Final Exam

The final exam is available here.

Note that the wording on problem 2 is changed slightly. One must assume that f is homogeneous in part a, and that I is a homogeneous ideal in part b) [see problem 1 for a definition of homogeneous].

Homework #6 correction

I had the numbers wrong in the additional problem that was assigned. The correct problem is now listed in the homework (I changed the 'b' vector from [-323,1035,654] to [-323,389,8]).

Homework #5 correction

There was a slight problem with question #2. Here is the corrected version:

2. a) Using a computer algebra system, show that the graph with edges
{(1,2),(1,4),(1,5),(1,6),(2,3),(2,5),(2,7),(3,4),(3,6),(3,9),(5,6),(6,8),(6,9),(7,8),(8,9)}

is 3-colorable. Use the Grobner basis you find to give an explicit 3-coloring. How many colorings are there?
b) Show that if you add edge (1,3) to the graph, it is still 3-colorable, and that the coloring is now unique up to permutation of the colors.

Homework #3 Comment

Please take a look at the comment on the homework #3 post.

HW Extension

It has come to my attention that the monomial ideals problem that I assigned is much easier with one more lemma under your belt. So, what I propose is that I extend the homework deadline to Thursday of this week, and add the lemma as an exercise that makes the rest of your homework much easier. It will also be useful at various points in the course.

Lemma:
Let I be an ideal of k[x_1,...,x_n]. Suppose that I has the property that for any f in I, every term of f also is in I. Show that I is a monomial ideal.

One can view this as the converse to Lemma 3 in section 2.4 of IVA; it is assigned as problem 1 in the exercises in that section. In some texts, this is the definition of a monomial ideal.

So, to recap, homework is now due on Thursday Feb 25th, and proving the preceding lemma is added to the list of assigned problems. This lemma should be extremely useful in the problem on sums, intersections, and radicals of monomial ideals, especially the radical problem. One can fumble through the sums and intersections problems without it, but trying the radical problem without it requires significantly more mathematical stamina.