Consider $C(X,\mathbb{R})$ be the ring of all continuous functions from a topological space $X$ to $\mathbb{R}$. I have proved the following facts:
$C([0,1],\mathbb{R})$ is not Noetherian.
$C(\mathbb{R},\mathbb{R})$ is not Noetherian.
Now I am interested in general topological spaces. Is there any necessary and sufficient condition for the topological space $X$ such that $C(X,\mathbb{R})$ is not Noetherian? Partial answers are also welcome.
Define an equivalence relation $\sim$ on $X$ by $x\sim y$ iff $f(x)=f(y)$ for all $f\in C(X,\mathbb{R})$. Then I claim $C(X,\mathbb{R})$ is Noetherian iff the quotient $X/{\sim}$ is finite. In one direction, if $X/{\sim}$ is finite, then $C(X,\mathbb{R})$ is finite-dimensional over $\mathbb{R}$ and hence Noetherian. Conversely, suppose $X/{\sim}$ is infinite. Let $F:X\to \mathbb{R}^{C(X,\mathbb{R})}$ be the product of all the maps $X\to \mathbb{R}$; then the image $F(X)$ is infinite, being in bijection with the set $X/{\sim}$. Since $F(X)$ is infinite and Hausdorff, there is a countably infinite discrete subset $\{x_n\}$ of $F(X)$ (you can easily construct such a sequence by induction with a little care). Let $C_N=F^{-1}(\{x_n:n\geq N\})\subset X$ and $$I_N=\{f\in C(X,\mathbb{R}):f(x)=0\text{ for all }x\in C_N\}.$$ Then $(I_N)$ is an ascending chain of ideals in $C(X,\mathbb{R})$. Furthermore, the $I_N$ are all distinct: since $\{x_n\}$ is discrete and $F(X)$ is completely regular, there exists a function $f_N:F(X)\to [0,1]$ such that $f_N(x_{N-1})=1$ and $f_N(x_n)=0$ for all $n\geq N$. The function $f_N\circ F$ is then in $I_N\setminus I_{N-1}$.