Congruences for Coefficients of Power Series Expansions of Rational Functions. Bachelor Thesis

Transcription

1 Congruences for Coefficients of Power Series Expansions of Rational Functions Bachelor Thesis Author Marc Houben Supervisor prof. dr. Frits Beukers

2 Acknowledgements First and foremost I would like to thank Frits Beukers for providing a lot of his time every week for answering my questions and listening to my progress. Writing my thesis under his supervision has been a pleasant and rewarding experience. I would also like to thank my friends from our seminar group for supporting me the past five months and providing useful feedback.

3 Contents Introduction 3 2 General Definitions 4 2. Basic Definitions and Conventions Polynomials Rational Functions Linear Recurrent Sequences 8 4 Algebraic Preliminaries 9 5 One Variable Case 5. Definitions and statement of result Proof of the first part of Theorem Multiple Variables 7 6. Definitions and sufficiency Restricting to subcones and necessity Characterization for a special type of rational functions

4 Introduction In this thesis we study what we will call Gaussian rational functions. A Gaussian rational function in n variables is a rational function for which the coefficients of its power series expansion around 0 satisfy certain congruences. The starting point for the definition of these congruences may be seen as the famous Euler s Theorem, as originally formulated in 763 by Leonhard Euler. In particular, we know that for any integer a, natural number r and prime number p, a mpr a mpr mod p r ) ) So, for example, the coefficients a k of the power series expansion of the rational function f given by satisfy the congruences f := ax ax = a k x k 2) k 0 a mp r a mp r mod p r ) 3) Our question will roughly be: Which rational functions satisfy the congruences 3)? The thesis can be seen as consisting of three parts. In the first part, consisting of sections 2, 3 and 4, we will state some basic definitions, results in the theory of linear recurrent sequences, and a give a small summary of the basics of algebraic number theory that we will need along the way. In the second part, consisting of section 5, we will give a characterization of Gaussian rational functions in one variable. In the third part, consisting of section 6, we will attempt to generalize our results to give a characterization for Gaussian rational functions in n variables. The results up to section 5 are not new. In particular, our characterization of Gaussian rational functions in one variable will be based on a result by Minton []. The aim of section 5 is mainly a translation of this result into a generalizable form. We also provide a slightly simpler alternative to the technique used in the proof of Minton. The results in section 6 are mostly new. Here we will provide ideas that work towards a characterization in the general case of n variables that can be applied to give a complete characterization for a special type of rational functions. This special characterization will not completely exhaust the theory we have developed, and there will be room for a possible extension of results for a wider range of applications. 3

5 2 General Definitions This section mainly consists of definitions and notational conventions that we will use throughout the thesis. In the last subsection we introduce rational functions, the main object of our study. 2. Basic Definitions and Conventions Convention 2. The set of natural numbers N is the set of positive integers N = {, 2, }. n usually denotes an arbitrary element of N, unless stated otherwise. Convention 2.2 Given an element k = k,, k n ) Z n we write k 0 if k i 0 for all i n. We will write k > 0 if k 0 and k i > 0 for at least one i. For d N, we will write k = d if k 0 and k + + k n = d. Definition 2.3 We call an indexed sequence of numbers {a i } i I an n-sequence if I = Z n 0. A -sequence will simply be called a sequence. Convention 2.4 In order to avoid tedious notation, we will often abbreviate an n-sequence of numbers {a k } k 0 with {a k }. Convention 2.5 When we talk about the numerator and denominator of rational numbers q Q, we will usually assume that q is in reduced form. That is, we have written q = a b with a Z, b N such that gcda, b) =. We would like to say that two rational numbers are congruent modulo a prime power p r if the numerator of their difference is a multiple of p r. This is captured by the following definition. Definition 2.6 Congruence for rational numbers) Let q = a /b and q 2 = a 2 /b 2 be rational numbers. Let p be a prime number, let r Z 0 and let p,, p m be the prime numbers dividing b b 2. We say that q is congruent to q 2 modulo p r, written q q 2 mod p r ), if p p i for all i m and q q 2 p r Z[/p,, /p m ]. Definition 2.7 v p function) Let p be a prime number, and let 0 k = k,, k n ) Z n. We define v p k) to be the largest integer r such that p r k i for all i n. Definition 2.8 Let m N and let a,, a m Z. We say that the set {a,, a m } is a complete residue system modulo m if, for any i, j {,, m}, a i a j mod m) if and only if i = j. Definition 2.9 Let α C and k Z 0. We define the generalized binomial coefficient as follows: ) α αα ) α k + ) := 4) k k! 4

6 Lemma ) 2.0 Let p,, p m be prime numbers and let k Z 0. If α Z[/p,, /p m ] then also Z[/p,, /p m ]. α k Proof. Write α = a/b with a Z and b = p r prm m. We have ) α = aa b) a bk )) k b k k! 5) Let p be a prime number such that p p i for all i m. Write k = c r p r + c r p r + + c 0 such that c 0,, c r Z 0 all strictly less than p. Now let 0 j r be an integer. Since p b, for any integer l the set {l, l b,, l bp j )} is a complete residue system modulo p j. It follows in particular that v p c j p j i= i + cj p j ) + + c 0 Since this holds for any integer 0 j r we have ) k v p k!) = v p i i= r = v p v p j=0 i= r c j p j v p a bi + cj p j + + c 0 ) ) 6) c j p j c j p j j=0 i= = v p k i= a bi )) i= i + cj p j ) + + c 0 8) a bi + cj p j + + c 0 ) ) ) ) = v p aa b) a bk )) 7) 9) Since this holds for any prime p not dividing b, we conclude using 5) that α k) Z[/p,, /p m ]. Definition 2. Almost all) Let S be a countable set. We will say that a condition P s) on elements of S holds for almost every s S, if it holds for all but finitely many s S. Definition 2.2 Given an integral domain I, we denote by I[[x,, x n ]] the set of formal power series in the variables x,, x n with coefficients in I. 0) ) 5

7 2.2 Polynomials Definition 2.3 Given an integral domain I and a polynomial P = a 0 + a x + + a d x d I[x], we define the reciprocal polynomial of P to be P := a 0 x d + a x d + + a d. Lemma 2.4 Let F be a field, and let P F [x]. Assume that P 0) 0. Then P is separable if and only if P is separable. Proof. Let d := deg P. Note that P x) = x d P ) x. It follows that α 0 is a root of P if and only if α is a root of P. Also, P 0) 0 and we assumed P 0) 0. We conclude that P has multiple zeros in its splitting field) if and only if P has multiple zeros in its splitting field). Definition 2.5 θ-operator) Given a polynomial P Z[x,, x n ] in n variables, we define θ i P, for i n, to be the polynomial defined by θ i P x,, x n ) := x i x i P x,, x n ). That is, θ i can be seen as the operator which acts by formally partially differentiating with respect to the i-th variable, followed by multiplication with this variable. In the case n =, we will abbreviate θ by θ. Definition 2.6 We denote by Z[x,, x n ] 0 the set of all polynomials in Z[x,, x n ] with non-zero constant term: Z[x,, x n ] 0 := {Q Z[x,, x n ] Q0) 0} 2) 6

8 2.3 Rational Functions Definition 2.7 Let n N. A rational function f = P/Q in the variables x,, x n is the quotient of two polynomials P, Q Z[x,, x n ]. We will denote the set of rational functions in the variables x,, x n by Qx,, x n ). If Q0) 0, i.e. the constant term of Q is non-zero, then f = P/Q has a power series expansion around 0, which means that we can write fx,, x n ) = k 0 a k x k := k,,k n 0 a k,,k n x k xkn n 3) For coefficients a k Z[/Q0)] see Proposition 2.8). The power series is uniquely determined if we demand that it converges for x = x,, x n ) small enough. However, the properties regarding the convergence of the power series will not be of our interest. We will often identify a rational function with its power series, so that we can regard rational functions as elements of Q[[x]]. Proposition 2.8 Let f = P/Q be a rational function and suppose that Q0) 0. Write the power series expansion of f as in 3). Then a k Z[/Q0)] for all k Z n 0. Hence in particular, if Q0) = then a k Z for all k Z n 0. Proof. If Q is constant then the statement is clearly true, so assume that Q is not constant. We can write f = P Q0) Q = P Q0) Q Q0) Where Q := Q0) Q is a non-zero polynomial with integer coefficients such that Q 0) = 0. The right hand side of 4) can be expanded as a geometric series: f = P = P ) ) + Q Q 2 Q0) Q Q0) Q0) + + 5) Q0) Q0) n Because Q has constant term equal to zero, the monomials that Q0)) consists of all have at least degree n. Therefore the expansion in 5) well-defines a power series, which converges for all x such that Q x) <. Since Q 0) = 0 and Q is continuous at 0, we conclude by uniqueness of power series Q0) expansions that this coincides with the power series expansion of f. Now since P and Q have integer coefficients, and Q0) is also an integer, we see immediately from 5) that the power series expansion of f has coefficients in Z[/Q0)]. Convention 2.9 We say a rational function P/Q Qx,, x n ) is written in reduced form if P, Q Z[x,, x n ] and P, Q have no common divisors in Q[x,, x n ] except for units). Q 4) 7

9 3 Linear Recurrent Sequences Definition 3. A rational linear recurrent sequence is a sequence {q n } n 0 in Q for which there exists an r N and coefficients c,, c r Q such that q n = c q n + c 2 q n c r q n r for all n > r. This last relation is called a recurrence relation. Remark 3.2. Note that in our definition above, for a rational linear recurrent sequence to be determined by its recurrence relation, we need initial conditions q,, q r. Note also that there is no restriction on q 0, the value of which will not be relevant for our purposes. Definition 3.3 The characteristic polynomial G of a rational linear recurrent sequence with coefficients c,, c r is given by Gx) = x r c x d c r. Proposition 3.4 Let {u n } be a rational linear recurrent sequence with coefficients c,, c r, and suppose that the characteristic polynomial G is separable and that G0) 0. Denote by ω,, ω r K the distinct roots of G in its splitting field K. Then there exist uniquely determined α,, α r K such that u n = r α i ωi n 6) i= for all n. Lemma 3.5 Suppose that {u n } is as in Proposition 3.4, and that α i Q for all i r. If ω i and ω j are conjugate, then α i = α j. Proof. Let H be the Galois group of K/Q, and suppose that ω i and ω j are conjugate. By a basic result in Galois theory there exists a σ H be such that σω i ) = ω j. Note that since u n Q, it is fixed under this σ, hence r r ) r u n = α i ωi n = σ α i ωi n = α i σω i ) n 7) i= Since the α i K are uniquely determined by {u n } Proposition 3.4), we conclude that α i = α j. i= i= 8

10 4 Algebraic Preliminaries This section is a summary of results and basic definitions in algebraic number theory, which will be used in the proof of Theorem 5.2. In particular the result found in Theorem 4.7 will be of great importance. Some statements are given without proof. Instead we will refer to [4] and [5]. Definition 4. A number field K is a finite field extension of Q. Since any finite field extension is algebraic, a number field K consists of algebraic numbers, hence it can be seen as a subset of C. In the remaining part of this section we will assume that K denotes a number field. Definition 4.2 An algebraic integer is an element ω C which is a root of some monic polynomial p Z[x]. We will denote by A C the set of all algebraic integers. Proposition 4.3 A is a subring of C. Proof. Can be found in [4] on page 68 Proposition 6..5). Lemma 4.4 A Q = Z. Proof. Let q = k/l Q be a rational number. We assume Convention 2.5) that gcdk, l) =. Suppose that q is an algebraic integer. Then there exists an n N and a,, a n Z such that k l ) n ) k n + a + + a n = 0 8) l It follows that k n = a k n l + + a n l n, hence l k n. We conclude that l =. Hence A Q Z. Conversely, an integer k Z is the root of the monic polynomial x k Z[x], so Z A Q. Definition 4.5 We denote by O K the set of algebraic integers of K, i.e. O K := A K. By Proposition 4.3, O K is a ring. We will call it the ring of integers of K. Proposition 4.6 Let α K. Then there exists a non-zero d Z such that dα is an algebraic integer. Proof. Since α is algebraic it is a zero of some polynomial with integer coefficients. That is, there exist a 0,, a n Z, a n 0 such that a n α n + a n α n + + a 0 = 0. Multiplying this by an n we find a n α) n + a n a n α) n + + a n n a 0, hence a n α is an algebraic integer. Definition 4.7 A prime P of K is a prime ideal of O K. Proposition 4.8 For any non-zero ideal A O K, O K /A is finite. Proof. Can be found in [4] on page 76 Proposition 2.2.3). Since every finite integral domain is a field, Proposition 4.8 has the following corollary: 9

11 Corollary 4.9 Every prime of K is maximal. Lemma 4.0 Suppose A, B O K are ideals such that A B, then there is an ideal C O K such that A = BC. Proof. Can be found in [4] on page 79 Proposition 2.2.7). Proposition 4. Every non-zero proper ideal A of O K can be written uniquely up to ordering) as a finite) product of primes of K. Proof. This is Proposition 2.2.8, page 80 in [4]. Lemma 4.2 Let A O K be an ideal. Then A Z {0}. Proof. Can be found in [4] on page 76 Lemma 2). Corollary 4.3 Let P be a prime of K. Then P Z = p) ideal in O K generated by p) for a unique prime number p N. We will say that P is a prime above p. Lemma 4.4 For every prime number p, there exist finitely many primes P above p. Proof. By Proposition 4., we can uniquely factorize p) = P e P g eg as a product of primes of K. It follows by Lemma 4.0 that P,, P g are precisely the primes above p. In particular there are finitely many such primes. Definition 4.5 Let P be a prime of K. We call O K /P the residue field of P. Lemma 4.6 Let p be a prime number. For every prime P above p, the residue field of P is a finite field of characteristic p. Proof. It follows directly from Proposition 4.8 and Corollary 4.9 that O K /P is a finite field. Since p P its characteristic is p. Theorem 4.7 Let K be a number field and let α O K. Suppose that, for almost every prime P of K, the residue ᾱ of α in the residue field κ of P satisfies ᾱ p = ᾱ. Then α Z. Proof. This follows from the Frobenius Density Theorem. A proof can be found in [5] on p.34 Thm 5.2). 0

12 5 One Variable Case 5. Definitions and statement of result In this section we will give a characterization for the so-called Gaussian rational functions in one variable. Part of our result is based on a proof for rational linear recurrent sequences by Minton, see []. In one variable the approach using rational linear recurrent sequences and our approach by rational functions are almost equivalent. We however choose the viewpoint of rational functions, because then there is a straightforward generalization to the multivariable case. We begin with the definition of Gauss sequences in one variable. Definition 5. Gauss sequences) Let {q n } n 0 be a sequence of rational numbers. Let p be a prime number. We say that {q n } is Gauss at p if for all m, r N, we have see Definition 2.6) q mp r q mp r mod p r ) 9) Furthermore, we call the sequence {q n } Gauss if it is Gauss at almost every prime p, and strictly Gauss if it is integral and Gauss at every prime p. We will call a power series f = a 0 + a x + Q[[x]] strictly) Gauss if the corresponding sequence of coefficients {a n } n 0 is strictly) Gauss. Remark 5.2. Using the v p -function as defined in 2.7, we can rewrite condition 9) as q k q k/p mod p v ) pk) for all k pz. Example 5.3 Let the power series f Z[[x]] be given by f = 2 x x x 2 = 2 + x + 3x2 + 4x 3 + 7x 4 + x 5 + 8x x 7 20) Notice how each term in the sequence of power series coefficients is the sum of the previous two terms. This Fibonacci-like sequence also known as the Lucas sequence) will, among other sequences of the same type, turn out to be strictly Gauss. 9 Definition 5.4 We call a rational function f = P/Q, P, Q Z[x], Q0) 0 in one variable of type L if fx) = or P = θq Definition 2.5). Remark 5.5. The set of non-constant functions of type L is closed under addition, since if Q, Q 2 Z[x] then θq + θq 2 = x d dx Q ) Q2 + Q x d dx Q ) 2 = θq Q 2 ) 2) Q Q 2 Q Q 2 Q Q 2 The following Theorem is the main result of this section. Theorem 5.6 Let P, Q Z[x] such that Q0) 0. Let f Q[[x]] be the power series expansion of P/Q. Suppose that Q is separable. Then f is Gauss if and only if it is a Q-linear sum of functions of type L.

13 Remark 5.7. Note that, for a rational function of type L, the degree of its numerator is always equal to the degree of its denominator. Therefore, Theorem 5.6 implies in particular that if a rational function P/Q in one variable is Gauss, then deg P deg Q. The implication = of Theorem 5.6 will be shown in more general form in section 6. The implication = will be of our concern for the rest of this section. Our proof will be an alternative and simpler version of the strategy used in []. Example 5.8 Consider f as in Example 5.3. Note that: f = 2 x x 2x2 = 2 x x2 x x 2 = 2 θ x x2 ) x x 2 22) Which is indeed a Q-linear sum of rational functions of type L Proof of the first part of Theorem 5.6 As stated earlier, the result in [] regards rational linear recurrent sequences, but we are interested in a result for rational functions. The relation between the two is shown by Lemma 5.9. Lemma 5.9 Let P, Q Z[x], Q0) 0 such that deg P deg Q. Let f = c 0 + c x + Q[[x]] be the power series expansion of P/Q. Then the corresponding sequence of coefficients {c n } n 0 is a rational linear recurrent sequence, with characteristic polynomial G = Q /Q0) see Definition 2.3). Proof. It follows from Proposition 2.8 that c n Q for all n N. Write Q = b 0 + b x + + b d x d. Note that: P = Qf 23) = b 0 c k x k + + b d c k x k+d 24) = k 0 k 0 k 0 b 0 c k x k + + k d b d c k d x k 25) = k>db 0 c k + b c k + + b d c k d )x k + Q 2 x) 26) where Q 2 x) is a polynomial of degree d. Since deg P deg Q = d we find that for all k > d: c k = b 0 b c k + + b d c k d ) 27) We conclude that {c n } is indeed a rational linear recurrent sequence, with characteristic polynomial G = x d + b b 0 x d + + b d b 0 = Q Q0) 28) 2

14 In Lemma 5.9 we assumed that deg P deg Q. In order to be able to extend our result to the case that deg P > deg Q we will need the following definition. Definition 5.0 We call a sequence of rational numbers {v n } vanishing if there exists a k N such that v n = 0 for all n > k. The smallest such k N is called the support of {v n }. Lemma 5. Let P, Q, f and {c n } n 0 be as in Lemma 5.9, but without the restriction that deg P deg Q. Then we can write {c n } = {u n } + {v n } for a rational linear recurrent sequence {u n } with characteristic polynomial Q /Q0) and a vanishing sequence {v n }. Proof. Let n := deg P and d := deg Q. By Euclidean division on polynomials we can write P = QV + R for polynomials V, R Q[x] such that deg R < deg Q. Now we see that f = P Q = QV + R Q = V + R Q 29) By Lemma 5.9 the sequence of power series coefficients {u n } of R/Q is a rational linear recurrent sequence with characteristic polynomials Q /Q0). Since V is a polynomial, its sequence of power series coefficients {v n } is vanishing. The desired result follows from the fact that {c n } = {u n } + {v n }. The main mechanism of our proof of Theorem 5.6 is the following result: Theorem 5.2 Suppose {u n } n 0 is a rational linear recurrent sequence with separable characteristic polynomial G, G0) 0. Let {v n } be a vanishing sequence and let {c n } = {u n } + {v n }. Let ω,, ω r be the roots of G in its splitting field K and write u n as in Proposition 3.4 Equation 6). If {c n } is Gauss then α i Q for all i r. Proof. Since ω,, ω r, α,, α r K, it follows from Proposition 4.6 that there exist non-zero integers d ω,, d ωr, d α,, d αr Z such that {d ω ω,, d ωr ω r, d α α,, d αr α r } O K. Now define d ω := r d ωi, d α := i= r d αi, ν i := d ω ω i, γ i := d α α i 30) i= Note that the ν i and γ i are all algebraic integers. Now consider the following matrix: ν ν 2 ν r Ω := ν 2 ν2 2 νr ν r ν2 r νr r 3) By the Vandermonde determinant formula we have detω) = ν i ν j ) 32) i<j n 3

15 Now suppose H is the Galois group of the field extension K/Q, and let σ H. Since σ permutes the roots ω,, ω n of G, it also induces a permutation on the ν i s. Thus we see, either directly from 32), or by noting that σ permutes the columns of Ω, that σ fixes detω) 2 = i<j n ν i ν j ) 2. Since this holds for any σ H, detω) 2 Q. Since detω) 2 is also an algebraic integer, we conclude by Lemma 4.4 that detω) 2 is in fact an integer. Now let s be the support of {v n }. p G0)d ω detω) 2. Let p be a prime number at which {c n } is Gauss, and such that Note that since G is separable we have detω) 0. We also assumed G0) 0 and constructed d ω such that it is non-zero. Therefore p G0)d ω detω) holds for almost every prime number p. Since {c n } is Gauss at almost every prime p, we conclude that the assumptions we make on p above hold for almost all prime numbers p. Now let d N such that d > s. Since c n = u n for all n > s, we have Furthermore, by Fermat s Little Theorem we have u dp u d mod p) 33) u p d u d mod p) 34) Subtracting equations 33) and 34) and filling in u n = r i= α iωi n we find r r ) p α i ω dp i α i ωi d 0 mod p) 35) i= i= If we consider this congruence modulo po K ideal in O K generated by p) then it still holds in particular pz po K ), hence r r ) p α i ω dp i α i ωi d 0 mod po K ) 36) i= i= Now multiply this equation by d dp ω d p α. We obtain r r ) p d p αα i d ω ω i ) dp d α α i d ω ω i ) d 0 mod po K ) 37) i= i= By Fermat s Little Theorem, d p α d α mod p), so in combination with 30) this reduces to r r ) p γ i ν dp i γ i νi d 0 mod po K ) 38) Therefore i= r i= i= γ i γ p i ) νdp i 0 mod po K ) 39) Now let P be a prime above p in K. Denote by ᾱ the residue of α in O K /P. It follows from 39) that r i= γ i γ p i ) νdp i = 0 40) 4

16 Since this holds for every d > s we find that ν s+)p γ γ p ). + + γ r γ p r ) ν s+r)p ν s+)p r. ν s+r)p r = 0. 4) 0 If we now define the matrix Then we have ν p ν p r detγ) =.. ν r )p ν r r )p ν s+)p ν s+)p r ν s+2)p Γ := ν s+2)p r.. ν s+r)p ν r s+r)p r k= ν s+)p k = i<j r ν p i νp j ) r k= ν s+)p k = detω) p r i= ν i s+)p 42) 43) We will now show that detγ) 0. Assume first to the contrary that detω) P. Then also detω) 2 P. But since detω) 2 is an integer, we have detω) 2 0 mod p). This contradicts p detω) 2. We conclude that detω) p P. Suppose now that ν i = 0 for some i i r). We know that ν i = d ω ω i, p d ω and that P is a prime ideal, thus ω i = 0. But then 0 = G ω i ) = G0). This contradicts our assumption that p G0). We conclude that ν i 0 for all i. Using once again that P is a prime ideal, it now follows that indeed detγ) 0. Now let i r. Combined with equation 4) we find γ i = γ p i. This holds for every prime P above almost every prime number p. So by Proposition 4.4, the residue γ i of γ i in P equals γ p i for almost every prime P of O K. Hence, by Theorem 4.7, γ i Z. We conclude by 30) that α i Q for all i r. We would now like to translate the result of Theorem 5.2 into a result for rational functions, which will form the promised proof of the implication = in Theorem 5.6. The translation will be established by combining Lemma 5. with Lemma 3.5. Proof of = in Theorem 5.6. Let P, Q Z[x], Q0) 0, f := P/Q, Q separable, and assume that f is Gauss. Let {c n } n 0 be the sequence of power series coefficients of f and write {c n } = {u n } + {v n } as in Lemma 5.. The characteristic polynomial of {u n } is G := Q /Q0). Since Q is separable, it follows from Lemma 2.4 that G is separable. It should also be clear that G0) 0, since we assume Q0) 0. If we again write u n as in Proposition 3.4, Theorem 5.2 tells us that α i Q for all i r. It follows from Lemma 3.5 that after relabelling the roots ω i of G we can write u n = β ω n + + ωk n ) ) + + βl ωk n l ωn k l 44) 5

17 For rational numbers β,, β l and k 0 = 0, k,, k l, k l = r Z 0, such that for every i l, {ω ki +,, ω ki } is a complete set of conjugate algebraic numbers, which are precisely the roots of the minimal polynomial m i := k i j=k i + x ω j ) Q[x] Now define S i n) := ω n k i ωn k i. Using this abbreviation, 44) reduces to u n = l β i S i n) 45) Define f i x) := n 0 S in)x n to be the generating function of the sequence {S i n)} n 0. Note that i= f i x) = n 0 k i j=k i + ω n j x n = k i j=k i + ki ω j x = j=k i + ω j x + ω j x ω j x = k i k i + k i j=k i + ω j x ω j x 46) Now if we define g j := ω j x, and G i := k i j=k i + g j, 46) transforms into see Remark 5.5) f i = k i k i k i j=k i + θg j g j = k i k i θg i G i 47) It follows that the generating function U of {u n } is given by U = l i= β i k i k i θg ) i G i 48) Now note that G i = m i Q[x], hence after rescaling G i to have integer coefficients) we see that the generating function of {u n } is a Q-linear sum of functions of type L. If we were able to prove that v m = 0 for all m N, then it follows that the generating functions of {u n } and {c n } are the same up to a possible constant. This implies the desired result, namely that f is a Q- linear sum of functions of type L. In order to show that this actually holds, we will need the = -part of Theorem 5.6 which we will prove in section 6), namely that every Q-linear combination of functions of type L is Gauss. Assuming this result, we find that U as in 48) is Gauss, hence so is the sequence {u n }. Now let m N and let p be a prime number greater than the support of {v n } such that {u n } is Gauss at p. Since we assume {c n } = {u n } + {v n } to be Gauss it follows that u m u mp mod p) and u m + v m u mp mod p) 49) From which it follows that v m 0 mod p). Since this holds for infinitely many primes p, we conclude that v m = 0. 6

18 6 Multiple Variables In section 5 we gave a characterization of Gaussian rational functions in one variable with separable denominator. In this section we work towards giving a similar characterization for rational functions in multiple variables. For now, this number of variables will be denoted by an arbitrary but fixed n N. The section consists of three parts. In the first part we will show sufficient conditions for a rational function in n variables to be Gauss, in the second part we will be concerned with finding necessary conditions, and in the third part we will apply our results to give a complete characterization for a special type of rational functions. 6. Definitions and sufficiency Definition 6. General Gauss sequences) Let {q k } k Z n 0 be an n-sequence of rational numbers. Let p be a prime number. We say that {q k } is Gauss at p if for all m Z n 0 and r N we have q mp r q mp r mod p r ) 50) We call the n-sequence {q k } Gauss if it is Gauss at almost every prime p. We call it strictly Gauss if it is integral and Gauss at every prime p. We will call a power series f = k 0 a kx k Q[[x,, x n ]] strictly) Gauss if the corresponding n- sequence of coefficients {a k } is strictly) Gauss. Remark 6.2. Alternatively, we can write condition 50) as q k q k/p mod p v pk) ) for all k pz n 0. Definition 6.3 Let {a k }, {b k } be n-sequences of rational numbers, and let q Q. We define sum of and scalar multiplication with such sequences in the straightforward way: {a k } + {b k } := {a k + b k } and q{a k } := {qa k }. Proposition 6.4 Q-vector space. The set G of Gauss n-sequences, together with the operations in Definition 6.3, is a Proof. Suppose {a k }, {b k } G are Gauss n-sequences, and let q Q. Let P, P 2 be the finite set of prime numbers for which condition 50) does not hold for {a k } and {b k } respectively, and let P 3 be the set of primes p such that p divides the denominator of q. Then {a k } + {b k } satisfies condition 50) for all primes p such that p P P 2, and q{a k } satisfies 50) for all primes p such that p P P 3. Since P P 2 and P P 3 are both finite sets, we conclude that {a k } + {b k } G and q{a k } G. Proposition 6.5 Let p,, p m be prime numbers, Z := Z[/p,, /p m ]. Let f Z[[x,, x n ]] such that f0) =. Then there exist a k,,k n = a k Z such that f = k>0 x k) a k 5) 7

19 We will first illustrate the method of obtaining the a k in Proposition 6.5 by considering the case n =. Then we can write f = + f x + f 2 x 2 + for f i Z. Note that f x) f = f ) f x) l = + f x + f 2 x 2 + ) f x + ) = + g 2 x 2 + g 3 x ) l l 0 For coefficients g i Z see Lemma 2.0). Using the same argument we find f x) f x 2 ) g 2 = + g 2 x 2 + g 3 x 3 + ) ) g2 x 2 ) l = + h 4 x 4 + h 5 x ) l l 0 For coefficients h i Z. Continuing this process we find a k Z note that a k = 0 is allowed) such that f k>0 xk ) a k =. For the case of general n we can use a similar argument, by factoring out the terms of a given degree in each step. Convention 6.6 Let d Z 0. If a power series k 0 a kx k satisfies a k = 0 for all k Z n 0 k < d, then we will say that it is Ox d ). We may abbreviate such a power series with Ox d ). such that Proof of Proposition 6.5. Let d N. We say that a power series g Z[[x,, x n ]] satisfies condition P d if g0) =, and g only contains terms of degree at least d i.e. is Ox d )). Suppose g satisfies condition P d. We can then write g k Z) g = + g k x k + Ox d+ ) 54) Define h := g x k ) g k. Then we implicitly use Lemma 2.0 for the second equality) k =d h = = k =d k =d + g k x k x k ) g k + Ox d+ ) 55) k =d + g k x k k =d k =d g k x k ) + Ox d+ ) 56) = + Ox d+ ) 57) So h satisfies condition P d+. Applying this inductively on f we obtain coefficients a k Z such that f x k) a k = 58) k>0 For the rest of this section we will assume that Z denotes a ring of the form Z[/p,, /p m ] for prime numbers p,, p m m Z 0 ). 8

20 Theorem 6.7 Let l n be a positive integer, and let f,, f l Z[[x,, x n ]] be power series such that f i 0) = for all i l. Let f,,f l ) x,,x l ) be the Jacobian matrix. Then the power series f Z[[x,, x n ]] defined by f := x x l f,, f l ) f f l x,, x l ) 59) is Gauss. Furthermore, if Z = Z then f is strictly Gauss. Proof. First consider the special case f i = x k i have, for any i, j l, So if we define K l to be the l l matrix k ij ) we find: f = x x l f,, f l ) f f l x,, x l ) = )l detk l ) xk x k l = ) l detk l ) f f l i l) with k i = k i,, k in ) Z n 0. Then we x j f i x j = k ij x k i 60) m,,m l x m k + +m l k l 6) If detk l ) = 0 then we see immediately that f is strictly Gauss, so assume that detk l ) 0. Then we have, for any rational numbers r,, r l, s,, s l : r k + + r l k l = s k + + s l k l = r,, r l ) = s,, s l ) 62) Now denote by a k the power series coefficients of f, 6) and 62) tell us in particular that, given k Z n 0, a k is either equal to 0 or equal to ) l detk l ), the latter holding precisely when k = m k + + m l k l for some m,, m l N. Let p be a prime number and let k pz n. If there do not exist m,, m l N such that m k + + m l k l = k, then there certainly do not exist m,, m l N such that m k + + m l k l = k/p. It follows that a k = a k/p = 0, so in particular a k a k/p mod p v pk) ). Now suppose there do exist m,, m l N such that m k + + m l k l = k. If p m i for all i then a k = a k/p = ) l detk l ), so again a k a k/p mod p v pk) ). The only case left to consider is p m i for some i. Assume without loss of generality that p m. Then we have the following: v p detk l )) = v p detk,, k l )) = v p m detk,, k l )) = v p detm k,, k l )) 63) = v p detm k + + m l k l,, k l )) v p m k + + m l k l ) 64) So a k = ) l detk l ) 0 a k/p mod p v pk) ). We conclude that f is indeed strictly Gauss. Now consider f as in the general case, i.e. f is given by 59) with f,, f l Z[[x,, x n ]], f i 0) = for all i. Consider the following differential form Ω: df Ω := x x l df l = x x l f,, f l ) f f l f f l x,, x l ) dx dx l = f dx dx l 65) Using Proposition 6.5 we know that we can write f i = x i) k aki i) for coefficients aki i) Z for all i. Using this we find k i >0 df i = a ki i) dxk i f i x k i k i >0 It follows in fact from 62) that a k/p = 0, but whether it is equal to ) l detk l ) or equal to 0 is irrelevant in this case. 66) 9

21 Now for any k Z n 0, define the power series gk) by g k) := x k. Combining this with 66) we can rewrite Ω as: df Ω = x x l df l f f l 67) l = x x l a ki i) dxk x k dxkl x k l 68) = ) l k >0,,k l >0 i= k >0,,k l >0 i= = ) l k >0,,k l >0 i= l dg k ) a ki i) x x l l a ki i) g k ) x x l g k ) g k l) dgk l) g k l) 69) g k),, g k ) ) x,, x l ) dx dx l 70) Note that, in view of equation 6), the sum in 70) well-defines a power series: The term of the sum corresponding to k,, k l is Ox k + + k l ), hence each of the coefficients of the power series that the sum defines can be computed as a finite sum of elements of Z. Now the special case above tells us that the power series defined by x x l g k ) g k l) g k),, g k ) ) x,, x l ) is strictly Gauss for any k,, k l Z n 0. Multiplying this by l i= a k i i) we obtain a power series that is Gauss, or even strictly Gauss if Z = Z. The same will hold after we sum over all k,, k l. Combined with 65) we conclude that f is Gauss, and strictly Gauss if Z = Z. Definition 6.8 We say a rational function f = P/Q, P, Q Z[x,, x n ], Q0) 0 is of type L if there is some subset of the variables X = {x i,, x il } {x,, x n } and there are polynomials f,, f l Z[x,, x n ] 0 such that f = x i x il f,, f l ) f f l x i,, x il ) 7) Here we use the convention that if X =, then f =. Remark 6.9. Note that the more general Definition 6.8 for rational functions of type L in n variables coincides with Definition 5.4 for n =. The following Proposition says that every Q-linear sum of functions of type L is Gauss, which has as a Corollary the promised implication = in Theorem 5.6. Proposition 6.0 Suppose that h,, h m are rational functions in n variables of type L and let q,, q m Q. Then f := q h + + q m h m 72) is Gauss. Proof. By Proposition 6.4, it suffices to prove that every function h of type L is Gauss. If h = then clearly h is Gauss. Now let X = {x i,, x il } be a non-empty subset of {x,, x n } and let 20

22 f,, f l Z[x,, x n ] 0. Applying Theorem 6.7 to g i := f i /f i 0) after a suitable relabelling of our variables we find that x i x il f,, f l ) f f l x i,, x il ) = x i x il g,, g l ) g g l x i,, x il ) 73) is Gauss. Proof of = in Theorem 5.6. This follows directly from Proposition 6.0 with n =. The following example illustrates an application of Proposition 6.0 for a two-variable rational function. Example 6. Define f := + xy 2 + x 2 y x 3 y 3 and f 2 := + xy 2. Check that 3f 2 = 3f f 2 + f θ x f 2 2θ y f 2 ) + f 2 θ y f 2θ x f ) + θ x f )θ y f 2 ) θ x f 2 )θ y f ) 74) From which it follows that = 3f f 2 + f θ x f 2 2θ y f 2 ) + f 2 θ y f 2θ x f ) + θ x f )θ y f 2 ) θ x f 2 )θ y f ) 75) f 3f f 2 = + θy f 2θ x f + θ xf 2 2θ y f 2 + θ ) xf )θ y f 2 ) θ x f 2 )θ y f ) 76) 3 f f 2 f f 2 = + y f y 2x f x + x f 2 x 2y f ) 2 y + xy f f x y 77) 3 f f 2 f f 2 Hence by Proposition 6.0, f = f 2 x f 2 y + xy 2 + x 2 y x 3 y 3 78) is Gauss. 9 Example 6. involves some tedious computational work, and it is probably not clear how 74) was obtained. Luckily enough, the result that /f as given by 78) is Gauss, can be obtained in a much more elegant way, that avoids the use of magic. The key to this is the observation that, defining X := x 2 y and Y := xy 2, f reduces to f = + X + Y XY. Before we can show that this indeed implies that /f is Gauss, we will need to do some work. The starting point for this is Theorem 6.2. This Theorem roughly says that if a variable x i appears only as a single power in the polynomial Q, i.e. we can write Q = P + x m i R where both P and R do not depend on x i, then P /Q is Gauss, and that we can obtain more Gaussian rational functions by repeatedly removing variables in this way. That is, if we were able to write P = P 2 + x m 2 j R 2 for some other variable x j such that P 2 and R 2 do not depend on x j, then P 2 /Q is also Gauss, etcetera for a possible P 3. Theorem 6.2 Let 0 k n and suppose that Q Z[x,, x n ] 0 satisfies the following condition: Defining P n := Q, there exist polynomials P k,, P n, R k,, R n Z[x,, x n ] and natural numbers m k+,, m n N such that for every k l < n, we have that P l, R l Z[x,, x l ] and P l+ = P l + x m l+ l+ R l. Then P k /Q is Gauss. 2

23 ) Proof. Let everything as above. Note that for every k l < n, we have P l = θ l+ m l+ P l+. Therefore P k Q = P n P n 2 P n P n P k P k+ = ) ) θn m n P n θ k+ m k+ P k+ 79) P n P k+ Now let i,, i r N such that k < i < < i r n. Since P ib does not depend on x ia for a < b, we see that x i x ir P i,, P ir ) r P i P ir x i,, x ir ) = θ ij P ij 80) P ij Hence, looking back at 79), we see that P k /Q is a Q-linear sum of rational functions of type L. We conclude, by Proposition 6.0, that P k /Q is Gauss. Example 6.3 Define Q := x 4 ++y 9 4+5z z 7 )+z 3. Then, by applying Theorem 6.2 for a suitably chosen ordering of the variables), P i /Q is Gauss for each of the following polynomials: P := + x 4 + z 3, P 2 := + x 4, P 3 := + z 3, P 4 :=, P 5 := + y z z 7 ) + z 3 8) It looks like, given a polynomial Q, we can sometimes filter out some terms T of Q using the approach in Theorem 6.2 to find that T/Q is Gauss. This gives rise to the following interesting Corollary: Corollary 6.4 Suppose that Q = k a kx k Z[x,, x n ] 0 is linear in each variable, i.e. has at most degree in x i for all i n. Then for each term T = a k x k appearing in Q, T/Q is Gauss. Proof. Define the ring of operators Θ := Q[ θ,, θ n ] = Q[θ,, θ n ]. We will first show that for any D Θ, DQ/Q is Gauss. Define P n := Q, and inductively for all 0 l < n: j= P l := θ l+ ) P l+, R l := x l+ θ l+ P l+ 82) Then the polynomials P l, R l satisfy the conditions in Theorem 6.2 for k = 0 and m = = m n =. It follows by the Theorem that θ l ) θ l+ ) θ n )Q) /Q is Gauss for any l n. Now note that for any i n and 0 l n, θ i ) 2 P l = 0, because P l is linear in x i. We can run the argument above for any ordering of our variables x i, so for any i,, i r n, we have that θ i ) θ ir )Q) /Q is Gauss. We conclude Proposition 6.4) that DQ/Q is Gauss for any D Θ. It remains to prove the following statement: For any term T occurring in Q, there exists a D Θ such that DQ = T. We will show this by induction on the number of variables n in Q. If n = 0 then the statement is clearly true. Suppose that the statement holds in the case of n variables. Since Q is linear in each variable, we can write any term as T = ax k xkn n, where k,, k n {0, }. If k n = 0, then T appears in P n. By our induction hypothesis, there exists a D 0 Q[θ,, θ n ] Θ such that D 0 P n = ax k xk n n, so choose D := D 0 θ n ). 9 22

24 If k n =, then T appears in x n Q n. In this case we find D Θ such that D Q n = ax k xk n n, so choose D := D θ n. In both cases, we end up with T = DQ. Example 6.5 Suppose that Q is given by Q = + x + y xy. Then Corollary 6.4 tells us that for any polynomial P Z[x, y] such that P is linear in both x and y i.e. P = a + bx + cy + dxy for a, b, c, d Z), P/Q is Gauss. In fact, if one runs the arguments in Theorem 6.2 and 6.4 carefully, one can even obtain that P/Q is strictly Gauss. We will not work this out here, because our main focus will be on rational functions that are not necessarily strictly Gauss. 9 To return to our problem of determining why /f as given by 78) is Gauss in an elegant way, note that Example 6.5 tells us in particular that / + X + Y XY ) is Gauss. We would therefore like that the Gaussness is not destroyed after substituting X = x 2 y and Y = xy 2. This gives rise to Proposition 6.6. Proposition 6.6 Let f = k 0 a k,,k n)x k xkn n Q[[x,, x n ]] be a power series. Let K = k ij ) i,j n be an n n matrix consisting of non-negative integer entries, ) and suppose that detk) 0. Then f is Gauss if and only if g := f t k t k n n,, t k n t knn n is Gauss seen as a power series in t,, t n ). 2 Proof. The power series expansion of g looks as follows: g := k 0 b k t k = m 0 a m,,m n)t m k + +m nk n t m k n + +m nk nn n = m 0 a m t Km 83) = Let p be a prime number such that f is Gauss at p, and such that p detk). Note that, since detk) 0, almost all prime numbers are of this form, so if we are able to prove that g is Gauss at p then we are done. Let k pz n 0. If there does not exist an m Zn 0 such that k = Km, then b k = 0. But then there certainly does not exist an m Z 0 such that k/p = Km. So we find b k 0 b k/p mod p v pk) ). Now suppose that there does exist an m Z n 0 such that k = Km. Denoting by adj K) the adjugate of K we obtain: m = adj K) k 84) detk) Since p detk), it follows that v p k) v p m). In particular we find m pz n 0. Since f is Gauss at p, we have a m a m/p mod p v ) pm). Since b k = a m and b k/p = a m/p, we also have b k b k/p mod p v ) pk). We conclude that g is Gauss at p. = Since K has integer entries, it follows immediately that for any m Z n 0, v pm) v p Km). So letting p be a prime number at which g is Gauss, m pz 0, k := Km, we find that b k b k/p mod p v pm) ). Since b k = a m and b k/p = a m/p, we conclude that f is Gauss at p. 2 To avoid ambiguity: g is obtained from f by substituting x i t k i t k ni n for all i n. 23

25 Example 6.7 Define Q := + X + Y XY, and let a, b, c, d Z 0 be such that ad bc 0. We saw in Example 6.5 that /Q is Gauss in the variables X, Y. It follows by Proposition 6.6 that / + x a y c + x b y d x a+b y c+d ) is Gauss in the variables x, y. In particular, in the case that a = d = 2 and b = c =, we see that /f as given by 78) is Gauss. 9 This is a good point to look back at our original problem, which was that of giving a characterization of Gaussian rational functions in n variables. We saw that in the case of n =, under the condition of separable denominator, the rational functions given by Proposition 6.0, i.e. the Q-linear sums of functions of type L, are precisely the ones are Gauss. One might hope that something similar holds in the case of n variables. All the results in this section up to Proposition 6.6 that produce Gaussian rational functions, followed from Proposition 6.0. Indeed, all the rational functions that we proved to be Gauss are Q-linear sums of functions of type L. However, Proposition 6.6 gives us a seemingly new method for finding Gaussian rational functions, and the natural question arises: Are these functions actually new? That is, given Proposition 6.0, does Proposition 6.6 produce Gaussian rational functions that are not Q-linear sums of functions of type L? The answer to this question is no. We can prove this using Lemma 6.9. Before stating the Lemma, we will need a definition. Definition 6.8 Let M = a i,j ) i,j n be an n n matrix with integer entries a i,j Z and let m {,, n}. Let A, B {,, n} such that A = B = m. Write A = {i,, i m }, B = {j,, j m } such that i < < i m and j < < j m. We define M A,B) to be the m m submatrix a ik,j l ) k,l m of M. Now define N := n m) and let A = {A,, A N } be )) an enumeration of the subsets of {,, n} that have cardinality m. We define M [A] := det to be the N N matrix whose entries M Ai,A j ) i,j N are the determinants of all m m submatrices of M under the ordering of A. Lemma 6.9 Let M be an n n matrix, let m {,, n}, let N := n m) and let A = {A,, A N } be an enumeration of the subsets of {,, n} of cardinality m. If detm) 0 then detm [A] ) 0. Proof. Let Λ m R n := {v v m v,, v m R n } m R n be the m-th exterior power of R n. Denoting by {e,, e n } the standard basis for R n, a basis for Λ m is given by B := {e i e im i < < i m n}. We can label the elements b,, b N of B with respect to A in the following way: For j N, write A j = {i,, i m } such that i < < i m, and then define b j := e i e im. Now let T : R n R n be the linear map corresponding to the matrix A with respect to the standard basis {e,, e n }). Since detm) 0, T has an inverse T. T induces a linear map Λ m T : Λ m R n Λ m R n by setting Λ m T v v m ) := Av Av m. Now the matrix of Λ m T with respect to the basis {b,, b N } is precisely M [A]. Denoting by I N : Λ m R n Λ m R n the identity map, we have I N = Λ m T T ) = Λ m T )Λ m T ). We conclude that Λ m T is invertible, hence detm [A] ) 0. 24

26 Proposition 6.20 Suppose that f is a rational function. Let K = k i,j ) i,j n be an n n matrix with non-negative integer entries k i,j Z 0 such that detk) 0. Then f is Q-linear sum of rational ) functions of type L in the variables x,, x n ) if and only if g := f t k t k n n,, t k n t knn n is a Q-linear sum of rational functions of type L in the variables t,, t n ). Proof. Let m n, Q,, Q n Z[x,, x n ] 0, let N := n m) and let A = {A,, A N } be an enumeration of the subsets of {,, n} of cardinality m. For r N, write A r = {j,, j m } such that j < < j m. We define J r x) J x) r := x j x jm Q j Q jm and J t) r by Q j,, Q jm ) x j,, x jm ), J t) r := t j t jm Q j Q jm Q j,, Q jm ) t j,, t jm ) 85) Note that, by the chain rule, for any r N we have Q j J r t) = t j t jm Q j,, Q jm ) Q j Q jm t j,, t jm ) = t j t jm x. Q j Q jm.... Q jm x Q j x n Q jm x n x t j..... x n t j x t jm x n t jm 86) Using the Cauchy-Binet formula we can work out the determinant of the product of the m n and n m matrices on the right hand side of 86) see Definition 6.8). We obtain From which it follows that J t) r = By Lemma 6.9, det K [A] ) 0, so we obtain J x). J x) N We see from 88) and 89) that every J r t) and every J r x) desired result follows. N l= J t). = K [A] J t) N = det ) x) K Ar,Al ) J l 87) J x). J x) N det )adj ) K [A] K [A] 88) J t). J t) N 89) can be written as a Q-linear in fact Z-linear) sum of J x),, J x) N,, J t). The can be written as a Q-linear in fact Z[/ detk [A] )]-linear) sum of J t) N 25

27 6.2 Restricting to subcones and necessity At this point the only Gaussian rational functions that we know of are given by Proposition 6.0. In an attempt to characterize Gaussian rational functions in n variables, one might try to reduce the problem of determining whether a rational function in n variables is Gauss to a simpler problem; for instance that of determining whether a rational function in n variables is Gauss. The motivation behind this approach is, of course, that we have a characterization for Gaussian rational functions in one variable. The following Proposition illustrates a special case of functions for which this approach works. Proposition 6.2 Let P, Q, R Z[x] such that Q is separable and Q0) 0. Then f := P/Q yr) is Gauss in the variables x, y) if and only if P/Q is Gauss in the variable x). Proof. = Suppose that f is Gauss with power series coefficients a k Z[/Q0)]. m Z 0, r N and almost every p prime Then in particular, for every a mp r,0) a mp r,0) mod p r ) 90) We can see 90) as the congruence conditions restricted to the coefficients of our power series corresponding to pure powers of x. Now note that we can expand f as follows: P f = Q yr = P Q y R = P + y R ) R 2 Q Q Q) + y2 + 9) Q In particular, we see that the part of the power series expansion of f containing the pure powers of x is precisely the power series expansion of P/Q. We conclude that P/Q must be Gauss. = Suppose that P/Q is Gauss. By Theorem 5.6, we can write P Q = θ x Q θ x Q l q 0 + q + + q l Q Q l for rational numbers q 0,, q l Q and Q,, Q l Z[x] 0. Define g := Q yr. Note that: f = P g = Q P g Q = θ ) y)g θ x Q θ x Q l q 0 + q + + q l g Q Q l ) 92) 93) It follows from application of Theorem 6.7 that θ y g)θ x Q i ) gq i = yx gq i g, Q i ) y, x) 94) is Gauss for every i. We conclude that every term in the expansion of the right hand side of 93) is Gauss, hence so is f. Later on we will be able to prove a stronger version of Proposition 6.2. Theorem This will be in the form of 26

28 The key to the implication = in Proposition 6.2, was examining the congruences restricted to the subspace of Z 2 0 generated by, 0), corresponding to the pure powers of x in the power series. We were able to split the power series into two parts, with the first part containing all the terms with powers of x, and the second part the rest. The first part turned out to be the power series expansion of a rational function namely P/Q), which we could conclude to be Gauss. The rest of this section will be an attempt to generalize this method. Definition 6.22 We define the support suppf) of a power series f = k 0 a kx k C[[x,, x n ]] as the powers k of x such that x k has non-zero coefficient: For a collection of power series f,, f l we define: suppf) := {k Z n 0 a k 0} 95) suppf,, f l ) := l suppf i ) 96) Definition 6.23 We call a subset X R n 0 a cone if it is closed under R 0-linear sums. That is, for any x, x 2 X and λ, λ 2 R 0 we have λ x + λ 2 x 2 X. Definition 6.24 For a subset S R n 0, we define ConeS) to be the cone generated by S. That is, i= ConeS) := m N{λ a + + λ m a m λ,, λ m R 0, a,, a m S} 97) Remark We can see ConeS) as the smallest cone containing S. In fact, ConeS) is the intersection of all cones containing S. Definition 6.26 We define the cone of a collection of power series f,, f l C[[x,, x n ]], which we will denote by Conef,, f l ), to be the cone generated by suppf,, f l ). Definition 6.27 We say a subset L R n 0 is positive-linearly dependent if there is a k N for which there exist pairwise distinct a 0,, a k L and not necessarily distinct) λ,, λ k R 0 such that a 0 = λ a + + λ k a k 98) If this is not the case we say that L is positive-linearly independent. Definition 6.28 Let X be a cone. We say a set B R n 0 X = ConeB)) and B is positive-linearly independent. is a basis for X if B generates X i.e. Lemma 6.29 Let S R n 0 that B S. be a finite subset and let X := ConeS). Then there is a basis B for X such 27