Algebraicity of some Hilbert-Kunz multiplicities (modulo a conjecture)
aa r X i v : . [ m a t h . A C ] J u l Algebraicity of some Hilbert-Kunzmultiplicities (modulo a conjecture)
Paul Monsky
Brandeis University, Waltham MA 02454-9110, USA. [email protected]
Abstract
Let F be a finite field of characteristic 2 and h be the element x + y + xyz of F [[ x, y, z ]]. In an earlier paper we made a precise conjecture as to the values of thecolengths of the ideals ( x q , y q , z q , h j ) for q a power of 2. We also showed that if theconjecture holds then the Hilbert-Kunz series of H = uv + h is algebraic (of degree2) over Q ( w ), and that µ ( h ) is algebraic (explicitly, + √ ). In this note, assumingthe same conjecture, we use a theory of infinite matrices to rederive this result, andwe extend it to a wider class of H ; for example H = g ( u, v ) + h . In a follow-uppaper, under the same hypothesis, we will show that transcendental Hilbert-Kunzmultiplicities exist. X In this section we develop some general results about Hilbert-Kunz seriesand multiplicities for characteristic 2 power series. (There are similar results,implicit in [5], in all finite characteristics but they are harder to prove.)
Definition 1.1 X is the vector space of functions I → Q where I = [0 , ∩ Z [ ] . If f = 0 is in the maximal ideal of F [[ u , . . . , u r ]] , char F = 2 , then φ f in X is the function iq → q − r deg( u q , . . . , u qr , f i ) ; here q denotes a power of 2and deg is colength in F [[ u , . . . , u r ]] . Note that φ f is well-defined. Definition 1.2 α in X is convex if for all i and q with < i < q , α (cid:16) iq (cid:17) ≥ α (cid:16) i − q (cid:17) + α (cid:16) i +1 q (cid:17) . Note that φ f (0) = 0, φ f (1) = 1, φ f is convex and φ f is Lipschitz. The first twoassertions are clear. If we set J = ( u q , . . . , u qr ) then multiplication by f inducesa map of ( J, f i − ) / ( J, f i ) onto ( J, f i ) / ( J, f i +1 ), yielding convexity. Finally, asLipschitz constant we can take the Hilbert-Kunz multiplicity, µ , of f .1 efinition 1.3 Suppose that α in X is convex Lipschitz with α (0) = 0 and α (1) = 1 . Then µ ( α ) = lim n →∞ α (2 − n ) · n , while S α is the element P α (2 − n )(2 w ) n of Q [[ w ]] . (The convexity of α shows that n → n α (2 − n ) isnon-decreasing. Since α is Lipschitz, the function is bounded and the limitexists.) Remarks
When α = φ f , µ ( α ) and S α (2 r − w ) are just the Hilbert-Kunz mul-tiplicity and Hilbert-Kunz series of f . Note that if α is as in Definition 1.3then µ ( α ) = lim w → − (1 − w ) S α ( w ) . For convexity shows that the co-efficientsof the power series (1 − w ) S α ( w ) are ≥ . So the limit is the value of this powerseries at 1. And we note that α (1)+ (cid:16) α (cid:16) (cid:17) − α (1) (cid:17) + (cid:16) α (cid:16) (cid:17) − α (cid:16) (cid:17)(cid:17) + · · · converges to µ ( α ) . We next define a bilinear product X × X → X and show that if f = 0and g = 0 are in the maximal ideals of F [[ u , . . . , u r ]] and F [[ v , . . . , v s ]], then φ f φ g = φ h , where h is the element f ( u ) + g ( v ) of F [[ u , . . . , u r , v , . . . , v s ]].(There is a similar construction, implicit in [5], in any finite characteristic.) Definition 1.4
Suppose α and β are in X . We define α β ( t ) by inductionon the denominator of t in I , according to the following procedure:Let α and α be the elements t → α (cid:16) t (cid:17) and t → α (cid:16) t (cid:17) of X ; define β and β similarly. Then:(1) α β (0) = 0 α β (1) = ( α (1) − α (0))( β (1) − β (0)) (2) If ≤ t ≤ α β ( t ) = α β (2 t ) + α β (2 t ) (3) If ≤ t ≤ α β ( t ) = α β (1) + α β (1) + α β (2 t −
1) + α β (2 t − t = 0, or 1 the two definitions of α β ( t ) given by theabove scheme coincide, so that α β is a well-defined element of X . α is constant then α β = 0, while if α is the identity function t , α β = ( β (1) − β (0)) t . In particular, t t = t .Now let T and T X → X be the maps taking α to t → α (cid:16) t (cid:17) and t → α (cid:16) t (cid:17) .Replacing t by t in (2) above and by t in (3) above gives: Theorem 1.5 If γ = α β then: T ( γ ) = ( T ( α ) T ( β )) + ( T ( α ) T ( β )) T ( γ ) = γ (cid:16) (cid:17) + ( T ( α ) T ( β )) + ( T ( α ) T ( β ))We now recall some notation used in both [1] and [5]. By an F [ T ]-module we’llmean a finitely generated F [ T ]-module annihilated by a power of T . Γ is the2rothendieck group of the set of isomorphism classes of such modules. Thereis a multiplication on Γ making it into a commutative ring; if V and W are F [ T ]-modules, a representative of their product is V ⊗ F W , with T acting by( T V ⊗ id) + (id ⊗ T W ). There is a Z -basis λ , λ , . . . of Γ with the followingproperty. If V is an F [ T ]-module then the class of V in Γ is P c i λ i where c i = ( − i dim( T i V /T i +1 V ). Because char F = 2, the multiplicative structureof Γ is very simple; λ i λ j = λ k where k is the “Nim-sum” of i and j . Definition 1.6 If α is in X , n ≥ and q = 2 n , then L n ( α ) is the element P q − (cid:16) α (cid:16) i +1 q (cid:17) − α (cid:16) iq (cid:17)(cid:17) ( − ) i λ i of Γ Q = Γ ⊗ Z Q . Note that L ( α ) = ( α (1) − α (0)) λ . If α = φ f and V is the F [ T ]-module F [[ u q , . . . , u qr ]] with T acting by multiplication by f , then q r L n ( α ) = P dim( T i V /T i +1 V )( − ) i λ i ; this is precisely the class of V in Γ.Suppose now that q = 2 n and 0 ≤ i < q . Since the Nim-sum of i and q is q + i , λ i λ q = λ q + i giving: Lemma 1.7 L n +1 ( α ) = L n ( α ) + λ q L n ( α ) Theorem 1.8 If γ = α β , L n ( γ ) = L n ( α ) · L n ( β ) . Proof
We argue by induction on n . Since γ (1) − γ (0) = ( α (1) − α (0))( β (1) − β (0)) the result holds for n = 0. Suppose that it’s true for a given n . Lemma1.7, Theorem 1.5 and the induction hypothesis show that L n +1 ( γ ) = L n ( α ) L n ( β ) + L n ( α ) L n ( β ) + λ q ( L n ( α ) L n ( β ) + L n ( α ) L n ( β )). But thisis ( L n ( α ) + λ q L n ( α )) · ( L n ( β ) + λ q L n ( β )) which is L n +1 ( α ) · L n +1 ( β ) byLemma 1.7. ✷ Theorem 1.9
Suppose h = f ( u , . . . , u r ) + g ( v , . . . , v s ) . Then:(1) For each n , L n ( φ h ) = L n ( φ f ) · L n ( φ g ) (2) φ h = φ f φ g Proof
With q = 2 n , let V be as in the paragraph following Definition 1.6.As we’ve seen V represents the element q r L n ( φ f ) of Γ. Replacing f by g weget a W representing the element q s L n ( φ g ) of Γ. Then q r + s L n ( φ f ) · L n ( φ g ) isrepresented by F [[ u , . . . , u r , v , . . . , v s ]] / ( u q , . . . , v qs ) with T acting by multi-plication by f ( u , . . . , u r ) + g ( v , . . . , v s ) = h . Since this F [ T ]-module repre-sents q r + s L n ( φ h ) we get (1). Suppose now that φ h ( t ) = φ f φ g ( t ) for some t = i n . Choose such a t with i as small as possible. Then i = 0, and theco-efficients of λ i − in L n ( φ h ) and L n ( φ f φ g ) differ. Theorem 1.8 then showsthat L n ( φ h ) = L n ( φ f ) L n ( φ g ), contradicting (1). ✷ heorem 1.10 If α and β are Lipschitz with Lipschitz constant m , then γ = α β is Lipschitz with Lipschitz constant m . Proof
We show that if 0 ≤ j < q then (cid:12)(cid:12)(cid:12) γ (cid:16) j +12 q (cid:17) − γ (cid:16) j q (cid:17)(cid:12)(cid:12)(cid:12) ≤ m q , arguingby induction on q . Note first that α , α , β and β are all Lipschitz withLipschitz constant m . We claim that when j < q the values of α β (andof α β ) at j +1 q and jq differ by at most m q . (When q = 1, j = 0, and thisis clear. When q > α and β (and α and β ) haveLipschitz constant m , together with the induction hypothesis.) Theorem 1.5then shows that γ (cid:16) j +12 q (cid:17) and γ (cid:16) j q (cid:17) differ by at most m q + m q = m q . Theargument is similar when j ≥ q , but now we make use of the values of α β (and of α β ) at j +1 − qq and j − qq . ✷ Lemma 1.11
Let δ r , r ≥ , be the class of F [ T ] /T r in Γ ; note that δ r = λ − λ + λ · · · + ( − ) r − λ r − . Then for α in X the following are equivalent:(1) α is convex.(2) For each n , L n ( α ) = P q − c i ( − ) i λ i with c ≥ c ≥ · · · ≥ c q − .(3) For each n , L n ( α ) is a linear combination of δ , . . . , δ q with the co-efficients of δ , . . . , δ q − ≥ . Proof
Since the c i in (2) is α (cid:16) i +1 q (cid:17) − α (cid:16) iq (cid:17) , (1) and (2) are equivalent.Suppose (2) holds. If we set c q = 0, then the formula for δ r given above showsthat L n ( α ) = P q − ( c i − c i +1 ) δ i . Since c − c , · · · , c q − − c q − are all ≥ ✷ Lemma 1.12
Suppose ≤ r, s ≤ q . Then, in Γ , δ r δ s is a linear combinationof δ , . . . , δ q with non-negative integer co-efficients. Furthermore δ r δ q = rδ q . Proof
Let V and W be the F [ T ]-modules F [ T ] /T r and F [ T ] /T s representing δ r and δ s . Writing V ⊗ F W (with T acting by T V ⊗ id + id ⊗ T W ) as a directsum of cyclic F [ T ]-modules we get the first assertion. The second is an easycalculation. ✷ Theorem 1.13 If α and β in X are convex, then so is α β . Proof
By Lemma 1.11, L n ( α ) and L n ( β ) are each linear combinations of δ , . . . , δ q with the co-efficients of δ , . . . , δ q − ≥
0. By Lemma 1.12 the sameis true of L n ( α ) · L n ( β ). Theorem 1.8 and Lemma 1.11 then show that α β is convex. ✷ Theorem 1.14
Suppose that α in X is convex Lipschitz with α (0) = 0 and α (1) = 1 . Suppose further that S α = P α (2 − n )(2 w ) n lies in a finite extension, L , of Q ( w ) . (We extend the imbedding of Q [ w ] in Q [[ w ]] to their fields offractions.) Then µ ( α ) is algebraic over Q of degree ≤ [ L : Q ( w )] . In fact there s a valuation ring containing Q [ w ] in L whose maximal ideal contains w − and whose residue class field contains a copy of Q ( µ ( α )) . Proof
Take H irreducible in Q [ W, T ] so that H ( w, (1 − w ) S α ) = 0. Then forany z in the open unit disc, H ( z, (1 − z ) S α ( z )) = 0. The remarks following Def-inition 1.3 show that H (1 , µ ( α )) = 0. Since H (1 , T ) = 0, µ ( α ) is algebraic over Q . Let g be Irr( µ ( α ) , Q ). Then ( W − , g ( T )) is a maximal ideal in Q [ W, T ] /H and we take a valuation ring in L that contains Q [ w, (1 − w ) S α ] = Q [ W, T ] /H ,and whose maximal ideal contracts to the above maximal ideal. ✷ Let f be the element x + y + xyz of Z/ x, y, z ], defining a nodal cubic.The values of φ f at q are known, and in particular, µ ( f ) = . In [2] weconjectured a precise value for all φ f (cid:16) iq (cid:17) , and showed that the conjectureimplied that µ ( uv + f ) is + √ . In this section we’ll rework this result usinginfinite matrix techniques from [3]; this approach will give rise to more generaltheorems. Definition 2.1 , t and ǫ will denote the elements t → , t → t and t → t − t of X . Definition 2.2
For m = 0 , , , . . . and t in I , φ m ( t ) is defined by inductionon the denominator of t as follows:(1) φ m (0) = φ m (1) = 0 (2) If ≤ t ≤ , φ m ( t ) = φ m +1 (2 t ) + (8 m + 6) t for m even, and φ m − (2 t ) + ǫ (2 t ) + (8 m + 6) t for m odd.(3) If ≤ t ≤ , φ ( t ) = φ (2 t −
1) + 6(1 − t ) (4) If ≤ t ≤ , φ m ( t ) = φ m − (2 t −
1) + ǫ (2 t −
1) + (8 m + 6)(1 − t ) for m = 0 even, and φ m +1 (2 t −
1) + (8 m + 6)(1 − t ) for m odd. When t = 0, or 1, the two definitions of φ m ( t ) given by the above schemeevidently coincide. So the φ m are well-defined elements of X . Replacing t by t in (2) and by t in (3) and (4) we get the “magnification rules”:(1) 8 T ( φ ) = φ + 3 t T ( φ ) = φ + 3(1 − t )(2) When m = 0 is even,8 T ( φ m ) = φ m +1 + (4 m + 3) t T ( φ m ) = φ m − + ǫ +(4 m +3)(1 − t )(3) When m is odd,8 T ( φ m ) = φ m − + ǫ + (4 m + 3) t T ( φ m ) = φ m +1 + (4 m + 3)(1 − t )Note also that 4 T ( ǫ ) = ǫ + t and that 4 T ( ǫ ) = ǫ + (1 − t ).5 onjecture 2.3 If f = x + y + xyz , then φ f = t + φ with φ as above. In [2] we presented evidence for a conjecture easily seen to be equivalent tothis. We noted in particular that both sides agree at all q and at each i . Theorem 2.4 If E = ǫ φ then lim n →∞ E (2 − n )2 n = + √ . Suppose now that Conjecture 2.3 holds. Then t + E = ( t + ǫ ) t + φ ) = φ uv φ f = φ uv + f . So Theorem 2.4 tells us that the Hilbert-Kunz multiplicityof uv + x + y + xyz is lim n →∞ (2 − n + E (2 − n )) 2 n = + √ , an observationmade in [2]. We now give a proof of Theorem 2.4 using the techniques of [3]. Lemma 2.5
Let T : X → X be T . Set E k = ǫ φ k − . Then:(1) T ( E ) = E + E + 6 t (2) T ( E k ) = E k − + E k +1 + (8 k − t + ( ǫ ǫ ) for k > (3) T ( ǫ ǫ ) = 4( ǫ ǫ ) + 4 t , and T ( t ) = 16 t Proof
Suppose k is even. Then T ( E k ) = 32 T ( ǫ φ k − ) = (4 T ( ǫ ) T ( φ k − ))+(4 T ( ǫ ) T ( φ k − )). The magnification rules following Definition2.2 show that this is ( ǫ + t ) φ k − + ǫ +(4 k − t )+( ǫ +1 − t ) φ k +(4 k − − t )). Expanding out we get ( ǫ φ k − )+(4 k − t +( ǫ φ k )+(4 k − t +( ǫ ǫ ) = E k − + E k +1 + (8 k − t + ( ǫ ǫ ). The other parts of the lemma are derivedsimilarly. ✷ Lemma 2.6
Let S be the power series P E (2 − n )(32 w ) n . Then (1 − w )(1 − w )(1 − w ) S = 4 w (1 − w ) + (2 w − w ) √ − w . Proof
Let l : X → Q be evaluation at 1, so that l ( E k ) = 0 for each k ,and l ( ǫ ǫ ) = 0, while l ( t ) = 1. Then E (2 − n )32 n is l ( T n ( E )) and S is just P l ( T n ( E )) w n . If we take Y to be the subspace of X spanned by ǫ ǫ and t , Lemma 2.5 shows that we are in the situation of Example 5.12 of [3]. Thefinal line of that paper is the desired result. ✷ Theorem 2.4 is now easily proved. Lemma 2.6 shows that the value, λ , of(1 − w ) S at w = is (cid:16) · (cid:17) (cid:16) · + q (cid:17) = + √ . Furthermore, S − λ − w is holomorphic in the disc | w | < . It follows that S (cid:16) w (cid:17) − λ − w isholomorphic in | w | <
4, and so the co-efficients in its power series expansion →
0. So E (2 − n ) · n − λ →
0, the desired result. ✷ We conclude this section by showing that the φ m of Definition 2.2 are convexand Lipschitz. Lemma 2.7 φ m (cid:16) q (cid:17) ≤ m +43 q for even m and m +33 q for odd m . roof When q = 2, φ m (cid:16) q (cid:17) = m +34 q . We argue by induction. Suppose q ≥ m is even, φ m (cid:16) q (cid:17) = φ m +1 (cid:16) q (cid:17) + m +38 q . By the induction hypothesis this is ≤ m +724 q + m +38 q = m +43(2 q ) . If m is odd, φ m (cid:16) q (cid:17) = φ m +1 (cid:16) q (cid:17) + q − q + m +38 q .By the induction hypothesis this is ≤ m q + m +48 q = m +33(2 q ) . ✷ Lemma 2.8 φ m (cid:16) − q (cid:17) ≤ m +43 q for odd m and m +33 q for even m . Proof q = 2 is clear. Suppose q ≥
2; we argue by induction. If m is odd, φ m (cid:16) − q (cid:17) = φ m +1 (cid:16) − q (cid:17) + m +38 q , while if m = 0 is even, φ m (cid:16) − q (cid:17) = φ m +1 (cid:16) − q (cid:17) + q − q + m +38 q , and we continue as in the proof of Lemma2.7. Finally, φ (cid:16) − q (cid:17) = φ (cid:16) − q (cid:17) + q . By the induction hypothesis thisis ≤ q + q = q . ✷ Lemma 2.9 φ m (cid:16) q +12 q (cid:17) and φ m (cid:16) q − q (cid:17) are ≤ φ m (cid:16) (cid:17) . Proof If m is odd, 8 (cid:16) φ m (cid:16) (cid:17) − φ m (cid:16) q +12 q (cid:17)(cid:17) = m +3 q − φ m +1 (cid:16) q (cid:17) . By Lemma 2.7this is ≥ m +3 q − m +83 q ≥
0. Also 8 (cid:16) φ m (cid:16) (cid:17) − φ m (cid:16) q − q (cid:17)(cid:17) ≥ m +3 q − φ m − (cid:16) − q (cid:17) .By Lemma 2.8 this is ≥ m +3 q − m − q ≥
0. The argument for even m issimilar. ✷ Theorem 2.10
The φ m are convex and Lipschitz. Proof
To prove convexity, we show that if 0 < j < q , then 2 φ m (cid:16) j q (cid:17) − φ m (cid:16) j − q (cid:17) − φ m (cid:16) j +12 q (cid:17) ≥
0, arguing by induction on q . The case q = 1 isimmediate. When j < q the induction assumption tells us that 2 φ s (cid:16) jq (cid:17) − φ s (cid:16) j − q (cid:17) − φ s (cid:16) j +1 q (cid:17) ≥ s ; this and the fact that ǫ and t are con-vex gives the result. When j > q , the induction assumption tells us that2 φ s (cid:16) j − qq (cid:17) − φ s (cid:16) j − − qq (cid:17) − φ s (cid:16) j +1 − qq (cid:17) ≥
0; this and the convexity of ǫ and 1 − t give the result. Finally the case j = q is handled by Lemma 2.9. Note also thatLemmas 2.7 and 2.8 show that | φ m (cid:16) q (cid:17) − φ m (0) | and | φ m (cid:16) − q (cid:17) − φ m (1) | areeach ≤ m +43 q . Since φ m is convex, it follows that it is Lipschitz with Lipschitzconstant m +43 . ✷ We generalize the calculations of Section 2 to show:
Theorem 3.1
Suppose β lies in a finite dimensional subspace of X stableunder T and T , and is convex Lipschitz. Set E = β φ with φ as in efinition 2.2. Then the power series S t + E ( w ) is algebraic over Q ( w ) , and µ ( t + E ) is algebraic over Q . Proof
Since β and φ are convex Lipschitz, the same is true of t + E . Inview of Theorem 1.14 we only need to prove the result for S . We shall mimicthe proof of Theorem 2.4. Take β , . . . β l , , t spanning a space stable under T and T . We are free to modify each β j by a linear combination of 1 and t andso may assume β j (0) = β j (1) = 0. Then T ( β j ) = (a linear combination of β i )+ a multiple of t , while T ( β j ) = (a linear combination of β i ) + a multiple of(1 − t ). Since T ( β j )(1) = T ( β j )(0) = β j (cid:16) (cid:17) we get: T ( β j ) = X r i,j β i + c j tT ( β j ) = X s i,j β i + c j (1 − t )with the r i,j , the s i,j and the c j all in Q .We proceed in several steps:I) Let R and S be the elements | r i,j | and | s i,j | of M l ( Q ). We define aninfinite matrix V with rows and columns indexed by the positive integersas follows. V is built up out of l by l blocks. The initial diagonal block is S while all succeeding diagonal blocks are matrices of zeroes. The blocksjust below the diagonal blocks are alternately R and S , as are the blocksjust to the right of the diagonal blocks. All other entries are zero.II) Let φ m be as in Definition 2.2. If m ≥ ≤ j ≤ l let E j + lm = β j φ m ; note that E = β φ in accord with the statement of thetheorem. Y ⊂ X is the subspace spanned by t and the β j ǫ , and wedefine y , y , . . . in Y as follows. If 1 ≤ j ≤ l , y j = 6 c j t . If m > y j + lm − (8 m + 6) c j t = P r i,j ( β i ǫ ) for odd m and P s i,j ( β i ǫ ) for even m . Note that 4 T ( β j ǫ ) = T ( β j ) ǫ + t ) + T ( β j ) ǫ + 1 − t ), so that Y is stable under T .III) With notation as above we claim that 8 T ( E j ) = P v i,j E i + y j . Thisamounts to:(1) If 1 ≤ j ≤ l , 8 T ( E j ) = P s i,j E i + P r i,j E i + l + y j (2) If m is odd, 8 T ( E j + lm ) = P r i,j E i + lm − l + P s i,j E i + lm + l + y j + lm (3) If m > T ( E j + lm ) = P s i,j E i + lm − l + P r i,j E i + lm + l + y j + lm Note that the left hand side of (3) is 8 T ( β j φ m ) = ( P r i,j β i + c j t ) φ m +1 + (4 m + 3) t ) + ( P s i,j β i + c j (1 − t )) φ m − + ǫ + (4 m + 3)(1 − t )). Expanding out and using the definition of y j + lm we get (3). Similarcalculations give (1) and (2).IV) Now set s = 2 l . It’s convenient to view the matrix V of I as built upout of s by s blocks. Set D = ( S RR ) and B = ( RR ) in M s ( Q ). Then thediagonal blocks of V are a single D followed by B ’s. If we take A = ( S )and C = ( S ), then the blocks just below the diagonal blocks are all A ’s,8hile those just to the right of the diagonal blocks are all C ’s. And allother entries are zero.The proof of Theorem 3.1 is now easy. III and IV tell us that we are in thesituation of Theorem 5.11 of [3] with T = 8 T and s , A , B , C , D as above.(Note that the y j are all in Y , that Y is finite-dimensional and stable under T , and that the condition of Lemma 5.10 of [3] on the sequence y , y , . . . istrivially satisfied.) Let l : X → Q be evaluation at 1 so that each l ( E j ) = 0.Then Theorem 5.11 of [3] shows that P l ( T n ( E )) w n = P E (2 − n )(8 w ) n isalgebraic over Q ( w ). So the same is true of − w + P E (2 − n )(2 w ) n = P (2 − n + E (2 − n ))(2 w ) n = S t + E ( w ). ✷ Definition 3.2 g = 0 in the maximal ideal of F [[ u , . . . , u r ]] is “strongly ra-tional” if φ g lies in a finite dimensional subspace of X stable under T and T . The following is shown in [4] and [5]:
Theorem 3.3 (1) If F is finite and r = 2 , g is strongly rational.(2) If g is strongly rational, the Hilbert-Kunz series of g lies in Q ( w ) , and µ ( g ) is rational.(3) If g ( u , . . . , u r ) and h ( v , . . . , v s ) are strongly rational, then so are g ( u ) + h ( v ) , g ( u ) h ( v ) , and all powers of g ( u ) . Remark
Much of the above is easy to prove. (1) however makes use of aresult on the finiteness of the number of ideal classes in certain -dimensionalrings. And the proof of (3) for g ( u ) + h ( v ) (or rather the generalization of thisresult to arbitrary finite characteristic p ) isn’t easy. But when p = 2 there’san immediate proof. Namely suppose that V and V are finite dimensionalsubspaces of X containing φ g and φ h and stable under T and T . Then thespace spanned by and V V is finite dimensional and stable under T and T . Furthermore it contains φ g φ h = φ g ( u )+ h ( v ) . If g is strongly rational, Theorem 3.1 tells us that S t +( φ g φ ) is algebraic over Q ( w ) and that µ ( t + ( φ g φ )) is algebraic. Now t + ( φ g φ ) = φ g t + φ ).This gives: Theorem 3.4
Suppose that Conjecture 2.3 holds; that is to say that t + φ = φ x + y + xyz . Then if g in F [[ u , . . . , u r ]] is strongly rational, the Hilbert-Kunzseries of g ( u , . . . , u r ) + x + y + xyz is algebraic over Q ( w ) , and the Hilbert-Kunz multiplicity is algebraic. In particular using Theorem 3.3 we find that ifwe assume Conjecture 2.3 then these algebraicity results hold for P g i ( u i , v i ) + x + y + xyz whenever F is finite over Z/ .
9n Theorem 3.1 it is possible in theory, once the r i,j , the s i,j and the c j areknown, to get a polynomial relation between w and S t + E and compute µ ( t + E ) by using the methods of [3]. This is daunting in practice but we’ll give oneinteresting partial result. Let M be the smallest subspace of X/ ( Q + Q · t ) thatcontains the image of β and is stable under T and T ; our hypotheses showit to be finite dimensional. If J and J are maps M → M let Ψ J ,J ( x, w ) bethe 2-variable polynomial det | xI − w ( J + xJ )( J + xJ ) | . Theorem 3.5
In the situation of Theorem 3.1, P E (2 − n )(8 w ) n lies in thesplitting field over Q ( w ) of Ψ T ,T ( x, w ) . Proof
We adopt the notation of Theorem 3.1 and its proof. P E (2 − n )(8 w ) n = P l ( T n ( E )) w n , and Theorem 5.11 of [3] shows that this power series lies in acertain extension L of Q ( w ) constructed from the matrices A , B and C . Wesaw in [3] that L ⊂ a splitting field over Q ( w ) of det | xI s − w ( Ax + Bx + C ) | .This last matrix is xI l − wx ( R + xS ) − w ( S + xR ) xI l . So our determinant is just x l det I l − w ( R + xS ) − w ( S + xR ) xI l . Since R and S give the action of T and T on M , this last determinant isΨ T ,T ( x, w ). ✷ Suppose β = φ g with g = u + u v + v . The methods of [4] show that M isfive dimensional, that the action of 4 T on M is given by β → β → β → β , β → β →
0, and that the action of 4 T is given by β → β → β → β , β → β →
0. A Maple calculation then shows that Ψ T , T ( x, w ) = − x Ψ ∗ where Ψ ∗ is the reciprocal polynomial w ( x + 1) − (2 w + w )( x + x ) − (2 w − w − w )( x + x ) + (2 w − w + 2 w − w − x . In an algebraicclosure of Q ( w ) let ρ , σ and τ be the roots of Ψ ∗ having positive ord; the other3 roots are ρ − , σ − and τ − . The Galois group of Ψ ∗ over Q ( w ) has order 48and consists of those permutations of the roots that permute the sets { ρ, ρ − } , { σ, σ − } , { τ, τ − } among themselves.Now Theorem 3.5 shows that P E (2 − n )(32 w ) n is in a splitting field of Ψ ∗ over Q ( w ). But as we saw in [3], the field L attached to the matrices A , B and C | xI s − w ( Ax + Bx + C ) | .In our case L ⊂ the degree 8 extension of Q ( w ) corresponding to the subgroupof the Galois group that stabilizes the set { ρ, σ, τ } . Let u = w ( ρ − ρ − )( σ − σ − )( τ − τ − ) and u = w ( ρστ + ρ − σ − τ − ). Using Galois theory we findthat u is in Q ( w ), that u has degree 4 over Q ( w ), and that u and u generatethe degree 8 extension of Q ( w ) mentioned above.So P E (2 − n )(32 w ) n lies in Q ( w, u , u ). A short calculation shows that u =( w − ( w + 1) ((1 − w ) − w ). One can also write down an irreducibleequation for u over Q ( w ) but it’s messy. (Some of the primes of Q [ w ] thatramify in Q ( w, u ) are (1 − w + 2 w ), (1 − w − w ) and (4 + 8 w − w − w − w − w + 81 w + 108 w )). Now the only fields between Q ( w )and Q ( w, u , u ) are Q ( w ), Q ( w, u ) = Q ( w, q (1 − w ) − w ), Q ( w, u ) and Q ( w, u , u ). So P E (2 − n )(32 w ) n , and consequently the conjectured Hilbert-Kunz series of u + u v + v + x + y + xyz , generates one of these 4 extensions of Q ( w ). I think it generates the full degree 8 extension, but verifying this wouldbe a very nasty computation.Now consider the integral closure of Q [ w ] in Q ( w, u , u ). There is just oneprime ideal in this ring lying over (1 − w ), and the argument of Theorem1.14 shows that µ ( t + E ), the putative Hilbert-Kunz multiplicity of u + u v + v + x + y + xyz lies in the residue class field of this ideal.The residue-class field is a degree 8 extension of Q generated by the images,¯ u and ¯ u of u and u . Q (¯ u ) is just Q ( q (13)(157)(2039)), while Q (¯ u ) isa degree 4 extension of Q with discriminant 2 · · · · · · · · Q (¯ u , ¯ u ) are Q , Q (¯ u ), Q (¯ u )and Q (¯ u , ¯ u ). So µ ( t + E ) generates one of these 4 extensions of Q . My beliefis that it generates the full degree 8 extension. References [1] C. Han, P. Monsky, Some surprising Hilbert-Kunz functions, Math. Z. 214(1993), 119–135.[2] P. Monsky, Rationality of Hilbert-Kunz multiplicities: a likely counterexample,Michigan Math. J. 57 (2008), 605–613.[3] P. Monsky, Generating functions attached to some infinite matrices, Preprint(2009), arXiv:math.CO/0906.1836.[4] P. Monsky, P. Teixeira, p -Fractals and power series I, J. Algebra 280 (2004),505–536.[5] P. Monsky, P. Teixeira, p -Fractals and power series II, J. Algebra 304 (2006),237–255.-Fractals and power series II, J. Algebra 304 (2006),237–255.