Let $I \subset k[x_1,\ldots,x_n]$ be an ideal in a polinomial ring over a field $k$. Assume that $I$ is quasihomogeneous (that is, $I$ is not homogeneous with the usual grading, but it is homogeneous if we set ${\rm deg}(x_i) = \omega_i$ for some $\omega_1,\ldots,\omega_n$ positive integers).
We can consider a minimal graded free resolution of $I$ as $k[x_1,\ldots,x_n]$-module.
$0 \longrightarrow F_p \longrightarrow \cdots \longrightarrow F_0 \longrightarrow I \longrightarrow 0 $
The $i$-th Betti number $\beta_i(I)$ is the rank of $F_i$.
Consider $J$ the homogenization of $I$ with respect to a new variable $x_0$. Now $J$ is a homogeneous ideal with the usual grading. If one considers its minimal graded free resolution as $k[x_0,\ldots,x_n]$-module one gets the Betti numbers $\beta_i(J)$.
Question: Is it true that $\beta_i(I) \leq \beta_i(J)$ for all $i$?
Comments:
- It is clearly true for $i = 0$ because dehomogenizing a minimal set of generators of $J$ provides a set of generators of $I$ (not necessarily minimal)
- If true, I also wonder if the proof that $\beta_i(I) \leq \beta_i(in(I))$ for $in(I)$ an initial ideal of $I$ can be adapted to prove this statement.
