Show that Delta_1 intersect Delta_2, Delta_1 union Delta_2 are both simplicial complexes too.
A simplicial complex is a collection (A) of subsets of {1..n}, (B) closed under shrinkage. (A) is obvious for both union and intersection, so we turn to (B).
Let F be a subset of {1..n}, and G a subset of F.
If F is in the intersection, then F is in Delta_1 and Delta_2, so G is too, hence G is in the intersection.
If F is in the union, then F is in one of Delta_1 or Delta_2, so G is in that one too, hence G is in the union.
2. Show H_{Delta_1 union Delta_2} = H_{Delta_1} + H_{Delta_2} - H_{the intersection}.
We computed H_Delta = \sum_{F in Delta} (t/(1-t))^|F|.
On the left hand side we sum over each F in the union, once.
On the right we sum over each F in the union either 1+0-0, 0+1-0, or 1+1-1 times, depending on whether it is in Delta_1, Delta_2, or both.
3. Take the two simplicial complexes from the last homework. Compute their Hilbert series (meaning, that of the associated Stanley-Reisner ideals) using the formula we have for general monomial ideals. Confirm that it matches the answer we get from the formula specifically for SR ideals.
The first one is I generated by {e, af, bd}. The usual monomial formula is 1/(1-t)^6 (1 -t-t^2-t^2 + t^3+t^3+t^4 - t^5). The fact that the LCMs are all products says that this is a complete intersection, i.e. the numerator factors, so we could also say 1/(1-t)^6 (1-t)(1-t^2)(1-t^2). Then that simplifies to 1/(1-t)^3 (1+t)^2.
Meanwhile, the formula specifically for S-R ideals gives us
1 + 5(t/(1-t)) + 8(t/(1-t))^2 + 4(t/(1-t))^3, for the 1 empty face, 5 vertices, 8 edges, and 4 triangles.
Multiply both sides by (1-t)^3, and the first calculation gives (1+t)^2, whereas the second gives (1-t)^3 + 5t(1-t)^2 + 8t^2(1-t) + 4t^3, and these are indeed the same.
The second ideal is generated by abc, so the general monomial formula is 1/(1-t)^4 (1-t^3) = 1/(1-t)^3 (1+t+t^2). The S-R formula is 1 + 4(t/(1-t)) + 6(t/(1-t))^2 + 3(t/(1-t))^3. Multiplying again by (1-t)^3, we get 1+t+t^2 vs. (1-t)^3 + 4t(1-t)^2 + 6t^2(1-t) + 3t^3, which again match.
4. Let I be a homogeneous ideal. But let's homogenize it again, anyway! Relate H_I and H_{homog(I)}.
Pick a homogeneous Gr\"obner basis. Then homogenizing it does exactly nothing to the basis; it only puts it into a ring with one more variable.
We can compute the Hilbert series from the leading terms of the Gr\"obner basis, as 1/(1-t)^{# variables} * something depending on those terms.
So the only difference between the two calculations is the number of variables.
Hence H_{homog(I)} = 1/(1-t) H_I.
