# Remark on the dimension of quotient by prime homogeneous ideal.

There's a remark near the end of a section I'm reading about Hilbert polynomials that I don't fully understand.

Let $K$ be a field, and let $A=K[X_0,\dots,X_N]$ be a polynomial ring, which is graded in the standard way (the elements of degree $n$ are the homogeneous polynomials of degree $n$). Let $\mathfrak{a}$ be a homogeneous prime ideal of $A$, and $d$ the dimension of the corresponding projective variety. Moreover, let $\chi(n,\alpha)=\dim_K A_n/\mathfrak{a}_n$. Here $A_n$ is the $K$-space of homogeneous elements of $n$ in $A$, and likewise for $\mathfrak{a}_n$.

The remark states that there is some $c_d\in\mathbb{N}$ so that $$\chi(n,\mathfrak{a})=c_d\frac{n^d}{d!}+c_{d-1}n^{d-1}+\cdots+c_0.$$

Can somebody clarify why such $c_d$ exists and how this equality is found? Thank you.

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One can show that $$\chi(n,(0))=\dim_K A_n=\binom{N+n}{N}$$ since this is the number of monomials of degree $n$. This is a combinatorial statement which is discussed here.

Now, we can choose a graded, free resolution of $M:=A/\mathfrak{a}$

$$\cdots\xrightarrow{\quad\phi_2\quad} F_1 \xrightarrow{\quad\phi_1\quad} F_0 \xrightarrow{\quad\phi_0\quad} M \xrightarrow{\quad\quad} 0$$ with $$F_i = \bigoplus_{j=1}^{r_i} A(-a_{ij})$$ where $A(-a)$ is the ring $A$, degree-shifted by $a\in\mathbb{Z}$, i.e. $A(-a)_n=A_{n-a}$. It is easy to see that $$\dim_K A(-a)_n = \binom{N+n-a}{N}$$ and it is then an easy calculation using the exactness to show that $$\chi(n,\mathfrak{a})=\dim_K M_n = \sum_{i\ge 0} (-1)^i \cdot \sum_{j=1}^{r_i} \binom{N+n-a_{ij}}{N}.$$

For $n > \max_{ij} (a_{ij}-N)$, the binomial coefficient $$\binom{N+n-a_{ij}}{N} = \frac{(n+N-a_{ij})(n-1+N-a_{ij})\cdots(1+n-a_{ij})}{N!}$$ is a polynomial expression in $n$. Let us denote this Polynomial by $P_M(n)$. We now want to show that this polynomial has degree equal to the dimension $d$ of the variety defined by $\mathfrak{a}$.

To do so, we perform induction on the dimension of $M$. If $\dim(M)=0$, then $M$ is Artinian and a finite-dimensional, graded $K$-vector space. This means $P_M(n)=0$ for large enough $n$ and hence, $P_M=0$. In the induction step, we know that $$\mathrm{codim}(\mathfrak{a})=N+1-\dim(M)\le N,$$ and thus there exists an $x\in A$ not contained in any prime which is minimal over $\mathfrak{a}$. We set $\mathfrak{b}:=\mathfrak{a}+(x)$ and $N:=M/(x)$, so $\mathrm{codim}(\mathfrak{b})=\mathrm{codim}(\mathfrak{a})+1$. Consequently, $\dim(N)=\dim(M)-1$.

The sequence $$0\to M(-1)\xrightarrow{\,\cdot x\,} M\to N\to 0$$ is exact (in every degree) and this implies $P_M-P_{M(-1)}=P_N$. In other words, $P_N$ is the first difference of $P_M$ and thus, $\deg(P_N)=\deg(P_M)-1$.

Note: The polynomial $P_M$ is usually called the Hilbert Polynomial of the module $M$, while your notion of $\chi(n,\mathfrak{a})$ is often referred to as $H_M(n)$ and called the Hilbert Function. It is important to note that the Hilbert Function itself is not always a Polynomial, but it does agree with $P_M$ for large values of $n$.

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Many thanks rattle. If you do find the time, I would like to see the verification that $\deg\chi(n,\mathfrak{a})=\dim (M)$. –  Buble Mar 5 '12 at 5:20