# Dimension of the space of homogeneous polynomials in a quotient ring

Consider the ring $$k[x_1,x_2,x_3,x_4]/(x_1x_3,x_1x_4,x_2x_3,x_2x_4)$$. How do I find the dimension of the vector space of homogeneous representatives of the quotient ring of each degree?

I've only managed to exhibit bases for low degrees:

0: $$1$$; dimension 1

1: $$x_1,x_2,x_3,x_4$$; dimension 4

2: $$x_1^2,x_2^2,x_3^2,x_4^2,x_1x_2,x_3x_4$$; dimension 6

3: $$x_1^3,x_2^3,x_3^3,x_4^3, x_1^2x_2,x_1x_2^2, x_3^2x_4,x_3x_4^2$$; dimension 8

So my conjecture is that the dimension of polynomials of degree $$k>0$$ is $$2k+2$$. Is it true? How to prove it?

The $$k$$-vector subspace of $$k[x_1,x_2,x_3,x_4]$$ of homogeneous polynomials of degree $$n$$ is spanned by the monomials of degree $$n$$, i.e. terms of the form $$x_1^{d_1}x_2^{d_2}x_3^{d_3}x_4^{d_4}$$ with $$d_1+d_2+d_3+d_4=n$$. The kernel of the quotient map is the $$k$$-vector space spanned by the monomials that satisfy either $$d_1d_3>0,\qquad d_1d_4>0,\qquad d_2d_3>0,\qquad\text{ or }\qquad d_2d_4>0.$$ Hence the $$k$$-vector subspace of $$k[x_1,x_2,x_3,x_4]/(x_1x_2,x_1x_3,x_2x_3,x_2x_4)$$ of homogeneous polynomials of degree $$n$$ is spanned by the monomials $$x_1^{d_1}x_2^{d_2}x_3^{d_3}x_4^{d_4}$$ that satisfy $$d_1+d_2+d_3+d_4=k,\qquad d_1d_3=d_1d_4=d_2d_3=d_2d_4=0.\tag{1}$$ For $$k=0$$ clearly $$d_1=d_2=d_3=d_4=0$$ is the unique solution. For $$k\neq0$$ we must have $$d_i\neq0$$ for some $$i$$. If $$d_1\neq0$$ or $$d_2\neq0$$ then $$d_3=d_4=0$$, and if $$d_3\neq0$$ or $$d_4\neq0$$ then $$d_1=d_2=0$$. So the number of solutions to $$(1)$$ is the same as the number of solutions to $$d_1+d_2=k,\quad d_3=d_4=0\qquad\text{ or }\qquad d_3+d_4=k,\quad d_1=d_2=0,$$ which is precisely twice the number of solutions to $$a+b=k$$ with $$a,b\in\Bbb{N}$$. This is of course precisely twice $$k+1$$, as you conjecture.

In the quotient, the product of any pair of variables vanishes except the products $$x_1x_2$$ and $$x_3x_4$$. This means a basis is given by the monomials of the form $$x_1^n,\qquad x_2^n,\qquad x_3^n,\qquad x_4^n,\qquad x_1^mx_2^n,\qquad x_3^mx_4^n.$$ Of course the first four are also of the form $$x_1^mx_2^n$$ or $$x_3^mx_4^n$$, simply with $$m=0$$ or $$n=0$$.
So for a given degree $$k$$, you want to count the number of monomials $$x_1^mx_2^n$$ and $$x_3^mx_4^n$$ with $$m+n=k$$. There are $$k+1$$ choices for $$m$$, and then $$n=k-m$$. So there are $$k+1$$ of monomials of each type, so a total of $$2k+2$$ monomials.
Here's a silly overkill way to find the Hilbert function. We compute a graded resolution for the ring $$M := k[x_0,x_1,x_2,x_3]/(x_0x_2, x_0x_3, x_1x_2, x_1x_3)$$ and then use additivity.
First let's find a resolution for the ideal $$I := (x_0 x_2, x_0 x_3, x_1 x_2, x_1 x_3)$$. Let $$S = k[x_0, x_1, x_2, x_3]$$. $$I$$ has four generators $$g_1 := x_0 x_2, g_2 := x_0 x_3, g_3 := x_1 x_2, g_4 := x_1 x_3$$, all of degree $$2$$, so our resolution begins $$S(-2)^{\oplus 4} \overset{\varphi_0}{\to} I \to 0$$ defined by $$e_i \mapsto g_i$$, where the $$e_i$$ are the standard basis vectors. We see that $$\varphi_0$$ has kernel generated by $$x_1 e_1 - x_0 e_3, x_3 e_1 - x_2 e_2, x_1 e_2 - x_0 e_4, x_3 e_3 - x_2 e_4 \, .$$ We continue our resolution by defining a map $$\varphi_1: S(-3)^{\oplus 4} \to S(-2)^{\oplus 4}$$ onto these generators for $$\ker(\varphi_0)$$. (The twist by $$-3$$ is necessary to make the map have degree $$0$$.) This can be written as left multiplication by the matrix $$\begin{pmatrix} x_1 & x_3 & 0 & 0\\ 0 & -x_2 & x_1 & 0\\ -x_0 & 0 & 0 & x_3\\ 0 & 0 & -x_0 & -x_2 \end{pmatrix} \, .$$ Denoting the standard basis vectors of $$S(-3)^{\oplus 4}$$ by $$f_1, f_2, f_3, f_4$$, then $$\varphi_1$$ has kernel generated by $$x_3 f_1 - x_1 f_2 - x_2 f_3 + x_0 f_4$$. So the final map in our sequence is $$\varphi_2: S(-4) \to S(-3)^{\oplus 4}$$, $$g_1 \mapsto x_3 f_1 - x_1 f_2 - x_2 f_3 + x_0 f_4$$. Thus we have constructed the free resolution $$0 \to S(-4) \underset{\varphi_2}{\overset{\begin{pmatrix} x_3\\ -x_1\\ -x_2\\ x_0\end{pmatrix}}{\longrightarrow}} S(-3)^{\oplus 4} \xrightarrow[\varphi_1]{\begin{pmatrix} x_1 & x_3 & 0 & 0\\ 0 & -x_2 & x_1 & 0\\ -x_0 & 0 & 0 & x_3\\ 0 & 0 & -x_0 & -x_2 \end{pmatrix}} S(-2)^{\oplus 4} \underset{\varphi_0}{\to} I \to 0$$ We can tack on $$M$$ to the righthand side in order to obtain a resolution for $$M$$: $$0 \to S(-4) \to S(-3)^{\oplus 4} \to S(-2)^{\oplus 4} \to S \overset{\pi}{\to} M \to 0$$ where $$\pi: S \to S/I = M$$ is the quotient map.
Recall that $$S(a)$$ has Hilbert function $$H_{S(a)}(d) = \binom{3 + a + d}{3}$$. (In general for $$S = k[x_0, \ldots, x_r]$$ we have $$H_{S(a)}(d) = \binom{r + a + d}{r}$$ by "stars and bars".) By additivity, then \begin{align*} H_M(d) &= \binom{d+3}{3} - 4 \binom{3 - 2 + d}{3} + 4 \binom{3 - 3 + d}{3} - \binom{3 - 4 + d}{3}\\ &= \binom{d+3}{3} - 4 \binom{d+1}{3} + 4 \binom{d}{3} - \binom{d-1}{3} \, . \end{align*} This gives $$H_M(0) = 1, H_M(1) = 4, H_M(2) = 6$$ and $$H_M(3) = 8$$, and reduces to $$H_M(d) = 2d+2$$ for $$d \geq 4$$.