I aim to assess the soundness of my approach. In an exercise, the question is posed regarding whether $\mathbb{R}[x]$ is Noetherian, and my response is negative. To support this, I employ the definition that a ring $R$ is Noetherian if each ascending chain of ideals in $R$ eventually becomes stable.
Considering the family $\{I_n\}_{n\in\mathbb{N}}$, where
$$I_n = \{f\in\mathbb{R}[x] \mid \text{such that } f(a) = 0 \ \forall a\in\mathbb{R}\setminus \{0,1,…,n\}\}.$$
I can easily demonstrate that $I_n$ is an ideal for all $n\in\mathbb{N} $.
Furthermore, if $f \in I_n$, then $f(a) = 0$ for all $a \in \mathbb{R}\setminus \{0,1,…,n\}$, and consequently, $f(a) = 0$ for all $a \in \mathbb{R}\setminus \{0,1,…,n, n+1\}$.
This establishes $I_n \subset I_{n+1}$ for all $n \in \mathbb{N}$, prompting me to observe the following ascending chain:
$$I_0 \subset I_1 \subset I_2 \subset \cdots. $$
However, this chain does not stabilize, indicating tha $\mathbb{R} [x]$ is not noetherian.
Is my solution correct, or am I missing something? Please provide feedback.
Best Answer
To expand on my comment, suppose $f \in \mathbb{R}[x]$ vanishes at every point outside of a finite set $S \subset \mathbb{R}$. Then, the set of elements of $\mathbb{R}$ which are sent to zero are $f^{-1}(\{0\}) \subset \mathbb{R}$, which is a closed set. Hence, since $\mathbb{R} \setminus S$ is dense in $\mathbb{R}$, we must have $f^{-1}(\{0\}) = \mathbb{R}$, where we view $f: \mathbb{R} \to \mathbb{R}$ as a continuous function. Hence $f$ is the zero function which implies $f$ is the zero polynomial.
Alternatively, any polynomial $f \in \mathbb{R}[x]$ is equal to its Taylor expansion around any point $$f(x) = \sum_{n \geq 0} \frac{f^{(n)}(a)}{n!}(x - a)^n$$ so if $f$ is uniformly zero on $\mathbb{R} \setminus S$, all of its derivatives would also vanish on $\mathbb{R} \setminus S$, so taylor expanding this polynomial around any point of $\mathbb{R} \setminus S$ would reveal it to be the zero polynomial.
As such, your ascending chain is just the trivial chain $(0) \subset (0) \subset \cdots$ which stabilizes immediately.
In fact, the division algorithm tells us that $\mathbb{R}[x]$ is a PID. In particular if we had an ascending chain of ideals $I_1 \subset I_2 \subset \cdots$ we would have $I_n = (g_n)$ where $g_n \in \mathbb{R}[x]$ and for any $n$, $g_{n + 1}$ divides $g_n$ in $\mathbb{R}[x]$. Must this stabilize?