Lecture 2: Gröbner Basis and SAGBI Basis

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1 Lecture 2: Gröbner Basis and SAGBI Basis Mohammed Tessema Suppose we have a graph. Suppose we color the graph s vertices with 3 colors so that if the vertices are adjacent they are not the same colors. How can we use algebra to answer questions like this? Can we color a graph with 4 colors? Can we classify graphs that can be colored with a certain number of colors? Dang vu Giang wrote a paper Gröbner basis and convex polytopes on planar graphs. A planar graph, is a graph where the edges do not cross. Using Gröbner basis he can show all planar graphs are 4-colorable. 1 Polynomial Rings 1.1 Polynomial Rings in one variable Let N = {0, 1, 2,... }. Let k be a field and x an indeterminate/variable. By a polynomial in x with coefficients in k we mean f(x) = a 0 + a 1 x + a 2 x a n x n, where a i k, n N. We define k[x] to be the set of all polynomials in the above form. Then (k[x], +, ) is a ring called the polynomial ring in one variable. Example 1 Q[x], R[x], C[x] Given f(x) = a 0 + a 1 x + a 2 x a n x n, a n 0, x n is the leading monomial a n x n is the leading term n is the degree of the polynomial If deg(c) = 0, then c is the non-zero constant. Note: deg(0) = Properties: If f, g k[x] we have: 1. deg(f + g) max{deg(f), deg(g)} 1

2 2. deg(fg) = deg(f) + deg(g) 3. deg(f g) = deg(f) deg(g) Division Algorithm: Given f, g k[x], g 0 there exists unique q, r k[x] such that where r = 0 or deg(r) < deg(g). Exercise 1 Prove the division algorithm. f = gq + r Definition 1 An ideal is a subset I of k[x] satisfying the following: 1. If f, g I then f + g I 2. If f I and h k[x] then fh I Example 2 Consider Q[x]. Let I = {xf(x) : f(x) Q[x]} = polynomials with 0 constant term. Definition 2 A set S k[x], generates an ideal I if every element of I is obtained as follows: m f = g i h i, where g i S, h i k[x], m N i=1 We say I is finitely generated if S <. If I is generated by a single polynomial f, we call it a principal ideal. Proposition 1 Every ideal I k[x] is principal. Exercise 2 Prove this proposition. Hint: Use the division algorithm. 1.2 Polynomial ring in several variables Let k be a field and x 1, x 2,... x n be variables. By a monomial in the x i s we mean n x α i i, i=1 where α i N. The polynomial ring k[x 1,..., x n ] is the set of all k-linear combinations of monomials. So we have m f = a i x α i, where x α i = n j=1 xα ij i, α i = (α ij ) n j=1 N n. i=1 Theorem 1 (Hilbert s Basis Theorem) Every ideal I k[x 1,..., x n ] is finitely generated. We will see a proof of this later using Gröbner Basis. 2

3 1.3 Algebraic Varieties Let k be an algebraically closed field (think k = C). Let f k[x 1,..., x n ]. We say v = (λ 1,... λ n ) k n is called a zero (solution) to f if f(λ 1,... λ n ) = 0. If S k[x 1,... x n ], then is called the algebraic variety defined by S. V (S) = {v k n : f(v) = 0, for all f S} Example 3 1. Let f = n i=1 a ix i, a i k. V (f) = {v k n : a i v i = 0} = {v k n : (a 1,... a n ) v = 0} = (a 1,..., a n ), which is a hyperplane 2. f = a i x α i. V (f) is called a hypersurface. Theorem 2 Let S k[x 1,..., x n ]. Let I =< S >. Then V (S) = V (I) Corollary 1 If S k[x 1,..., x n ], there exists f 1,..., f m S such that V (S) = V (f 1,..., f m ) Theorem 3 (Hilberts Nullstellensatz- weak form) Let I k[x 1,..., x n ]. Then V (I) = if and only if I =< 1 >= k[x 1,... x n ] 2 Introduction to Gröbner Basis 2.1 Monomial Orders Definition 3 By a monomial order on monomials x α, α N n we mean a total order, on the monimials satisfying 1. x α x β then x α x γ x β x γ for all α, β, γ N n 2. x α x 0 = 1 (1 is the least element.) Example 4 1. The dictionary (lexicographic order): x α lex x β if and only if α β = 0 or the first non-zeo entry of α β is positive. and 2. Graded lex order: x α glex x β if α > β or if α = β and x α lex x β So we have x 2 y 3 lex xy 75 x 2 y 3 glex xy 75 3

4 Theorem 4 (Robbiano) If n 2 then there are uncountable cardinality of distinct monomial orders equal to R = c. Proof: Pick a transcendental γ R. Let u = (1, γ, γ 2,..., γ n 1 ). Let A, B N n. Then A u B if and only if Au Bu. Show that u is a nomial order. Let u = (1, γ,..., γ n 1 ), v = (1, τ,..., τ n 1 ) where γ, τ R + are transcendental. Claim: u v. Then (A B) u > 0 so we have A u > B u, so A u B. We also have that (A B) v < 0, so that B v > A v, so that B v A. 2.2 Monomial Ideals and Dickson s Lemma A monomial ideal is an ideal generated by monomials in k[x 1,..., x n ]. Properties: 1. I is a monomial ideal if and only if for each f = k i x α i I, each x α i I when k i In fact, each monomial in I is a k-basis for I. 3. The Stanley-Reisner ring is k[x 1,..., x n ]/I where I are the square free monomial ideals Lemma 1 (Dickson s Lemma) Every monomial ideal in k[x 1,..., x n ] is finitely generated. To prove this induct on n (See Cox Froberg) Consequences of Dickson s Lemma: Recall: A monomial order is a total order on N n with (0,... 0) being the least element. Given any monomial, x α, there can be infinitely many monomial x β x α. Example 5 x 2 y lex xy n, n N. But if x α 1 x α 2... it eventually terminates. Proof: I =< x α i > i=1=< x α 1,..., x αs, where x α 1 is the smallest. If x β I then x β = x α i x γ. So we have x β x α i x α 1 Definition 4 Let be a monomial order, f = k i x α i k[x 1,..., x n ]. The initial of f in (f) is defined to be the largest monomial in f with respect to. Given I k[x 1,..., x n ] we have In (I) =< in (f) : f I\{0} > Definition 5 Let I be an ideal and G = {g 1,..., g s } be the finite subset of I. WE say G is a Gröbner basis for I if In (I) =< in (g 1 ),..., in (g s ) > Theorem 5 HBT: Every ideal I k[x 1,..., x n ] is finitely generated. 4

5 Proof: Let G be a Gröbner basis for I. Claim I =< G >. Let f I, in (f) < in(g i ) >. We have in (f) = in (g i )x β, g i G and f 1 = f cx β g i where c = lc(f). lc(g i ) Clearly, f i I and in (f) > in (f 1 ). If this is zero we are done. Otherwise we can repeat this process. Inductively define f 2, f 3,... where in (f) > in (f 1 ) > in (f 2 ) >.... This will terminate at 0. Thus we can write f as a polynomial in {g i } s i=1. Recall: Gröbner basis depends on. Question: If 1 2 does this imply that in 1 (I) in 2 (I)? Theorem 6 There are only finitely many distinct initial ideals for I k[x 1,..., x n ]. From the above theorem there is a Gröbner basis for I that works for any. We call such a Gröbner basis a universal Gröbner basis. Division Algorithm: Let G = {g 1,..., g s } be a Gröbner basis for some I k[x 1,..., x n ]. Then, for f I, there exists a unique remainder r k[x 1,..., x n ] such that f = s i=1 g ih i +r where r = 0 or no monomial in r is divisible by in(g i ) for some i = 1,..., s. Q: If H = {g 1,..., g r }, f k[x 1,..., x n ] is there a polynomial r k[x 1,..., x n ] where f = g i h i + r, r =.... Let T = {g 1,..., g s } be a finite subset of k[x 1,..., x n ]. Let I =< T >. Pick f, g T, in (f) = x α, in (f) = x β, lcm{x α, x β } = x γ. Definition 6 The S-polynomial of f and g, S(f, g) = xγ lt(f) f xγ lt(g) g Buchberger criterion: T will be a Gröbner basis for I if and only if for all f, g I, S(f, g) = 0 or in (S(f, g)) is divisible by in (g i ) for some i. Applications of Gröbner Basis: Ideal membership: We can easily check if f I or not. Solving a system of polynomial equations. Given f 1,..., f s k[x 1,..., x n ]. We can compute V (f 1,..., f s ) = V (I) = V (G). Coloring of graphs: Given a graph (V, E), E V V. Let V = {x 1,..., x n }. Plan to color the graph by 3 colors in such a way that no two adjacent vertices have the same color. Let the colors be the three distinct cube roots of unity, {ζ, ζ 2, ζ 3 = 1}. Since each x i can be potentially assigned as 1, ζ, ζ 2, we can define f i = x 3 i 1, i = 1, 2,..., n over C[x 1,..., x n ]. Let x i x j E, i j. We have x 3 i x 3 j = (x i x j )(x 2 i + x i x j + x 2 j). 5

6 Let g ij = x 2 i + x i x j + x 2 j if x i x j E. Now consider I =< f i, g ij >, where i = 1,... n and we include all edges. Therefore the graph is three colorable if and only if I C[x 1,..., x n ]. REU problem 1 1. Can we find some class of graphs that can be three colorable? 2. Similarly four coloring of graphs: Let V = {x 1,..., x n } with colors {1, ζ, ζ 2, ζ 3 }, where g ij = x 3 i + 3x 2 i x j + 3x i x 2 j + x 3 j. Theorem 7 Every planar graph is four colorable. 6