Problem Set 1 Solutions

Transcription

1 Math 918 The Power of Monomial Ideals Problem Set 1 Solutions Due: Tuesday, February 16 (1) Let S = k[x 1,..., x n ] where k is a field. Fix a monomial order > σ on Z n 0. (a) Show that multideg(fg) = multideg(f)+ multideg(g) for non-zero polynomials f, g S. Proof. Say multideg(f) = α 0 multideg(g) = β 0. Then we can write f = a 0 x α 0 + α I g = b 0 x β 0 + a α x α β I b β x β where I I are some index sets a 0, b 0, a α, b β are in the field k. Since f g are non-zero, we know that a 0 b 0 are non-zero. Furthermore, by the definition of multidegree, α 0 > σ α β 0 > σ β for all α I for all β I. We have fg = a 0 b 0 x α 0 +β 0 + a0 b β x α0+β + b 0 a α x α+β 0 + a α b β x α+β. β I α I α I,β I Since > σ is a monomial order, relative ordering of terms is preserved when we multiply monomials. In particular, α 0 + β 0 > σ α 0 + β > σ α + β α 0 + β 0 > σ α + β 0 > σ α + β for all α I for all β I. Therefore, since a 0 b 0 0, we must have that multideg(f g) = multideg(f)+ multideg(g) (b) A special case of a weight order is constructed as follows. Fix u Z n 0. Then, for α, β in Z n 0, define α > u,σ β if only if u α > u β, or u α = u β α > σ β, where denotes the usual dot product of vectors. Verify that > u,σ is a monomial order. Proof. We first show that > u,σ is a total ordering. Let α, β Z n 0. Assume that α β. Since Z 0 is totally ordered with the usual definition of >, exactly one of the following cases must be true: (i) u α > u β (ii) u α < u β (iii) u α = u β. By definition of > u,σ, if case (i) holds then α > u,σ β. Similarly, if (ii) holds then β > u,σ α. In the case (iii), since > σ is given to be a total order, exactly one of the following cases holds: α > σ β so α > u,σ β; β > σ α so β > u,σ α; or α = σ β so α = u,σ β. Therefore, exactly one of α > u,σ β or β > u,σ α or α = u,σ β holds. We conclude that > u,σ is a total ordering.

2 2 To demonstrate the second requirement for a monomial ordering, let α, β Z n 0 such that α > u,σ β. Let γ Z n 0. If u α > u β, then u (α + γ) = u α + u γ > u β + u γ = u (α + γ) which shows that α + γ > u,σ β + γ. In the case that u α = u β, then α > σ β. Note that u (α + γ) = u α + u γ = u β + u γ = u (α + γ). However, since > σ is a monomial ordering, we must have α + γ > σ β + γ. Thus again, α + γ > u,σ β + γ. Finally, to show that > u,σ is a well-ordering, we apply the Corollary to Dickson s Lemma verify that α u,σ 0 for all α Z n 0. Since α Z n 0, it is true that u α 0 = u 0. If u α > 0 then we are done. If the dot product is zero, then we must have α σ 0 since > σ is a well-ordering itself so α u,σ 0 yet again. (c) A particular example of a weight order is the elimination order which was introduced by Bayer Stillman. Fix an integer 1 i n let u = (1,..., 1, 0,..., 0), where there are i 1 s n i 0 s. Then the ith elimination order > i is the weight order > u,grevlex. Prove that > i has the following property: if x α is a monomial in which one of x 1,..., x i appears, then x α > i x β for any monomial x β involving only x i+1,..., x n. Does this property hold for the graded reverse lexicographic order? Solution: We first prove the desired result then compare the elimination order with the graded reverse lexicographic order. Proof. By the definitions of u, x α x β, it is clear that u α > 0 yet u β = 0. Thus, by definition, x α > i x β. This property does not hold for the graded reverse lexicographic order. For example, let i = 1 S = k[x 1, x 2 ] where x 1 > grevlex x 2. Then x 3 2 > grevlex x 1. (2) Let I be a non-zero ideal in k[x 1,..., x n ]. Let G = {g 1,..., g t } F = {f 1,..., f r } be two minimal Gröbner bases for I with respect to some fixed monomial order. Show that {LT (g 1 ),..., LT (g t )} = {LT (f 1 ),..., LT (f r )}. Proof. Since both F G are minimal Gröbner bases for I, we have that the leading coefficient of each f i g j must equal 1. Consider f 1. Since G is a Gröbner basis for I f 1 I, there is some g i such that LT (g i ) divides LT (f 1 ). Renumber if necessary so that i = 1. Then, since g 1 I F is a Gröbner basis for I, there must exist some f j such that LT (f j ) divides LT (g 1 ). We conclude that LT (f j ) divides LT (f 1 ). But, since F is given to be minimal, LT (f 1 ) is not in the ideal generated by the leading terms in F {f 1 }. We conclude that j = 1 so LT (f 1 ) = LT (g 1 ). We repeat this argument starting with f 2. We again have that there exists some g l such that LT (g l ) divides LT (f 2 ). Since F is a minimal Gröbner basis LT (f 1 ) = LT (g 1 ), we know that l 1. We may relabel, if necessary, to assume that l = 2. Arguing as above yields LT (f 2 ) = LT (g 2 ). Continuing in this fashion, we see that this procedure must stop at which point t = r, after relabeling, LT (f i ) = LT (g i ) for i = 1,..., t.

3 (3) Suppose that I = (g 1,..., g t ) is a non-zero ideal of k[x 1,..., x n ] fix a monomial order on Z n 0. Suppose that for all f in I we obtain a zero remainder upon dividing f by G = {g 1,..., g t } using the Division Algorithm. Prove that G is a Gröbner basis for I. (We showed the converse of this statement in class.) Solution: Below are two possible proofs for this exercise. Proof. We argue by contradiction suppose that G is not a Gröbner basis for I. Clearly, (LT(g 1 ),..., LT(g t )) in(i). Thus, we must have in(i) (LT(g 1 ),..., LT(g t )). Let f I be a non-zero polynomial such that LT(f) (LT(g 1 ),..., LT(g t )). Apply the Division Algorithm to divide f by G. Then, since LT(f) is not divisible by LT(g i ) for any i, the first step of the algorithm yields that LT(f) is added to the remainder column. This is a contradiction to the hypothesis that when we divide f by G we obtain a zero remainder. Therefore, G must be a Gröbner basis for I. 3 Proof. We saw in class that G is a Gröbner basis if only if for all pairs i j, the remainder on division of the S-polynomial S(g i, g j ) is zero. By definition, S(g i, g j ) = LCM(LM(g i), LM(g j )) LT(g i ) g i LCM(LM(g i), LM(g j )) g j. LT(g j ) Since I = (g 1,..., g t ), we see that each S-polynomial S(g i, g j ) is in I. Thus, by assumption, when we divide S(g i, g j ) by G we obtain a zero remainder. We conclude that G is a Gröbner basis for I. (4) Consider the ideal I = (xy + z xz, x 2 z) k[x, y, z]. For what follows, use the graded reverse lexicographic order with x > y > z. You are not permitted to use a computer algebra system for this exercise. Be sure to show all of your work. (a) Apply Buchberger s Algorithm to find a Gröbner basis for I. Is the result a reduced Gröbner basis for I? Solution: Start by letting g 1 = xy xz + z, g 2 = x 2 z G = {g 1, g 2 }. Then S(g 1, g 2 ) = x2 y xy g 1 x2 y x 2 g 2 = x 2 z + xz + yz. Applying the Division Algorithm to divide S(g 1, g 2 ) by G yields S(g 1, g 2 ) = zg 2 + xz + yz z 2. We let g 3 = xz + yz z 2 (the remainder from dividing S(g 1, g 2 ) by G) append this to G. Thus, G = {g 1, g 2, g 3 }. We then calculate S(g 1, g 3 ) = xyz xy g 1 xyz xz g 3 = y 2 z xz 2 + yz 2 + z 2. Applying the Division Algorithm to divide S(g 1, g 3 ) by G yields S(g 1, g 3 ) = zg 3 y 2 z + 2yz 2 z 3 + z 2. We let g 4 = y 2 z + 2yz 2 z 3 + z 2 (the remainder from dividing S(g 1, g 3 ) by G) append this to G. Thus, G = {g 1, g 2, g 3, g 4 }. We show that G is a Gröbner basis for I

4 4 by demonstrating that S(g 1, g 4 ), S(g 2, g 3 ), S(g 2, g 4 ) S(g 3, g 4 ) have zero remainders when divided by G. The end results are: S(g 1, g 4 ) = xy2 z xy g 1 xy2 z y 2 z g 4 = xyz 2 xz 3 + xz 2 + yz 2 = z 2 g 1 + zg 3 S(g 2, g 3 ) = x2 z x 2 g 2 x2 z xz g 3 = xyz + xz 2 z 2 = zg 1 S(g 2, g 4 ) = x2 y 2 z g x 2 2 x2 y 2 z y 2 z g 4 = 2x 2 yz 2 x 2 z 3 + x 2 z 2 y 2 z 2 = 2xz 2 g 1 + (z 3 + z 2 )g 2 2z 2 g 3 + zg 4 S(g 3, g 4 ) = xy2 z xz g 3 xy2 z y 2 z g 4 = y 3 z + 2xyz 2 y 2 z 2 xz 3 + xz 2 = 2z 2 g 1 + (z 2 + z)g 3 (y + z)g 4. Note that G is not a reduced Gröbner basis for I. For example, the monomial xz is a term of g 1 LT (g 3 ) = xz. So, xz is in the ideal generated by the leading terms in G {g 1 }. (b) Use your answer from part (a) to determine if f = xy 3 z z 3 + xy is in I. Solution: Dividing f by the Gröbner basis G found in part (a) yields f = (y 2 z + yz 2 + z 3 + 1)g 1 + (z 3 + 1)g 3 + zg 4 + ( yz 4 + z 5 3yz 3 2z 3 yz + z 2 z). Since the remainder r = yz 4 + z 5 3yz 3 2z 3 yz + z 2 z is non-zero, f is not in the ideal I. (5) Consider the affine variety V = V(x 2 +y 2 +z 2 4, x 2 +2y 2 5, xz 1) in C 3. Use a computer algebra system Gröbner bases to find all the points of V. Solution: Let I = (x 2 + y 2 + z 2 4, x 2 + 2y 2 5, xz 1) C[x, y, z]. Using CoCoA working with lexicographic order with x > lex y > lex z, we find that a Gröbner basis for I is G = {g 1, g 2, g 3 } where g 1 = y 2 z 2 1 g 2 = x 2z 3 + 3z g 3 = 2z 4 + 3z 2 1 Thus V = V(g 1, g 2, g 3 ). Note that g 3 depends on z alone. Using the quadratic formula we see that 1 g 3 = 0 z = 1, 1,, Setting z = 1, we see that Setting z = 1, we see that g 2 = 0 x = 1 g 1 = 0 y = 2, 2. g 2 = 0 x = 1

5 5 g 1 = 0 y = 2, 2. Setting z = 1, we see that 2 Setting z = 1 2, we see that Therefore, { V = g 2 = 0 x = 2 g 1 = 0 y = 3 3 2, 2. g 2 = 0 x = 2 g 1 = 0 y = (1, ± 2, 1), ( 1, ± 2, 1), 3 3 2, 2. ( ) ( 2, 3 ± 2, 1 2, )} 3 2, ± 2, 1. 2