Hadamard products of symbolic powers and Hadamard fat grids

In this paper we address the question if, for points $P, Q \in \mathbb{P}^{2}$, $I(P)^{m} \star I(Q)^{n}=I(P \star Q)^{m+n-1}$ and we obtain different results according to the number of zero coordinates in $P$ and $Q$. Successively, we use our results to define the so called Hadamard fat grids, which are the result of the Hadamard product of two sets of collinear points with given multiplicities. The most important invariants of Hadamard fat grids, as minimal resolution, Waldschmidt constant and resurgence, are then computed.


Introduction
In the last few years, the Hadamard products of projective varieties have been widely studied from the point of view of Projective Geometry and Tropical Geometry.The main problem in this setting is the behaviour of the Hadamard product between varieties with many points with zero coordinates.
The paper [8], where Hadamard products of general linear spaces are studied, can be considered the first step in this direction.Successively, the second author, with Calussi, Fatabbi and Lorenzini, in [5] and [6], address the Hadamard products of linear varieties not necessarily in general position.Moreover, they show that the Hadamard product of two generic linear varieties V and W is projectively equivalent to a Segre embedding and that the Hilbert function of the Hadamard product V W of two varieties, with dim(V ), dim(W ) ≤ 1, is the product of the Hilbert functions of the original varieties V and W .
An important result contained in [8] concerns the construction of star configurations of points, via Hadamard product.This construction found a generalization in [10] where the authors introduce a new construction using the Hadamard product to present star configurations of codimension c of P n and which they called Hadamard star configurations.Successively, the first author and Calussi introduce a more general type of Hadamard star configuration; any star configuration constructed by their approach is called a weak Hadamard star configuration.In [2] they classify weak Hadamard star configurations and, in the case c = n, they investigate the existence of a (weak) Hadamard star configuration which is apolar to the generic homogeneous polynomial of degree d.The use of Hadamard products in this context permits a complete control both in the coordinates of the points forming the star configuration and the equations of the hyperplanes involved on it.This fact leads to the question if other interesting set of points can be constructed by Hadamard product.Recent results in this direction can be found in [3], where the second author along with C. Capresi and D. Carrucoli built Gorenstein set of points in P 3 , with given h−vector, by creating, via Hadamard products, a stick figure of lines to which they apply the results of Migliore and Nagel [21], and also in [9] where the third author along with E. Carlini, M.V. Catalisano and G. Favacchio found a relation between star configurations where all the hyperplanes are osculating to the same rational normal curve (contact star configurations) and Hadamard products of linear varieties.
This paper addresses another question, proposed by the second author during the CMO Workshop "Ordinary and Symbolic Powers of Ideals" (May 14-19, 2017, Oxaqua, Mexico) stating Question 1.1.Is it true that for P, Q points in P 2 , I(P ) m I(Q) n = I(P Q) m+n−1 ?
The answer of this question is given in Theorem 4.2.Successively this result is used in Section 5 to define the so called Hadamard fat grids, which are the result of the Hadamard product of two sets of collinear points with given multiplicities.Finally, we compute algebraic invariants of Hadamard fat grids, as minimal degree, minimal resolution, Waldschmidt constant and resurgence.We also point out that the results of Section 5 enlarge the known literature on the minimal graded resolution of sets of fat points in P 2 with all the same multiplicities supported on a complete intersection (see for instance [11,18]).
The authors thank G. Favacchio for some useful discussions on a preliminary version of the paper.

Preliminary results on join and Hadamard products of ideals
be a polynomial ring over an algebraically closed field.Let m = x 0 , . . ., x N be the homogeneous irrelevant ideal.
Let I j (y j ) be the image of the ideal I j in K[x, y] under the map x → y j .Then the join I 1 * I 2 * • • • * I r , as defined in [23], is the elimination ideal and their Hadamard product I 1 I 2 • • • I r , as defined in [8], is the elimination ideal We define the r-th secant of an ideal I ⊂ K[x] to be the r-fold join I with itself: Similarly we define the r-th Hadamard power of an ideal I ⊂ K[x] to be the r-fold Hadamard product of I with itself: Remark 2.1.The Hadamard product of points in a projective space is a coordinate-wise product as in the case of the Hadamard product of matrices.Let p, q ∈ P N be two points with coordinates [p 0 : p 1 : • • • : p N ] and [q 0 : q 1 : • • • : q N ] respectively.If p i q i = 0 for some i, the Hadamard product p q of p and q, is defined as p q = [p 0 q 0 : p If p i q i = 0 for all i = 0, . . ., n then we say p q is not defined.This definition extends to the Hadamard product of varieties in the following way.Let X and Y be two varieties in P N .Then the Hadamard product X Y is defined as X Y = {p q : p ∈ X, q ∈ Y, p q is defined}.
Thus the defining ideal of the Hadamard product X Y of two varieties X and Y , that is, the ideal I(X Y ), equals the Hadamard product of the ideals I(X) I(Y ) (see [8,Remark 2.6]).Definition 2.2.If I is a homogeneous ideal of S, the m-th symbolic power of I is defined as where Ass(I) denotes the set of the associated primes of I.If I is a radical ideal then In this paper we will always deal with ideals of fat points.Given distinct points p 1 , . . ., p s ∈ P N and nonnegative integers m i (not all 0), let Z = m 1 p 1 + • • • + m s p s denote the scheme (called a set of fat points) defined by the ideal I Z = s i=1 I(P i ) m i , where I(P i ) is the ideal generated by all homogeneous polynomials vanishing at P i .When all the m i are equal, we say that Z is a homogeneous set of fat points.For ideals of this type, the m-th symbolic power can be simply defined as If I m is the regular power of an ideal I, then there is clearly a containment I m ⊆ I (m) and indeed, for 0 = I R, I r ⊆ I (m) holds if and only if r ≥ m [1, Lemma 8.1.4].A much more difficult problem is to determine when there are containments of the form I (m) ⊆ I r .The results of [12] and [20] show that I (m) ⊆ I r holds whenever m ≥ N r, where N is the codimension of the ideal.
As a stepping stone, the second author and B. Harbourne ([4]) introduce an asymptotic quantity which we refer to as the resurgence, namely ρ(I) = sup{m/r : I (m) ⊆ I r }.
In particular, if m > ρ(I)r, then one is guaranteed that I (m) ⊆ I r .There are still, however, very few cases for which the actual value of ρ(I) is known, and they are almost all cases for which ρ(I) = 1.For example, by Macaulay's unmixedness theorem it follows that ρ(I) = 1 when I is a complete intersection (see Proposition 2.4).And if I is a monomial ideal, it is sometimes possible to compute ρ(I) directly.In this paper we will show that if I is the defining ideal of a Hadamard fat grid, ρ(I) = 1 even if I is a not complete intersection (see Proposition 5.18 and Corollary 5.19).
We mention few results which will be useful for the rest of the paper.
Proof.The proof is analogous to [22,Lemma 2.6].A polynomial f belongs to (∩J l ) K if and only if f (y From the previous lemma we get the following Corollary 2.6.Let I, J be two ideals in K[x 0 , . . ., x N ] with primary decomposition respectively Remark 2.7.The right-hand term in (1) is not in general a minimal primary decomposition of I J since it can contains some redundant term.As a matter of fact consider the ideals in K[x 0 , x 1 , x 2 ] with primary decompositions which does not have four ideals as expected.This is due to the fact that Proof.Let u = (u 0 , . . ., u N ) an integer vector and denote by Then the proof follows from [23, Lemma 2.3] and [22,Lemma 2.6].Definition 2.9.Let H i ⊂ P n , i = 0, . . ., n, be the hyperplane x i = 0 and set In other words, ∆ i is the i−dimensional variety of points having at most i + 1 nonzero coordinates.Thus ∆ 0 is the set of coordinates points and ∆ n−1 is the union of the coordinate hyperplanes.Note that elements of ∆ i have at least n − i zero coordinates.We have the following chain of inclusions: Let R be the ring K[x 0 , x 1 , . . ., x n ].Given a vector of nonnegative integers I = (i 0 , . . ., i n ), we denote by X I the monomial Theorem 2.11.Let I be an ideal in C[x 0 , . . ., x n ] and consider a point is a generating set for I(P ) I.Moreover, if f 1 , . . ., f s is a Gröbner basis for I, then f P 1 , . . ., f P s is a Gröbner basis for I(P ) I.

Preliminary results on ACM sets of fat points in
In this section we recall some known results and a standard technique used for sets of fat points in P 1 × P 1 since they are the main tools to find a minimal graded free resolution of special sets of fat points in P 2 called Hadamard fat grids (see Definition 5.1 in Section 5).Indeed, we prove that a Hadamard fat grid inherits some properties from an arithmetically Cohen-Macaulay set of a fat points in P 1 × P 1 , such as that its defining ideal is generated by product of linear forms.We should refer the reader to Chapters 4 and 6 in [17] for more details on arithmetically Cohen-Macaulay sets of fat points in P 1 × P 1 .
Let Z be a finite set of points in P 1 × P 1 , and let R/I Z denote the associated N 2 -graded coordinate ring.When R/I Z is Cohen-Macaulay, that is depth R/I Z = dimR/I Z = 2, then Z is called an arithmetically Cohen-Macaulay (ACM for short) set of points.Remark 3.1.For the ease of the reader, we now recall a standard argument that relates sets of (fat) points in P 1 × • • • × P 1 (n-times) and sets of (fat) points in P 2n−1 in the case n = 2 (see for more details Section 3, Theorem 3.2, Corollaries 3.3 and 3.4 in [13] and, for instance, also [14,15,16]).We observe that in our case in P 1 a point is also a hyperplane, and this allows us to use hyperplane sections and related constructions for our study. Let ] be the coordinate ring for P 1 × P 1 , which we shall also view as the coordinate ring for P 3 .Let Z ⊂ P 1 × P 1 be a finite set of fat points.Since I Z defines both a set of fat points in P 1 × P 1 and a union of linear varieties (fat lines) in P 3 , by abuse of notation, we denote by Z also the subvariety of P 3 defined by this ideal.
In P 1 × P 1 a complete intersection of type (r, 0) and (0, s), or simply an (r, s)-grid, can be viewed as 2 families of linear forms, and we denote by A 1,i the linear combinations of x 1,0 and x 1,1 , and by A 2,i the linear combinations of x 2,0 and x 2,1 .Set Let T be the polynomial ring in r + s variables a 1,1 , . . ., a 1,r , a 2,1 , . . ., a 2,s .We form the monomial ideal in T given by the intersection of ideals of the form (a 1,i , a 2,j ) m ij corresponding to the components of X.This intersection defines a height 2 monomial ideal, J ⊂ T .Following Theorem 2 in [13], a set of fat points Z in P 1 × P 1 is ACM if and only if J ⊂ T is CM, where T is the ring previously defined.Since Z can be viewed as an ACM set of lines in P 3 (it is 1-dimensional) we can still find a suitable hyperplane section in order to get a set X of points in P 2 that shares the same Betti numbers as Z.In particular, with the above described method, we will show that a Hadamard fat grid in P 2 share the same graded Betti numbers as an ACM set of fat points in P 1 × P 1 (see Theorem 5.6).
In the sequel, it useful to consider in Z×Z and in N×• • •×N the partial (lexicographic) ordering induced by the usual one in Z and in N, respectively; we will denote it by "≤".
Consider an (r, s)-grid, and let For each i = 1, . . ., r and j = 1, . . ., s, set For i = 1 . . ., r, let S be the set of the s-tuples of type The next Lemma 3.2 shows that the set S of tuples associated to M and N is totally ordered.It will be useful to find the graded Betti numbers of a Hadamard fat grid (see Section 5).
We have Lemma 3.2.For i = 1 . . ., r and for h = 0 . . ., m 1 + n s − 2, the set S is a totally ordered set of tuples with usual lex ordering.
Proof.Let r = 1.From construction, it is max Since 1 for all j 1 = j 2 , and since Hence S is totally ordered for r = 1.
Suppose r > 1 and the lemma true for all the sets S of s-tuples with r < r and prove it for S. Let 1 ≤ k ≤ r an integer such that m k + n j 1 − 1 ≤ m k + n j 2 − 1 for all j 1 = j 2 .From our construction, we have that k = r.
Define for i = 1 . . ., r where By induction S is totally ordered.We observe that and by construction and for all i = 1 . . ., r = k, we have Thus the set S is totally ordered.
Finally, we define A Z to be the set of (m -tuples, that is, the set of ( r i=1 m i + rn s − r)-tuples that one gets by rearranging the elements of ÃZ in descending order.
We now recall how to compute the graded Betti numbers of I(Z) when Z is an ACM set of fat points in P 1 × P 1 .Let A Z = (α 1 , . . ., α m ) be the tuple associated to Z. Define the following two sets from A Z : where we take α −1 = 0.With this notation, we get that a minimal bigraded free resolution of an ACM set of points in P 1 × P 1 is given by Theorem 3.3.[17, Theorem 6.27] Suppose that Z is an ACM set of fat points in P 1 × P 1 with A Z = (α 1 , . . ., α m ).Let C Z and V Z be constructed from A Z as above.Then a bigraded minimal free resolution of I(Z) is given by The following result will be one of the main tool that allows us to find that the Betti numbers of a given Hadamard fat grid in P 2 whose support is a complete intersection.Proposition 3.4.Let Z be a set of fat points in P 1 × P 1 whose support is a complete intersection of type (r, s) (or (r, s)-grid)) and whose multiplicities are of type m i + n j − 1 for i = 1, . . ., r and j = 1, . . ., s.Then Z is ACM.
Proof.From Theorem 6.21 in [17] and Lemma 3.2, we get the conclusion.

Hadamard product of symbolic powers of ideals of points
We focus now our attention on the Hadamard product I(P ) m I(Q) n , where P, Q ∈ P 2 .Hence, we work on the polynomial ring S = K[x] = K [x 0 , x 1 , x 2 ], over an algebraically closed field and we still denote by m = x 0 , x 1 , x 2 the irrelevant ideal.Since we will multiply, by Hadamard, only two ideals, we change the name of the variables involved in the process, with respect to our original definition.More precisely we will use the extra variables y = (y 0 , y 1 , y 2 ) and z = (z 0 , z 1 , z 2 ) for the Hadamard product of the ideals I and J: I J = (I(y) + J(z) We set H = (x 0 − y 0 z 0 , x 1 − y 1 z 1 , x 2 − y 2 z 2 ).We start with some preliminary results.
Lemma 4.1.Let P = [p 0 : p 1 : p 2 ] be a point, then Proof.For i) we can use Theorem 2.11.Hence the generators of I(P ) m t are the Hadamard transformations, with respect to P , of the monomials in m t .According to Definition 2.10, such transformations of the monomials in m t are the same monomials scaled by a constant P I , hence I(P ) m t = m t .
For ii) we can assume, without loss of generality, that p 0 = 0 where the last equality follows from the fact that y 0 annihilates y 0 z 0 .Hence m t ⊆ I(P ) m t .Similarly, if the point P has p 0 = p 1 = 0 one has The following result gives a positive answers to Question 1.1 when the points P and Q have non-zero coordinates.Theorem 4.2.Let P and Q be two points in P 2 \ ∆ 1 .Then for m, n ≥ 1 one has Proof.From Proposition 2.3 we know that For this aim, we define Hence we have the following sequence of equalities Denote by I the ideal in the last equality in the previous formula, i.e.
Since the elements in Ĥ can be seen as ( 6) we can substitute the ideal Ĥ with the ideal generated by i − y i z i = 0, k i − z i y i = 0, k i − y i z i = 0, By definition, k i −y i z i = 0 generates I(P ) i −y i z i = 0 generates I(P ) m n .By Lemma 4.1 (i), we know that A simple calculation shows that any term of a form f of degree ∈ m n (z ) which follows that any term containing y i r i z i t i is zero as well.A similar calculation for terms containing k(3) follows that the term containing y i r i z i t i is zero.Therefore we conclude that any term of a form f of degree d in Ĩ containing k is zero.Hence we can simply cancel k (4) i from ( 7).We have that Therefore, and, by Proposition 2.3, the last ideal is equal to I(P Q) m+n−1 and the theorem is proved.
Remark 4.3.If we remove the condition P, Q ∈ P 2 \ ∆ 1 in Theorem 4.2 we are able to proof only the following inclusion To prove this inequality is enough to apply the proof of Theorem 4.2 using part ii) of Lemma 4.1.
We observe that if P Q is defined, and m = n = 1 then it follows from the definition of Hadamard product that I(P ) I(Q) = I(P Q), and hence Question 1.1 has an affirmative answer.Theorem 4.2 shows that Question 1.1 has an affirmative answer, when we consider points which are not in the coordinates lines.When one of the point is taken in a coordinate line, the formula in Question 1.1 is no more valid, as stated in the following: Proof.(a): without loss of generality let P = [p 0 : p 1 : p 2 ] with p i = 0 and assume that Q ∈ ∆ 0 has exactly one non-zero coordinate, that is Q = [q 0 : 0 : 0].Assume that m = 1.
Since P Q = Q, therefore we only need to show that I(P ) I(Q) n = I(Q) n .We have that Note that in general we have We have that 2 / ∈ I(P Q) m+n−1 therefore we have that I(P Q) m+n−1 = I(P ) m I(Q) n .If m = 1 we have: (b): Let m ≤ n.Without loos of generality assume that P = [p 0 : 0 : p 2 ] and Q = [q 0 : q 1 : 0].We have that I(P Q) = (x 1 , x 2 ).
Now again without loss of generality assume that P = [p 0 : p 1 : 0] and Q = [q 0 : q 1 : 0].One can see that x m 2 ∈ I(P ) m I(Q) n .Since x m 2 ∈ I(P Q) m+n−1 therefore the equality fails.Similarly one can show that

Hadamard Fat Grids
In this section we introduce and study a particular set of fat points in P 2 , that we call Hadamard fat grid, whose support is a complete intersection.In particular, in Theorem 5.7 we describe a graded minimal free resolution of a Hadamard fat grid using the results from Section 3, enlarging the known literature on the minimal graded resolution of homogeneous sets of fat points in P 2 supported on a complete intersection.We also compute the Waldschmidt constant and the resurgence of the ideal defining a Hadamard fat grid (see Proposition 5.17 and Corollary 5. 19).
Let P M = {P 1 , . . ., P r } and Q N = {Q 1 , . . ., Q s } be two sets of collinear points in P 2 \ ∆ 1 with assigned positive multiplicities, respectively, M = {m 1 , . . ., m r } and N = {n 1 , . . ., n s }.In terms of ideals we have Then the set of fat points defined by I(P M ) I(Q N ), is called a Hadamard fat grid and it is denoted by HF G(P M , Q N ).
According to Theorem 4.2 and Corollary 2.6, the ideal of HF G By Lemma 3.1 in [8], we know that the Hadamard product Z S of a collinear set Z by a point S ∈ P 2 \ ∆ 1 is still a collinear set lying on the line S L, where Z ⊂ L. Hence, if we denote by P and Q the lines in which the sets P M and Q N lie respectively, one has P i Q j ∈ P Q j for all i = 1, . . ., r and for all j = 1, . . ., s.
And similarly, P i Q j ∈ P i Q for all j = 1, . . ., s and for all i = 1, . . ., r.
This shows that HF G(P M , Q N ) has the structure of a planar grid.Specifically, it is a set of fat points whose support is a complete intersection of type (r, s) in P 2 .
Example 5.2. Figure 1 shows a Hadamard fat grid for r = 4 and s = 5; Figure 1(a) shows the geometric structure, with all lines involved in the grid, while in Figure 1(b) we represent the multiplicities of each point of the grid.
Remark 5.3.From Figure 1(b) we see that the multiplicities of the points in the grid have a specific behaviour.If we assume that m i ≤ m i+1 , for i = 1, . . ., r − 1 and n j ≤ n j+1 for each j = 1, . . ., s − 1, then the multiplicities of P i Q j and P i Q j+1 differ of n j+1 − n j for all i = 1, . . ., r. Similarly the multiplicities of P i Q j and P i+1 Q j differ of m i+1 − m i for all j = 1, . . ., s. Hence the Hadamard fat grids are a subclass of all possible fat grids in the plane.
From the rest of the paper we assume that s ≥ r and the multiplicities are ordered in non-decreasing order, that is m i ≤ m i+1 , for i = 1, . . ., r − 1 and n j ≤ n j+1 for each j = 1, . . ., s − 1.We denote by To short the notation, set m ij = m i + n j − 1 for i = 1, . . ., r and j = 1, . . ., s.
Lemma 5.4.Let Z be a set of fat points in P 1 ×P 1 whose support is a complete intersection of type (r, s) (or (r, s)-grid)) and whose multiplicities m ij are the same as a Hadamard Fat grid HF G(P M , Q N ).Then Z share the graded same Betti numbers as a set of fat points Y in P 2 .
Proof.From Lemma 3.4, Z is an ACM set of fat points in P 1 × P 1 and from Remark 3.1, its ideal I Z defines a set of fat lines in P 3 , and it still continues to be ACM and 1-dimensional.Hence, after a "proper hyperplane section" we get a set of fat points Y in P 2 that has the same graded Betti numbers as Z.That is, if is a proper hyperplane section, we have the former of which is ACM.
Lemma 5.5.Let X and X be two sets of fat points in P 2 whose support is an (r, s)−grid and with the same multiplicities m ij as a HF G(P M , Q N ).Then X and X share the same graded Betti numbers.
Proof.Following again the same method as Theorem 3.2 in [13], Remark 3.1 and Lemma 5.4, we construct the polynomial ring T in the r + s new variables and form a height 2 monomial ideal J in T given by the intersection of ideals of the form (a 1,i , a 2,j ) m ij corresponding to the components of X.Thus, starting from the scheme defined by the ideal J, after suitable sequences of proper hyperplane sections we can construct a set (fat) of lines in P 3 that corresponds to an ACM set of fat points Z in P 1 × P 1 and since, from Proposition 3.4, Z is ACM then, from Theorem 3.2 in [13], T /J is CM .Analogously, we can construct a set of lines in P 3 that corresponds to an ACM set of fat points Z in P 1 × P 1 and since Z is ACM then T /J is CM.Again, starting from the scheme defined by the ideal J, after two other suitable sequences of hyperplane sections, we get two sets of fat points X and X in P 2 that both share the same graded Betti numbers as Z and Z ( and as J).
We show the main two results of this section.
Theorem 5.6.Let X be a Hadamard fat grid HF G(P M , Q N ) in P 2 and Z be an ACM set of fat points in P 1 × P 1 supported on an (r, s)-grid with the same multiplicities m ij as the Hadamard fat grid X.Then X and Z share the same Betti numbers.
Proof.From Lemma 5.4 we can construct a set of fat points Y an (r, s)-grid in P 2 that preserves the same multiplicities on m ij and the same Betti numbers as Z.From Lemma 5.5, X and Y (and Z) share the graded same Betti numbers.
We now are able to compute a minimal free graded resolution of a Hadamard fat grid.We will use the results from previous Section 3, to prove the main result of this section.
Theorem 5.7.Let X = HF G(P M , Q N ) be a Hadamard fat grid in P 2 .Then a graded minimal free resolution of I(X) is given by Proof.We can construct A X = (α 1 , . . ., α m ) associated to X using the method described as Section 3. Applying Theorem 3.3 and Theorem 5.6, we get the conclusion.
From Corollary 3.4 in [13] and Theorem 5.7, we have that Corollary 5.9.If X is a Hadamard fat grid X = HF G(P M , Q N ) in P 2 , then its homogeneous ideal is minimally generated by products of linear forms of type H i and V j .Because of the subadditivity of α, this limit exists (see Lemma 2.3.1 of [7]).Moreover, α(I) > 0 (see Lemma 2.3.2 of [7]).
We need some preliminary result.We can have a similar argument for the other summation as follows. Since One can also observe that deg V ns−ns+t−1 1 = 0 for t = 1.Hence for t = 1 we have, For t = 1, . . ., m r + n s , let G = {g 1 , g 2 , . . ., g t } be a set of minimal generators of I(P M , Q N ).We claim that g t ≥ g ns−n s−r+1 for 1 ≤ t ≤ n s − n s−r+1 .
be a homogenous polynomial, of degree d, of the form f = |I|=d a I X I and consider a point P ∈ P n \ ∆ n−1 .The Hadamard transformation of f by P is the polynomial (3) f P = |I|=d a I P I X I .

( 5 )
I(P ) m = I(P ) * m m and I(Q) n = I(Q) * m n .Thus, I(P ) m I(Q) n = [I(P ) * m m ] [I(Q) * m n ] can be computed by applying successively the definition of join of two ideals and of Hadamard products of ideals.

Proposition 4 . 4 .
Let P and Q be two points in P 2 .(a) If P ∈ P 2 \ ∆ 1 and P Q is defined then for m = 1 and n ≥ 1 Question 1.1 has an affirmative answer.(b) If P, Q ∈ ∆ 1 \∆ 0 then Question 1.1 has no affirmative answer but for m = n = 1.

Theorem 5 . 1 •• V b 1 +k 1 •= 1
10.A minimal set of generators of the idealI(P M , Q N ) consists of m r + n s generators of types H a 1 −k • • H ar−k r • • V bs+k s for k = 0, . . ., m r + n s − 1 where we adopt the convention that H a i −k i if a i − k ≤ 0 and V b j +k j = 1 if b j + k ≤ 0.That is, a minimal set of generators is of type(8)