The A-hypergeometric System Associated with a Monomial Curve

We make a detailed analysis of the A-hypergeometric system (or GKZ system) associated with a monomial curve and integral, hence resonant, exponents. We characterize the Laurent polynomial solutions and show that these are the only rational solutions. We also show that for any exponent, there are at most two linearly independent Laurent solutions, and that the upper bound is reached if and only if the curve is not arithmetically Cohen--Macaulay. We then construct, for all integral parameters, a basis of local solutions in terms of the roots of the generic univariate polynomial associated with A. We determine the holonomic rank r for all integral exponents and show that it is constantly equal to the degree d of X if and only if X is arithmetically Cohen-Macaulay. Otherwise there is at least one exponent for which r = d + 1.


Introduction
In this paper we make a detailed analysis of the A-hypergeometric system (or GKZ system) associated with a monomial curve and integral, hence resonant, exponents. We describe all rational solutions and show in Theorem 1.10 that they are, in fact, Laurent polynomials. We also show that for any exponent, there are at most two linearly independent Laurent solutions, and that the upper bound is reached if and only if the curve is not arithmetically Cohen-Macaulay. We then construct, for all integral parameters, a basis of local solutions in terms of the roots of the generic univariate polynomial (0.5) associated with A. We also determine in Theorem 3.7 the holonomic rank r(α) for all α ∈ Z 2 and show that d ≤ r(α) ≤ d + 1, where d is the degree of the curve. Moreover, the value d + 1 is attained only for those exponents α for which there are two linearly independent rational solutions and, therefore, r(α) = d for all α if and only if the curve is arithmetically Cohen-Macaulay.
In order to place these results in their appropriate context, we recall the definition of the A-hypergeometric systems. These were introduced in a series of papers in the mid 1980's by the Gel'fand school, particularly Gel'fand, Kapranov, and Zelevinsky (see [7,9], and the references therein). Let A = {ν 1 , . . . , ν r } ⊂ Z n+1 be a finite subset which spans the lattice Z n+1 . Suppose, moreover, that there exists a vector λ = (λ 0 , . . . , λ n ) ∈ Q n+1 such that λ, ν j = 1 for all j = 1, . . . , r, i.e. the set A lies in a rational hyperplane. Let A also denote the (n + 1) × r matrix whose columns are the vectors ν j . Let L ⊂ Z r be the sublattice of elements v ∈ Z r such that A · v = 0. Given α ∈ C n+1 , the A-hypergeometric system with exponent (or parameter) α is: ν ji x j ∂ϕ ∂x j = α i ϕ ; i = 1, . . . , n + 1 (0. 2) where D v is the differential operator in C r : 1991 AMS Subject Classification: Primary 33C70, Secondary 14M05, 33D20. 1 The A-hypergeometric system is holonomic (with regular singularities) and, consequently, the number of linearly independent solutions at a generic point is finite [7]. Let r(α) denote the holonomic rank of the system, i.e. the dimension of the space of local solutions at a generic point in C r . If we drop the assumption that A lies in a hyperplane, then the regular singularities property is lost but, as Adolphson [2] has shown, the system remains holonomic. The singular locus is described by the zeroes of the principal A-determinant ( [11]). We set R := C[ξ 1 , . . . , ξ r ]/I A , where I A is the toric ideal (0.3) When n = 1, we can assume without loss of generality that A = 1 1 · · · 1 1 0 k 1 · · · k m d , where 0 < k 1 < · · · k m < d. Note that the condition that the columns of A generate the lattice Z 2 is equivalent to gcd(k 1 , . . . , k m , d) = 1. The homogeneous ideal I A defines a monomial curve X A ⊂ P m+1 of degree d whose homogeneous coordinate ring is R. X A is normal if and only if d = m + 1. Recall that X A is said to be arithmetically Cohen-Macaulay if and only if the ring R is Cohen-Macaulay.
The system associated with (0.4) admits very interesting solutions. Let denote the generic polynomial with exponents 0, k 1 , . . . , k m , d. It is not hard to see that the powers ρ s (x), s ∈ Z, of the roots of f (x; t), viewed as functions of the coefficients, are algebraic solutions of the A-hypergeometric system with exponent (0, −s). This fact was observed by Mayr [17] who constructed series expansions for these functions. These have more recently been refined by Sturmfels [22]. The total sum p s (x) := ρ s 1 (x) + · · · + ρ s d (x) (0.6) will then be a rational solution with the same exponent. Similarly, one can show that the local residues give algebraic solutions with exponent (−a, −b) and, again, the total sum of residues will be a rational solution.
In §1 we describe explicitly all rational solutions of the A-hypergeometric system associated with a monomial curve. Since for A as in (0.4), the principal A-determinant factors into powers of x 0 , x d , and the discriminant ∆(f ), we know a priori what the possible denominators of a rational solution may be. However, we show in Theorem 1.10 that there are no rational solutions whose denominator involves ∆(f ) and therefore, every rational solution must be a Laurent polynomial. This is a somewhat surprising result which is peculiar to the case n = 1 (see Example 1.11). One may give explicit formulas for these Laurent polynomials in terms of hypergeometric polynomials in fewer variables. When applied to the sum of powers of roots one recovers the classical Girard formulas. One also obtains similar expressions for total residues in terms of hypergeometric polynomials.
We show that for any α ∈ Z 2 the dimension of the space of rational A-hypergeometric functions with parameter α is at most 2. Moreover, the value 2 may be reached for only finitely many values of α and this happens if and only if the ring R is not Cohen-Macaulay.
In §2 we exhibit a family of algebraic A-hypergeometric functions defined in terms of the roots of the polynomial f (x; t). These are the building blocks for the construction, in §3, of local bases of solutions and the determination of the holonomic rank for all integral exponents. It becomes necessary to consider four possibilities for the exponent α. These cases admit combinatorial descriptions (see (1.9)) and correspond to the existence of a polynomial solution; a one-dimensional space of rational -non-polynomial-solutions; a two-dimensional space of rational solutions; or no rational solution for the given exponent. For u ∈ N m+2 , the derivative D u maps A-hypergeometric functions to A-hypergeometric functions while changing the exponent from α to α − A · u. A careful analysis of the kernel and image of this operator together with Corollary 5.20 of [2] leads to the determination of the holonomic rank for all values of α. We show, in particular, that and that r(α) = d + 1 exactly for those parameters α ∈ Z 2 for which the dimension of the space of rational solutions is 2. Hence, r(α) = d for all α ∈ Z 2 if and only if the curve X A is arithmetically Cohen-Macaulay. These results allow us to clarify the relationship between the holonomic rank and vol(P ), the normalized volume of the convex hull P ⊂ R n+1 of A and the origin (which equals the degree d in the case of curves). It was originally claimed in [9, Theorem 2] that r(α) = vol(P ) in all cases, but it was pointed out by Adolphson that, for resonant exponents, the proof required the assumption that the ring R be Cohen-Macaulay (see [10]). In [2,Corollary 5.20], Adolphson showed that r(α) = vol(P ), for α a semi-nonresonant exponent, without any additional assumptions on R. The first explicit example where the equality fails is given in [23] and described in Example 1.8.i). In the forthcoming monograph [19], Saito, Sturmfels and Takayama prove (0.8) using Gröbner deformation methods and show that the inequality r(α) ≥ vol(P ) holds without restrictions on n.
Very little seems to be known about the problem of finding rational solutions of differential equations beyond the case of linear differential operators in one variable. In this case, Singer [20] has shown that one can determine in a finite number of steps whether a given equation has a rational solution and find a basis for the space of such solutions. Abramov and Kvasenko ( [1]) have further studied the problem of effectively finding rational solutions for such operators. In our case one could, in principle, use non-commutative elimination to obtain linear operators in one variable with coefficients that depend rationally on the other variables and apply Singer's decision procedure to characterize the rational solutions. This can be done in small examples using non-commutative Gröbner bases packages such as kan ( [24]) but we have not been able to obtain any general results in this manner.
Gel'fand, Zelevinsky, and Kapranov have constructed series solutions for (0.1)-(0.2), associated with regular triangulations of the polytope P . When the exponent α is non-resonant for the triangulation, it is possible to obtain in this manner vol(P )-many independent solutions. There are, however, very interesting cases in which the exponents are integral and, therefore, automatically resonant. For example, it has been observed by Batyrev in [4] that the period integrals of Calabi-Yau hypersurfaces in toric varieties satisfy an A-hypergeometric system with exponents α = (−1, 0, . . . , 0). In this case, series solutions have been obtained by Hosono, Lian, and Yau in [14,15] (see also [3,5,13]). Very recently, Stienstra [21] has generalized the Γ-series construction of Gel'fand, Zelevinsky, and Kapranov to obtain series solutions in the case of resonant exponents under a maximal-degeneracy assumption. In particular, if α = 0 and P admits a unimodular triangulation -which implies that R is Cohen-Macaulay-all solutions of (0.1)-(0.2) may be obtained in this manner. This method also yields all solutions of interest in the context of toric mirror symmetry.
A key result, in the curve case, is Theorem 1.9 which asserts that for m ≥ 1, there are no rational solutions with integral exponents in the Euler-Jacobi cone. This corresponds to the classical vanishing theorem for the total sum of residues, a statement which has a generalization as the Euler-Jacobi theorem (see [16] for the most general form of this result). It is interesting to note then that Euler-Jacobi vanishing is a consequence of the fact that residues satisfy the A-hypergeometric system. We also point out that while the characterization of Laurent solutions follows from formal arguments, the proof of the Euler-Jacobi vanishing involves transcendental methods.

Rational solutions
The polynomial solutions of a general A-hypergeometric system admit a very simple description. Given α ∈ Z r , we define the hypergeometric polynomial As usual, we set Φ A (α; x) := 0 if α ∈ A · N r . The following result, whose verification is left to the reader, is Proposition 2.1 in [18]: is the unique, up to scaling, polynomial solution of the Ahypergeometric system with exponent α. Moreover, for any u ∈ N r , where D u stands for the partial derivative ∂ |u| /∂x u .
The purpose of this section is to describe the rational solutions of the A-hypergeometric system associated with a matrix A as in (0.4). Note that for m = 0 the system restricts to the homogeneity equations (0.2). Therefore we may assume throughout that m ≥ 1. To simplify our notation, we will index all (m + 2)-tuples by 0, k 1 , . . . , k m , d. Let e 0 , e k 1 , . . . , e d denote the standard basis of Z m+2 . For i = 1, . . . , m we have (1. 3) The following observation will be useful in the sequel: Suppose ϕ is a local holomorphic solution of (0.1)-(0.2), polynomial with respect to any of the variables x 0 , x k 1 , . . . , x d . Then ϕ is a Laurent polynomial.
Proof: Since ϕ satisfies the equations (0.1), it follows from (1.3) that for all ℓ ∈ N, and, consequently, if ϕ is polynomial in any of the variables, it must be so in all of the variables x k i , i = 1, . . . , m, and we may write: where u varies in a finite subset of N m and each ϕ u (x 0 , x d ) is homogeneous in each variable with respective degrees β 0 , β d satisfying But, because of (1.4), for r, s ∈ N sufficiently large, D r 0 D s d ϕ = 0, and consequently one, and therefore both, of β 0 , β d must be an integer. Hence ϕ is a Laurent polynomial. ⋄ Note that in the one-dimensional case, an A-hypergeometric Laurent polynomial may not contain a non-zero term of the form c u x u with u k i < 0. This follows from the fact that the singular locus Σ of the hypergeometric system is given by the zeroes of the principal A-determinant, i.e.
where ∆(f ) is the discriminant of the generic polynomial (0.5). Alternatively, if a solution contains a non-zero term c u x u with u k i < 0, being in the kernel of the differential operator it must also contain non-trivial terms of the form c v x v with v k i = u k i −j d for all positive integers j and this is clearly impossible. A similar argument shows that a Laurent solution may not contain terms of the form: c u x u with both u 0 < 0 and u d < 0.
Thus, any Laurent solution must be of the form where L 0 (x) has as denominators only powers of x 0 and L d (x) has as denominators only powers of x d . We note that the study of L 0 (x) and L d (x) is completely symmetric. Indeed, let ℓ j = d − k m−j+1 , j = 1, . . . , m, and is a Laurent solution of the A-hypergeometric system and exponent α = (α 1 , α 2 ) (although α should be viewed as a column vector we will, for simplicity of notation, always write exponents as row vectors), the functionR(y 0 , y ℓ 1 , . . . , y ℓ m , y d ) obtained from R by substituting: is a solution of theÂ-hypergeometric system and exponentsα = (α 1 , dα 1 − α 2 ). Lemma 1.3. For α ∈ A · N m+2 , the only Laurent solutions of the A-hypergeometric system are the constant multiples of the hypergeometric polynomial (1.1).
Proof: Suppose there is a Laurent solution L(x) of exponent α containing a term of the form x u /x r d , with u ∈ N m+2 , u d = 0 and r > 0 (we will always assume that monomials are written in reduced form). Then A·(u−re d ) = α. Let v ∈ N m+2 be such that A·v = α, then A · u = A · (v + r e d ), and the operator D u − D r d D v being in the hypergeometric system, must vanish on L. This means that L must also contain a term x w whose derivative Since v d ≥ 0, this is clearly impossible. Arguing by symmetry, we see that there cannot be a solution containing a term of the form x u /x r 0 , with u 0 = 0 and r > 0. ⋄ We will denote by B the matrix B = 1 1 · · · 1 0 k 1 · · · k m and x ′ the vector consisting of the first m + 1 variables (x 0 , x k 1 , . . . , x k m ). Similarly, let C be the matrix Given α ∈ Z 2 we define where α ′ (r) = α + r (1, d), and Note that both sums are finite. This follows from the fact that Φ B (α ′ (r); The statement for (1.8) follows by symmetry. In fact, we should observe that the change of variables (1.6) transforms Ψ A d (α; x) into ΨÂ 0 (α; x). The following subsets of Z 2 will play an important role in the description of A-hypergeometric functions: Note that via the change of variables (1.6), and denoting for α = (α 1 , α 2 ) ∈ Z 2 ,α = (α 1 , dα 1 − α 2 ), we have forÂ as in (1.5): It is clear from the definitions (1.7) and (1. and, On the other hand, the importance of the sets E 0 (A) and E d (A) stems from the fact that according to Lemma 1.3, it is only for α ∈ E 0 (A) (respectively α ∈ E d (A)) that the Laurent polynomial Ψ A 0 (α; x) (respectively Ψ A d (α; x)) may be -and as the following result shows is-A-hypergeometric. Note also that there are no A-hypergeometric Laurent polynomials with exponent α ∈ J(A) and it will be a consequence of Theorem 1.10 that there are no rational A-hypergeometric functions with exponent α ∈ J(A).
iii) The functions Ψ A d (α; x) and Ψ A 0 (α; x) span the space of Laurent solutions of the Ahypergeometric system with parameter α.
Proof: Clearly, i) is an immediate consequence of ii) and, because of symmetry, it suffices to show (1.12) Thus, it remains to prove (1.12) for the partial derivative D d . We have The last equality follows since α ∈ A · N m+2 implies that Suppose now that L(x) is a Laurent solution with exponent α and write L(x) = L d (x) + L 0 (x). If we decompose further: x r d then, the polynomials A r (x ′ ) must be solutions of the B-hypergeometric system and exponent α ′ (r). Thus, is in the hypergeometric system and must vanish on L. This means that A symmetric argument shows that if the component L 0 (x) is non-trivial then it must be a constant multiple of Ψ A 0 (α; x), which proves part iii). ⋄ We state for emphasis: Proof: The ring R is a particularly simple example of an affine semigroup ring whose properties have been extensively studied (see for example [6,Chapter 6], [12], [25]). In fact, Proposition 1.6 is a special case of Theorem 2.6 in [12] which gives necessary and sufficient conditions for an affine semigroup ring which, like R does, admits a system of monomial parameters. Their condition (ii) is easily seen to be equivalent, in our notation, to E(α) = ∅. ⋄ 1.7. Remarks: i) When the curve X A is normal, i.e. d = m + 1, the sets defined in (1.9) have a very simple description: the image I(A) coincides with the "cone" (properly speaking semigroup) The complement J(A) of these three sets, i.e. the set of α ∈ Z 2 for which there are no A-hypergeometric Laurent polynomials of exponent α is the Euler-Jacobi cone: iii) The conditions (1.10) and (1.11) are far from being sharp. It is easy to see, for example, x) are non-trivial. Hence, there is for each α at most one, up to constant multiple, Laurent solution of exponent α. Note also that the toric ring R is always Cohen-Macaulay but it is normal if and only if d = 2. iv) It is not hard to prove (see for example [2,Lemma 3.12]) that there exists v ∈ A · N m+2 such that v + C ⊂ I(A). Thus, for α ∈ v + C, there is a unique A-hypergeometric Laurent polynomial and it is given by (1.1). Moreover, when this observation is combined with the inequalities in iii), it follows that the set E(A) is finite. i) This is the "running example" in [23]. Let The exponent α = (1, 2) is the unique element in C such that α ∈ I(A) and α 1 < α 2 < 3α 1 .
, is A-hypergeometric and the only A-hypergeometric Laurent polynomials of exponent β are the multiples of the ii) Consider the system associated with the matrix A = 1 1 1 1 1 0 6 7 13 14 and α = (2, 18) ∈ I(A). We have It follows from (1.10) and (1.11) that there are no A-hypergeometric Laurent polynomials whose exponent α is in the Euler-Jacobi cone (1.13). In fact, as the following result shows, there are no rational solutions with exponent in that region. Theorem 1.9. The A-hypergeometric system associated with the matrix (0.4) has no rational solutions whose exponent α lies in the Euler-Jacobi cone.
Given α ∈ Z 2 , we will denote by R(α) the vector space of rational A-hypergeometric functions of exponent α. Before proving Theorem 1.9 we note the following consequence. Proof: Suppose ϕ ∈ R(α). For ℓ ∈ N sufficiently large, β = α − ℓ (1, k 1 ) lies in the Euler-Jacobi cone. Then, D ℓ k 1 ϕ is a rational solution in H(β), and thus it is identically zero by Theorem 1.9. Hence, ϕ is polynomial in x k 1 and by Proposition 1.2 it must be a Laurent polynomial. ⋄ 1.11. Example: Theorem 1.10 is not true for n > 1. It fails already in the simplest two-dimensional situation: Consider the hypergeometric system associated with the matrix The lattice L has rank one; in fact L = Z · (1, −1, −1, 1) T , and the system (0.1)-(0.2) is equivalent to Gauss' classical hypergeometric equation. The function 1/(x 1 x 4 − x 2 x 3 ) is a solution with parameters (−2, −1, −1).
Proof of Theorem 1.9: The proof will be by induction on m. We begin by considering the case m = 1 and write, for simplicity, k 1 = k. Note that in this case the lattice L has rank 1 and is generated by ω := (d − k)e 0 − de k + ke d . In particular, for appropriate values of α, we can write the A-hypergeometric functions in terms of classical hypergeometric functions (see [9, §3.1]). The discriminant of the generic polynomial x 0 + x k t k + x d t d is, up to factors which are powers of x 0 and x d , Suppose now that R(x) = P (x)/Q(x) ∈ R(α) with α ∈ EJ , i.e. dα 1 < α 2 < 0. Note that both P and Q are bihomogeneous relative to the Z 2 -degree defined by A. We can then write P (x) = x v P 1 (z), where P 1 is a polynomial and P 1 (0) = 0. Thus, up to a constant, with c 0 = 0, and A · u = α.
Since R is in the kernel of the differential operator the coefficients in (1.14) satisfy the relation: for all j ∈ Z. Setting j = −1 we get On the other hand, ku k + du d = α 2 < 0 and du 0 + (d − k)u k = dα 1 − α 2 < 0, which means that in either case, u k < 0. Therefore the left-hand side of (1.16) is never zero and, consequently neither is the right-hand side. This implies that u 0 ≥ 0 and u d ≥ 0 and for all j ≥ 0 for some constant c. But a function with such an expansion may not be rational. Indeed, Stirling's formula implies that, asymptotically, c j ∼ µ j −α 1 λ j / √ j, for some constant µ. On the other hand, for a rational function R whose denominator is a power of (1 − λ z), we would have c j ∼ p(j) λ j with p a polynomial. This completes the proof of Theorem 1.9 in the case m = 1.
Assume now that Theorem 1.9, and its consequence Theorem 1.10, are valid for m − 1, m ≥ 2, and consider the A-hypergeometric system associated with the matrix A in (0.4). Let R(x) be a rational solution with parameter α in the Euler-Jacobi cone; that is, α = (α 1 , α 2 ) with dα 1 < α 2 < 0. Suppose we can write R(x) = P (x)/(x r d Q(x)) ; r > 0 and assume that x d does not divide P (x) or Q(x). Let again x ′ = (x 0 , x k 1 , . . . , x k m ) and write By inductive assumption, since A ℓ (x ′ ) is a rational B-hypergeometric function, it must be a Laurent polynomial as described by Theorem 1.4. (Note that even though we may have ℓ = gcd(k 1 , . . . , k m ) > 1 we may easily reduce to the system associated with a matrix where k i has been replaced by k i /ℓ.) Set x d = 0 and consider the non-trivial rational function A −r (x ′ ) = P (x ′ , 0)/Q(x ′ , 0) which has B-exponent α ′ (r) = (α 1 + r, α 2 + d r). It is clear that α ′ (r) ∈ E(B) and hence there may be, up to constant, at most one Laurent solution with exponent α ′ (r). We distinguish the two possible cases.
Then it contains a non-zero term of the formx u Successive applications of the fact that R is in the kernel of the operator (1.15), with k m in the place of k, yields that R must also contain a non-zero term whose derivative D Then it contains a non-zero term of the formx u By symmetry we can then assume that R(x) = P (x)/Q(x) and neither x d nor x 0 divide Q. Thus R(x) is written as in (1.17) with r = 0. For each ℓ ≥ 0, A ℓ (x ′ ) is a solution of the B-hypergeometric system with exponent (α 1 − ℓ, α 2 − d ℓ). By inductive hypothesis, these must be Laurent polynomials and it is easy to check that the only possible denominators are powers of x k m . Thus, only powers of x k m may appear in the denominator of the above expansion for R, which implies that q 0 (x ′ ) must be of the form x d k m and, therefore, Q(x) has bidegree (d, k m d). On the other hand, a symmetric argument would imply that Q(x) must contain a term of the form x e k 1 and hence Q(x) should have bidegree (e, k 1 e). Since m > 1, this implies that Q(x) has degree 0, but then R(x) is a polynomial solution which is impossible since, being in the Euler-Jacobi cone, α ∈ A · N m+2 . ⋄ As we have noted before, given s ∈ Z, the sum (0.6) p s (x) = ρ s 1 (x) + · · · + ρ s d (x), of the powers of the roots of the generic polynomial (0.5) is a rational A-hypergeometric function with exponent (0, −s). By Theorem 1.10 it must be a Laurent polynomial and, therefore expressible in terms of Ψ A 0 and Ψ A d . In fact, Corollary 1.12. For s > 0, (

1.19)
Proof: It suffices to consider the normal case, d = m + 1, and then set the appropriate variables equal to zero. For s > 0, it follows from (1.11), that p s (x) must be a multiple of Ψ A d ((0, −s); x). It is easy to see that the value of the multiple must be s by specialization to the case when f (x; t) = t d + t d−1 . The statement for s < 0 follows by symmetry after observing that with the change of variables (1.6), the polynomialf (y; τ ) associated with the matrixÂ in (1.5) is given byf (y; τ ) = τ d · f (x; τ −1 ) and, consequently, its roots are the inverse of those of f . ⋄.

Remarks:
i) Note that each term in the right hand side of (1.18) is of total degree zero and, therefore, we may express p s (x) as a polynomial in x d−j /x d = (−1) j σ j (ρ 1 , . . . , ρ d ), j = 1, . . . , d, where σ j is the j-th elementary symmetric polynomial. This yields the classical Girard formulas.
ii) As we have also noted in the introduction, the total sum of the local residues (0.7) gives a rational A-hypergeometric function with exponent (−a, −b) and hence, as in Corollary 1.12, it must be a multiple of We end this section with a result that should be seen as a complement to (1.10) and (1.11) and which will be of use in §2. Proof: It suffices to prove the first statement and then deduce ii) by symmetry. Suppose β ∈ Z 2 with β 1 = α 1 − r, r > 0 and s(β) = s(α) then, if Ψ A d (α; x) = 0 we have by (1.12): The result now follows from the fact that Ψ A d ((0, −s); x) = s p s (x). ⋄

Algebraic solutions
In this section we introduce a family ψ ρ of local algebraic solutions of the A-hypergeometric system associated with a monomial curve. These solutions, which are given in terms of the roots ρ(x) of the generic polynomial (0.5), will play a central role in §3 when we compute the holonomic rank and construct a basis of local solutions for all exponents α ∈ Z 2 .

Proof: By Theorem 1.4 iii), Ψ A (α) is a linear combination λΨ
. Moreover, if α 1 = 0 the result follows from (1.18) and (1.19). Therefore, computing derivatives with respect to x d , the result follows for s(α) = dα 1 − α 2 > 0, α 1 < 0 where Ψ A 0 (α; x) = 0. By symmetry i) also holds for α 2 > 0, which implies λ = 1 since, because of Proposition 1.14, Ψ A d ((0, −s(α))) = 0. A similar argument shows that if Ψ A 0 (α; x) = 0 then µ = 1. The second assertion is an immediate consequence of i) and iii) in Theorem 1.  Before giving the proof of Theorem 2.4, we first recall the construction by Gel'fand, Zelevinsky, and Kapranov [9] of Γ-series solutions for the A-hypergeometric system and the expressions obtained by Sturmfels [22] for the roots of f (x; t) in terms of them. We will begin by considering the normal case, and eventually we will specialize coefficients to study the general case. We only need to consider the coarsest triangulation of the polytope P , the convex hull of A and the origin, i.e. the one consisting of the single simplex P . As before, we let L stand for the integral kernel of A, that is, the sublattice of elements v ∈ Z d+1 such that A · v = 0. Given u ∈ Q d+1 we define the formal power series where, for any rational number u and any integer v, we write if u is a negative integer and u ≥ −v, otherwise.
If u has no negative integer coordinates, or is of the form (0, . . . , 0, 1, −1, 0, . . . , 0), the series x u 0 0 · · · x u d d is a formal solution of the A-hypergeometric system with parameters A · u ∈ Q 2 (see [ It follows from [9, Proposition 2] that there exists an open set V ⊂ C d+1 \ Σ of the form for some positive real constant M , where all these series converge, and, according to [22,Theorem 3.2], locally on V, they define the holomorphic d roots of the generic polynomial d j=0 x j · t j . Given a positive integer s we consider the powers ρ s i (x) and write We now consider the A-hypergeometric system associated with the matrix (0.4) and recall that we are assuming that gcd(k 1 , . . . , k m , d) = 1. Let J denote the complement of {0, k 1 , . . . , k m , d} in {0, 1, . . . , d}, and V J the (m + 2)-dimensional subspace of C d+1 obtained by setting x j = 0, j ∈ J. Note that V ∩ V J is non-empty.
The same is true of any of its derivatives D u θ b , u ∈ N m+2 , with respect to variables x i , i ∈ J.
Proof: Note that if ρ s = (ρ s 1 , . . . , ρ s d ) T and θ = (θ 1 , . . . , θ d ) T , then θ = M −1 · ρ s where M is the non-singular matrix M = (ξ a i ), i, a = 1, . . . , d. In particular, θ b (x) ∈ H((0, −s))(U). We now claim that for any set of indices a 1 , . . . , a s such that a 1 + · · · + a s = b + ℓd, for some ℓ ∈ N, for some non-zero constant λ. Indeed, forgetting for the moment the coefficients, suppose that x w = x w (1) · · · x w (s) is a monomial appearing in the product Then A · w = (0, −s) T and therefore, x w must differ from the monomial inside the bracket (2.12) by a monomial of the form x v with v ∈ L. This means that all the monomials in the power series of θ b (x) appear in the Γ-series of (2.12). But, on the other hand, since we already know that θ b (x) is A-hypergeometric, if a monomial, such as the one in (2.12), appears in its expansion then the whole Γ-series must appear and with the appropriate coefficients.
Consequently, multiplication of the monomial in the bracket (2.12) by x v yields a term in the Γ-series which does not involve any variables from the index set J. On the other hand, it is easy to check that all coefficients γ(u i , v i ) are non-zero and therefore the restriction The statement about the derivatives of θ b (x) follows from the fact that for b < d, the exponents of x 0 and x n in the bracket in (2.12) are not integers, while the exponents with which any of the other variables x k i appears in the Γ-series cannot be bounded since for any ℓ ∈ N, the element Thus, θ d (x) = 0 if and only if ρ s 1 (x) + · · · + ρ s d (x) = 0. The assertion for α = (0, s) with s a positive integer follows from symmetry. In view of the definition (2.6), the statement for α 1 < 0 follows from that for α 1 = 0 using the assertion in Lemma 2.5 about the derivatives of the Γ-series θ b (x).

Bases of solutions and holonomic rank
In this section we will determine the holonomic rank of the A-hypergeometric system associated with a monomial curve for all integral exponents and exhibit explicit bases of local solutions constructed in terms of the roots of the generic polynomial (0.5).
Four different scenarios need to be considered: • The exponent α ∈ I(A): In this case r(α) = d and we construct in Theorem 3.1, d − 1 local solutions which, together with the hypergeometric polynomial Φ A (α) define a basis of solutions.
The holonomic rank equals d and we have from Theorem 2.4 a basis of algebraic solutions.
In Theorem 3.7 we determine the holonomic rank r(α) for all α ∈ Z 2 . Our starting point is a result of Adolphson ([2,Corollary 5.20]) which states that even without assuming that the ring R is Cohen-Macaulay, the equality r(α) = vol(P ) holds for so-called seminonresonant exponents α. In our particular case, this condition is equivalent to α being in the Euler-Jacobi cone (1.13).
Proof: Since Ψ A (α) = d j=1 ψ j (α) = 0 it follows that χ(α) satisfies the equations (0.2) with exponent α. In order to check that the higher-order equations (0.1) are satisfied as well we show, first of all, that if α 1 > 0, then We claim that the second summand is identically zero. In fact, where, we recall where t ∈ C * , and iii) For M sufficiently large and j = 1, . . . , m, Proof: In view of (3.7), the first assertion follows from (3.6) together with the fact that ρ j (t * x) = t −1 ρ j (x), j = 1, . . . , d. Proof: In view of Theorem 2.4, it will be enough to show that χ(α) is not an algebraic function. In fact, we will show that its orbit under the monodromy action of π 1 (C m+2 \ Σ) is infinite.
It is straightforward to check that R(α) is algebraic and invariant under the monodromy action, i.e. R(α) is a rational function. Since we have just shown that the function d j=1 ψ j (α) log ρ j is not algebraic, the proof is complete. ⋄ 3.6. Remark: Note that since the function χ(α) is not algebraic, any rational A-hypergeometric function R with exponent α in the Euler-Jacobi cone must be a linear combination of ψ 1 (α), . . . , ψ d (α). On the other hand, with similar arguments as in the proof of Theorem 3.5, it is possible to show the existence of a loop γ whose action on the roots is a cyclic permutation. It is then easy to see that R must be a constant multiple of d j=1 ψ j (α), and therefore it must vanish. This gives an alternative proof of Theorem 1.9.
Proof: Note, first of all, that the lower bound follows from Theorem 2.4 (for α ∈ E 0 (A) ∪ E d (A)), Theorem 3.1 (for α ∈ I(A)), and Theorem 3.5 (for α ∈ J(A)). Suppose now that α is in the Euler-Jacobi cone EJ . Then, as we observed before, α is semi-nonresonant in the sense of Adolphson and it follows from [2, Corollary 5.20] that r(α) = d. For any α ∈ Z 2 , there exist u ∈ N m+2 such that α −A.u lies in EJ , and, because of Theorem 1.9, for any such u the kernel of the linear map contains R(α). We will determine the dimension of H(α) by computing the kernel and the image of D u for suitable u. Suppose first that α ∈ J(A). For u = ℓ e k 1 , ℓ >> 0, we have α − A.u ∈ EJ and, it follows from Proposition 1.2 and Corollary 1.5 that ker(D u ) = R(α) = {0}. Therefore, D u is a monomorphism, which implies that dim H(α) ≤ d. Since it is at least d, we deduce dim H(α) = d. Suppose now that α ∈ J(A), then dim(R(α)) = 1 or 2. We can again choose u = ℓ e k 1 , ℓ >> 0, so that β := α −A.u ∈ EJ . As the kernel of D u is precisely R(α) it will be enough to show that, for some ℓ sufficiently large, the dimension of the image of D u is d − 1. From (2.6) and Theorem 3.5, we deduce that the functions ψ j (β) generate a subspace of the image of dimension at least d − 1. The proof will be completed by showing that the function χ(β) defined in (3.6) is not in the image D u (H(α)).
Consider first the case α ∈ E 0 (A). Choosing ℓ = s d, we factor D u = D sd k 1 )s, 0). It is enough to show that χ(β) ∈ D k 1 s d (H(α ′ )). Note that α ′ ∈ E 0 (A) as well and, therefore, we may assume without loss of generality that α 1 < 0 and β = α − ℓe d ∈ EJ for some sufficiently big ℓ. Let χ(α) be as in (3.6). Since β ∈ EJ , it follows from Lemma 3.3 that D ℓ d (χ(α)) = χ(β). Therefore, if φ ∈ H(α) is such that D ℓ d (φ) = χ(β) we must have φ = χ(α) + F where F depends polynomially on x d . On the other hand, because of iii) in Proposition 3.4 and the fact that φ is hypergeometric, we have for all M large enough. This implies that F is polynomial on x k 1 , . . . , x k m as well. But, it follows from i) in Proposition 3.4 that F (t * x) = t α 2 F (x) + t α 2 log t Ψ(α)(x) which is impossible since the fact that the action of t does not affect x 0 implies that F (t * x) is polynomial in t. By symmetry, the result also holds for α ∈ E d (A). Thus, it remains to consider the case α ∈ I(A). For ℓ large enough, so that α ′ 1 = α 1 − ℓ(d − k 1 ) < 0, we have α ′ = (α ′ 1 , α 2 ) ∈ E 0 (A) and an argument similar to the one above yields the result.  is an isomorphism. ⋄