1. For I a monomial ideal in R = C[x_1,...,x_n], let the multigraded Hilbert series be the sum of all the monomials in R that aren't in I.
a) Compute this for R = C[x,y] and I = 0, as a ratio of polynomials.
b) Generalize our inclusion-exclusion formula for Hilbert series to multigraded Hilbert series.
a) Every monomial is uniquely the product of some x^i and some y^j, so the sum of all monomials = (1 + x + x^2 + ...)(1 + y + y^2 + ...) = 1 / (1-x)(1-y).
b) At each step of inclusion-exclusion, we want to take out or add back the monomials that are multiples of some \product_i (x_i)^(e_i) for some exponents (e_i).
Adding all those monomials up (to give the term), we get the prefactor \product_i 1/(1-x_i) times the monomial \product_i (x_i)^(e_i).
Now the same inclusion-exclusion argument gives the formula
H = \product_i 1/(1-x_i) \sum_{S subset of 1..n} (-1)^|S| \product_i (x_i)^{max_{j in S} e_{ij}}.
(When all the x_i are set equal to t, we recover the singly graded Hilbert series, and the maxes get added together.)
2. Let f,g be two polynomials, and fix a term order <. Assume that gcd(init(f),init(g)) = 1. Show that the reduction algorithm reduces the S-polynomial of f and g to zero (using f and g alone). (Moral: one needn't bother fiddling with the S-polynomial of f and g in this case.)
Write f = init(f) + f', g = init(g) + g'.
By the assumption, S(f,g) = init(g) f - init(f) g = (g-g')f - (f-f')g = f'g - g'f.
As we apply the reduction algorithm, we add multiples of f and g; each step along the way looks like pg - qf. Since the initial terms only decrease during reduction, init(qf) is at most init(g'f) = init(g')init(f) < init(g)init(f). Hence init(q) < init(g).
Do we ever get stuck? If p=0 and q=0, we're done. If p=0 but q not, we can use f to reduce, similarly if q=0 but p not.
If p,q are not both zero, we can talk about the initial terms of pg and qf.
We claim they are different, for otherwise init(p) init(g) = init(q) init(f), so init(g) divides init(q) init(f). Then by the gcd=1, we learn init(g) divides init(q), but that's impossible because init(g) > init(q).
Since the leading monomials of pg and qf are different, one of them is the leading monomial of pg-qf. So the algorithm is not stuck; the leading monomial of pg-qf is a multiple of either the leading monomial of g or f, depending.
(This was definitely the hardest homework question so far!)
3. If f and g are monomials, what is their S-polynomial?
It's lcm(f,g) - lcm(f,g) = 0.
4. Compute a Gröbner basis for I = < x^2 - y, x^3 - z > with respect to lex order.
1&2: x^3 - x^3 -> xy - z, call this 3.
1&3: x^2 y - x^2 y -> y^2 - xz, call this 4.
2&3: x^3 y - x^3 y -> yz - x^2 z -> yz - xy^2 -> yz -yz = 0.
1&4: x^2 z - x^2 z -> yz - xy^2 -> yz - yz = 0.
2&4: x^3 z - x^3 z -> z^2 - x^2 y^2 -> z^2 - z^2 = 0.
3&4: xyz - xyz -> z^2 - y^3, call this 5.
1&5, 2&5, 4&5: gcd = 1 so we don't need to test.
3&5: xy^3 - xy^3 -> z y^2 - x z^2 -> z y^2 - z y^2 -> 0.
In all, x^2 - y, x^3 - z, xy - z, y^2 - xz, z^2 - y^3.
Can any of the generators in your list be safely left out?
Generator #2 reduces to 0 using #1 and #3. So not only is it in the ideal they generate, but it isn't necessary for reducing anybody else -- #1 and #3 are on the job.
5. Let p = x^2 + 3xy - y^2. Find a term order that makes x^2 the leading term, another that makes -y^2 the leading term, and show that 3xy isn't be the leading term under any term order.
Lex order with x>y makes x^2 the leading term, x < y makes y^2 the leading term.
Either x < y or x > y. If x < y, then xy < y^2. If x > y, then x^2 > xy. So no matter what, xy is not the leading term.
