The limit as p -> infinity of the Hilbert-Kunz multiplicity of sum(x_i^(d_i))
aa r X i v : . [ m a t h . A C ] J u l The limit as p → ∞ of the Hilbert-Kunzmultiplicity of P x d i i Ira M. Gessel and Paul Monsky
Brandeis University, Waltham MA 02454-9110, [email protected] and [email protected]
Abstract
Let p be a prime. The Hilbert-Kunz multiplicity, µ , of the element P x d i i of( Z /p )[ x , . . . , x s ] depends on p in a complicated way. We calculate the limit of µ as p → ∞ . In particular when each d i is 2 we show that the limit is 1 + the coefficientof z s − in the power series expansion of sec z + tan z . Suppose s ≥ d , . . . , d s are positive integers, and h is the element P x d i i of A = ( Z /p )[ x , . . . , x s ]. Let e n ( h ) be the colength of the ideal generated by h and the x qi where q = p n . For fixed p , Hilbert-Kunz theory tells us that e n = µq s − + O ( q s − ) for some µ > µ is the Hilbert-Kunz multiplicity of h .When s = 2, µ = min( d , d ) and so is independent of p , but the dependenceon p is subtle when s ≥ µ (and in fact all of the e n ) when s = 3. Thisresult was extended to s > µ → a limit as p → ∞ , and a formulafor the limit. The formula has been discovered by others since, but perhapsbecause their arguments were more complicated, they haven’t presented themfor publication.Much of the interest of the above result lies in an elegant expression for thelimit when each d i is 2; the limit is 1 + the coefficient of z in the power seriesexpansion of sec z + tan z . This was conjectured by the second author, and thefirst used Eulerian polynomials to provide a proof. At the request of severalcolleagues we’re here writing down our old proofs. The limits of Hilbert-Kunzmultiplicities have been studied in other situations; see Trivedi [4].1e begin with some easy results. When any d i is 1, e n = q s − irrespective of p , so the limit of µ is 1. Assume from now on that each d i >
1. The followingfunction was studied in [3].
Definition 1.1 D p ( k , . . . , k s ) = length A/ ( P x i , x k i , . . . , x k s s ) . Evidently, D p ( k , . . . , k s ) is the length of( Z /p )[ x , . . . , x s ] / (cid:16) ( x + · · · + x s ) k , x k , . . . , x k s s (cid:17) . Since ( Z /p )[ x , . . . , x s ] is a free module of rank p s − over ( Z /p )[ x p , . . . , x ps ], itfollows that D p ( pk , . . . , pk s ) = p s − D p ( k , . . . , k s ). Lemma 1.2
Suppose d i u i ≤ p ≤ d i v i . Then for n > , each of the e n ( h ) /q s − lies between dp − s D p ( u , . . . , u s ) and dp − s D p ( v , . . . , v s ) , where d is the prod-uct of the d i . Proof
For n > e n is bounded below by the colength of the ideal generatedby h and the x qdiuip i . Since A is free of rank d over ( Z /p )[ x d , . . . , x d s s ] thiscolength is dD p (cid:16) qu p , · · · , qu s p (cid:17) = (cid:16) qp (cid:17) s − dD p ( u , . . . , u s ). Dividing by q s − , weget the lower bound, and the upper bound is derived similarly. ✷ Now the A -module (cid:16)P x i , x k , . . . , x k s s (cid:17) / (cid:16)P x i , x k +11 , x k , . . . , x k s s (cid:17) is annihi-lated by x , and so may be viewed as a ( Z /p )[ x , . . . , x s ]-module. As such, itis cyclic, generated by x k and annihilated by x + · · · + x s and by each of x k , . . . , x k s s . So its length is at most D p ( k , . . . , k s ) . Lemma 1.3
If each of u , . . . , u s is < p then D p ( u + 1 , u + 1 , . . . , u s + 1) − D p ( u , . . . , u s ) ≤ sp s − . Proof
The argument preceding the lemma shows that D p ( u +1 , u , . . . , u s ) − D p ( u , u , . . . , u s ) ≤ D p ( u , . . . , u s ). Since each u i is < p , this is ≤ p s − . Com-bining this with s − ✷ Theorem 1.4 e ( h ) p s − and µ differ by at most dsp . Consequently, lim p →∞ ( µ ) =lim p →∞ (cid:16) e ( h ) p s − (cid:17) , provided the latter limit exists. Proof
Set u i = ⌊ pd i ⌋ . Since d i >
1, each u i < p . If we let v i be u i + 1, thenLemmas 1.2 and 1.3 show that all the e n ( h ) /q s − , n >
0, lie in an interval oflength ≤ ( sp s − )( dp − s ) = dsp . But µ is in the closure of this interval. ✷ Theorem 1.5
Suppose for each p we are given integers a , . . . , a s ≥ with a i = pd i + O (1) . Then there is a k such that for all p the difference between µ and dp − s D p ( a , . . . , a s ) is at most kp . roof Fix N large and let u i = a i − N , v i = a i + N . Then d i u i ≤ p ≤ d i v i .And when p is large, u i ≥ v i ≤ p . Now dp − s D p ( a , . . . , a s ) and each e n /q s − , n >
0, lie between dp − s D p ( u , . . . , u s ) and dp − s D p ( v , . . . , v s ). Theargument of Theorem 1.4 shows they lie in an interval of length ≤ N (cid:16) dsp (cid:17) .The closure of this interval contains µ . ✷ When a , . . . , a s are ≤ p , Theorem 2.20 of [3] gives the following formula for D p ( a , . . . , a s ). Let γ be ⌊ P ( a i − ⌋ . Then D p ( a , . . . , a s ) is the sum as λ runs over Z of the coefficients of the t γ + λp in the polynomial Π (cid:16) − t ai − t (cid:17) . In thenext section we’ll combine this result with Theorem 1.5 to calculate the limitof µ as p → ∞ . For λ in Z , C λ = P ( ǫ · · · ǫ s ) (cid:16) ǫ d + · · · + ǫ s d s − λ (cid:17) s − , wherethe sum extends over the s -tuples ǫ , . . . , ǫ s with each ǫ i in {− , } and ǫ d + · · · + ǫ s d s > λ . For each p choose integers a , . . . , a s ≥ P a i ≡ s (2), and a i = pd i + O (1). Suppose ǫ , . . . , ǫ s are in {− , } and λ is in Z . Let a in Q be ǫ d + · · · + ǫ s d s − λ ; this is independent of p . Let α be (cid:16) P ( ǫ i a i − (cid:17) − pλ .Since P a i ≡ s (2), α is in Z . Evidently α = pa + O (1).We fix ǫ · · · ǫ s and λ , and study how the coefficient of t α in (1 − t ) − s dependson p . When a < α < p and the coefficient is 0. When a = 0, α is O (1) and the coefficient is O (1); since s ≥ O ( p s − ). Now suppose a >
0. Then for large p , α > (cid:16) α + s − s − (cid:17) . Since α + s − pa + O (1), we get s − (cid:16) pa (cid:17) s − + O ( p s − ) = − s ( s − a s − p s − + O ( p s − ). Lemma 2.2
Let γ = P ( a i − . Then the coefficient of t γ in Π (cid:16) − t ai − t (cid:17) is − s ( s − C p s − + O ( p s − ) with C as in Definition 2.1. Proof Π (cid:16) − t ai − t (cid:17) = (1 − t ) − s Π(1 − t a i ). Multiplying the second product out weexpress our coefficient in terms of coefficients of (1 − t ) − s . Explicitly it is (thecoefficient of t γ in (1 − t ) − s ) − (the sum of the coefficients of the t γ − a i )+(the sumof the coefficients of the t γ − a i − a j ) − · · · . The paragraph before the lemma, with λ = 0, tells us the behavior of each term as p → ∞ and gives the result. ✷ More generally:
Lemma 2.3
The coefficient of t γ − λp in Π (cid:16) − t ai − t (cid:17) is − s ( s − C λ p s − + O ( p s − ) . urthermore C λ = C − λ . Proof
The argument of Lemma 2.2 gives the first result. Since the coefficientsof t γ + N and t γ − N in Π (cid:16) − t ai − t (cid:17) are equal, the second result follows. ✷ Note that when λ ≥ s , ǫ d + · · · + ǫ s d s ≤ s ≤ λ and so C λ = 0. By Lemma 2.3, C λ = 0 when | λ | ≥ s . Theorem 2.4 As p → ∞ , µ → d (2 − s )( s − ( P C λ ) = d (2 − s )( s − ( C + 2 P λ> C λ ) . Proof
Since C λ = C − λ , the sums are equal. By Theorem 1.5 it sufficesto show that p − s D p ( a , . . . , a s ) → − s ( s − P C λ as p → ∞ . But this followsimmediately from Lemmas 2.2, 2.3 and the result from [3] quoted at the endof the Introduction. ✷ Example 2.5
Supppose s = 4 and each d i is . Then C = (cid:16) + + + (cid:17) − (cid:16) + + − (cid:17) = , while C λ = 0 for λ = 0 . So µ → · · = . In fact, µ = (cid:16) p +2 p +32 p +2 p +1 (cid:17) if p ≡ and (cid:16) p − p +32 p − p +1 (cid:17) if p ≡ . From now on we assume each d i is 2. Definition 2.6 If a is an integer, f s ( a ) = a s − − (cid:16) s (cid:17) ( a − s − + (cid:16) s (cid:17) ( a − s − − · · · , where we make the convention that c s − = 0 when c < . Theorem 2.7 As p → ∞ , µ → s − · s − ( f s ( s ) + 2 f s ( s −
4) + 2 f s ( s −
8) + 2 f s ( s −
12) + · · · ) . This may also be written as s − · s − · P f s ( a ) , the sum extending over all a ≡ s (4) . Proof C = (cid:16) s (cid:17) s − − (cid:16) s (cid:17) (cid:16) s − (cid:17) s − + (cid:16) s (cid:17) (cid:16) s − (cid:17) s − − · · · = s − · f s ( s ); similarly C λ = s − · f s ( s − λ ). Now apply Theorem 2.4, noting that d · − s ( s − = s − . ✷ Example 2.8
Supppose s = 5 . f (5) = 5 − · + 10 · = 230 , while f (1) = 1 = 1 . So by Theorem 2.7, µ → · (232) = . In fact, if p > , µ = p +1524 p +12 . Proceeding as in Example 2.8, the second author calculated the limit of µ foreach s ≤
10, finding that in each case the limit was 1+ the coefficient of z s − in the power series expansion of sec z + tan z . In the next section we’ll useEulerian polynomials to show that this holds for all s ; this insight is due tothe first author. 4 The case h = P x i Definition 3.1
For n ≥ , A n = (1 − T ) n +1 (1 + 2 n T + 3 n T + · · · ) . For example, A = 1 + 11 T + 11 T + T . Lemma 3.2 A n is a polynomial, and A n (1) = n ! . Proof
Let ∆ be the operator f → f ( T + 1) − f ( T ) on Z [ T ]. The n -folditerate of ∆ evidently takes T n to the constant n !. It follows that (1 − T ) n (1 +2 n T + 3 n T + · · · ) = (a polynomial in T ) + n !(1 − T ) . Multiplying by 1 − T andevaluating at T = 1 we get the result. ✷ Euler [1], [2] evaluated these Eulerian polynomials at −
1. The values at i areless familiar but we’ll show how to derive them by an easy method. As diver-gent series are out of fashion, we’ll proceed formally. Let O be the completelocal ring C [[ T, z ]]. If u is in the maximal ideal of O , e u will denote the element P n ≥ u n n ! of O . Lemma 3.3 In O , (cid:16)P n ≥ A n ( T )(1 − T ) n z n n ! (cid:17) (1 − T e z ) = e z − . Proof P n ≥ A n ( T )(1 − T ) n +1 z n n ! = P n ≥ z n n ! (1 n + 2 n T + 3 n T + · · · ) = e z + T e z + T e z + · · · . Multiplying by (1 − T e z )(1 − T ) we find that (cid:16)P n ≥ A n ( T )(1 − T ) n z n n ! (cid:17) (1 − T e z ) = e z (1 − T ). Subtracting off the n = 0 term we get e z (1 − T ) − (1 − T e z ) =e z − ✷ Lemma 3.4 In C [[ z ]] , P n ≥ A n (i)(1+i) n · z n n ! = − e i z e i z − i . Proof
There is a continuous ring automorphism of O taking T to T and z to z (1 − T ). Applying this to Lemma 3.3 we find that (cid:16)P n ≥ A n ( T ) z n n ! (cid:17) (1 − T e z (1 − T ) ) = e z (1 − T ) −
1. Now this is an identity in C [ T ][[ z ]]. Applying thecontinuous ring homomorphism C [ T ][[ z ]] → C [[ z ]] that takes T to i and z to z we get the result. ✷ Theorem 3.5 If s ≥ , A s − (i)(1+i) s − · s − is the coefficient of z s − in the powerseries expansion of sec z + tan z . Proof
Multiplying both sides of Lemma 3.4 by 1+i and adding 1 we find thatin C [[ z ]], 1 + P n ≥ A n (i)(1+i) n − z n n ! = − ie i z e i z − i . If by sin z , cos z , sec z , tan z we meanthe Taylor series expansions of these functions, then − ie i z e i z − = (1+sin z ) − i cos z cos z − i(1 − sin z ) .Since z cos z and cos z − sin z are each sec z + tan z , 1 + P ∞ A n (i)(1+i) n − z n n ! = sec z + tan z in C [[ z ]], and we compare the coefficients of z s − . ✷ emma 3.6 P a f s ( a ) T a − = (1 + T ) s A s − , where f s ( a ) is as in Definition2.6. Proof f s ( a ) = a s − − (cid:16) s (cid:17) ( a − s − + (cid:16) s (cid:17) ( a − s − − · · · . So P f s ( a ) T a − =(1 s − +2 s − T +3 s − T + · · · ) − (cid:16) s (cid:17) (1 s − T +2 s − T +3 s − T + · · · )+ (cid:16) s (cid:17) (1 s − T +2 s − T + 3 s − T + · · · ) − · · · = (1 s − + 2 s − T + 3 s − T + · · · )(1 − T ) s =(1 + T ) s A s − . ✷ Theorem 3.7
Let c = A s − (i)(1+i) s − . Then P f s ( a ) , the sum extending over all a ≡ s (4) , is s − (cid:16) A s − (1) + c + ¯ c (cid:17) . Proof
Let P = P j f s ( j − s ) T j . Since 4 divides j if and only if j − s ≡ s (4),our sum is ( P (1) + P ( −
1) + P (i) + P ( − i)). Now P = P f s ( j ) T j +3 s which is T s +1 (1+ T ) s A s − by Lemma 3.6. Thus P (1) = 2 s − A s − (1) and P ( −
1) = 0.Furthermore, P (i) = ( − i) s − (1 + i) s A s − (i). Since ( − i) s − (1 + i) s (1 + i) s − =( − i) s − (2i) s − = 2 s − , P (i) = · s − · A s − (i)(1+i) s − = 2 s − · (cid:16) c (cid:17) . Conjugating wefind that P ( − i) = 2 s − · (cid:16) ¯ c (cid:17) . ✷ Theorem 3.8
Suppose h = P s x i . Then as p → ∞ , the Hilbert-Kunz multi-plicity of h → the coefficient of z s − in sec z + tan z . Proof
Theorems 2.7 and 3.7 show that µ → s − (cid:16) A s − (1) + c + ¯ c (cid:17) . But A s − (1) = ( s − c ( s − and ¯ c ( s − are each thecoefficient of z s − in sec z + tan z . ✷ References [1] L. Euler, Remarques sur un beau rapport entre les s´eries des puissances tantdirectes que r´eciproques, Acad´emie des sciences de Berlin, Lu en 1749, OperaOmnia Serie I, Bd. 15, 70–90.[2] L. Euler, Institutiones calculi differentialis. . . , volume 2, chapter 7.[3] C. Han, P. Monsky, Some surprising Hilbert-Kunz functions, Math. Z. 214(1993), 119–135.[4] V. Trivedi, Hilbert-Kunz multiplicity and reduction mod p , Nagoya Math. J.185 (2007), 123–141., Nagoya Math. J.185 (2007), 123–141.