# Computation of basic Hilbert functions

I want to compute the Hilbert function for the ring $$M:=\frac{k[x,y,z,w]}{(x,y)\cap(z,w)}$$ and compare it to the Hilbert function for the ring $$N:=k[x,y].$$

I tried computing the bases for each $M_i$ and $N_i$:

\begin{align} &M_1=\{x,y,z,w\} & &N_1=\{x,y\}\\ &M_2=\{x^2,xy,y^2,z^2,zw,w^2\}& & N_2=\{x^2,xy,y^2\}\\ &M_3=\{x^3,x^2y,xy^2,y^3,z^3,z^2w,zw^2,w^3\}& &N_3=\{x^3,x^2y,xy^2,y^3\}\\ &\vdots & & \vdots \end{align}

Did I compute these bases correctly? Since any elements in $(x,y)\cap(z,w)$ are identified with $0$ in $M$, it's like there are two copies of $k[x,y]$ insides of $M$. That is, there will be twice as many basis elements in $M_i$ than there are in $N_i$. So then the Hilbert functions should be: $$H_M(t)=\begin{cases} 0&\text{ if t\leq 0}\\ 2(t+1)&\text{ if t>0} \end{cases} \;\;\;\;\;\; H_N(t)=\begin{cases} 0&\text{ if t\leq 0}\\ t+1&\text{ if t>0} \end{cases}$$

Does this work?

(I'm not looking for a general theory to compute Hilbert functions yet... I was just introduced to them).

• There are a number of ways to compute this Hilbert series, and in short, multiplying by $2$ is not correct. As a combinatorialist, my favorite way is to use Stanley-Reisner theory to compute this. However, my guess is that you might find this dissatifying. – Ken Duna Jan 28 '17 at 17:08
• So what's an example of an element of some $M_i$ not in the proposed basis above? And is my $H_N$ correct? – FTem Jan 28 '17 at 17:19
• I apologize, I had just woke up when I wrote that. You are correct. I computed the Hilbert Series in a high powered way and got $\frac{1 + 2t - t^2}{(1-t)^2}$. At first glance, that seemed like it had not chance of being what you had, but it is. Your intuition is good (in fact that is what caused me to re-evaluate my initial answer). There is only one small problem with your function and I'll address that in a true response. – Ken Duna Jan 28 '17 at 18:04
• Cool!! Do you have any resources on how to compute this using a more formal approach now? (How did you get that function?) – FTem Jan 28 '17 at 18:14
• Your intuition is correct, but this isn't enough. The ideal $(x,y)\cap(z,w)$ equals $(xz,xw,yz,yw)$ (why?). Now the non-zero monomials in the factor ring $x^ay^bz^cw^d$ must have the form $x^ay^b$, respectively $z^cw^d$ since $c\ge1$ implies $a=b=0$, respectively $d\ge1$ implies $a=b=0$. – user26857 Jan 29 '17 at 8:13

The only real error is $H_N(0) = H_M(0) = 1$. This is because the $0^{th}$ graded piece of each of these rings is $k$, which is a one dimensional $k$ vector space.

To explain my initial poor comment about you being wrong:

Using Stanley-Reisner theory, you can see very quickly that the Hilbert Series (the generating function of $H_M(t)$) is $$\frac{1 + 2x - x^2}{(1-x)^2}_.$$

At first glance, it seems like this means your answer is off. But once your start computing coefficients, you notice a pattern.

$$\frac{1+2x-x^2}{(1-x)^2} = (1+2x-x^2)(1+2x + 3x^2 + 4x^3 + \dots)$$

The constant term is $1$.

The coefficient of $x$ is $2 + 2 = 4$.

For $n \geq 2$: The coefficient of $x^n$ in this expression is $$(n+1) + 2n - (n-1) = 2n+2 = 2(n+1).$$

As to how I got that Hilbert Series:

At the risk of self-promotion: For my senior project in my undergrad degree, I wrote A Beginner's guide to Stanley-Reisner rings which you can find on my webpage. It is not great writing by any means (I was just beginning as a mathematician), but it might make what I write next make more sense.

Another way to write your ideal is $(xw, xz, yw,yz)$. This is the Stanley-Reisner ideal of the one-dimensional simplicial complex whose facets are $\{x,y\}$ and $\{w,z\}$. Once you have this, you are home free because the Hilbert-Series of the Stanley-Reisner ring is $$\frac{h_0 + h_1 x + \dots + h_d x^d}{(1-t)^d}_.$$

Where $d$ is the dimension of the simplicial complex plus one, and the $h_i$'s come from the $h$-vector of the complex (these numbers come from the combinatorics of the complex). $d$ is also the Krull dimension of the Stanley-Reisner ring.