Dimension of the space of homogeneous polynomials in a quotient ring

Question

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?

Answer 1

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.

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Earlier version of this answer:

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.

Answer 2

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 $d \geq 4$.