g=x^2y and g = xyz, a similar analysis shows that the operators

    C · x^py^p D_x,p-2D_y,p-1D_z,3,   resp.   C' · y^pz^p, D_x,p-3D_y,p-1D_z,4

for suitably chosen non-zero constants C, C'  ∈𝔽_p, have the required property.



Old proof for above claim. 

If p≡ 1  3, then (f)=1 and therefore (g,f)=1
by part (ii) of Proposition <ref>, so hereafter we can assume that p≡ 2 3, and we
keep the notations introduced along Lemma <ref>. Moreover, using Theorem <ref>, it is enough to show that the result holds
when g∈{x^3,x^2y,xyz}; indeed, since g is a monomial of degree 3 g can only be one of the following ten monomials:

    x^3,y^3,z^3,x^2y,x^2z,xy^2,xz^2,xyz,y^2z,yz^2.

In the below table, we plan to show that any fraction of the form g/f can be transformed by a K-algebra isomorphism that just permutes the variables into the fraction g_0/f, where g_0∈{x^3,x^2y,xyz}; this will show, using Theorem <ref>, that it is enough to prove that the result holds
when g∈{x^3,x^2y,xyz}.



g_0         g     K–algebra isomorphism 
 x^3         y^3    x⟼ y, y⟼ x, z⟼ z

x^3         z^3    x⟼ z, y⟼ y, z⟼ x

x^2y         x^2z    x⟼ x, y⟼ z, z⟼ y

x^2y         xy^2    x⟼ y, y⟼ x, z⟼ z

x^2y         xz^2    x⟼ z, y⟼ x, z⟼ y

x^2y         y^2z    x⟼ y, y⟼ z, z⟼ x

x^2y         yz^2    x⟼ z, y⟼ y, z⟼ x




In this way, thanks to our previous argument hereafter we can assume, without loss of generality, that g∈{x^3,x^2y,xyz}.
First of all, if g=x^3, then the matrix A appearing in the statement of Lemma <ref> contains the following minor of size 3:

    [ 0 0 C; 0 C 0; B 0 0 ].

This shows that the rank of A is 3 and therefore (g,f)=1
by Lemma <ref>. Secondly, if g=xyz, then
A contains the following minor of size 3:

    [ B 0 0; 0 B 0; 0 0 B ].

Hence again (g,f)=1 by Lemma <ref>. Finally, if g=x^2y then the rank of A is 2, so we can not simply appeal to Lemma <ref>; however, an inspection of the calculation done along the proof of Lemma <ref>
shows that x∈ I_1 (gf^p-1). This shows that g=x^2y∈ (x)⊆ I_1 (gf^p-1), hence (g,f)=1 by part (i) of Proposition <ref>.



Notice that, in the example considered in
Lemma <ref>, (f)=2> (g,f)=1. From here, one
might ask whether, in general, (g,f)≤ (f); however, this
is not the case, as the below example shows. The unjustified
calculations were done with Magma <cit.>.


Let R=k [x,y,z,w], g=y and f=xy^p+1
+yz^p+1+zw^p+1; when p∈{2,3,5}, (g)=1,
(f)=2, but (g,f)=4.

For any prime p, what is easy to show in this example is that (g,f)≥ 2; indeed, notice
that

    gf^p-1=∑_0≤ i,j,k≤ p-1
     i+j+k=p-1p-1i,j,k(y^iz^jw^k)^p · (x^i y^p-kz^p-1-iw^k).

We claim that, whereas y^p∈ I_1 (gf^p-1), g=y∉ I_1 (gf^p-1). Indeed, if in the
above expansion we pick j=k=0 and i=p-1, then one gets that
gf^p-1=(y^p)^p (x^p-1)+…, and this choice is the only one
that makes the basis element x^p-1 appearing in this expansion. This
shows that y^p∈ I_1 (gf^p-1); moreover, notice that, if one
choices a i,j,k as above where i<p-1, then the coefficient
of the corresponding basis element is made up by monomials that are divisible
by either z or w. This shows that y^p is the smallest possible
power of y that belongs to I_1 (gf^p-1), hence
g=y∉ I_1 (gf^p-1) and therefore (g,f)≥ 2, as
claimed.

Moreover, again about Lemma <ref>, we want to
single out that the assumption p≠ 2,3 can not be removed, as the following examples show.


Let p=2, let R=k[x,y,z], f=x^3+y^3+z^3 and
g=xyz; we claim (g,f)=2. Indeed, on the one hand,
gf^p-1=(x^2)^2· (yz)+(y^2)^2· (xz)+(z^2)^2· (xy), so g=xyz∉ I_1(gf^p-1)=(x^2,y^2,z^2); this shows, by part (i)
of Proposition <ref>, that
(g,f)≥ 2. On the other hand,

    gf^p^2-1=     (x^2)^4· (x^2yz)+(y^2)^4· (xy^2z)
    +(z^2)^4· (xyz^2)+(xy)^4 · (x^3z)+
    (xy)^4 · (y^3z)
       +(xz)^4 · (x^3y)+
    (xz)^4 · (yz^3) +(yz)^4 · (xyz^3),

and g^p f^p^2-p=x^8 (yz)^2+y^8 (xz)^2+z^8 (xy)^2; these
last two computations show that

    g^p f^p^2-p∈ (x^2,y^2,z^2,xy,xz,yz)^[p^2]=I_2 (gf^p^2-1)^[p^2],

and therefore Lemma <ref> ensures
(g,f)=2, as claimed.

Now, assume that p=3 (g and f are the same); in this case, one can check that J:=I_1 (gf^2)=(x^2+2xy+y^2+2xz+2yz+z^2) and g=xyz∉ J. One way to check it is the following; denote
by V(J) the hypersurface defined by J. This hypersurface contains the point (1,1,1), which is a point which does not belong to V(xyz). This shows that xyz∉ J. 

The above argument shows that (g,f)≥ 2 and, actually, one can check either by hand or by computer that (g,f)=2.


We conclude this section with an example showing that the
level of a pair of polynomials is, in general, not finite. This in fact answers a question raised in <cit.>.


Let R=k[x,y] with char k = p, and let f=x^p+1+y^p+1 and
g=x.
Then (g,f)=∞.
In particular, no δ∈𝒟_R
exists with δ(g/f)=g^p/f^p.


Let e ≥ 2 be an arbitrary even integer. We will show that (g,f) > e. By Lemma <ref>, this is equivalent to showing that I_e(g^pf^p^e-p)^[p^e]⊈I_e(gf^p^e-1)^[p^e].

First we show that I_e(gf^p^e-1) is a monomial ideal. Indeed, we have 

    gf^p^e-1 = ∑_i = 0^p^e-1p^e-1i x^i(p+1)+1y^(p^e-1-i)(p+1).

By the description of I_e in Remark <ref>, to find generators of I_e(gf^p^e-1), we take out p^e-th powers. If for two indices i and j the corresponding terms above differ by a p^e-th power, then they both contribute to the same generator. But this happens only if the exponents for x and y are congruent modulo p^e. From i(p+1) + 1 ≡ j(p+1) + 1 p^e we obtain i ≡ j p^e since p+1 is a unit modulo p^e. But if 0 ≤ i,j ≤ p^e - 1 and i ≡ j p^e, then i = j. So we see that the terms occurring in gf^p^e-1 are independent over Frac(R^p^e). Hence the generators for I_e(gf^p^e-1) that we get from Remark <ref> are monomials, and so I_e(gf^p^e-1) is a monomial ideal. It follows that also I_e(gf^p^e-1)^[p^e] is a monomial ideal. 

Now we show that g^pf^p^e-p∉ I_e(gf^p^e-1)^[p^e]. Since the latter is a monomial ideal, it is sufficient to find a monomial that occurs in g^pf^p^e-p with non-zero coefficient which is not in this ideal. For this, set m := x^p^e - p^2 + py^p^e+1 - p.
We claim that this monomial occurs in g^pf^p^e-p with non-zero coefficient. We have

    g^pf^p^e-p = ∑_i = 0^p^e-pp^e-pi x^i(p+1)+py^(p^e-p-i)(p+1).

We see that our monomial m occurs for index i = (p^e - p^2)/(p+1), which is an integer because e is even. To evaluate the binomial coefficient for this value of i, we can look at the p-adic digits of the numbers involved. We have p^e - p = (p-1)p^e-1 + (p-1)p^e-2 + … + (p-1)p, and we have i = (p-1)p^e-2 + (p-1)p^e-4 + … + (p-1)p^2. Using Lucas's theorem <cit.>, we find that the binomial coefficient evaluates to 1, so in particular it is non-zero. 

Now we need to show that m ∉ I_e(gf^p^e-1)^[p^e]. This ideal is generated by monomials which are also p^e-th powers, and m is an element of this ideal if and only if at least one of these monomials divides m. The largest p^e-th power dividing m is y^(p-1)p^e. Hence, it is enough to show that y^(p-1)p^e∉ I_e(gf^p^e-1)^[p^e], or equivalently, that y^p-1∉ I_e(gf^p^e-1). In view of Remark <ref>, we look at terms in the product gf^p^e-1 that contribute something of the form y^n to I_e(gf^p^e-1). A term does this if and only if the exponent for x is strictly lower than p^e. In Equation (<ref>) above, this happens for terms with index i for which i(p+1) + 1 ≤ p^e - 1, which is equivalent to

    i ≤p^e - 2/p+1 = p^e - p - 2/p + 1,

where we used again that e is even. But for such indices i, the exponent for y is given by

    (p^e - 1 - i)(p+1) ≤ p^e+1 + p^e - p - 1 - p^e + p + 2 = p^e+1 + 1.

So the contribution of these terms to I_e(gf^p^e-1) is at least y^p. Thus the lowest exponent n such that y^n ∈ I_e(gf^p^e-1) is n = p, and in particular y^p-1∉ I_e(gf^p^e-1). 



This is the old version of the proof. There are some mistakes in it. 

Notice that

    gf^p-1=∑_i=0^p-1p-1i
    (x^i y^p-1-i)^p · (x^i+1y^p-1-i).

From this equality we deduce that g=x∉ I_1(gf^p-1)=(x^p,x^iy^p-1-i: 0≤ i≤ p-2), hence (g,f)≥ 2. Now, assume that e≥ 2 is even and write

    gf^p^e-1=∑_i=0^p^e-1p^e-1i
    x^ip+(i+1)(y^p)^p^ey^p^e-(i+1)(p+1).

Notice that, if one chooses i=p^k-1 (0≤ k≤ e-1), then one
obtains that gf^p^e-1 contains in its support

    p^e-1p^k-1(y^p)^p^e· (x^p^k+1+p^k-p
    y^p^e-(p^k+1+p^k)) if k≤ e-2, and p^e-1p^e-1-1(xy^p-1)^p^e· (x^p^e-1-1
    y^p^e-p^e-1).

Moreover, if i=p^e-1, then one obtains that gf^p^e-1 contains in its support (x^p)^p^e (x^p^e-p).
On the other hand, observe that for any integer i one has ip+i+1<p^e if and only if
i<(p^e-1)/(p+1). As we assume e is even, (p^e-1)/(p+1) is an integer so the
given condition is equivalent to
i≤ ((p^e-1)/(p+1))-1. In this case, one can also notice that
p^e-(i+1)(p+1)≥ 0 if and only if i≤ ((p^e-1)/(p+1))-1. Combining
these two remarks, it follows that, if one chooses
i≤ ((p^e-1)/(p+1))-1, then one
obtains that gf^p^e-1 contains in its support

    p^e-1i(y^p)^p^e· (x^ip+p+1
    y^p^e-(i+1)(p+1)).

Now, let 1≤λ≤ p-1 be an integer. 
Since e is assumed to be even,
i= λ (p^e-1)/(p+1) is an integer as well. One
obtains that gf^p^e-1 contains in its support

    p^e-1i(x^λ-1y^p-λ-1)^p^e· (x^p^e-(λ-1)
    y^p^e-(λ+p+1)).

On the other hand, if one writes in a unique way

    gf^p^e-1=∑_0≤ i,j≤ p^e-1h_i,j^p^e x^i y^j,

then, since gf^p^e-1 is homogeneous and (gf^p^e-1)
=1+(p+1)(p^e-1), it follows, for any 0≤ i,j≤ p^e-1 that
h_i,j is homogeneous and

    (h_i,j)    =1+(p+1)(p^e-1)- (x^i y^j)/p^e
       ≥1+(p+1)(p^e-1)-2(p^e-1)/p^e
    =1+(p-1)(p^e-1)/p^e≥ p-2.

Summing up, all our above computations show that, for any e≥ 2,

    I_e (gf^p^e-1)⊆
    (x,y^2), if p=2,
     (y^p,x^jy^p-2-j: 1≤ j≤ p-2),
     otherwise.

Hence

    I_e (gf^p^e-1)^[p^e]⊆ J:=
    (x^2^e,y^2^e+1), if p=2,
     (y^p^e+1,x^p^e jy^p^e+1-2p^e-jp^e: 1≤ j≤ p-2),
     otherwise.

So, in order to prove that g^p f^p^e-p∉ I_e (gf^p^e-1)^[p^e], it is enough
to check that g^p f^p^e-p∉ J; moreover, since J is a monomial
ideal, it is enough to check that there is at least one non–zero term
of g^p f^p^e-p that does not belong to J. In this case, one can
check, by a matter of degrees (keeping in mind that
e≥ 2), that x^p^e-1+p^e-2-1 y^p^e+1-p^e-1 is an element
in the support of g^p f^p^e-p such that
x^p^e-1+p^e-2-1 y^p^e+1-p^e-1∉ J. This finally shows that
g^p f^p^e-p∉ I_e (gf^p^e-1)^[p^e] and therefore
(g,f)=∞, as claimed.






 §.§ Some open questions




The following questions are open, to the best of our knowledge.




  * Does an algorithm exist which, on input polynomials f and g, decides whether (g,f)<∞? 


  * Under which conditions one can ensure that (g,f)≤ (f)?



















  * In <cit.>, it is shown that, if R
is an F–finite ring of characteristic p≥ 3, f∈ R, and
e is the largest F–jumping number of f that lies inside
(0,1), then (f)=⌈ 1-log_p (1-e)⌉. Is it possible
to obtain a similar result for (g,f)?




alpha


