ON HILBEFIT POLYNOMIAL OF CERTAIN DETERMINANTAL IDEALS

Let X=(Xij) be an m(1) by m(2) matrix whose entries Xij, 1≤i≤m(1), 
1≤j≤m(2); are indeterminates over a field K. Let K[X] be the polynomial ring in 
these m(1)m(2) variables over K. A part of the second fundamental theorem of 
Invariant Theory says that the ideal I[p

Given any two bi-vectors a and a' of lengths p and p' respectively, we define a a' to mean that p > p' and an a(k,t) a'(k,i) for k [1,2] and i [I,p'].We note that this defines a partial order on the set vet (2) of all bl- vectors Given any d N, by a bl-tableau of depth d, we mean a mapping T: [l,d]  vet (2), which to every e [l,d], associates Tie] vec (2).
A bl-tableau T of depth d is said to be standard if len(T[e]) is positive for each e [I,d] and TIll T [2] ..

. T[d].
Given any m N*(2), a vet(2) and a bi-tableau T of depth d, we say that (I) T is bounded by m if Tie] m for e 1,2,...,d (il)T is predominatpd by a if a Tie] for e 1,2,...,d.The area of a bl-tableau T of depth d is denoted by are(T) and is defined as Finally for any element V in an overrlng of Q and any integer r, we set and Let X (Xlj) be any m(1) by m(2) matrix whose entries Xij i m(1), J m(2), are Indetermlnates over a field It.Let K[X] be the polynomlal ring in these m(1)m(2) varlables.Clearly to every bl-vector a of length p and bounded by m, there corresponds a uniquely determined p x p minor of X, say mor(X,a), formed by the rows with row numbers a(l,l), a(l,2),***,a(l,p) and by the columns with column numbers a(2,1), a(2,2),...,a (2,p).By convention we define mor(X,a) if len(a) O.
Further, to a bl-tableau T of depth d, there corresponds a product of minors of X, which we call a monomlal in minors of X.A monomlal in minors of X is said to be standard If the corresponding bl-tableau Is standard.
Using straightening formula, DeConclnl, Eisenbud, and Procesl [2], proved that the standard monomlals in minors of X form a base for the vector space K[x] over K.
Abhyankar proves this by enumerating the number of standard bl-tableaux with certain conditions.
In fact, Abhyankar [3] proves that the number of standard bl-tableaux of area V, bounded by m and predominated by a given fixed bl-vector a of length p Is and FD(m,p,a), D Z, are iuteger valued funct[,s defined in section 2.
t|e further poves that If a is a given b[-vector of length p, bounded by m and if lip,a] is the idea| in K[X] generated by all minors of X corresponding to the bi- vectors b not predom,ated by a, then lip,a] is a homogeneous prime ideal in K[X] and H(V) is the HIbert function as well as the Hilbert polynomial of lip,a] in K[X].In particular, if a(k,i) i for k [|,2] and i e [;,p], then lip,a] is the ideal in K[X] generated by (p+l) by (p+|) minors of X and hence it follows by the above theorem that it is a prime ideal in K[E].This forms a part of the second fundamental theorem of Invarlant theory and was originally proved by Pascal in 1888 and then reproved by Mount [4], Eagon and Hochster [5] and others.Now H(V) is a polynomial in V with rational coefficients and of degree C.
Further C! times the coefficient of V C in H(V) is Fo(m,p,a) which equals the order of the irreducible variety defined by lip,a] in the (m(1) m(2)-l) dimensional projective space over K (Refer to remark (20.18) of [I]).From this it follows that F0(m,p,a) is a positive integer (see theorem (2.1) of section 2).
Further arithmetic genus of (-I)C[(Fo FI+ F2-...+ (-I)CF C) -I] 0. Thus FD(m,p,a), D Z determine lip,a] is important geometric characters of the variety defined by lip,a] from Zariskl and Samuel [6].Hence it is interesting to study the properties of FD(m,p,a), D e Z.For any m e N*(2), p N*, we put vec(2) --the set of all bl-vectors, vec [2,p] the set of all bl-vectors of length p and vec(2,m,p) {a vec[2,p]: a m}.
This paper settles both these problems in an affirmative manner.

THE MAIN RESULT.
We follow the usual convention that the sum over an empty family Is zero.For any p N and A Z, we let J(p) all subsets of [l,p] and J(p,A) all subsets of [l,p] with cardlnality A.
Let there be given any m N (2), p e N b vec[2,p+l], u J(P),U e [O,p] and We define the sets M[p,b,k,u] and M(p,b,k,U) as follows: bl[p,b,k,u] a e vec [2,p]: a(k,t b(k,i for all i e [l,p], a(k' i) e [b(k' i)+l b(k' i+l)-l] for all t e u and a(k',J) b(k',J), for all J e [1,p]\u and M(p,b,k,U) -U M[p,b,k,u], u e J(p,U) Given any m e N (2), p e N b e vec [2,p+l] and k e [1,2], let m e N (2) be such that (I) m*(k')ffim(k') and m*(k)=m(k)+l holds.Further let p N and a e vec(2,m ,p be such that either In the above recurslve formula, by interchanging the summations over I and U and by noting the fact that U I In what follows, we shall prove some results which enable one to answer the first problem posed by Abhyankar, in an affirmative manner [Refer to problem (I) at the end of sect [on 2].# then {b(k',J)}, j 1,2 p, is a sequence of consecutive integers i.e. b(k',J+1)=b(k',J)+l, for all J if,p-l] and consequently M(p,b,k,U) for every U ) I.
Finally, if U > and u J(p) is any set of cardinally U, then there Is an i e u such that i [I, p-l].
e(i) for i 1,2,...,p.Further let for every E e Z, HE(m,p,a det G(m,p,a,e).Then for every D Z, we define eeZ(p,E) FD(m,p,a) () ,(m,p,a)" EZ In view of the usual properties of binomial coefficients and determinants, we may assume that 0 e i (m(k)-a(k,i)) for i 1,2,...,p.Hence the above sum is S.(;.UDPIKAR essentially fi_nlte. p,a) is a positiv,_-integer, for all D c Z, FD(n,p,a) is a ,negatlve integer and {D c Z: FD(m,p,a) O}_[O,C(ra,p,a)l where C(m,p,a) mtn{C[m,p,a,k], C[m,p,a,k']}.
of [1].le further proposes the following problems.Given any m e N (2), p N and a vee(2,m,p), does there exist a nonnegattve Integer C*(m,p,a) such that Problem (I) {D e Z: FD(m,p,a) O} [O,C (m,p,a)]?Problem (II) If so, then is tt possible ttmt C (m,p,a) C(m,p,a)?

(N
2a) p p and m (k) a (k, 0 and (,i) b(,i) for all e [1,2] and i [l,p] and b(,p+l) m()+l for all E [1,2] holds or (m e N (2), p e N a e vec[2,p], U e [O,p] and D e Z, we set U Su(m,p,a,D) FD_I(m,p,a)-We now state a useful recursive formula in the followlng.THEOREM 3.1.Let there be given any m p] a M(p,b,k,U) I PROOF.Refer to Theorem (9.8*), section 9 of [1].

THEOREM 3. 2 .
Let k e [1,2], p e N and b e vec[2, p+l] be given.Then we have the ollow[ng.(I) M(p,b,k,0) M[p,b,k,#] is a nonempty set consisting of the unique element b e vec[2,p] obtained from b by putting (,i) b(,i) for all e [1,2] and i [1,p].(II) If M(p,b,k,l) of cardlnality zero, we have M(p,b,k,0) M[p,b,k,#].The rest follows from the definition of M[p,b,k,]. (II) Let u i {i} where i is an arbitrarily chosen element of [I, p-l].Then by definition of U[P,b,k 1,u1] we have, b,k,U), then we must have a(k',i) [b(k',l) + I, b(k',i + I) I], which is a contradiction since {b(k',J)}, J 1,2, p, is a sequence of consecutive integers.Hence M(p,b,k,U) for every U > if M(p,b,k,l) .Hence the theorem., , LEMMA 3.1.Let m N (2), p N b vec[2,p + I] and k [1,2] be given.* * * * * * Let m e N (2), p e N and a e vec(2,m ,p be such that (I) is satisfied and either

(
2a)  or (2b) is satisfied.Let be the unique element of vec[2,p] obtained from b by putting (,i) --b(,i) for all e [1,2] and i e if,p].
We shall use this form of the recursive formula later [n this section.