MATH 495/800 Assignment 3. Summer 2006. Due 9:30am Wednesday June 21st. Late policy: -10% for each day late. Problems marked * are for MATH 800 students. Michael Monagan Problems from the book. 2.7 #2,3,7 2.8 #5a,11 2.9 - 3.1 #1,2,4,5*,6(b+c),7 3.3 #6,8,14 For exercises #2 and #3 in 2.7 you may use Maple as a calculator but you must show all steps of Buchberger's algorithm. For the exercises in 2.8, 3.1, and 3.3 use Maple to compute Groebner bases. For 3.1 #7c, check your answer using the lexdeg([t],[x,y,z]) monomial ordering in Maple Additional exercises. 1: Section 2.4 exercise 5 concerned the ideal where f1=x^2*y-z, and f2=x*y-1. In parts b) and d) you found two polynomials r and g in I for which division by {f1,f2} resulted in a non-zero remainder. Apply Buchberger's algorithm to compute a reduced Grobner basis G for using lex order with x>y>z. To simplify the work involved, you may apply the "very useful lemma." Now verify that r divided by G and g divided G have remainder 0. 2: Using Maple, determine which of the following ideals are the same? a) < y^3-z^2, x*z-y^2, x*y-z,x^2-y > b) < x*y-z^2, x*z-y^2, x*y-z, x^2-y > c) < x*z-y^2, x+y^2-z-1, x*y*z-1 > d) < y^2-x^2*y, z-x*y, y-x^2 > 3: By hand, compute a REDUCED Grobner basis for the following linear system firstly using lexicograhical ordering (plex in Maple) with x > y > z and secondly the grevlex ordering (tdeg in Maple) also with x > y > z. S = { x + y + z = 1, x - 2*y - z = 2, y + 2*z = 5 } You may use apply proposition 4 from section 2.9 to skip S-polynomial calculations. You should get the same Groebner basis. Why? Some other stuff for interest only. 1: Let f = x^4*y^5 + x^5*y^2 + x^3*y^9. Show that there is no monomial ordering on k[x,y] s.t. LM(f) = x^4*y^5. Notice that if you consider the prime power monomial orderings, this statement must imply that p1^4*p2^5 < max( q1^5*q2^2, r1^3*r2^9 ) for all primes p1, p2, q1, q2, r1, r2. 2: Let I be an ideal in k[x1,...,xn] which is non-trivial. Let T = {all possible monomial orderings on k[x1,...,xn]}. I gave in class an example of an infinite class of monomial orderings for k[x,y], the prime power orderings, so T is infinite when n > 1. Since we know that a Groebner basis for I depends on the monomial ordering this suggests that there might be an infinite number of different reduced Groebner bases for a given ideal. Surprise. This is not the case! Let R = {all reduced Groebner bases for I for each monomial ordering in T} Theorem: R is finite. It follows from the above that a finite set F in k[x1,...,xn] exists which is a Groebner basis for I for EVERY monomial ordering -- just take F to be the union of R. Such a Groebner basis is called a universal Groebner basis. Some examples of universal Groebner bases. F = {x,y} for and G = {x-y^2,x*y-x,y^3-y^2,x^2-x} for . So by adding enough "redundant" polynomials in I to a given Groebner basis we can make it a Groebner basis for all monomial orderings.