(and the Frobenius problem) - a gentle introduction

Transcription

1 GRÖBNER BASES (and the Frobenius problem) - a gentle introduction Niels Lauritzen

2 c Niels Lauritzen NIELS LAURITZEN DEPARTMENT OF MATHEMATICAL SCIENCES UNIVERSITY OF AARHUS DK-8000 ÅRHUS C DENMARK niels@imf.au.dk URL: Version 0.99 (August 2004)

3 Chapter 1 Gröbner bases Mathematics owes many of its roots to the classical business of solving equations. This is often overlooked and sometimes ridiculed in the setting of more abstract mathematics. A good example is linear algebra. Here you can rush through the abstract definitons of a vector space, homomorphism, dimension and whatnot in two hours. Does this make you an expert in linear algebra? Nope! The fun begins when you start to solve systems of linear equations. You get to ask your own questions and appreciate (slowly) the abstract concepts. Recall the beautiful method of Gauss elimination for solving systems of linear equations. Take as an example x y z = 0 x + y z = 0 x + y 1 = 0. We eliminate x by subtracting the first equation from the second and third. This gives the simpler system: x y z = 0 2y = 0 2y + z 1 = 0. Definition We call two systems of equations equivalent if they have the same set of solutions. Can we mimic Gauss elimination for systems of algebraic (polynomial) equations in several variables such as 1

4 2 CHAPTER 1. GRÖBNER BASES x 3 x y 2 = 0 x 2 y 3 = 0? The point of Gröbner basis theory is that with some ingenuity one actually can generalize Gauss elimination to the more general setting of algebraic equations. These insights were given by Buchberger and Hironaka around Today the theory of Gröbner bases is a vibrant field with many applications in pure mathematics, computer science, statistical models (HMM) in bioinformatics ($). 1.1 Reducing (solving) equations Let us restrict ourselves to the case of three variables x, y, z. A polynomial consists of a sum of terms like for example P = x 3 x y 2 + xyz 17. To proceed with the theory we need to have an ordering on these terms. We simply fix the lexicographical ordering given by x > y > z. In this setting we have x 3 > xyz 17 > x > y 2. I hope you can see the idea from the above example, because I do not want to write it out (it easily becomes boring). The idea is just like the usual ordering on entries in an encyclopedia, where for example ABC > BAA. We have simply replaced letters with powers of the variables. Given a polynomial f and a term ordering, we let τ f denote the largest term in f. As an example we have τ P = x 3. A fundamental trick for solving equations is the following generalization of Gauss elimination. Theorem The system of (algebraic) equations f(x, y, z) = 0 g(x, y, z) = 0 is equivalent to f(x, y, z) = 0 (1.1) g(x, y, z) + λ(x, y, z)f(x, y, z) = 0, (1.2) and f(x, y, z) = 0 g(x, y, z) = 0 λ(x, y, z)g(x, y, z) + µ(x, y, z)f(x, y, z) = 0, where λ(x, y, z) and µ(x, y, z) are polynomials.

5 1.2. A KEY EXAMPLE 3 Proof. Exercise. We will need the following fundamental definition for reducing systems of equations. Definition The equation f can be reduced by the equation g if the largest term τ g in g divides a term τ f in f. In this case f reduces to modulo g. A set of equations f f (τ f /τ g )g f 1 (x, y, z) = 0 f r (x, y, z) = 0 is called reduced if f i cannot be reduced by f j for i j. Example In the lexicographical ordering x > y, x 2 + y 2 1 reduces to xy + xz + y 2 1 modulo x y z. Example Use Theorem to transform the system of equations x y z = 0 x + y z 2 = 0 x 2 + y 2 1 = 0 into an equivalent reduced system of equations. Exercise Prove that every system of equations can be transformed into a reduced one using Theorem A key example Consider the system of equations x 3 y 2 x = 0 y 3 x 2 = 0. Reducing the first equation by the second we get x 3 y 2 x x(x 2 y 3 ) = xy 3 x y 2.

6 4 CHAPTER 1. GRÖBNER BASES This gives the equivalent system of equations xy 3 x y 2 = 0 x 2 y 3 = 0 This really sucks! We have transformed our system of equations into an equivalent system, which by no means is easier to solve. We are missing a crucial device. This device is called the S-polynomial The S-polynomial The least common multiple of xy 3 and x 2 is x 2 y 3. We form the S-polynomial S(xy 3 x y 2, x 2 y 3 ) = x(xy 3 x y 2 ) y 3 (x 2 y 3 ) = x 2 xy 2 + y 6. This leads to the equivalent system of equations xy 3 x y 2 = 0 x 2 y 3 = 0 x 2 xy 2 + y 6 = 0 Notice now that the third equation can be reduced by the second. We continue our reduction steps: xy 3 x y 2 = 0 x 2 y 3 = 0 xy 2 y 6 + y 3 = 0 x + y 2 y 7 + y 4 = 0 x 2 y 3 = 0 xy 2 y 6 + y 3 = 0 x + y 2 y 7 + y 4 = 0 x 2 y 3 = 0 y 9 2y 6 y 4 + y 3 = 0 HEUREKA! Now we have an equation involving only the variable y. We may now solve for y, insert in the previous equations and then solve for x. Think about it! We have discovered the equation y 9 2y 6 y 4 +y 3 = 0 buried in the original equations x 3 y 2 x = 0 and y 3 x 2 = 0.

7 1.3. THE DIVISION ALGORITHM 5 But, wait, the above system is not reduced since x in the first equation divides x 2 in the second equation. Below I have sketched an algorithm (the division algorithm in several variables) showing that the second equation reduces to the equation 0 = 0 using the first and third equation. This means that it can be left out completely and we end up with a nice reduced system of equations. x + y 2 y 7 + y 4 = 0 y 9 2y 6 y 4 + y 3 = The division algorithm We may formalize the reduction process in the socalled division algorithm for getting a remainder using a system of equations. Below it is sketched how the division algorithm shows that the remainder f F is 0, where f = x 2 y 3 and F = (x y 7 + y 4 + y 2, y 9 2y 6 y 4 + y 3 ). The underlined terms are the largest terms. I hope you can spot the idea. Writing it out formally only slows the perception of the human brain. x 2 y 3 : (x y 7 + y 4 + y 2, y 9 2y 6 y 4 + y 3 ) x 2 xy 7 + xy 4 + xy 2 xy 7 xy 4 xy 2 y 3 xy 7 y 14 + y 11 + y 9 xy 4 xy 2 + y 14 y 11 y 9 y 3 xy 4 + y 11 y 8 y 6 xy 2 + y 14 2y 11 y 9 + y 8 + y 6 y 3 xy 2 + y 9 y 6 y 4 y 14 2y 11 2y 9 + y 8 + 2y 6 + y 4 y 3 y 14 2y 11 y 9 + y 8 y 9 + 2y 6 + y 4 y 3 y 9 + 2y 6 + y 4 y Definition of a Gröbner basis We will give a first approximation to the definition of a Gröbner basis.

8 6 CHAPTER 1. GRÖBNER BASES Definition A system f 1,...,f r of equations is called a Gröbner basis if every S-equation S(f i, f j ) reduces to 0 = 0 using f 1,..., f r. Given this definition there is an obvious algorithm for finding a Gröbner basis equivalent to a given system of equations. We have sketched it by example. It is called Buchberger s algorithm. The fundamental question is now why it stops? In principle it could run indefinitely and produce infinitely many new S-equations Enter ideals We need an algebraic object containing all the possible equations that we can get from the system f 1,...,f r of equations. Naturally this is the set f 1,...,f r := {λ 1 f λ r f r λ 1,...,λ r polynomials}. In more mathematical terms this is known as the ideal generated by f 1,...,f r in the ring k[x, y, z] of polynomials. We will state the main theorem in the theory of Gröbner bases, which says that our provisional definition of a Gröbner basis is equivalent to something which is easier to handle mathematically (but a bit more difficult conceptually). Theorem A system f 1,..., f r of equations is a Gröbner basis if and only if for every f f 1,...,f r there exists j = 1,...,r such that τ fj divides τ f. Mathematically this is a much cleaner definition to work with. It has the disadvantage of completely hiding the basic idea of solving equations. The proof of the main theorem may be found in for example [1] or [2]. Now we can sketch a proof why the obvious algorithm for computing Gröbner bases terminates. Suppose that we at a certain step have the system F = (f 1,...,f r ) of equations and that we are adding a non-zero reduction of S = S(f i, f j ) F using F to the system F. Then (Exercise 5) S f 1,...,f r and none of τ fj divides τ S. Consider the set T = {(a, b, c) N 3 x a y b z c = τ f, for some f f 1,...,f r }. Then we may find finitely many vectors v 1,..., v N T such that if w T, then w = v j + r for some j = 1,..., N and r N 3. This is a special case of a surprising result due to the American mathematician Dickson (1914):

9 1.5. LEXICOGRAPHIC ORDERING 7 Lemma Let S be a subset of N n. Then we may find finitely many vectors v 1,..., v N N n such that if w S, then w = v j + r for some j = 1,...,N and r N n. Now why does this result imply that the algorithm terminates? Well, for every new S that we add to the system of equations, we actually get a new vector w from τ S in T with the property that w is not a sum of any of the previous vectors with a vector in N 3. So at a certain point (according to Dickson) all the S-equations that we add have to reduce to zero (Exercise 6). 1.5 Lexicographic ordering The mathematical super hero Gauss already knew the inner workings of lexicographical ordering: Dein e duobos terminis Ma α b β c γ...et Ma α b β c γ... priori ordinem altiorem tribuemus quam posteriori, si fit vel α > α, vel α = α et β > β, vel α = α, β = β et γ > γ, vel etc. i.e. se e differentiis α α, β β, γ γ etc. prima, quae non-evascit, positiva evadit. For our purposes we note the following (Exercise 3) Theorem Lexicographic ordering is a well ordering on terms. C. F. Gauss, Maple breaks down! This example is due to Anders Nedergaard Jensen. Try it out on maple. Is Mathematica capable of doing it? Solve f(x, y, z) = 0 for f = x 4 + y 4 + z 4 4xyz + x + y + z?

10 8 CHAPTER 1. GRÖBNER BASES Leads to the system of equations 4x 3 4yz + 1 = 0 4y 3 4xz + 1 = 0 4z 3 4xy + 1 = 0. Reduced equation in z is z 32 z z z z z z z z z z z z z z z z 18 = Exercises In what follows R will denote the ring of polynomials in 3 variables x, y, z. We fix the lexicographic ordering given by x > y > z. 1. Solve the system of equations in 1.2 aiming to get an equation in x instead. In other words, work with the lexicographic ordering, where y > x. 2. Prove Theorem Prove that lexicographic ordering is a well ordering i.e. for every non-empty set T of terms there is a smallest term in T. 4. We say that f can be reduced by a set of polynomials F = {f 1,...,f r } if f can be reduced by some g F. In this case we write f F f (τ/τ g )g. A natural thing to do is to continue the process of reducing modulo g: f F f 1 F f 2 F (a) Let ǫ F (f) denote the largest term in f with can be reduced by some g F. Prove that ǫ F (f) is strictly bigger than ǫ F (f ), where f F f. (b) Prove that the successive reduction modulo F stops.

11 1.7. EXERCISES 9 5. Suppose that F = (f 1,..., f r ) is a system of equations and that we are adding a non-zero reduction of S = S(f i, f j ) F using F to F. Prove that and that none of τ fj divides τ S. S f 1,...,f r 6. Prove in greater detail (read: Convince yourself) that Buchbergers algorithm terminates. 7. Try to run the example in 1.6 on Maple and Mathematica.

12 10 CHAPTER 1. GRÖBNER BASES

13 Chapter 2 The Frobenius problem Consider positive integers 0 < a 1 a r such that gcd(a 1,...,a r ) = 1. Which numbers can be represented as a sum of a 1,...,a r? In other words for which positive integers N do we have an integral solution to a 1 x a r x r = N, where x 1,...,x r 0. Let us look at a few examples. Example (i) Let a 1 = 3 and a 2 = 7. Then clearly 1, 2, 4, 5, 8, 11 is not a sum of 3 and 7. But since 12, 13 and 14 is a sum of 3 and 7 we conclude that every N 12 is a sum of 3 and 7. (ii) Let a 1 = 5 and a 2 = 7. Then 1, 2, 3, 4, 6, 8, 9, 11, 13, 16, 18, 23 is not a sum of 5 and 7. But since 24, 25, 26, 27 and 28 is a sum of 5 and 7 we conclude that every N 24 is a sum of 5 and 7. It can be shown that there is a largest number N(a 1,..., a r ) (the Frobenius number) which is not a sum of a 1,...,a r. 2.1 Exercises 1. Prove that the Frobenius number exists. 2. Prove that N(a 1, a 2 ) = a 1 a 2 a 1 a 2, where 0 < a 1 < a 2 and gcd(a 1, a 2 ) = 1. This result is due to Sylvester (1884). 11

14 12 CHAPTER 2. THE FROBENIUS PROBLEM 3. Suppose that 0 < a 1 a 2 a r are natural numbers. Consider the ideal I = x 1 t a 1,..., x r t ar. Let G be a Gröbner basis of I over the lexicographic ordering given by t > x 1 > > x r. Prove that the Frobenius problem: can N be written as a sum of a 1,...,a r can be solved if and only if the remainder of t N divided by G is of the form x m 1 1 x mr. In this case r a 1 m a r m r = N. 4. Use Maple and the previous exercise to compute N(11, 13, 17). 5. Compute N(238, 569, 791). 6. Let r k denote the smallest positive integer k (mod a 1 ) such that x 2 a x r a r = k has a solution x 2,...,x k N. Let r = max(r 1, r 2,...,r a1 1). Prove that N(a 1,..., a r ) = r a 1. This result is due to Brauer and Shockley (1962).

15 Appendix A Gröbner basis commands in Maple It is quite easy to perform Gröbner basis computations using Maple. Simply log into your account and type the command maple in an X-window. Then you fire up a text based version of maple sufficient for our purposes. Below the author succeeded in computing a Gröbner basis for the ideal x t 3, y t 4, z t 5 over the lexicographical term ordering < given by t > x > y > z. Notice that input from the user is preceeded by > below. You can also find the computation of the Gröbner basis of (x 3 x y 2, x 2 y 3 ) over the lexicographical term ordering x > y below. This probably caused you a great deal of headache. Maple does this without any sloppy errors. The last line of input divides t 131 by the Gröbner basis gb. The output is the monomial z 23 y 4. How does this relate to the Frobenius problem? [187]frodo: /DMF> maple \ˆ/ Maple 6 (IBM INTEL LINUX22)._ \ / _. Copyright (c) 2000 by Waterloo Maple Inc. \ MAPLE / All rights reserved. Maple is a registered trademark of < > Waterloo Maple Inc. Type? for help. > with(groebner); [fglm, gbasis, gsolve, hilbertdim, hilbertpoly, hilbertseries, inter_reduce, is_finite, is_solvable, leadcoeff, leadmon, leadterm, normalf, pretend_gbasis, reduce, spoly, termorder, testorder, univpoly] > gb := gbasis([x-tˆ3, y-tˆ4, z-tˆ5], plex(t, x,y,z)); 13

16 14 APPENDIX A. GRÖBNER BASIS COMMANDS IN MAPLE bytes used= , alloc=917336, time= gb := [y - z, -y + x z, x y - z, -z + y x, x - y z, -x + t z, t y - z, t x - y, -x + t ] > normalf(tˆ131, gb, plex(t, x, y, z)); bytes used= , alloc= , time= z y > gbasis([xˆ3 - x - yˆ2, xˆ2 - yˆ3], plex(x,y)); [-2 y + y - y + y, x + y - y + y ]

17 Bibliography [1] Cox, Little, and O Shea, Ideals, Varieties and Algorithms, Springer Verlag, [2] Niels Lauritzen, Concrete Abstract Algebra, Cambridge University Press,