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.