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This is a slight extension of a remark a read a few days ago.

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 $\chi(n,\alpha)=\dim_K A_n/\mathfrak{a}_n$ (also known as the Hilbert function). Here $A_n$ is the $K$-space of homogeneous elements of $n$ in $A$, and likewise for $\mathfrak{a}_n$.

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

Taking $d$ generic linear forms, say $f_1,\dots,f_d$, why does $$ \chi(n,\mathfrak{a}+(f_1,\dots,f_d))=c_d? $$


One theorem I have read is that if $F$ is a homogeneous polynomial of degree $j$ where $F$ is not a zero divisor modulo $\mathfrak a$, i.e., if $G\in A$ and $FG\in\mathfrak{a}$, then $G\in\mathfrak{a}$, then $\chi(n,\mathfrak{a}+(F))=\chi(n,\mathfrak{a})-\chi(n-j,\mathfrak{a})$. I'm wondering how it can be extended to an ideal generated by more than one linear form, to get the above equality. Thank you.

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By induction! We claim that $\chi(n,\mathfrak{a}+(f_1,\ldots,f_k))$ has the same leading coefficient as $\chi(n,\mathfrak{a})$ for any $k\le d$ and generic (linear) choice of the $f_i$. For $k=1$, the statement follows from the result you quoted:

$$\begin{align*} \chi(n,\mathfrak{a}+(f))&=\chi(n,\mathfrak{a})-\chi(n-1,\mathfrak{a}) \\&= \sum_{k=0}^d c_k \frac{n^k}{k!} - c_k\frac{(n-1)^k}{k!} \\&=\sum_{k=0}^d \frac{c_k}{k!} \cdot \left( n^k - \sum_{j=0}^k \binom{k}{j} (-1)^{k-j} n^j \right) \\&=\sum_{k=0}^d \frac{c_k}{k!} \cdot\sum_{j=1}^{k} \binom{k}{j} (-1)^{k-j} n^{j-1} \\&= n^{d-1}\cdot\frac{c_d}{(d-1)!} + \left<\mathrm{terms\ of\ lower\ degree}\right> \end{align*} $$

On the other hand, for $k> 1$, set $\mathfrak{b}:=\mathfrak{a}+(f_1,\ldots,f_{k-1})$. By induction hypothesis, the leading coefficients of $\chi(n,\mathfrak{a})$ and $\chi(n,\mathfrak{b})$ coincide. By the case $k=1$, the leading coefficients of $\chi(n,\mathfrak{b})$ and $\chi(n,\mathfrak{b}+(f_k))=\chi(n,\mathfrak{a}+(f_1,\ldots,f_k))$ also coincide, so we are done.

Remark: $d!$ times the leading coefficient of the Hilbert Polynomial is also referred to as the degree of the projective variety defined by $\mathfrak{a}$. It is a nice exercise to check that for $\mathfrak{a}=(f)$, it actually agrees with the degree of $f$. It is, thus, very reassuring to see that the degree remains invariant under intersection with a generic hyperplane - and that's precisely what it means to add a generic, linear polynomial to $\mathfrak{a}$.

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  • $\begingroup$ Thanks again rattle, I appreciate the explanation. $\endgroup$
    – Buble
    Mar 10, 2012 at 1:32

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