Term Orders And Weights

Let $I \subset \mathbb{C}[x_1, \dots, x_n]$ be an ideal.

- For $t \in \mathbb{R}$, let $w(t) = (t^n, t^{n-1}, \dots, t^1) \in \mathbb{R}^n$. Show that for sufficiently large $t$, $\text{in}_{w(t)}(I) = \text{in}_\prec(I)$ where $\prec$ is the lexicographic term order with $x_1 \succ \cdots \succ x_n$.
- Describe a weight vector $w$ such that $\text{in}_w(I) = \text{in}_{\prec'}(I)$ where $\prec'$ is the
*graded reverse lexicographical order*with $x_1 \succ' \cdots \succ' x_n$ i.e. $x^a \succ' x^b$ if either $\deg(x^a) > \deg(x^b)$, or $\deg(x^a)=\deg(x^b)$ and the right most non-zero entry of $a-b$ is negative.

### Partial Solution:

- Let $G=\{g_1,\ldots , g_N\}$ be a GrÃ¶bner basis for $I$ with respect to the lexicographic term order, then we have that $\langle \text{in}_\prec (g_1) , \ldots, \text{in}_\prec(g_N)\rangle = \text{in}_\prec (I)$. Since $\{g_i\}$ is a finite set of polynomials, there exists a sufficiently large $t$ such that $\text{in}_{w(t)}(g_i) = \text{in}_\prec(g_i)$ for all $i$ with respect to the weight vector $w(t) = (t^n, t^{n-1}, \dots, t^1)$, which implies $\text{in}_{w(t)}(I) = \text{in}_\prec(I)$.
- Let $w(t) = (1-t^1, 1-t^{2}, \ldots, 1- t^n)$, then for sufficiently small $t$, we have $\text{in}_{w(t)}(I) = \text{in}_{\prec'}(I)$.