The Hilbert-Kunz function for Binomial Hypersurfaces

In this article, I give an iterative closed form formula for the Hilbert-Kunz function for any binomial hypersurface in general, over any feild of arbitrary positive characteristic. I prove that the Hilbert-Kunz multiplicity associated to 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.


Introduction
Let ( , n) be a Noetherian local ring of dimension and of prime characteristic > 0. Let be an n-primary ideal and = where > 0 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 , ( ) = ( ) + ( −1 ) .
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 = [ 1 , . . . , ] where is a field of arbitrary prime characteristic > 0, 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 1-dimensional case. In the example of the 1-dimensional case, I prove the well known 2 Algebra fact that for any 1-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)).

Stating the Problem
For the rest of this paper we will assume that = [ 1 , . . . , ] where is a field of arbitrary prime characteristic > 0 and is an arbitrary ideal in . Let = / . Put m := m + where m = ( 1 , . . . , ). We will assume that ⊆ m; otherwise,m = . Consider ) .
We are presently interested in the case when is a hypersurface, that is, when the ideal is generated by a single polynomial.
It is easy to check that the set S equipped with the lexicographic order < lex is in fact a totally ordered set, and this set contains many elements. Let us denote these many elements of the set S arranged in the < lex order by There are many elements in the set M. For any U = ( 1 , 2 , . . . , ) ∈ S, let us denote by X −1−U the element of M given by The set S is in one-to-one correspondence with the set M via the map U → X −1−U . Let us denote by X −1−U the image of the element U under this correspondence.
(ii) the proof of this follows from Remark 2.
ideal +1 belong to . Due to degree reasons, X −1−U = B +1, ( +1) does not belong to . Therefore X −1−U is the only generator of the ideal +1 which does not belong to .
It follows from Lemmas 4, 5, and 6 above that the following chain is a filtration of ideals such that ( +1 / ) = 1. 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 ( /(m [ ] + )). 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. Proof. Fix any ∈ {0, . . . , − 1}, it follows from Lemma 5 that ( +1 / ) = 1. Therefore, reducing modulo the ideal , we have (( +1 + )/( + )) ≤ 1. Let +1 be as in Lemma 5. It follows from Lemma 6 that X −1−U generates +1 . Therefore reducing modulo the ideal , it follows that the quotient ( +1 + )/( + ) is zero if and only if X −1−U ∈ + and is generated by the single element 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: This is the key checking condition for computing the length ( /(m [ ] + )).

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 1 , . . . , , 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.

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 1 , . . . , as follows.
(ii) On the set of all monomials in the variables 1 , . . . , , ⊳ is the degree lexicographic order with respect to the order ⊳ defined on the variables 1 , . . . , .
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 [ − 1], . . ., and so on till the least initial term 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 ∈ M, we need to check whether or not ∈ + . Theorem 10 below provides a way for doing this checking.
Definition 9. A monomial in the variables 1 , . . . , of the type 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 [ ] −1 where is a nonzero scalar and [ ] is a term of such that [ ] divides . Say , then we can conclude that the term [1] divides and we are done.
If not, then look at the portion (say, 1 ) of the polynomial which consists of sum of all terms of the type [ ] −1 where is a nonzero scalar, is a combination monomial in and such that ∉ , and [ ] is a term of which divides . Clearly then 1 contains the sum 1 If ⋅ 1 equals + l for some nonzero scalar , then condition (ii) of the theorem holds and with ℎ = 1 and we are done.
So assume that the product ⋅ 1 contains at least one term (say, 1 ) 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 1 and [ ] = 1 for some nonzero scalar and some term [ ] of . Hence = 1 [ ] −1 for some nonzero scalar and some term [ ] of which divides 1 . But since 1 is a combination monomial in and not belonging to , it follows from the equation that is a term of the polynomial 1 , hence a contradiction.

The Case of Binomial Hypersurfaces
In this section, we will study the case where the polynomial contains only 2 terms; that is, = [2] + [1]. The affine variety defined by the ideal where = ( ) = ([2] + [1]) is called a binomial hypersurface.

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.  [1] belongs to . Let 0 denote the least positive integer for which this happens. Then the polynomial given by where 1 is a nonzero scalar in the ground field and + +1 = 0 (mod ) for each ∈ {1, . . . , 0 − 1}, 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 [1] does not divide . Then by Theorem 10, there exists a polynomial ℎ whose terms are of the type [ ] −1 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 ⋅ ℎ 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 [

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. − the power of in the term [2] . (12) Note that some of the Δ 's can be negative, some can be positive, and some can be 0. Without loss of generality, we can assume that Δ 1 ≤ ⋅ ⋅ ⋅ ≤ Δ . Let be the integer such that 0 ≤ ≤ and, the ordered In other words, we can say that the ordered set 1 ⊳ ⋅ ⋅ ⋅ ⊳ 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 ∈ {1, . . . , }, let ,min and ,max denote the minimum and the maximum powers, respectively, of the variable that appears in the expression of the polynomial . We can similarly define ,min , ,max , ,min , and ,max for all ∈ {1, . . . , } and for all ∈ {1, . . . , }. Observe that since each is a zero difference variable, we have ,min = ,max for all ∈ {1, . . . , }.

The Number max,
Definition 12 (of max, ). Let be an arbitrary element of the set M (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

4.2.
3. An Explicit Account of max, . Corollary 15 below gives an explicit account of the number max, , for any monomial which is divisible by [7]. The number max, is important to us because of Theorem 11 which provides us with a way to count the required length ( /(m [ ] + )).
It is now easy to see that for any ≥ 1, the power of any zero difference variable or any positive difference variable in The following lemma and its corollary give an explicit account of max, for any monomial in M which [2] divides. But for stating the lemma, we need some notation. Let ∈ {1, . . . , } and let be any positive integer. If ≥ , then let denote the least positive integer ≥ ( − , ) which is divisible by . Let be defined by the equation And let This is equivalent to saying that the maximum value of the positive integer for which the value of max, is the same (call it 0 max ).
The main result of this subsubsection is Lemma 24 below which provides an account of the length count for a fixed value of max, . But for stating the lemma, we need to introduce some notation first.
For an arbitrary element max in max [2] and for every ∈ {1, . . . , }, let The symbol Min in the above definition of Min , max stands for "minimum convergent. " The reason behind this notation is explained in Remark 19 below.
Proof. The proof is by reverse induction on , the base case of induction being = . Observe that if is any monomial in F , , 1 ,..., −1 such that the power of the variable in it satisfies the inequality 0 ≤ ≤ − 1, then neither where for each ∈ {1, . . . , − 1}, is a fixed integer such that , it follows that monomials in the set F , , 1 ,..., −1 are not divisible by [1].
Observe now that if is any monomial in F , , 1 ,.. such a monomial satisfies ∉ + . The total number of such monomials in F , , 1 ,..., −1 is −Min , 0 max − ,min . And if is any monomial in F , , 1 ,..., −1 such that the power of the variable in it satisfies the inequality 0 ≤ < ,min , then neither [2] nor [1] divides and hence ∉ + . The total number of such monomials in F , , 1 ,..., −1 is ,min . Hence the total number of monomials in the set F , , 1 ,..., −1 for which ∉ + equals This proves the base case of induction in Case 1.
This proves the base case of induction in this subcase.
This proves the base case of induction in this subcase. ). Hence such a monomial satisfies ∉ + . The total number of such monomials in F , , 1 ,..., −1 is ,max − ,min . And if is any monomial in F , , 1 ,..., −1 such that the power of the variable in it satisfies the inequality 0 ≤ < ,min , then neither [2] nor [1] divides and hence ∉ + . The total number of such monomials in F , , 1 ,..., −1 is ,min . Hence the total number of monomials in the set F , , 1 ,..., −1 for which ∉ + equals This proves the base case of induction in this subcase. Hence the base case of induction is proved in Case 2. Since there exists at least one ∈ {1, . . . , − 1} such that ,min ≤ < ,max , it follows that monomials in the set Algebra F , , 1 ,..., −1 are not divisible by [1]. 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 F 1 for which ∉ + equalsD 1, 0 max .
Proof. Let 1 be an arbitrarily fixed integer such that − Min It then follows from Lemma 20 that for any such fixed 1 , the total number of monomials in F 2, , 1 such that ∉ + equals 2 . Also if 1 is an arbitrarily fixed integer such that 0 ≤ 1 < 1,min , then it follows from Lemma 21 that for any such fixed 1 , the total number of monomials in F 2, , 1 such that ∉ + equals  , the total number of monomials in F 2, , 1 for which ∉ + equals̃2 , 0 max . It again follows from Lemma 22 that for any fixed integer 1 such that 1,min ≤ 1 < 1,max , the total number of monomials in F 2, , 1 for which ∉ + equals 2, 0 max . Hence in this case, the total number of monomials in the set F 1 for which ∉ + equals Lemma 24. The total number of monomials in the set P( 0 1 , 0 2 , . . . , 0 ) for which ∉ + equals S 0 max , where S 0 max , is defined inductively as follows: Proof. Fix an arbitrary integer 1 such that − 1 ≥ 1 ≥ 1,min . Let It is now easy to see that the total number of monomials in P 1 for which ∉ + equals̃1 , 0 max = S 0 max ,0 (the proof of this is similar to the proof of Lemma 23).

An Illustrative Example.
For this example as well as for the example given in Section 4.4 below, we will need the following notation.
For every ∈ Z such that 0 ≤ ≤̃2 ,last , we have Since for any ∈ Z such that 0 ≤ ≤̃2 ,last , we know that ∫ ( 1 ) := S 6, − 6 − , and = the total number of zero difference variables = 2 here, therefore we have For getting the formula for the Hilbert-Kunz function in this case, we need to compute ∫ ( 2 ) for the various values of under consideration. A straight forward computation tells us that for any ∈ Z such that 0 ≤ <̃1 ,last , we have Similarly, for any ∈ Z such that ≥̃1 ,last , we have  are given above.
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 .

Example of the One-Dimensional Case.
In this subsection, we will discuss the case of 1-dimensional binomial hypersurfaces and will observe that using the above formula for the Hilbert-Kunz function, we do get that in this case of 1-dimension, the associated Hilbert-Kunz multiplicity is an integer (e.g., see Chapter 6, 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 1-dimensional rings.
There are 3 possible cases: (1) 1 ⊳ 2 where 1 is a negative difference variable and 2 is a positive difference variable; (2) 1 ⊳ 2 where 1 is a negative difference variable and 2 is a zero difference variable; (3) 1 ⊳ 2 where 1 is a zero difference variable and 2 is a positive difference variable.
In Cases (2) and (3), it is easy to check (without using the formula above) following.
We will now discuss Case (1). Since 1 is a negative difference variable and 2 is a positive difference variable, then using our notation in Section 4.2.1, we can denote 1 by 1 and 2 by 1 . Then writing 1 ⊳ 2 is the same as writing 1 ⊳ 1 . In this case, there are 3 subcases. Subcase 1.1 (when 1 < Δ 1 ). 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