Proving that ${{X}_{m}}=\{P\in {{k}^{n}}\left| {{f}_{1}}(P)=...={{f}_{m-1}}(P)=0 \;\;\hbox{and} \;\; {{f}_{m}}(P)=1\} \right. $ is eventually empty.

Question

We consider a field $k$, $n \in \mathbb{N}$ and ${{f}_{1}},{{f}_{2}},...\in k[{{x}_{1}},...,{{x}_{n}}]$. For every integer $m\ge 2$, we define ${{X}_{m}}=\{P\in {{k}^{n}}\left| {{f}_{1}}(P)=...={{f}_{m-1}}(P)=0 \;\;\hbox{and} \;\; {{f}_{m}}(P)=1\}. \right.$ I need to show that there exists an integer $N$ such that ${{X}_{m}}=\varnothing $ for every $m \ge N$.
I encountered this problem in a course of commutative algebra, in the chapter of Noetherian rings.
First, I suspected that I should use the property "every increasing sequence of ideals is eventually constant", which characterizes a Noetherian Ring, but I am not very sure. Any help would be greatly appreciated.

Answer 1

For $m \geq 1$, let $ I_m $ be the ideal generated by $f_1, f_2, \dots, f_m$. These ideals form an ascending sequence $$I_1 \subseteq I_2 \subseteq I_3 \subseteq \cdots$$ which must stabilise since $k[x_1, \dots, x_n]$ is Noetherian. Let $N$ be minimal such that $I_{N-1} = I_N$. We find in particular that, for all $m \geq N$, we have $f_m \in I_{N-1}$,which can be rephrased as $f_m = \sum_{r=1}^{N-1} \lambda_r f_r$ for some $\lambda_r \in k[x_1, \dots, x_n]$.

Observe that if $P \in k^n$ satsifies $$f_1(P) = f_2(P) = \cdots f_{N-1}(P) = 0$$ then, for $m \geq N$, we have $$f_m(P) = \sum_{r=1}^{N-1} \lambda_r f_r(P) = \sum_{r=1}^{N-1} \lambda_r(P) \cdot 0 = 0\text{.}$$

Let $m \geq N$ and suppose for contradiction that there exists $P \in X_m$. By definition of $X_r$ we have $f_r(P)=0$ for all $1 \leq r \leq N-1$ which, by the preceding, implies $f_m(P)=0$. Yet $f_m(P)=1$, again from the definition of $X_m$. This is absurd. We conclude that $X_m = \emptyset$ for all $m \geq N$, as desired.