MATH 441/751/819 Assignment 3.  Spring 2012.
Due Thursday February 23rd 2:30pm.
Late policy: -20% for up to 24 hours late.  Zero for more than 24 hours late.
Problems marked * are for MATH 819 students only.
Enjoy, Michael Monagan

Problems from the book.
2.7 #1,2,3,7
2.8 #5a,11
2.9 #13a-b
3.1 #1,2,4,5*,6b*,7*
3.3 #6,8

Use Maple for 2.7 #1.
For exercises #2 and #3 in 2.7, assume x>y>z.
Do parts (a) and (b) by hand and check your results using Maple.
For part (c) use Maple as appropriate.
To reduce the number of steps, you may, if you wish, apply the
"very useful lemma" and proposition 4 of 2.9.

For the exercises in 2.8, 2.9, 3.1, and 3.3 use Maple to compute Groebner bases.
For exercise 13 of 2.9, don't hand in your answer.
For 3.1 #7c, check your answer using the lexdeg([t],[x,y,z]) ordering in Maple.
For exercises 6 and 8 of 3.3, plot the surfaces in Maple.

Additional exercises.

1: 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 >

2: By hand, compute a REDUCED Groebner basis for the following
   linear system using lexicographical ordering with x > y > z.
      S = { x + y + z = 1, x - 2*y - z = 2, y + 2*z = 5 }
   Use proposition 4 from section 2.9 to skip S-polynomial calculations.
   What would happen if we use the grlex ordering with x > y > z?

3: Lemma 3 of 2.7 says that if you have a Groebner basis G = {p,g1,...,gt}
   a polynomial ideal I with LT(p) in <LT(G-{p})>, then G-{p} is a Groebner
   basis for I.  The proof does not prove that G-{p} is a basis for I.
   I think this is okay because it follows from Corollary 6 of 2.5.
   As an exercise, prove that the remainder of p divided G-{p} is 0 to
   show p is in <g1,...,gt> hence that G-{p} is a basis for I.

For interest only.

   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 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.
   So by adding enough "redundant" polynomials in I to a given Groebner basis
   we can make it a Groebner basis for all monomial orderings.
   Some examples of universal Groebner bases are: F = {x,y} for <x,y> 
   and G = {x-y^2,x*y-x,y^3-y^2,x^2-x} for <x-y^2,x*y-x>.