Primary Decomposition of Ideals Arising from Hankel Matrices

Primary Decomposition of Ideals Arising from Hankel Matrices Paul Brodhead University of Wisconsin-Madison Malarie Cummings Hampton University Cora Seidler University of Texas-El Paso August 10 2000 Abstract Hankel matrices have many applications in various fields ranging from engineering to computer science. Their internal structure gives them many special properties. In this paper we focus on the structure of the set of polynomials generated by the minors of generalized Hankel matrices whose entries consist of indeterminates with coefficients from a field k. A generalized Hankel matrix M has in its j th codiagonal constant multiples of a single variable X j. Consider now the ideal I r (M) in the polynomial ring kx 1... X m+n 1 generated by all (r r)-minors of M. An important structural feature of the ideal I r (M) is its primary decomposition into an intersection of primary ideals. This decomposition is analogous to the decomposition of a positive integer into a product of prime powers. Just like factorization of integers into primes the primary decomposition of an ideal is very difficult to compute in general. Recent studies have described the structure of the primary decomposition of I 2 (M). However the case when r > 2 is substantially more complicated. We will present an analysis of the primary decomposition of I 3 (M) for generalized Hankel matrices up to size 5 5. 1 Introduction The properties of the ideals generated by the minors of matrices whose entries are linear forms are hard to describe unless the forms themselves satisfy some strong condition. Here we compute a primary decomposition for ideals in polynomial rings that are generated by minors of Hankel matrices. To be precise let k be a field and 19

let 2 m n be integers. A generalized Hankel Matrix is defined as r 11 X 1 r 12 X 2 r 13 X 3 r 1n X n r 22 X 2 r 23 X 3 r 24 X 4 r 2n+1 X n+1 M =. r mm X m r mm+1 X m+1 r mm+2 X m+2 r mm+n 1 X m+n 1 where the X i are indeterminates and the r ij are nonzero elements of a field k. In the present work we analyze the structure of a m n generalized Hankel matrix M with m 3. In particular we determine the minimal primary decomposition of ideals generated by the 3 3 minors of M. By I 3 (M) we denote the ideal in the polynomial ring F X 1... X m+n 1 which is generated by the 3 3 minors. We denote I 2 (M) the ideal in the polynomial ring F X 1... X m+n 1 which is generated by the 2 2 minors. Let I n (M) be the primary decomposition of ideals generated by n n minors of a generalized Hankel matrix M. In previous research the structure of I 2 (M) has been described. However little is known about the cases of minors with n 3. In our research we have analyzed I 3 (M) for 3 4 matrices for 4 4 matrices and 5 5 matrices. In Section 2 we describe the primary decomposition of ideals and definitions related to the understanding of I 3 (M). In Section 3 we give the structure of I 2 (M). In Section 4 we prove that I 3 (M) for a 3 4 matrix is prime. In Section 5 we give several examples and conjectures for I 3 (M) for a 4 4 matrix. In Section 6 we discuss the symmetry of I 3 (M) for some examples with 5 5 matrices. In Section 6 we have further thoughts over the project and possible future work. 2 Primary Decomposition of Ideals The primary decomposition of an ideal in a polynomial ring over a field is an essential tool in commutative algebra and algebraic geometry. The process of computing primary decompositions of ideals is analogous to the factorization of positive integers into powers of primes. Just like factoring an integer into powers of primes finding the primary decomposition of an ideal is generally very difficult to compute. In this section we will provide the reader with some basic properties of ideals and their primary decompositions. We will first introduce several basic terms and concepts associated to ideals followed by the definition of a primary decomposition and examples. Definition 2.1 Let R be a commutative ring and I be an ideal 1. An ideal I R is irreducible if it is not the intersection of strictly larger ideals. 2. R is Noetherian if every increasing chain of ideals I 1 I 2 I n eventually becomes constant. 3. I is primary if whenever ab I and a I then b n I for some positive integer n. 4. I k x 1 x n is prime if whenever f g k x 1 x n and fg I then either f I or g I. 5. Let I R be an ideal. The radical rad(i) is the ideal 20

I= {r R: r n I for some positive integer n }. Lemma 2.2 If I is primary then I is prime. Proof. See LA. Example 2.3 1. R = Z. The only primary ideals are those of the form (p n ) for a prime number p and the zero ideal. The radical of (p n ) is equal to (p) which is a prime ideal. 2. Let R = kx y z/(xy z 2 ) and let P = ( x z) R. Then P is prime because R/P = k y is a domain. Then xȳ = z 2 P 2 but x P 2. Futhermore ȳ rad(p 2 ) = P. Hence P 2 is not primary. Note A power of a prime need not be primary even though its radical is prime. Definition 2.4 A primary decomposition of an ideal I R is a decomposition of I as an intersection I = I 1 I r of primary ideals with pairwise distinct radicals which is irredundant. Corollary 2.5 If R is a Noetherian ring then every ideal has a primary decomposition. Thus we see that the intersection of ideals is similar to the factorization of integers into their primes since every integer has a prime factorizaion. However we don t get uniqueness of the decomposition in full generality. Example 2.6 Let R = k x y. Then (x 2 xy) = (x) (x y) 2 = (x) (x 2 y) Fortunately not all is lost since the set of radical ideals associated to each primary component is unique. This motivates the following definition. Definition 2.7 Let I = Q 1 Q n with P i = rad(q i ) and P j = rad(q j ) 1. The ideals P i are called the primes associated to I and the set {P i } is denoted by Ass(I). 2. If a P i does not contain any P j j i then Q i is called an isolated component. Otherwise Q i is called an embedded component. Example 2.8 Consider I = (z 2 zx) = (z) (z 2 x) Here (z) is an isolated component but (z 2 x) is embedded since (z x)= (z 2 x) contains (z) = (z) 21

Theorem 2.9 The isolated components of a primary decomposition are unique. Proof. See FR. We will close this section with an example of the computation of the primary decomposition of a monomial ideal. Example 2.10 Let I = (z 3 x 2 y yx 2 z) k x y z. Then I = (z 3 x 2 yx 2 z) (z 3 y yx 2 z) = (z 3 x 2 y) (z 3 x 2 ) (z 3 x 2 z) (z 3 y) (z 3 y x 2 ) (z 3 y z) Now observe that (z 3 x 2 ) (z 3 x 2 y) and (z 3 y) (z y) and (z 3 x 2 ) (z x 2 ) so we can delete (z 3 x 2 y) (z y) (z x 2 ). Thus we get the primary decomposition I = (z 3 x 2 ) (z 3 y) 3 Structure of I 2 (M) In recent studies uerrieri and Swanson computed the minimal primary decomposition of ideals generated by 2 2 minors of generalized Hankel matrices. They showed that the primary decomposition of I 2 (M) is either primary itself or has exactly two minimal components and sometimes also one embeded component. They also identified two integers s and t intrinsic to M which allow one to decide if I 2 (M) is prime. To define s and t we first need to to transform M into a special form by scaling the variables. The scaling of variables does not change the number of primary components or the prime and primary properties. So without loss of generality M becomes the following generalized Hankel matrix with all r ij units in F. We can define s as: X 1 X 2 X 3 X n X 2 X 3 X 4 X n+1 X 3 r 34 X 4 r 35 X 5 X n+2 M = X 4 r 45 X 5 r 46 X 6 X n+3. X m r mm+1 X m+1 r mm+2 X m+2 X m+n 1 s=min { j 4: i 3 such that r ij 1}. The integer t is defined in a similar way to s for a matrix obtained from rotating M 180 degrees and then rescaling the variables. So without loss of generality M is transformed to 22

M = X 1 r 12 X 2 r 13 X 3 r 1n 1 X n 1 X n X 2 r 23 X 3 r 24 X 4 r 2n X n X n+1 X m 2 r m 2m 1 X m 1 r m 2m X m r m 2m+n 4 X m+n 4 X m+n 3 X m 1 X m X m+1 X m+n 3 Xm + n 2 X m X m+1 X m+2 X m+n 2 Xm + n 1 with all r ij units in F We can define t as: t=max {j m + n 4: i m 2 such that j< n+i-1 and r ij r ij+1 } Now that we have s and t we can now describe the structure of I 2 (M) as shown in the following theorem. Theorem 3.1 Let. be ideals in the ring F X 1 X m+n 1. Q 1 = I 2 (M) + (X s X m+n 1 ) Q 2 = I 2 (M) + (X 1 X t ) Q 3 = I 2 (M) + (X1 m+n 4 Xm+n 1) m+n 4 I. Q 1 Q 2 and Q 3 are primary to the prime ideals (X s X m+m 1 ) (X 1 X m+n 2 ) and (X 1 X m+n 1 ) respectively. II. If s and t do not exist then I 2 (M) is a prime ideal. III. If s > t then I 2 (M) = Q 1 Q 2 is a primary decomposition. IV. If s t then I 2 (M) = Q 1 Q 2 Q 3 is an irredundant primary decomposition. 4 I 3 (M) for 3x4 Hankel Matrices In the primary decomposition of I 2 (M) we saw that each primary component Q i looks like I 2 (M) + J i for some ideal J i. In a similar way we have the same kind of breakdown for each primary component in the primary decomposition of I n (M). Proposition 4.1 Let M be a generalized Hankel matrix and let be a roebner basis of I n (M). If the primary decomposition of I n (M) is Q 1 Q 2... Q k then each Q i is of the form I n (M) + J i for some ideal J i. Furthermore if the set of generators for each Q i is {h 1 h 2... h mi } then the set of generators for each J i is precisely {h 1 h2... hmi } where each hj is the normal form of hj with respect to. 23

Proof. Since I n (M) = Q 1 Q 2... Q n we have that I n (M) is a sub-ideal of Q i for all i where 1 i n. Now suppose that Q i is the ideal generated by {h 1 h 2... h mi } and the röbner basis for I 3 (M) is = {g 1 g 2... g n }. Then taking the normal form of each Q i with respect to gives us {h 1 h2... hmi }. So we have Q i =<h 1 h 2... h mi > = <g 1 g 2... g n > + <h 1 h2... hmi > = I n (M)+ <h 1 h2... hmi >. Now we have that I n <h 1 h2... hmi > and In <h 1 h2... hmi >. Therefore we have that each Q i is precisely I n (M)+ <h 1 h2... hmi >. Thus each Ji is precisely <h 1 h2... hmi >. This finishes the proof. QED The last proposition is used in our algorithms for finding the primary decompostion of I 3 (M). Utilizing this proposition we now give the primary decomposition of I 3 (M) for any generalized 3 4 Hankel matrix M. Theorem 4.2 If M is any generalized 3 4 Hankel matrix then I 3 (M) is prime. Proof. By section two it is enough to consider the primary decomposition of a matrix of the following form: M = X 1 X 2 X 3 X 4 X 2 X 3 X 4 X 5 X 3 r 34 X 4 r 35 X 5 X 6 Now by considering r 34 and r 35 as variables SINULAR computed the primary decomposition of I 3 (M ). Our output was just I 3 (M ) itself namely the ideal generated by: r 35 x 1 x 2 5 r 35 x 2 x 4 x 5 x 1 x 4 x 6 + x 2 x 3 x 6 x 2 3x 5 + x 3 x 2 4 r 34 x 3 x 4 x 5 r 34 x 3 4 r 35 x 2 x 2 5 + r 35 x 3 x 4 x 5 + x 2 x 4 x 6 x 2 3x 6 r 34 x 1 x 4 x 5 r 34 x 2 x 2 4 x 1 x 3 x 6 + x 2 2x 6 x 2 x 3 x 5 + x 2 3x 4 r 34 x 1 x 2 4 r 34 x 2 x 3 x 4 r 35 x 1 x 3 x 5 + r 35 x 2 2x 5 x 2 x 3 x 4 + x 3 3 Hence I 3 (M ) is itself primary. Now by Lemma 1 this implies that I 3 (M ) is prime. Our goal was to show that I 3 (M ) is prime. However after one more SIN- ULAR computation we found that I 3 (M )=I 3 (M ). Therefore I 3 (M ) is prime. This finishes the proof. QED 24

5 I 3 (M) of a 4 4 Hankel Matrix As in the 3 4 Hankel matrix case we can assume for the 4 4 Hankel matrix that the first two rows and the first and last columns have coefficients equal to one. The remaining four coefficients r ij can assume any value. Thus we assume the 4 4 Hankel matrix takes on the following form: M = X 1 X 2 X 3 X 4 X 2 X 3 X 4 X 5 X 3 r 34 X 4 r 35 X 5 X 6 X 4 r 45 X 5 r 46 X 6 X 7 r34 X In the coefficient matrix R k (M) = 4 r 35 X 5 there are fifteen possible combinations where some r ij 1. According to many examples we computed it seems r 45 X 5 r 46 X 6 clear that the primary decomposition of I 3 (M) for these fifteen matrices breaks up into three cases. The primary decomposition I 3 (M) can be equal to one two or three ideal components. It is our belief that similar to previous sections there are three possible choices for the primary decomposition of I 3 (M): I 3 (M) = Q 1 I 3 (M) = Q 1 Q 2 I 3 (M) = Q 1 Q 2 Q 3 In the following subsections we present each case in further detail. 5.1 I 3 (M) = Q 1 Our analysis of 4 4 Hankel matrices shows only eight possible combinations of the coefficient matrix where there exists only one ideal component. These are the possible combinations of R k (M) where not all r ij = 1 : 1 r35 1 1 R k (M) = r34 r 35 1 r 46 where k = 1.. 8. r34 1 r 45 1 r34 1 r 45 r 46 r34 1 1 r 46 1 r35 r 45 r 46 The examples computed ran on SINULAR for specific values of r ij. r34 r 35 r 45 1 r34 r 35 r 45 r 46 25

Example 5.1 M = X 1 X 2 X 3 X 4 X 2 X 3 X 4 X 5 X 3 11X 4 3X 5 X 6 X 4 4X 5 7X 6 X 7 The output obtained by SINULAR for Q 1 = I 3 (M) + J 1 is: I 3 (M) is equal to: 1=x(5)^3+10663*x(4)*x(5)*x(6)+10664*x(3)*x(6)^2+10664*x(4)^2*x(7)- x(3)*x(5)*x(7) 2=x(4)*x(5)^2-10664*x(4)^2*x(6)-10664*x(3)*x(5)*x(6)+10664*x(2)*x(6)^2+ 10664*x(3)*x(4)*x(7)-x(2)*x(5)*x(7) 3=x(3)*x(5)^2+10664*x(3)*x(4)*x(6)-x(2)*x(5)*x(6)+10664*x(1)*x(6)^2+ 10663*x(3)^2*x(7)+x(2)*x(4)*x(7)-x(1)*x(5)*x(7) 4=x(4)^2*x(5)+10663*x(3)*x(4)*x(6)+10664*x(1)*x(6)^2+10664*x(3)^2*x(7)- x(1)*x(5)*x(7) 5=x(3)*x(4)*x(5)-x(2)*x(5)^2-10664*x(3)^2*x(6)+10664*x(1)*x(5)*x(6)+ 10664*x(2)*x(3)*x(7)-10664*x(1)*x(4)*x(7) 6=x(3)^2*x(5)+x(2)*x(4)*x(5)-2*x(1)*x(5)^2-x(2)*x(3)*x(6)+x(1)*x(4)*x(6)+ x(2)^2*x(7)-x(1)*x(3)*x(7) 7=x(4)^3-x(2)*x(5)^2-10664*x(3)^2*x(6)-x(2)*x(4)*x(6)+10665*x(1)*x(5)*x(6)+ 10665*x(2)*x(3)*x(7)-10665*x(1)*x(4)*x(7) 8=x(3)*x(4)^2-2*x(2)*x(4)*x(5)+x(1)*x(5)^2+x(2)^2*x(7)-x(1)*x(3)*x(7) 9=x(3)^2*x(4)-x(2)*x(4)^2-x(2)*x(3)*x(5)+x(1)*x(4)*x(5)+x(2)^2*x(6)- x(1)*x(3)*x(6) 10=x(3)^3-2*x(2)*x(3)*x(4)+x(1)*x(4)^2+3*x(2)^2*x(5)-3*x(1)*x(3)*x(5) and J 1 = 0. Now we show that I 3 (M) which is also Q 1 is equal to: 1=x(5)^3+10663*x(4)*x(5)*x(6)+10664*x(3)*x(6)^2+10664*x(4)^2*x(7)- x(3)*x(5)*x(7) 2=x(4)*x(5)^2-10664*x(4)^2*x(6)-10664*x(3)*x(5)*x(6)+10664*x(2)*x(6)^2+ 10664*x(3)*x(4)*x(7)-x(2)*x(5)*x(7) 3=x(3)*x(5)^2+10664*x(3)*x(4)*x(6)-x(2)*x(5)*x(6)+10664*x(1)*x(6)^2+ 10663*x(3)^2*x(7)+x(2)*x(4)*x(7)-x(1)*x(5)*x(7) 4=x(4)^2*x(5)+10663*x(3)*x(4)*x(6)+10664*x(1)*x(6)^2+10664*x(3)^2*x(7)- x(1)*x(5)*x(7) 5=x(3)*x(4)*x(5)-x(2)*x(5)^2-10664*x(3)^2*x(6)+10664*x(1)*x(5)*x(6)+ 10664*x(2)*x(3)*x(7)-10664*x(1)*x(4)*x(7) 6=x(3)^2*x(5)+x(2)*x(4)*x(5)-2*x(1)*x(5)^2-x(2)*x(3)*x(6)+x(1)*x(4)*x(6)+ x(2)^2*x(7)-x(1)*x(3)*x(7) 7=x(4)^3-x(2)*x(5)^2-10664*x(3)^2*x(6)-x(2)*x(4)*x(6)+ 10665*x(1)*x(5)*x(6)+10665*x(2)*x(3)*x(7)-10665*x(1)*x(4)*x(7) 26

8=x(3)*x(4)^2-2*x(2)*x(4)*x(5)+x(1)*x(5)^2+x(2)^2*x(7)-x(1)*x(3)*x(7) 9=x(3)^2*x(4)-x(2)*x(4)^2-x(2)*x(3)*x(5)+x(1)*x(4)*x(5)+x(2)^2*x(6)- x(1)*x(3)*x(6) 10=x(3)^3-2*x(2)*x(3)*x(4)+x(1)*x(4)^2+3*x(2)^2*x(5)-3*x(1)*x(3)*x(5) Since J 1 Q 1 = I 3 (M) + 0 Q 1 = I 3 (M). Also we see that I 3 (M) = I 3 (M) so by definition of prime we have that Q 1 is a prime ideal component hence I 3 (M) is prime. 5.2 I 3 (M) = Q 1 Q 2 There are five possible combinations of the coefficient matrix for there to exist two ideal components. These are the possible combinations of the coefficient matrix of R k (M) where r ij 1: R k (M) = { r34 1 1 1 r34 r 35 1 1 1 1 r 45 r 46 1 r35 r 45 1 1 r35 1 r 46 } where k = 1.. 5. Example 5.2 M = X 1 X 2 X 3 X 4 X 2 X 3 X 4 X 5 X 3 X 4 3X 5 X 6 X 4 4X 5 X 6 X 7 The output obtained from SINULAR for Q 1 = I 3 (M) + J 1 where J 1 is an ideal composed of the following polynomials: 1=x(4)*x(5)-10664*x(3)*x(6) 2=x(2)*x(5)-10664*x(1)*x(6) 3=x(4)^2-x(2)*x(6) 4=x(3)*x(4)-x(1)*x(6) 5=x(3)^2-3*x(1)*x(5) 6=x(2)*x(3)-x(1)*x(4) 7=x(1)*x(5)^2-5332*x(1)*x(4)*x(6)+7998*x(2)^2*x(7)-7998*x(1)*x(3)*x(7) 8=x(1)*x(3)*x(5)*x(6)+15995*x(1)*x(2)*x(6)^2-7997*x(2)^2*x(4)*x(7)+ 7997*x(1)^2*x(6)*x(7) 9=x(1)*x(2)*x(4)*x(6)-2*x(1)^2*x(5)*x(6)+15994*x(2)^3*x(7)- 15994*x(1)^2*x(4)*x(7) 10=x(1)*x(2)^2*x(6)^2+10663*x(1)^2*x(3)*x(6)^2+15994*x(2)^3*x(4)*x(7)- 15994*x(1)^2*x(2)*x(6)*x(7) The Q 1 is equal to: 27

1=x(4)*x(5)-10664*x(3)*x(6) 2=x(2)*x(5)-10664*x(1)*x(6) 3=x(4)^2-x(2)*x(6) 4=x(3)*x(4)-x(1)*x(6) 5=x(3)^2-3*x(1)*x(5) 6=x(2)*x(3)-x(1)*x(4) 7=x(5)^3-12441*x(3)*x(6)^2-7998*x(3)*x(5)*x(7)+2666*x(2)*x(6)*x(7) 8=x(3)*x(5)^2-5332*x(1)*x(6)^2+7998*x(2)*x(4)*x(7)+7997*x(1)*x(5)*x(7) 9=x(1)*x(5)^2-5332*x(1)*x(4)*x(6)+7998*x(2)^2*x(7)-7998*x(1)*x(3)*x(7) 10=x(1)*x(3)*x(5)*x(6)+15995*x(1)*x(2)*x(6)^2-7997*x(2)^2*x(4)*x(7)+ 7997*x(1)^2*x(6)*x(7) 11=x(1)*x(2)*x(4)*x(6)-2*x(1)^2*x(5)*x(6)+15994*x(2)^3*x(7)- 15994*x(1)^2*x(4)*x(7) 12=x(1)*x(2)^2*x(6)^2+10663*x(1)^2*x(3)*x(6)^2+15994*x(2)^3*x(4)*x(7)- 15994*x(1)^2*x(2)*x(6)*x(7) and the output for Q 2 = I 3 (M) + J 2 where J 2 is an ideal composed of the following polynomials: 1=x(5) 2=x(3)*x(6)^2+x(4)^2*x(7) 3=x(1)*x(6)^2-x(3)^2*x(7)+2*x(2)*x(4)*x(7) 4=-3*x(2)*x(5)*x(7) 5=-15994*x(2)*x(5)*x(6)-15995*x(1)*x(6)^2+15995*x(3)^2*x(7)+ x(2)*x(4)*x(7)+15994*x(1)*x(5)*x(7) 6=3*x(1)*x(5)*x(6) 7=x(2)*x(3)*x(6)-x(1)*x(4)*x(6)-x(2)^2*x(7)+x(1)*x(3)*x(7) 8=0 9=6*x(1)*x(5)^2+15993*x(2)*x(3)*x(6)-15993*x(1)*x(4)*x(6)- 15993*x(2)^2*x(7)+15993*x(1)*x(3)*x(7) 10=0 11=-3*x(2)^2*x(5)+3*x(1)*x(3)*x(5) Q 2 (M) is equal to: 1=x(5) 2=x(3)*x(6)^2+x(4)^2*x(7) 3=x(1)*x(6)^2-x(3)^2*x(7)+2*x(2)*x(4)*x(7) 4=x(4)^2*x(6)-x(2)*x(6)^2-x(3)*x(4)*x(7) 5=x(3)*x(4)*x(6)-x(3)^2*x(7)+x(2)*x(4)*x(7) 6=x(3)^2*x(6)-x(2)*x(3)*x(7)+x(1)*x(4)*x(7) 7=x(2)*x(3)*x(6)-x(1)*x(4)*x(6)-x(2)^2*x(7)+x(1)*x(3)*x(7) 8=x(4)^3-x(2)*x(4)*x(6)+x(2)*x(3)*x(7)-x(1)*x(4)*x(7) 9=x(3)*x(4)^2+x(2)^2*x(7)-x(1)*x(3)*x(7) 10=x(3)^2*x(4)-x(2)*x(4)^2+x(2)^2*x(6)-x(1)*x(3)*x(6) 11=x(3)^3-2*x(2)*x(3)*x(4)+x(1)*x(4)^2 28

We conclude that I 3 (M) = (I 3 (M)+J 1 ) (I 3 (M)+J 2 ) where each J i =<h 1.. hmi > if is a ro bner basis for I 3 (M). 5.3 I 3 (M) = Q 1 Q 2 Q 3 The last two possible combinations of R k for the 4 4 Hankel matrix have three ideal components for I 3 (M). The following are the possible R k (M) where not all r ij = 1 R k (M) = { 1 1 r 45 1 1 1 1 r 46 } where k = 1 2. Example 5.3 M = X 1 X 2 X 3 X 4 X 2 X 3 X 4 X 5 X 3 X 4 X 5 X 6 X 4 4X 5 X 6 X 7 Each Q i = I 3 (M) + J i. So I 3 (M) is the same for all Q i. Then I 3 (M) is: For Q 1 we have J 1 : 1=x(5)^2-x(4)*x(6) 2=x(4)*x(5)-x(3)*x(6) 3=x(3)*x(5)-x(2)*x(6) 4=x(2)*x(5)-x(1)*x(6) 5=x(4)^2-x(2)*x(6) 6=x(3)*x(4)-x(1)*x(6) 7=x(2)*x(4)-x(1)*x(5) 8=x(3)^2-x(1)*x(5) 9=x(2)*x(3)-x(1)*x(4) 10=x(2)^2-x(1)*x(3) Q1 is: 1=x(5)^2-x(4)*x(6) 2=x(4)*x(5)-x(3)*x(6) 3=x(3)*x(5)-x(2)*x(6) 4=x(2)*x(5)-x(1)*x(6) 5=x(4)^2-x(2)*x(6) 6=x(3)*x(4)-x(1)*x(6) 7=x(2)*x(4)-x(1)*x(5) 8=x(3)^2-x(1)*x(5) 9=x(2)*x(3)-x(1)*x(4) 10=x(2)^2-x(1)*x(3) 29

For Q 2 we have J 2 : 1=x(5)*x(6)-7998*x(4)*x(7) 2=x(3)*x(6)-7998*x(2)*x(7) 3=x(5)^2-7998*x(3)*x(7) 4=x(4)*x(5)-7998*x(2)*x(7) 5=x(3)*x(5)-7998*x(1)*x(7) 6=x(2)*x(5)-x(1)*x(6) 7=x(4)^2-x(2)*x(6) 8=x(3)*x(4)-x(1)*x(6) 9=x(3)^2-x(1)*x(5) 10=x(2)*x(3)-x(1)*x(4) 11=x(1)*x(6)^2-7998*x(2)*x(4)*x(7) 12=x(1)*x(4)*x(6)-7998*x(2)^2*x(7) Q2 is: 1=x(5)*x(6)-7998*x(4)*x(7) 2=x(3)*x(6)-7998*x(2)*x(7) 3=x(5)^2-7998*x(3)*x(7) 4=x(4)*x(5)-7998*x(2)*x(7) 5=x(3)*x(5)-7998*x(1)*x(7) 6=x(2)*x(5)-x(1)*x(6) 7=x(4)^2-x(2)*x(6) 8=x(3)*x(4)-x(1)*x(6) 9=x(3)^2-x(1)*x(5) 10=x(2)*x(3)-x(1)*x(4) 11=x(1)*x(6)^2-7998*x(2)*x(4)*x(7) 12=x(1)*x(4)*x(6)-7998*x(2)^2*x(7) And for Q 3 we have J 3 : 1=x(5) 2=x(3)*x(6)^2+x(4)^2*x(7) 3=x(1)*x(6)^2-x(3)^2*x(7)+2*x(2)*x(4)*x(7) 4=x(2)*x(3)*x(6)-x(1)*x(4)*x(6)-x(2)^2*x(7)+x(1)*x(3)*x(7) 5=x(3)^2*x(4)*x(7)-x(2)*x(4)^2*x(7)+x(2)^2*x(6)*x(7)- x(1)*x(3)*x(6)*x(7) 6=x(3)^3*x(7)-2*x(2)*x(3)*x(4)*x(7)+x(1)*x(4)^2*x(7) 7=x(2)*x(3)*x(4)^2*x(7)-x(1)*x(4)^3*x(7)-x(1)*x(3)^2*x(6)*x(7)+ x(1)*x(2)*x(4)*x(6)*x(7)+x(2)^3*x(7)^2-x(1)*x(2)*x(3)*x(7)^2 8=x(2)*x(4)^2*x(6)^2*x(7)-x(2)^2*x(6)^3*x(7)+x(3)*x(4)^3*x(7)^2- x(1)*x(4)^2*x(6)*x(7)^2 9=x(2)^2*x(4)^2*x(6)*x(7)-x(1)*x(3)*x(4)^2*x(6)*x(7)- x(2)^3*x(6)^2*x(7)-x(2)^2*x(3)*x(4)*x(7)^2-x(1)*x(2)^2*x(6)*x(7)^2+ x(1)^2*x(3)*x(6)*x(7)^2 10=x(2)^2*x(4)^3*x(7)-x(1)*x(3)*x(4)^3*x(7)-x(2)^3*x(4)*x(6)*x(7)+ x(1)^2*x(4)^2*x(6)*x(7)+x(2)^3*x(3)*x(7)^2-x(1)*x(2)*x(3)^2*x(7)^2 30

Q3 is: 1=x(5) 2=x(3)*x(6)^2+x(4)^2*x(7) 3=x(1)*x(6)^2-x(3)^2*x(7)+2*x(2)*x(4)*x(7) 4=x(4)^2*x(6)-x(2)*x(6)^2-x(3)*x(4)*x(7) 5=x(3)*x(4)*x(6)-x(3)^2*x(7)+x(2)*x(4)*x(7) 6=x(3)^2*x(6)-x(2)*x(3)*x(7)+x(1)*x(4)*x(7) 7=x(2)*x(3)*x(6)-x(1)*x(4)*x(6)-x(2)^2*x(7)+x(1)*x(3)*x(7) 8=x(4)^3-x(2)*x(4)*x(6)+x(2)*x(3)*x(7)-x(1)*x(4)*x(7) 9=x(3)*x(4)^2+x(2)^2*x(7)-x(1)*x(3)*x(7) 10=x(3)^2*x(4)-x(2)*x(4)^2+x(2)^2*x(6)-x(1)*x(3)*x(6) 11=x(3)^3-2*x(2)*x(3)*x(4)+x(1)*x(4)^2 Similar to the previous section I 3 (M) = (I 3 (M) + J 1 ) (I 3 (M) + J 2 ) (I 3 (M) + J 3 ) where J i =<h 1.. hmi > if is a röbner basis for I3 (M). 5.4 Section Conclusions All three cases of I 3 (M) for the 4 4 Hankel matrix are similar to I 2 (M). We see that each equality pertains to its respective subsection I 3 (M) = I 3 I 3 (M) = (I 3 (M) + J 1 ) (I 3 (M) + J 2 ) I 3 (M) = (I 3 (M) + J 1 ) (I 3 (M) + J 2 ) (I 3 (M) + J 3 ) where J i =<h 1.. hmi > if is röbner basis for I3 (M). In the subsections we presented Q i to show that some of the elements of a Q i are contained in Q j but not all where i j for the same I 3 (M). So from section 2 we have that these Q i are isolated ideal components. 6 5 5 Matrices In this section we will analyze I 3 (M) for 5 5 genralized Hankel matrices. We will demonstrate our results with an example. Example 6.1 Let A be the following matrix A = X 1 X 2 X 3 X 4 X 5 X 2 X 3 X 4 X 5 X 6 X 3 2X 4 3X 5 5X 6 X 7 X 4 7X 5 11X 6 13X 7 X 8 X 5 17X 6 19X 7 23X 8 X 9 31

Based on a Singular computation A has a primary decomposition I = Q 1 Q 2 Q 3 Q 4 Q 5 where Q 1... Q 5 are giving below. Q 1 = I 3 (M) + Q 2 = I 3 (M) + Q 3 = I 3 (M) + Q 4 = I 3 (M) + Q 5 = I 3 (M) + (x 1 (x 4 ( x 7 ( x 1 ( x 9 x 2 x 5 x 8 x 2 x 6 x 2 8 x 3 x 6 x 9 x 3 x 5 x 7 x 8 x 4 x 7 x 2 7 x 2 6 x 2 x 5 x 6 x 8 x 5 x 6 x 2 3 x 5 x 6 x 2 4 x 5 x 8 x 4 x 6 x 2 x 3 x 3 x 6 x 3 x 4 x 4 x 8 x 2 5 x 7 x 8 x 2 5 x 2 x 4 x 2 7 x 4 x 5 x 2 2 x 1 x 3 x 4 x 5 x 2 3 x 6 x 7 x 2 4 x 2 8 + 6954x 7 x 9 ) x 4 x 6 x 2 x 3 x 5 x 7 x 2 6 4384x 5 x 7 x 2 4 6x 3 x 5 + 6x 2 x 6 ) x 2 2 x 2 6 8206x 4 x 8 ) x 4 8 x 4 2 x 4 x 5 5466x 3 x 6 ) x 5 x 6 + 5726x 6 x 3 ) From the above table we notice that Q 1 begins with terms of single variables (X 1 X 2 X 3 ) Q 2 begins with (X 4 X 5 X 6 ) Q 3 with (X 7 X 8 X 9 ) Q 4 with (X 1 ) and Q 5 with (X 9 ). If we look at the placement of these terms we notice that those of Q 1 lie on or above the X 3 diagognal: X 1 X 2 X 3 X 5 X 2 X 3 X 5 X 3 X 5 X 5 X 5 the terms for Q 2 lie between the X 3 and X 7 diagonals: X 3 X 4 X 5 X 3 X 4 X 5 X 6 X 3 X 4 X 5 X 6 X 7 X 4 X 5 X 6 X 7 X 5 X 6 X 7 and the terms for Q 3 lie on or below the X 7 diagonal: X 5 X 5 X 5 X 7 X 5 X 7 X 8 X 5 X 7 X 8 X 9 32

We also notice that the term of Q 4 and Q 5 are placed at opposite ends of the X 5 diagonal. X 1 X 5 X 5 X 5 X 5 X 5 X 9 With these facts in mind suppose that possibly some symmetry exists. Let the X 5 diagonal be the line of symmetry. If we reflect or map terms to each other along this diagonal we have the following mapping φ : x 1 x 9 x 2 x 8 x 3 x 7 x 4 x 6 x 5 x 5 Using this mapping for the terms Q 1 and Q 3 we get the following: Q 1 = I 3 (M) + Q 3 = I 3 (M)+ (x 1 x 9 ( x 7 x 3 x 2 x 8 x 8 x 2 x 3 x 7 x 9 x 1 x 4 x 7 x 6 x 3 x 2 6 x 2 4 x 5 x 6 x 5 x 4 x 5 x 6 x 5 x 4 x 4 x 6 x 6 x 4 x 3 x 6 x 7 x 4 x 2 5 x 2 5 x 2 5 x 2 5 x 4 x 5 x 5 x 6 x 4 x 5 x 6 x 5 x 2 4 x 2 6 x 4 x 6 x 2 6 4384x 5 x 7 x 2 4 4384x 5 x 3 x 2 4 6x 3 x 5 x 6 x 4 x 2 6 6x 7 x 5 + 6x 8 x 4 8206x 4 x 8 ) 8206x 6 x 2 +6x 2 x 6 ) After carefully analyzing the above table it is clear that after the mapping every term in φ(q 1 ) Q 3 and every term in φ(q 3 ) Q 1. Please note that there are some variations of coefficients. After performing the same procedure for all Q i we have Q 1 Q 3 Q 2 Q 2 Q 4 Q 5 We believe that this same type of symmetry exists for all the different 5 5 matrices. The amount of symmetry may depend on the values of s and t which in turns depends on the amount and placement of coefficents. We hope to further investigate this in future research. 33

7 Future Work Throughout our research we analyzed the primary decomposition of I 3 (M) for M as a 3 4 4 4 or 5 5 Hankel matrix. One important result that we proved is the primary decomposition of I 3 (M 3 4 ). However much more work still needs to be done. We were on the verge of discovering the primary decomposition of I 3 (M 4 4 ). Considering that there are 15 possible primary decompositions for I 3 (M 4 4 ) (depending on the placement of four coefficients) our hypothesis was that there are 8 decompositions that are prime 5 decompositions that are the intersections of two ideals and 2 decompositions that are the intersections of three ideals. Towards the end of our reseach our hope was that we could prove this using SINULAR much in the same way as we did for I 3 (M 3 4 ). However at the time of this writing SINULAR was in its sixth day computing our input. So is our conjecture necessarily true? Other possiblities for future work consist of analyzing the patterns inherent in the primary decompostions of I 3 (M n m ) for nm 5. Specifically for I 3 (M 5 5 ) are the symmetries we discussed inherent in all the primary decompositions of I 3 (M 5 5 )? If so are these symmetries based on s s and t s? More generally assuming that these symmetries exist can they also be found in the primary decompositons of I 3 (M n m ) for any mxn matrix? A progressive result would be a theorem describing the primary decomposition of I 3 (M) for any Hankel matrix M. Now supposing we find the primary decomposition of I 3 (M n m ) for all n m Hankel matrices the best result possible would be a theorem describing the primary decomposition of I n (M) for any Hankel matrix M. This was our ultimate goal. However our present work shows how complicated this case is. To work on or expand on any of our questions would make for promising future research. References CLS D. Cox J. Little & D. O Shea Ideals Varieties and Algorithms 2nd Ed. Springer-Verlag New York 1997. FR S LA R. Fröberg An Introduction to röbner Bases John Wiley and Sons New York 1997. A. uerrieri & I. Swanson On the Ideal of Minors of Matrices of Linear Forms Preprint (2000) 1-9. R. Laubenbacher Computational Algebra and Applications SIMU 2000 class notes (2000) 1-48. 34