Algebraic Geometry I

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1 6.859/ Iteger Programmig ad Combiatorial Optimizatio Fall 2009 Lecture 14: Algebraic Geometry I Today... 0/1-iteger programmig ad systems of polyomial equatios The divisio algorithm for polyomials of oe variable Multivariate polyomials Ideals ad affie varieties A divisio algorithm for multivariate polyomials Dickso s Lemma for moomial ideals Hilbert Basis Theorem Gröber bases 0/1-Iteger Programmig Feasibility Normally, a ij x j = b i j=1 x j {0, 1} i = 1,... m j = 1,..., Equivaletly, a ij x j b i = 0 j=1 i = 1,... m 2 x j x j = 0 j = 1,..., Motivates study of systems of polyomial equatios Refresher: Polyomials of Oe Variable Some basics: Let f = a 0 x m a 1 x m 1 a m, where a 0 = 0. We call m the degree of f, writte m = deg(f). m We say a 0 x is the leadig term of f, writte LT(f) = a 0 x. m For example, if f = 2x 3 4x 3, the deg(f) = 3 ad LT(f) = 2x 3. 1

2 If f ad g are ozero polyomials, the The Divisio Algorithm: I: g, f deg(f) deg(g) LT(f) divides LT(g). Out: q, r such that f = q g r ad r = 0 or deg(r) < deg(g) 1. q := 0; r := f 2. WHILE r = 0 AND LT( g) divides LT(r) DO 3. q := q LT(r)/LT(g) 4. r := r (LT(r)/LT(g))g Polyomials of More tha Oe Variable Fields: A field cosists of a set k ad two biary operatios ad which satisfy the followig coditios: (a b) c = a (b c) ad (a b) c = a (b c), a b = b a ad a b = b a, a (b c) = a b a c, there are 0, 1 k such that a 0 = a 1 = a, give a k there is b k such that a b = 0, give a k, a = 0, there is c k such that a c = 1. Examples iclude Q, R, ad C. Moomials: A moomial i x 1,..., x is a product of the form with α 1,..., α Z. We also let α := (α 1,..., α ) ad set α 1 α 2 α x 1 x 2... x, x α := x 1 α 1 x 2 The total degree of x α is α := α 1 α. Polyomials: α 2 α... x. 2

3 A polyomial i x 1,..., x is a fiite liear combiatio of moomials, f = a α x α, If a α = 0, the a α x α is a term of f. The total degree of f, deg(f), is the maximum α such that a α = 0. α S where a α k for all α S, ad S Z is fiite. The set of all polyomials i x 1,..., x with coefficiets i k is deoted by k[x 1,..., x ]. We call a α the coefficiet of the moomial x α. Example: f = 2x 3 y 2 z 3 2 y 3 z 3 3xyz y 2 Four terms, total degree six Two terms of max total degree, which caot happe i oe variable What is the leadig term? Orderigs o the Moomials i k[x 1,..., x ] For the divisio algorithm o polyomials i oe variable, > x m1 > x m > > x 2 > x > 1. I Gaussia elimiatio for systems of liear equatios, x 1 > x 2 > > x. Note that there is a oe-to-oe correspodece betwee the moomials i k[x 1,..., x ] ad Z. A moomial orderig o k[x 1,..., x ] is ay relatio > o Z that satisfies 1. > is a total orderig, 2. if α > β ad γ Z, the α γ > β γ, 3. every oempty subset of Z has a smallest elemet uder >. Examples of Moomial Orderigs Lex Order: For α, β Z, α > lex β if the left-most ozero etry of α β is positive. We write x α > lex x β if α > lex β. For example, (1, 2, 0) > lex (0, 3, 4) ad (3, 2, 4) > lex (3, 2, 1). Also, x 1 > lex x 5 2x 3 3. Graded Lex Order: For α, β Z, α > grlex β if α > β or α = β ad α > lex β. For example, (1, 2, 3) > grlex (3, 2, 0) ad (1, 2, 4) > grlex (1, 1, 5). 3

4 Further Defiitios Let f = α a αx α be a ozero polyomial i k[x 1,..., x ] ad let > be a moomial order. The multidegree of f is multideg(f) := max {α Z : a α = 0}. > The leadig coefficiet of f is The leadig moomial of f is The leadig term of f is LC(f) := a multideg(f ). multideg(f ) LM(f) := x. LT(f) := LC(f) LM(f). Example Let f = 4xy 2 z 4z 2 5x 3 7x 2 z 2 ad let > deote the lex order. The multideg(f) = (3, 0, 0), LC(f) = 5, LM(f) = x 3 LT(f) = 5x 3. The Basic Algebraic Object of this Lecture A subset I k[x 1,... x ] is a ideal if it satisfies: 1. 0 I, 2. if f, g I, the f g I, 3. if f I ad h k[x 1,... x ], the h f I. Let f 1,..., f s k[x 1,... x ]. The { } s f 1,..., f s := h i f i : h 1,..., h s k[x 1,... x ] is a ideal of k[x 1,... x ]. (We call it the ideal geerated by f 1,..., f s.) i=1 A ideal I is fiitely geerated if I = f 1,..., f s, ad we say that f 1,..., f s are a basis of I. 4

5 Polyomial Equatios Give f 1,..., f s k[x 1,... x ], we get the system of equatios f 1 = 0,..., f s = 0. If we multiply the first equatio by h 1, the secod oe by h 2, ad so o, we obtai which is a cosequece of the origial system. h 1 f 1 h 2 f 2 h s f s = 0, Thus, we ca thik of f 1,..., f s as cosistig of all polyomial cosequeces of f 1 = f 2 = = f s = 0. Let f 1,..., f s k[x 1,... x ]. The we set V (f 1,..., f s ) := {(a 1,..., a ) k : f i (a 1,..., a ) = 0, i = 1,..., s} ad call V (f 1,..., f s ) a affie variety. If f 1,..., f s = g 1,..., g t, the V (f 1,..., f s ) = V (g 1,..., g t ). Let V k be a affie variety. The we set I(V ) := {f k[x 1,..., x ] : f(a 1,..., a ) = 0 for all (a 1,..., a ) V }. If V is a affie variety, the I(V ) is a ideal. Drivig Questios Does every ideal have a fiite geeratig set? Give f k[x 1,..., x ] ad I = f 1,..., f s, is f I? Fid all solutios i k of a system of polyomial equatios Fid a ice basis for f 1,..., f s. A Divisio Algorithm i k[x 1,..., x ] Goal: Divide f by f 1,..., f s. f 1 (x 1,..., x ) = = f s (x 1,..., x ) = 0. Example 1: Divide f = xy 2 1 by f 1 = xy 1 ad f 2 = y 1, usig lex order with x > y. This leads to xy 2 1 = y (xy 1) ( 1) (y 1) 2. Example 2a: Divide f = x 2 y xy 2 y 2 by f 1 = xy 1 ad f 2 = y 2 1, usig lex order with x > y. This evetually leads to x 2 y xy 2 y 2 = (x y) (xy 1) 1 (y 2 1) x y 1. 5

6 Theorem 1. Fix a moomial order o Z, ad let (f 1,..., f s ) be a ordered tuple of polyomials i k[x 1,..., x ]. The every f k[x 1,..., x ] ca be writte as f = a 1 a s f s r, where a i, r k[x 1,..., x ], ad either r = 0 or r is a liear combiatio, with coefficiets i k, of moomials, oe of which is divisible by ay of LT(f 1 ),..., LT(f s ). We call r a remaider of f o divisio by (f 1,..., f s ). If a i f i = 0, the 1. a 1 := 0;..., a s := 0; r := 0 2. p := f 3. WHILE p = 0 DO 4. i := 1 multideg(f) multideg(a i f i ). 5. WHILE i s AND o divisio occurred DO 6. IF LT(f i ) divides LT(p) THEN 7. a i := a i LT(p)/LT(f i ) 8. p := p (LT(p)/LT(f i ))f i 9. ELSE 10. i := i IF o divisio occured THEN 12. r := r LT(p) 13. p := p LT(p) More Examples Example 2b: Divide f = x 2 y xy 2 y 2 by f 1 = y 2 1 ad f 2 = xy 1, usig lex order with x > y. This leads to x 2 y xy 2 y 2 = (x 1) (y 2 1) x (xy 1) 2x 1. The remaider is differet from the oe i Example 2a! Example 3a: Divide f = xy 2 x by f 1 = xy 1 ad f 2 = y 2 1 with the lex order. The result is xy 2 x = y (xy 1) 0 (y 2 1) ( x y). Example 3b: Divide f = xy 2 x by f 1 = y 2 1 ad f 2 = xy 1 with the lex order. The result is xy 2 x = x (y 2 1) 0 (xy 1) 0. The secod calculatio shows f f 1, f 2, but the first does ot! 6

7 Moomial Ideals A ideal I is a moomial ideal if there is A Z such that I cosists of all fiite sums α A h αx α. We write I = x α : α A. Let I = x α : α A. The x β I iff x β is divisible by x α for some α A. x β is divisible by x α iff β = α γ for some γ Z. Thus, α Z cosists of the expoets of all moomials divisible by x α. If I is a moomial ideal, the f I iff every term of f lies i I. Dickso s Lemma Let A Z. The (α Z ) α A ca be expressed as the uio of a fiite subset of the α Z. A moomial ideal I = x α : α A ca be writte i the form I = x α(1),..., x α(s), where α(1),..., α(s) A. Hilbert Basis Theorem: Prelimiaries Let I k[x 1,..., x ] be a ideal other tha {0}. Let LT(I) = the set of leadig terms of elemets i I. LT(I) is a moomial ideal. There are g 1,..., g s I such that Hilbert Basis Theorem LT(I) = LT(g 1 ),..., LT(g s ). Theorem 2 (Hilbert 1888). Every ideal I k[x 1,..., x ] has a fiite geeratig set. That is, I = g 1,..., g s for some g 1,..., g s I. Hilbert Basis Theorem: Proof Let I = {0}. Recall that LT(I) = LT(g 1 ),..., LT(g s ). Claim: I = g 1,..., g s. 7

8 Let f I. If we divide f by g 1,..., g s, we get Suppose r = 0. Note that r I. f = a 1 g 1 a s g s r, where o term of r is divisible by ay of LT(g 1 ),..., LT(g s ). Claim: r = 0. Hece, LT(r) LT(I) = LT(g 1 ),..., LT(g s ). So LT(r) must be divisible by some LT(g i ). Cotradictio! Thus, f = a 1 g 1 a s g s, which shows I g 1,..., g s. Gröber Bases Fix a moomial order. A subset {g 1,..., g s } of a ideal I is called a Gröber basis if LT(I) = LT(g 1 ),..., LT(g s ). Equivaletly, {g 1,..., g s } is a Gröber basis of I iff the leadig term of ay elemet i I is divisible by oe of the LT(g i ). Note that every ideal I = {0} has a Gröber basis. Moreover, ay Gröber basis of I is a basis of I. Next Time Properties of Gröber bases Computatio of Gröber bases (Buchberger s Algorithm) Solvig 0/1-iteger programs Solvig (geeral) iteger programs 8

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