Lecture 15: Algebraic Geometry II

6.859/15.083 Integer Programming and Combinatorial Optimization Fall 009 Today... Ideals in k[x] Properties of Gröbner bases Buchberger s algorithm Elimination theory The Weak Nullstellensatz 0/1-Integer Programming The Structure of Ideals in k[x] Lecture 15: Algebraic Geometry II Theorem 1. If k is a field, then every ideal of k[x] is of the form f for some f k[x]. Moreover, f is unique up to multiplication by a nonzero constant in k. Proof: If I = {0}, then I = 0. So assume I = {0}. Let f be a nonzero polynomial of minimum degree in I. Claim: f = I. Clearly, f I. Let g I be arbitrary. The division algorithm yields g = q f + r, where either r = 0 or deg(r) < deg(f). I is an ideal, so q f I, and, thus, r = g q f I. By the choice of f, r = 0. But then g = q f f. Reminder: Gröbner Bases Fix a monomial order. A subset {g 1,..., g s } of an ideal I is a Gröbner basis of I if LT(g 1 ),..., LT(g s ) = LT(I). Equivalently, {g 1,..., g s } I is a Gröbner basis of I iff the leading term of any element in I is divisible by one of the LT(g i ). 1

Properties of Gröbner Bases I Theorem. Let G = {g 1,..., g s } be a Gröbner basis for an ideal I, and let f k[x 1,..., x n ]. Then the remainder r on division of f by G is unique, no matter how the elements of G are listed when using the division algorithm. Proof: First, recall the following result: Let I = x α : α A be a monomial ideal. Then a monomial x β lies in I iff x β is divisible by x α for some α A. Suppose f = a 1 g 1 + + a + a + r with r = r s g s + r = a 1 g 1 + sg s. Then r r I and, thus, LT(r r ) LT(I) = LT(g 1 ),..., LT(g s ). The lemma implies that LT(r r ) is divisible by one of LT(g 1 ),..., LT(g s ). This is impossible since no term of r, r is divisible by one of LT(g 1 ),..., LT(g s ). S-Polynomials Let I = f 1,..., f t. We show that, in general, LT(I) can be strictly larger than LT(f 1 ),..., LT(f t ). Consider I = f 1, f, where f 1 = x 3 xy and f = x y y + x with grlex order. Note that so x I. Thus x = LT(x ) LT(I). x (x y y + x) y (x 3 xy) = x, However, x is not divisible by LT(f 1 ) = x 3 or LT(f ) = x y, so that x LT(f 1 ), LT(f ). What happened? The leading terms in a suitable combination may cancel, leaving only smaller terms. ax α f i bx β f j On the other hand, ax α f i bx β f j I, so its leading term is in LT(I). This is an obstruction to {f 1,..., f t } being a Gröbner basis. Let f, g k[x 1,..., x n ] be nonzero polynomials with multideg(f) = α and multideg(g) = β. Let γ i = max(α i, β i ). We call x γ the least common multiple of LM(f) and LM(g). The S-polynomial of f and g is defined as γ x x S(f, g) = f g. LT(f) LT(g) γ

An S-polynomial is designed to produce cancellation of leading terms. Example: Let f = x 3 y x y 3 + x and g = 3x 4 y + y with the grlex order. Then γ = (4, ). Moreover, x 4 y 4 S(f, g) = x 3 y f x y 3x 4 y g 1 = x f y g 3 1 = x 3 y 3 + x 3 y 3 t Consider i=1 c if i, where c i k and multideg(f i ) = δ Z n + for all i. t If multideg( i=1 c t if i ) < δ, then i=1 c if i is a linear combination, with coefficients in k, of the S-polynomials S(f j, f k ) for 1 j, k t. Moreover, each S(f j, f k ) has multidegree < δ. Properties of Gröbner Bases II t c i f i = c jk S(f j, f k ) i=1 j,k Theorem 3. A basis G = {g 1,..., g s } for an ideal I is a Gröbner basis iff for all pairs i = j, the remainder on division of S(g i, g j ) by G is zero. Sketch of proof: s Let f I be a nonzero polynomial. There are polynomials h i such that f = i=1 h i g i. It follows that multideg(f) max(multideg(h i g i )). If <, then some cancellation of leading terms must occur. These can be rewritten as S-polynomials. The assumption allows us to replace S-polynomials by expressions that involve less cancellation. We eventually find an expression for f such that multideg(f) = multideg(h i g i ) for some i. It follows that LT(f) is divisible by LT(g i ). This shows that LT(f) LT(g 1 ),..., LT(g s ). 3

Buchberger s Algorithm Consider I = f 1, f, where f 1 = x 3 xy and f = x y y + x with grlex order. Let F = (f 1, f ). S(f 1, f ) = x ; its remainder on division by F is x. Add f 3 = x to the generating set F. S(f 1, f 3 ) = xy; its remainder on division by F is xy. Add f 4 = xy to the generating set F. S(f 1, f 4 ) = xy ; its remainder on division by F is 0. S(f, f 3 ) = y + x; its remainder is y + x. Add f 5 = y + x to the generating set F. The resulting set F satisfies the S-pair criterion, so it is a Gröbner basis. Buchberger s Algorithm The algorithm: In: F = (f 1,..., f t ) {defining I = f 1,..., f t } Out: Gröbner basis G = (g 1,..., g s ) for I, with F G 1. G := F. REPEAT 3. G := G 4. FOR each pair p q in G DO 5. S := remainder of S(p, q) on division by G 6. IF S = 0 THEN 7. UNTIL G = G Buchberger s Algorithm Proof of correctness: G := G {S} The algorithm terminates when G = G, which means that G satisfies the S-pair criterion. Proof of finiteness: The ideals LT(G ) from successive iterations form an ascending chain. 4

Let us call this chain J 1 J J 3. Their union J = J i is an ideal as well. By Hilbert s Basis Theorem, it is finitely generated: i=1 J = h 1,..., h r. Each of the h l is contained in one of the J i. Let N be the maximum such index i. Then J = h 1,..., h r J N J N+1 J. So the chain stabilizes with J N, and the algorithm terminates after a finite number of steps. Minimal Gröbner Basis Let G be a Gröbner basis for I, and let p G be such that LT(p) LT(G \ {p}). Then G \ {p} is also a Gröbner basis for I. A minimal Gröbner basis for an ideal I is a Gröbner basis G for I such that 1. LC(p) = 1 for all p G.. For all p G, LT(p) LT(G \ {p}). A given ideal may have many minimal Gröbner bases. But we can single one out that is better than the others: A reduced Gröbner basis for an ideal I is a Gröbner basis G for I such that 1. LC(p) = 1 for all p G.. For all p G, no monomial of p lies in LT(G \ {p}). Reduced Gröbner Basis Lemma 4. Let I = {0} be an ideal. Then, for a given monomial ordering, I has a unique reduced Gröbner basis. (One can obtain a reduced Gröbner basis from a minimal one by replacing g G by the remainder of g on division by G \ {g}, and repeating.) Elimination Theory Systematic methods for eliminating variables from systems of polynomial equations. For example, consider x 1 + x + 3x 3 + 4x 4 + 5x 5 + 15x 6 15 = 0, x 1 x 1 = 0,..., x 6 x 6 = 0. The reduced Gröbner basis with respect to lex order is G = {x 6 x 6, x 5 + x 6 1, x 4 + x 6 1, x 3 + x 6 1, x + x 6 1, x 1 + x 6 1}. So the original system has exactly two solutions: x = (1, 1, 1, 1, 1, 0) or x = (0, 0, 0, 0, 0, 1) 5

Given I = f 1,..., f s k[x 1,..., x n ], the l-th elimination ideal I l is the ideal of k[x l+1,..., x n ] defined by I l = I k[x l+1,..., x n ]. I l consists of all consequences of f 1 = f = = f s = 0 which eliminate the variables x 1,..., x l. Eliminating x 1,..., x l means finding nonzero polynomials in I l. Theorem 5. Let I k[x 1,..., x n ] be an ideal, and let G be a Gröbner basis of I with respect to lex order where x 1 > x > > x n. Then, for every 0 l n 1, the set G l = G k[x l+1,..., x n ] is a Gröbner basis of the l-th elimination ideal I l. Proof: It suffices to show that LT(I l ) LT(G l ). We show that LT(f), for f I l arbitrary, is divisible by LT(g) for some g G l. Note that LT(f) is divisible by LT(g) for some g G. Since f I l, this means that LT(g) involves only x l+1,..., x n. Any monomial involving x 1,..., x l is greater than all monomials in k[x l+1,..., x n ]. Hence, LT(g) k[x l+1,..., x n ] implies g k[x l+1,..., x n ]. Therefore, g G l. The Weak Nullstellensatz Recall that a variety V k n can be studied via the ideal This gives a map V I(V ). On the other hand, given an ideal I, I(V ) = {f k[x 1,..., x n ] : f(x) = 0 for all x V }. V (I) = {x k n : f(x) = 0 for all f I}. is an affine variety, by Hilbert s Basis Theorem. This gives a map I V (I). Note that the map V is not one-to-one: for example, V (x) = V (x ) = {0}. Recall that k is algebraically closed if every nonconstant polynomial in k[x] has a root in k. 6

Also recall that C is algebraically closed (Fundamental Theorem of Algebra). Consider 1, 1 + x, and 1 + x + x 4 in R[x]. They generate different ideals: However, V (I 1 ) = V (I ) = V (I 3 ) =. I 1 = 1 = R[x], I = 1 + x, I 3 = 1 + x + x 4. This problem goes away in the one-variable case if k is algebraically closed: Let I be an ideal in k[x], where k is algebraically closed. Then I = f, and V (I) are the roots of f. Since every nonconstant polynomial has a root, V (I) = implies that f is a nonzero constant. Hence, 1/f k. Thus, 1 = (1/f) f I. Consequently, g 1 = g I for all g k[x]. It follows that I = k[x] is the only ideal of k[x] that represents the empty variety when k is algebraically closed. The same holds when there is more than one variable! Theorem 6. Let k be an algebraically closed field, and let I k[x 1,..., x n ] be an ideal satisfying V (I) =. Then I = k[x 1,..., x n ]. (Can be thought of as the Fundamental Theorem of Algebra for Multivariate Polynomials: every system of polynomials that generates an ideal smaller than C[x 1,..., x n] has a common zero in C n.) The system f 1 = 0, f = 0,..., f s = 0 does not have a common solution in C n iff V (f 1,..., f s ) =. By the Weak Nullstellensatz, this happens iff 1 f 1,..., f s. Regardless of the monomial ordering, {1} is the only reduced Gröbner basis for the ideal 1. Proof: Let g 1,..., g s be a Gröbner basis of I = 1. Thus, 1 LT(I) = LT(g 1 ),..., LT(g s ). Hence, 1 is divisible by some LT(g i ), say LT(g 1 ). So LT(g 1 ) is constant. Then every other LT(g i ) is a multiple of that constant, so g,..., g s can be removed from the Gröbner basis. Since LT(g 1 ) is constant, g 1 itself is constant. 7

0/1-Integer Programming: Feasibility Normally, n a ij x j = b i j=1 x j {0, 1} i = 1,... m j = 1,..., n Equivalently, n f i := a ij x j b i = 0 i = 1,... m j=1 g j := x j x j = 0 j = 1,..., n An algorithm: In: A Z m n, b Z m Out: a feasible solution x to Ax = b, x {0, 1} n 1. I := f 1,..., f m, g 1,..., g n. Compute a Gröbner basis G of I using lex order 3. IF G = {1} THEN 4. infeasible 5. ELSE 6. Find x n in V (G n1 ) 7. FOR l = n 1 TO 1 DO 8. Extend ( x l+1,..., x n) to ( x l,..., x n) V (G l 1 ) Example: Consider x 1 + x + 3x 3 + 4x 4 + 5x 5 + 15x 6 = 15 x 1, x,..., x 6 {0, 1} The reduced Gröbner basis is G = {x 6 x 6, x 5 + x 6 1, x 4 + x 6 1, x 3 + x 6 1, x + x 6 1, x 1 + x 6 1} G 5 = {x 6 x 6 }, so x 6 = 0 and x 6 = 1 are possible solutions We get x = (1, 1, 1, 1, 1, 0) or x = (0, 0, 0, 0, 0, 1) 8

Structural insights: The polynomials in the reduced Gröbner basis can be partitioned into n sets: S n contains only one polynomial, which is either x n, x n 1, or x n x n. S i, for 1 i n 1, contains polynomials in x n,..., x i. Similar to row echelon form in Gaussian elimination. Allows solving the system variable by variable. Example: Consider x 1 + x + 3x 3 + 4x 4 + 6x 5 = 6, x 1,..., x 5 {0.1} The reduced Gröbner basis is { } x 5 x 5, x 4 x 5, x 4 x 4, x 3 + x 4 + x 5 1, x + x 5 1, x 1 + x 4 + x 5 1 The sets are S 5 = {x 5 x 5 } S 4 = {x 4 x 5, x 4 x 4 } S 3 = {x 3 + x 4 + x 5 1} S = {x + x 5 1} S 1 = {x 1 + x 4 + x 5 1} 0/1-Integer Programming: Optimization Modify the algorithm as follows: n Let h = y j=1 c j x j. Consider k[x 1,..., x n, y] and V (f 1,..., f m, g 1,..., g m, h). Use lex order with x 1 > > x n > y. The reduced Gröbner basis is either {1} or its intersection with k[y] is a polynomial in y. Every root of this polynomial is an objective function value that can be feasibly attained. Find the minimum root, and work backwards to get the associated values of x n,..., x 1. Example: { } min x 1 + x + 3x 3 : x 1 + x + x 3 = 3, x 1,..., x 3 {0, 1}. The reduced Gröbner basis is { } 1 7y + y, 3 + x 3 y, 4 + x + y, 1 x 1. The two roots of 1 7y + y are 3 and 4. The minimum value is y = 3, and the corresponding solution is (1, 1, 0). 9

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