Abstract

I give an iterative closed form formula for the Hilbert-Kunz function for any binomial hypersurface in general, over any field of arbitrary positive characteristic. I prove that the Hilbert-Kunz multiplicity associated with any binomial hypersurface over any field of arbitrary positive characteristic is rational. As an example, I also prove the well known fact that for 1-dimensional binomial hypersurfaces the Hilbert-Kunz multiplicity is a positive integer and give a precise account of the integer.

1. Introduction

Let be a Noetherian local ring of dimension and of prime characteristic . Let be an -primary ideal and where is an integer. The “Hilbert-Kunz function” of with respect to is defined as where th Frobenious power of , that is, the ideal generated by , .

The associated Hilbert-Kunz multiplicity is defined to be

Monsky showed in his paper [1] that the limit exists and is a real constant. Further he showed that Several authors have investigated . They showed that is a rational number for certain special kind of binomial hypersurfaces [2], cubic curves and surfaces [3] and full flag varieties, and elliptic curves [4].

A binomial hypersurface is defined in the beginning of Section 4. In this paper, we are interested in giving an iterative closed form formula for the Hilbert-Kunz function for any binomial hypersurface in general. Our methods work for any positive characteristic. Our work generalizes the work of Conca [2]. In [2], Conca computes the Hilbert-Kunz function of monomial ideals and also of those Binomial hypersurfaces whose terms defining the hypersurface are relatively prime. In [2], Conca also proves that the Hilbert-Kunz multiplicity associated with these special Binomial Hypersurfaces is always rational. In this paper, we prove that the Hilbert-Kunz multiplicity associated with any binomial hypersurface over any field of arbitrary positive characteristic is rational. The rationality result appears in Eto’s work (Theorem 2.2, [5]), where the multiplicity is interpreted as a volume of a polytope. Here we give a different proof for the rationality result. Our work also generalizes the work [6] of Watanabe, where he deals only with normal toric varieties.

The organization of this work is more or less clear from the table of contents. But to be precise, until Section 4 begins, the matter of this work holds true for any hypersurface over any field of positive characteristic and need not have to be a binomial hypersurface! Also, the filtration introduced in Section 2.1 is effective for any general ideal of a polynomial ring where is a field of arbitrary prime characteristic , not just when the ideal is generated by a single polynomial. The iterative closed form formula for the Hilbert-Kunz function for binomial hypersurfaces appears in Theorem 26 in Section 4.2.5. After that, in Section 4.4, we discuss the example of the -dimensional case. In the example of the -dimensional case, I prove the well known fact that for any -dimensional binomial hypersurface, the associated Hilbert-Kunz multiplicity is a positive integer and I give a precise account of the integer. In Section 4.5, I give an expression for the Hilbert-Kunz multiplicity for a general binomial hypersurface (see (67)).

2. Stating the Problem

For the rest of this paper we will assume that where is a field of arbitrary prime characteristic and is an arbitrary ideal in . Let . Put where . We will assume that ; otherwise, . Consider

Note that the rings and are isomorphic, where . Hence the Hilbert-Kunz function is given by We are presently interested in the case when is a hypersurface, that is, when the ideal is generated by a single polynomial.

2.1. A Filtration for Computing the Length

Let denote the set of all -tuples such that for each , . Let us arrange the elements of the set in the lexicographic order; that is, we say that if in the vector difference , the leftmost nonzero entry is . It is easy to check that the set equipped with the lexicographic order is in fact a totally ordered set, and this set contains many elements. Let us denote these many elements of the set arranged in the order by .

Let . There are many elements in the set . For any , let us denote by the element of given by . The set is in one-to-one correspondence with the set via the map . Let us denote by the image of the element under this correspondence.

For any and any , let us denote by the product , that is, the product in up to the th variable.

Given any and any , define

Definition 1. The ideals : for any , let where .
Declare to be the polynomial ring itself.

Remark 2. From the definition of the lexicographic order on , it follows easily that for any , either or there exists such that , , and .

Notation 1. For any , let .

Lemma 3. (i) For any , ;
(ii) for all and for all .

Proof. (i) It follows from Remark 2 that . Hence ;
(ii) the proof of this follows from Remark 2.

Lemma 4. For any , .

Proof. As (where ), it is enough to show that for all and for all .
Case  1 (when). In this case, clearly for all . And is either equal to or equal to . Hence in either case, . Therefore, .
Case  2 (when ). In this case, there exists such that , , and .
Clearly then, for all . Also, is either equal to or equal to . Hence in either case, . For , . Hence for all . Therefore, .

Notation 2. Let be any -module. Let denote the length of with respect to .

Lemma 5. Let . Put . Then .

Proof. Fix . We will show that for each , . We claim that for any , .
Case  1 (when ). In this case, we have . Hence for any and for any .
Case  2 (when ). It follows from Lemma 3 that for all . Therefore for all . This implies that for all and for all .
It remains to show that for all . It follows from the proof of Lemma 3(i) that . Hence for any , is either a multiple of or a multiple of . Therefore for all .

Lemma 6. For all , .

Proof. It follows from Lemma 3 that for all . It also follows from the same lemma that for all and for all . Therefore the generators (where ) of the ideal belong to . Due to degree reasons, does not belong to . Therefore is the only generator of the ideal which does not belong to . Hence .

It follows from Lemmas 4, 5, and 6 above that the following chain is a filtration of ideals such that .

Therefore reducing modulo the ideal , it follows that the following chain is a filtration of ideals, which is having the property that each successive quotient either is zero or is generated by a single element. The total number of successive quotients of the above mentioned filtration which are generated by a single element equals the length . The following corollary to Lemma 6 above gives a precise account of which successive quotient of the above mentioned filtration is generated by a single element and which successive quotient is zero.

Corollary 7. For any , , and if and only if .

Proof. Fix any , it follows from Lemma 5 that . Therefore, reducing modulo the ideal , we have . Let be as in Lemma 5. It follows from Lemma 6 that generates . Therefore reducing modulo the ideal , it follows that the quotient is zero if and only if and is generated by the single element otherwise.

Hence for knowing that which successive quotient of the above mentioned filtration is generated by single element and which one is zero, the only condition that needs to be verified is the following:  or not for every ?

This is the key checking condition for computing the length .

3. A Procedure for Doing the Key Check

We are interested in the case when the ideal is generated by a single polynomial, say . In this section, we will first define a term order on the set of all monomials in the variables , and then with respect to , we will arrange the terms of the polynomial , and with the help of all this notation, we will describe a way of doing the key checking.

3.1. The Term Order

To order the terms of the polynomial , let us put an order (denote it by ) on the set of all monomials in the variables as follows.(i)Set .(ii)On the set of all monomials in the variables , is the degree lexicographic order with respect to the order defined on the variables . Say, the polynomial has many terms. Let us denote the most initial (with respect to ) term of by , the next most initial term of by , , and so on till the least initial term . Hence we have For example, when and , then , , , , and .

Remark 8. Note here that the terms , of are assumed to be containing scalar coefficients.

3.2. The Main Theorem

Given any term of the polynomial , define . Recall the set . Let be an arbitrary element of the set . Say the monomial chosen above from the set equals for some . The key checking condition for the monomial says that “ or not.” Let us denote by the ideal . We call the ideal of convergence corresponding to . Given any , we need to check whether or not . Theorem 10 below provides a way for doing this checking.

Definition 9. A monomial in the variables of the type where and are terms of the polynomial and is any nonnegative integer, is called a combination monomial in and if contains no negative powers of any of the underlying variables . In particular, the monomial itself is a combination monomial in and .

Theorem 10. Let . if and only if one of the following mutually exclusive conditions holds.(i)The term of divides the monomial .(ii)The term of does not divide the monomial , but there exists a polynomial whose terms are of the type where is a nonzero scalar, is a combination monomial in and such that , is a term of which divides , such that the product equals where is a nonzero scalar.

Proof. If condition (i) of the theorem holds, the proof follows easily from the construction of the ideal . If condition (ii) of the theorem holds, then it is easy to see that .
Conversely, suppose . Note that due to degree reasons, the monomial can never belong to the ideal (the proof of this follows easily from the construction of the ideal ). Therefore the fact that implies that there exists a polynomial (say ) such that the product equals where is a nonzero scalar in the ground field .
Since the product contains a term of the type (for some nonzero scalar ), therefore the polynomial contains finitely many terms of the form where is a nonzero scalar and is a term of such that divides . Say is the part of the polynomial which has all its terms of the form where is some nonzero scalar and is some term of which divides . If at least one of the ’s appearing in the expression is equal to , then we can conclude that the term divides and we are done.
If not, then look at the portion (say, ) of the polynomial which consists of sum of all terms of the type where is a nonzero scalar, is a combination monomial in and such that , and is a term of which divides . Clearly then contains the sum . If equals for some nonzero scalar , then condition (ii) of the theorem holds and with and we are done.
So assume that the product contains at least one term (say, ) which is a combination monomial in and not belonging to . Since the product equals for some nonzero scalar , therefore there exists a nonzero term in such that is not a term of the polynomial and for some nonzero scalar and some term of . Hence for some nonzero scalar and some term of which divides . But since is a combination monomial in and not belonging to , it follows from the equation that is a term of the polynomial , hence a contradiction.

4. The Case of Binomial Hypersurfaces

In this section, we will study the case where the polynomial contains only terms; that is, . The affine variety defined by the ideal where is called a binomial hypersurface.

4.1. The Main Theorem for Binomial Hypersurfaces

In this subsection, we will prove a theorem which will reduce the “key checking condition” mentioned above to checking of a combinatorial condition. The theorem which does this job is the following.

Theorem 11. if and only if either divides or there exists a positive integer for which contains no negative powers of any of the underlying variables and belongs to . (Note here that by the notation , we mean multiplied times and similarly for the notation .)

Proof. If divides , then it follows from Theorem 10 above that .
Suppose there exists a positive integer for which has no negative powers and belongs to . Let denote the least positive integer for which this happens. Then the polynomial given by where is a nonzero scalar in the ground field and () for each , is having the property that the product equals where is a nonzero scalar. Hence . We now need to prove the other way round.
Suppose and does not divide . Then by Theorem 10, there exists a polynomial whose terms are of the type where is a nonzero scalar, is a combination monomial in and not belonging to , and is a term of which divides , such that the product equals where is a nonzero scalar.
It is easy to see that for the case of a binomial hypersurface, any combination monomial in and is either equal to itself or it is of the type for some positive integer or it is of the type for some positive integer . It is also elementary to observe that any combination monomial in and of the type always belongs to . So any combination monomial not belonging to is either equal to itself or of the type for some positive integer . It hence follows that the terms of the polynomial are either of the type for some positive integer or of the type . But we have assumed that the term does not divide , so cannot be a term of the polynomial . Hence the terms of the polynomial are only of the type for some positive integer .
Let denote the largest positive integer for which is a term of the polynomial . Clearly then has no negative powers of any of the underlying variables . Moreover, since the product equals (where is a nonzero scalar), it follows that belongs to .

4.2. A Formula for the Hilbert-Kunz Function for Binomial Hypersurfaces

In this subsection, we will give a closed form iterative formula for computing the Hilbert-Kunz function for any binomial hypersurface in general, over any field of positive characteristic. The notation and terminology remain the same as in the previous part of this paper.

4.2.1. Notation

Recall from Section 3.1 the variables and the term order on the set of all monomials in these variables. Let be the polynomial defining the binomial hypersurface where and are defined as in Section 3.1.

For any , let Note that some of the ’s can be negative, some can be positive, and some can be . Without loss of generality, we can assume that .

Let be the integer such that and, the ordered set equals the set . For , let and . Similarly, let denote the number of ’s for which (note that can also be ). Let be the elements of the set . For , let and . Finally, let and let be the elements of the set . Let us denote by the ordered set and for , let . In other words, we can say that the ordered set is the same as the ordered set . We call ’s the negative difference variables, ’s the zero difference variables, and ’s the positive difference variables. For any , let and denote the minimum and the maximum powers, respectively, of the variable that appears in the expression of the polynomial . We can similarly define , , , and for all and for all . Observe that since each is a zero difference variable, we have for all .

For any , let and . For any and any positive integer , let and be defined by the equation .

4.2.2. The Number

Definition 12 (of ). Let be an arbitrary element of the set (this set has been defined earlier) such that [7] divides . It follows from Theorem 11 that for those monomials which are divisible by [7], we need to keep an account of the largest positive integer for which the monomial has no negative power for any of the variables . Let denote this largest positive integer.

4.2.3. An Explicit Account of

Corollary 15 below gives an explicit account of the number , for any monomial which is divisible by [7]. The number is important to us because of Theorem 11 which provides us with a way to count the required length .

Remark 13. Let be any monomial in which divides; then it is of the form where for any , for every , and for every .
Hence for any positive integer , is of the form where for any , for every , and for every .
It is now easy to see that for any , the power of any zero difference variable or any positive difference variable in is . But the power of any negative difference variable in may be negative for some values of .

The following lemma and its corollary give an explicit account of for any monomial in which divides. But for stating the lemma, we need some notation. Let and let be any positive integer. If , then let denote the least positive integer which is divisible by . Let be defined by the equation . Let And let For any real number , let denote the smallest positive integer ≥. For any , let

Lemma 14. For any , if is such that , then the maximum value of the positive integer for which is ≥ equals

Proof. Observe that since , we have if and only if . So we need to compute the greatest positive integer , that is, ≤.
Case  1 (when  ). In this case, we have . The least positive integer which is divisible by is . Therefore the least positive integer is . Therefore the greatest negative integer is . And hence the greatest positive integer is .
Since , it follows that . The smallest positive integer is . Therefore the greatest positive integer is . As increases, decreases, the lower bound being equal to . Therefore the greatest positive integer equals .
Hence the greatest positive integer is equal to
This is equivalent to saying that the maximum value of the positive integer for which is ≥ equals
Case  2 (when ). In this case, we have 2 subcases.
Subcase  2.1 (when ). Here we have and . Therefore and hence . Therefore the least nonnegative integer is . The rest of the proof is similar to the proof of Case  1.
Subcase  2.2 (when ). Here we have, and . Therefore . Hence the least positive integer is . The rest of the proof is similar to the proof of Case  1.

The following corollary is immediate.

Corollary 15. Let be a monomial in which divides and let be a positive integer. Let and be of the form as given in Remark 13. For , let be the maximum value of the positive integer for which is .
Then equals the minimum of all s as ranges from to .

Let the set of all monomials in which are divisible by . Let .

Remark 16. It follows from Corollary 15 above that .

4.2.4. The Length Count for a Fixed

Remark 17. It is easy to verify that if is a positive integer such that contains no negative powers and is a positive integer such that , then contains no negative powers. It is also easy to verify that if is a positive integer such that belongs to and is a positive integer such that , then belongs to . Hence for any , if and only if either divides or belongs to .

Remark 18. Recall from Remark 13 as well as from Corollary 15 that given any monomial in , the value of depends only on the powers of the negative difference variables appearing in . It now follows that for each , if we fix numbers such that , then for any in the set the value of is the same (call it ).

The main result of this subsubsection is Lemma 24 below which provides an account of the length count for a fixed value of . But for stating the lemma, we need to introduce some notation first.

For an arbitrary element in and for every , let The symbol in the above definition of stands for “minimum convergent.” The reason behind this notation is explained in Remark 19 below.

Remark 19. Observe that if the power of the variable in the monomial is , then the power of in the monomial is . If is divisible by , then . Now if , then belongs to . That is, if , then belongs to . That is, if , then belongs to . So is the minimum possible power of the variable in the monomial for which belongs to .

Recall the ordered set of positive difference variables. Let and . For any , let And for any , let For any in , let and . For any in and any , let Recall the set from Remark 18. For any in this set , the value of is the same, namely, . Lemma 24 below gives an account of the length count for the set . But we will break up the proof of Lemma 24 in a few short steps as follows.

Lemma 20. For every , let where for each , is a fixed integer such that and at least one is .
Then the total number of monomials in the set for which equals .

Proof. The proof is by reverse induction on , the base case of induction being . Observe that if is any monomial in such that the power of the variable in it satisfies the inequality , then divides and . Hence such a monomial satisfies . The total number of such monomials in is . And if is any monomial in such that the power of the variable in it satisfies the inequality , then neither nor divides and hence . The total number of such monomials in is . Hence the total number of monomials in the set for which equals This proves the base case of induction. The rest of the proof follows by reverse induction and is left to the reader.

Lemma 21. For every , let where for each , is a fixed integer such that and at least one is .
Then the total number of monomials in the set for which equals .

Proof. The proof is by reverse induction on , the base case of induction being . Observe that if is any monomial in such that the power of the variable in it satisfies the inequality , then neither nor divides and hence . The total number of such monomials in is . Hence the total number of monomials in the set for which equals . This proves the base case of induction. The rest of the proof follows by reverse induction and is left to the reader.

Lemma 22. For every , let where for each , is a fixed integer such that .
Then the total number of monomials in the set for which equals (i) if there exists at least one such that ;(ii) if for every , we have and ;(iii) if for every , we have and there exists at least one such that .

Proof. The proof is by reverse induction on , the base case of induction being . Let us first prove the base case of induction.
Case  1. When there exists at least one such that .
Then since we have that for each , is a fixed integer such that , it follows that monomials in the set are not divisible by .
Observe now that if is any monomial in such that the power of the variable in it satisfies the inequality , then divides and . Hence such a monomial satisfies . The total number of such monomials in is . If is any monomial in such that the power of the variable in it satisfies the inequality , then divides but does not divide and . Hence such a monomial satisfies . The total number of such monomials in is . And if is any monomial in such that the power of the variable in it satisfies the inequality , then neither nor divides and hence . The total number of such monomials in is . Hence the total number of monomials in the set for which equals This proves the base case of induction in Case  1.
Case  2. When for every , we have and . In this case, there are 3 subcases.
Subcase  2.1 (). Observe that if is any monomial in such that the power of the variable in it satisfies the inequality , then divides and . Hence such a monomial satisfies . The total number of such monomials in is . And if is any monomial in such that the power of the variable in it satisfies the inequality , then neither nor divides and hence . The total number of such monomials in is . Hence the total number of monomials in the set for which equals This proves the base case of induction in this subcase.
Subcase  2.2 ( and ). Observe that if is any monomial in such that the power of the variable in it satisfies the inequality , then divides and . Hence such a monomial satisfies . The total number of such monomials in is . If is any monomial in such that the power of the variable in it satisfies the inequality , then divides but does not divide (since ) and . Hence such a monomial satisfies . The total number of such monomials in is . And if is any monomial in such that the power of the variable in it satisfies the inequality , then neither nor divides and hence . The total number of such monomials in is . Hence the total number of monomials in the set for which equals This proves the base case of induction in this subcase.
Subcase  2.3 ( and ). Observe that if is any monomial in such that the power of the variable in it satisfies the inequality , then divides and . Hence such a monomial satisfies . The total number of such monomials in is . If is any monomial in such that the power of the variable in it satisfies the inequality , then divides (since and for every ). Hence such a monomial satisfies . The total number of such monomials in is . If is any monomial in such that the power of the variable in it satisfies the inequality , then divides but does not divide (since ) and (since ). Hence such a monomial satisfies . The total number of such monomials in is . And if is any monomial in such that the power of the variable in it satisfies the inequality , then neither nor divides and hence . The total number of such monomials in is . Hence the total number of monomials in the set for which equals This proves the base case of induction in this subcase. Hence the base case of induction is proved in Case  2.
Case  3. When for every , we have and there exists at least one such that .
Since there exists at least one such that , it follows that monomials in the set are not divisible by . The rest of the proof of this case is similar to the proof of Case  1. This proves the base case of induction in Case  3.
This proves the base case of induction. The rest of the proof follows by reverse induction and is left to the reader.

Let

Lemma 23. The total number of monomials in the set for which equals .

Proof. Let be an arbitrarily fixed integer such that . It then follows from Lemma 20 that for any such fixed , the total number of monomials in such that equals . Also if is an arbitrarily fixed integer such that , then it follows from Lemma 21 that for any such fixed , the total number of monomials in such that equals . Now there are 3 cases.
Case  1 (if ). In this case, the total number of monomials in the set for which equals
Case  2 (if and ). It follows from Lemma 22 that for any fixed integer such that , the total number of monomials in for which equals . Hence in this case, the total number of monomials in the set for which equals
Case  3 (if and ). It follows from Lemma 22 that for any fixed integer such that , the total number of monomials in for which equals . It again follows from Lemma 22 that for any fixed integer such that , the total number of monomials in for which equals . Hence in this case, the total number of monomials in the set for which equals

Lemma 24. The total number of monomials in the set for which equals where is defined inductively as follows:

Proof. Fix an arbitrary integer such that . Let It is now easy to see that the total number of monomials in for which equals (the proof of this is similar to the proof of Lemma 23).
Fix an arbitrary integer such that . Let Since , therefore any monomial is divisible neither by nor by . Hence the total number of monomials in for which equals .
So if we define as the subset of consisting of all those monomials in for which , then the total number of monomials in the set for which equals . Similarly, if we define as the subset of consisting of all those monomials in it for which , then the total number of monomials in the set for which equals . Proceeding inductively, it is now easy to see that the total number of monomials in the set for which equals .

Definition 25. Given any in , let where is defined in the same way as is defined in Lemma 24 (just replace in Lemma 24 by ).

4.2.5. The Formula for the Hilbert-Kunz Function

We will now state the formula for the Hilbert-Kunz function for the general binomial hypersurface that we have considered above. But for stating the formula, we need some more notation, which we provide first. Set and for every , let ; Set , and for every , let ; Set , and for every , let . Recall that for any , let and for any and any positive integer , and are defined by the equation . Also recall the definitions of and from Section 4.2.3.

For every , let Clearly for every .

For any nonnegative integer such that belongs to the set , let and let

Theorem 26. The formula for the Hilbert-Kunz function of the general binomial hypersurface under consideration is

Proof. Observe that for any fixed and any such that , the number provides an account of the number of monomials in which satisfy the following properties simultaneously.(i).(ii)The power of each of the variables is .(iii)The power of the variable is where and .
Therefore for any such that , denotes the number of monomials in for which and the power of the variable is where .
Clearly then the number of monomials in for which and the power of the variable is where equals . Similarly, the number of monomials in for which and the power of the variable is where and equals . We can similarly account for the terms , , and in the formula (45) of the Hilbert-Kunz function.

4.3. An Illustrative Example

For this example as well as for the example given in Section 4.4 below, we will need the following notation.

Given any real number , let denote the smallest integer and let denote the largest integer . We call and the upper and the lower integral parts of .

Let where is a field of arbitrary prime characteristic , and let be the ideal in defined by where . The differences , and are therefore given by , respectively. Hence the ’s are the negative difference variables, the ’s are the zero difference variables, and the ’s are the positive difference variables. Here , , , and . Let be as defined in Section 4.2.3 above. Here, and . For any , let and be defined by the equation(s) . That is, For every such that , we have Using the formulae given in Lemma 24, it follows that Since for any such that , we know that and the total number of zero difference variables here, therefore we have For getting the formula for the Hilbert-Kunz function in this case, we need to compute for the various values of under consideration. A straight forward computation tells us that for any such that , we have where Similarly, for any such that , we have Clearly then is a rational function in and for all values of . Therefore, the formula for the corresponding Hilbert-Kunz function is where Since all the quantities involved in the above formula for the Hilbert-Kunz function are rational functions in the variables and , therefore the Hilbert-Kunz function is a rational function in and .

4.4. Example of the One-Dimensional Case

In this subsection, we will discuss the case of -dimensional binomial hypersurfaces and will observe that using the above formula for the Hilbert-Kunz function, we do get that in this case of -dimension, the associated Hilbert-Kunz multiplicity is an integer (e.g., see Chapter , Corollary 6.2 of [7]). In this case, this integer happens to be equal to the ordinary multiplicity because the Hilbert-Kunz multiplicity is equal to the ordinary multiplicity for -dimensional rings.

In the case of -dimensional binomial hypersurfaces, we have , where , and the terms and of are monomials in the variables and .

There are possible cases:(1) where is a negative difference variable and is a positive difference variable;(2) where is a negative difference variable and is a zero difference variable;(3) where is a zero difference variable and is a positive difference variable. We will first discuss Cases  (2) and (3).

In Cases  (2) and (3), it is easy to check (without using the formula above) following. The Hilbert-Kunz function evaluated at and The associated Hilbert-Kunz multiplicity which is an integer.

We will now discuss Case  (1).

Since is a negative difference variable and is a positive difference variable, then using our notation in Section 4.2.1, we can denote by and by . Then writing is the same as writing . In this case, there are subcases.

Subcase  1.1 (when ). It is an exercise to check that using the above formula for the Hilbert-Kunz function, we get that for large enough.

The Hilbert-Kunz function evaluated at equals where for any real number , and denote the upper and lower integral parts of , respectively.

Hence the associated Hilbert-Kunz multiplicity equals which is nothing but , which is an integer.

Subcase  1.2 (when). It is an exercise to check that using the above formula for the Hilbert-Kunz function, we get that for large enough the Hilbert-Kunz function evaluated at equals Hence the associated Hilbert-Kunz multiplicity equals , which is an integer.

Subcase  1.3 (when ). Let be the smallest positive integer such that if , then . Similarly, let be the smallest positive integer such that if , then . Clearly then .

It is an exercise to check that using the above formula for the Hilbert-Kunz function, we get that for large enough, the Hilbert-Kunz function evaluated at equals where for any real number , denotes the upper integral part of .

Hence the associated Hilbert-Kunz multiplicity equals , which is an integer.

4.5. Rationality of the Hilbert-Kunz Multiplicity

The Hilbert-Kunz multiplicity for the general Binomial hypersurface (whose Hilbert-Kunz function is given by (45) above) is by definition equal to the coefficient of in the formula given by (45). We can easily see from the above discussion that for any such that , is obtained iteratively by first computing for all , then , and so on till for all . Moreover, for any nonnegative integer such that , we know that belongs to the set and

So let us first give a formula for . I will rather give a formula for for any in in general and then one can put in it to get the required formula for .

Let be arbitrary. To compute , we first need to compute . I will compute only in the most general case, that is, when In this case, we have In all other cases, it is a tedious job to compute because we will have to take care of various complicated possibilities like for some , we may have and for some other , we may have . This makes the computation of highly tedious and complicated. Let us now compute for the most general case.

For , let denote the sum of all possible terms which are products of -many items either of the type or of the type where such that the following two properties hold.(i)At least one of the items in each such product is of the type ;(ii)not all of the items in any such product are of the type . Similarly, for any , let denote the sum of all possible terms which are products of -many items either of the type or of the type where such that the following two properties hold.(i)At least one of the items in each such product is of the type ;(ii)not all of the items in any such product are of the type . For , let denote the sum of all possible terms which are products of the type .

Then

Hence the formula for () is given by Since , we get the formula for from (64) above by putting in it.

Let us now write down the formula (45) for the Hilbert-Kunz function, in a different way which will help us see how the formula for the Hilbert-Kunz multiplicity will look like. But for that, we first need some notation. For each , let and for any positive integer , let . For each , let denote the coefficient of in the formula (45) for the Hilbert-Kunz function. For any , let , , and . For any integer such that and any integer such that , let sum of all possible terms which are products of many s and many s where the s belong to the set and the s belong to the set .

Let be arbitrary. Then for any integer such that , let And for any integer such that , let us define inductively as follows.

Let   and for any , let .

Now if we expand the formula ((45) above) of the Hilbert-Kunz function, assuming that , we get that the expression in (45) equals where the formulae for can be obtained from (64). The Hilbert-Kunz multiplicity is by definition the coefficient of in the expression (66) above. Computing the coefficient of in the expression (66) above, we get that the Hilbert-Kunz multiplicity equals

If we expand the above formula (expression (67)) of the Hilbert-Kunz multiplicity, then inside the expansion, we will get terms which are of the form:(i)finite (by “finite”, I mean depending upon ) sums of the form where is some positive integer and and are upper or lower integral parts of some real numbers which are of the form “polynomials in with rational coefficients”;(ii)polynomials in with rational coefficients (including constant terms which are rational numbers);(iii)products of the above mentioned types of terms.

It hence follows from the above discussion that the Hilbert-Kunz multiplicity associated with any general binomial hypersurface is always rational.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.