MATH 497 INTRODUCTION TO APPLIED ALGEBRAIC GEOMETRYHOMEWORK 3 SOLUTIONS

Assigned 9/14, due 9/18 in class.

Problem 1. Working with your group, implement the division algorithm (Chapter 2, Section 3,Theorem 3 in CLO).

Solution. The code is written below.

def polydivset(f,g):R = f.parent()n = len(g)p, r, a = f, 0, [R.zero() for i in range(n)]while p != 0:

i, flag = 0, Falsewhile i < n and not flag:

if R.monomial divides( g[i].lt(), p.lt() ) == True:q = R(p.lt()/g[i].lt())p = p - q*g[i]flag == True

else:i = i + 1

if flag == False:r, p = r + p.lt(), p - p.lt()

return a, r

Do Section 3, Exercises 5 and 6 with your division algorithm as well.In Chapter 2 Section 4, do Exercise 12. Note there is a typo, it should say “or u · α = u · β and

α >σ β.”

§3 : 1.) Compute the remainder on division of the given polynomial f by the order set F . Use thegrlex order, then the lex order in each case.a.) f = x7y2 + x3y2 − y + 1, F = (xy2 − x, x− y3)

Solution. The code is below12 R.<x, y> = PolynomialRing(QQ,order = ’deglex’);3 F = [x*yˆ2 - x, x - yˆ3];4 f = xˆ7*yˆ2 + xˆ3*yˆ2 - y +1;5 polydivset(f,F)6 ([xˆ6 + xˆ2, 0], xˆ7 + xˆ3 - y + 1)7

1

12 R.<x, y> = PolynomialRing(QQ,order = ’lex’);3 F = [x*yˆ2 - x, x - yˆ3];4 f = xˆ7*yˆ2 + xˆ3*yˆ2 - y +1;5 polydivset(f,F)6 ([xˆ6 + xˆ5*y + xˆ4*yˆ2 + xˆ4 + xˆ3*y + xˆ2*yˆ2 +7 2*xˆ2 + 2*x*y + 2*yˆ2 + 2, xˆ6 + xˆ5*y + xˆ4 + xˆ3*y +8 2*xˆ2 + 2*x*y + 2], 2*yˆ3 - y + 1)9

b.) Repeat part a with the order of the pair F reversed.

Solution. The code is below12 R.<x, y> = PolynomialRing(QQ,order = ’degrevlex’);3 F = [x*yˆ2 - x, x - yˆ3];4 f = xˆ7*yˆ2 + xˆ3*yˆ2 - y +1;5 polydivset(f,F)6 ([xˆ6 + xˆ2, 0], xˆ7 + xˆ3 - y + 1)7

12 R.<x, y> = PolynomialRing(QQ,order = ’invlex’);3 F = [x*yˆ2 - x, x - yˆ3];4 f = xˆ7*yˆ2 + xˆ3*yˆ2 - y +1;5 polydivset(f,F)6 ([xˆ6 + xˆ2, 0], -y + xˆ7 + xˆ3 + 1)7

§3 : 2.) Compute the remainder on division:a.) f = xy2z2 + xy − yz, F = (x− y2, y − z3, z2 − 1)

Solution. The code is below12 R.<x, y, z> = PolynomialRing(QQ);3 F = [x - yˆ2, y - zˆ3, zˆ2 - 1];4 f = x*yˆ2*zˆ2 + x*y - y*z;5 polydivset(f,F)6 ([-x*zˆ2, 0, xˆ2], xˆ2 + x*y - y*z)7

b.) Repeat part a with the order of the set F permuted cyclically.

Solution. The code is below2

12 R.<x, y, z> = PolynomialRing(QQ);3 F = [y - zˆ3, zˆ2 - 1, x - yˆ2];4 f = x*yˆ2*zˆ2 + x*y - y*z;5 polydivset(f,F)6 ([0, x*yˆ2, -x], xˆ2 + x*y - y*z)7

12 R.<x, y, z> = PolynomialRing(QQ);3 F = [zˆ2 - 1, x - yˆ2, y - zˆ3,];4 f = x*yˆ2*zˆ2 + x*y - y*z;5 polydivset(f,F)6 ([x*yˆ2, -x, 0], xˆ2 + x*y - y*z)7

§3 : 5.) We will study the division of f = x3 − x2y − x2z + x by f1 = x2y − z and f2 = xy − 1.a.) Compute using grilled order:

r1 = remainder of f on division by (f1, f2)

r2 = remainder of f on division by (f2, f1)

Your results should be different. Where in the division algorithm did the differenceoccur? (You may need to do a few steps by hand here.)

Solution. The results are computed below.12 R.<x, y, z> = PolynomialRing(QQ);4 F = [xˆ2*y - z, x*y - 1];5 f = xˆ3 - xˆ2*y - xˆ2*z + x;6 polydivset(f,F)7 ([-1, 0], xˆ3 - xˆ2*z + x - z)8

12 R.<x, y, z> = PolynomialRing(QQ);4 F = [x*y - 1, xˆ2*y - z];5 f = xˆ3 - xˆ2*y - xˆ2*z + x;6 polydivset(f,F)7 ([-x, 0], xˆ3 - xˆ2*z)8

b.) Is r = r1 − r2 in the ideal 〈f1, f2〉? If so, find an explicit expression r = Af1 + Bf2.If not, say why not.

3

Solution. r = x− x2y. This is clearly in the ideal since

r = 0 · f1 − x · f2c.) Compute the remainder of r on division by (f1, f2).Why could you have predicted

your answer before doing the division?

Solution. The remainder is 0 since r is in the ideal.

d.) Find another polynomial g ∈ 〈f1, f2〉 such that the remainder on division of g by(f1, f2) is nonzero.

Solution. Let g = yz − xy. Notice that −y(x2y − z) + xy(xy − 1) = −x2y2 + yz +x2y2 − xy = g, so g is in the ideal.

12 R.<x, y, z> = PolynomialRing(QQ);4 F = [xˆ2*y - z, x*y - 1];5 g = y*z + x*y;6 polydivset(g,F)7 ([0, -1], y*z - 1)8

12 R.<x, y, z> = PolynomialRing(QQ);4 F = [xˆ2*y - z, x*y - 1];5 g = y*z - 1;6 polydivset(g,F)7 ([0, 0], y*z - 1)8

e.) Does the division algorithm give us a solution for the ideal membership problem forthe ideal 〈f1, f2〉? Explain your answer.

Solution. No, it does not on its own since elements that are in the ideal may not yielda zero remainder.

§3 : 6.) Using the grlex order, find an element g of 〈f1, f2〉 = 〈2xy2 − x, 3x2y − y − 1〉 ⊂ R[x, y]whose remainder on division by (f1, f2) is nonzero.

Solution. Consider

g = −3

2x(2xy2 − x) + y(3x2y − y − 1) = −3x2y2 + 3x2

2+ 3x2y2 − y2 − y

12 R.<x, y, z> = PolynomialRing(QQ);4 F = [2*x*yˆ2 - x, 3*xˆ2*y - y - 1];5 g = 3xˆ2/2 - yˆ2 - y;6 polydivset(g,F)7 ([0, 0], 3/2*xˆ2 - yˆ2 - y)8

4

§4 : 12.) Another important weight order is constructed as follows. Let u = (u1, . . . , un) be in Zn≥0,and fix a monomial order >σ ( such as >lex or >grevlex) on Zn≥0. Then for α, β ∈ Zn≥0,define α >u,σ β if and only if

u · α > u · β or u · α = u · β and α >σ β.

We call >u,σ the weight order determined by u and >σ.a.) Use Corollary 6 to prove that >u,σ is a monomial order.

Solution. Since the dot product yields elements in Z, which is totally ordered by >and >σ is a total order by definition, it follows that >u,σ is a total order.Suppose u · α > u · β. Then consider

u · (α + γ) = u · α + u · γ > u · β + u · γ = u(β + γ).

Alternatively, suppose α >σ β. By definition, >σ satisfies this property.

b.) Find u ∈ Zn≥0 so that weight order >u,lex is the order >grlex .

Solution. If u = (1, 1, . . . , 1), then we get the graded lexicographic ordering.

Remark 0.1. Projection onto the diagonal results in the `1 norm, which is equivalentto ordering by degrees.

c.) In the definition of >u,σ, the order >σ is used to break ties, and it turns out that tieswill always occur in this case. More precisely, prove that given u ∈ Zn≥0, there areα 6= β in Zn≥0 such that u · α = u · β.

Solution. Let u = (u1, u2, . . . , un). Define α = (u2, 0, . . . , 0) and β = (0, u1, . . . , 0).Then u · α = u1u2 = u · β.

d.) A useful example of a weight order is the elimination order introduced by Bayer andStillman (1987b). Fix an integer 1 ≤ i ≤ n and let u = (1, ..., 1, 0, ..., 0), where thereare i 1s and n− i 0s. Then the ith elimination order >i is the weight order >u, grevlex.Prove that>i has the following property: if xα is a monomial in which one of x1, ..., xiappears, then xα >i x

β for any monomial involving only xi+1, ..., xn.

Solution. Let α and β be as described. Then u · α > 0 = u · β.

5