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Prove that the ring of continuous functions $f:\mathbb R\to\mathbb R$ is not Noetherian.

I know that to be Noetherian, every ideal is generated by finitely many elements or equivalently R satisfies the ascending chain condition.

So, if I can find an ideals that are contained in each other that don't terminate then it is not Noetherian.

My professor briefly touched on Noetherian rings so it is still a little bit confusing. How do I go about finding these ideals? Or should I show that every ideal is generated by finitely generated elements? Any help is much appreciated!

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    $\begingroup$ how about consider the ideal generated by all functions that are zero on a fixed set M. Then you can change M to get infinitely many ideals. $\endgroup$
    – Chen Jiang
    Oct 17, 2014 at 2:00

2 Answers 2

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Consider $I_n=\{f \in C(\mathbb{R}):f(x)=0, \forall x \geq n\}$ ($n >0$). Then clearly $I_n$'s form an infinite ascending chain of ideals.

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Consider the ideal $\mathfrak m=\{f:f(0)=0\}$. Then $e^{-x^{-2}}\in \bigcap \mathfrak m^n$, since $f^{(k)}(0)=0$ for all $k$, but there is no elemet $g\in\mathfrak m$ such that $e^{-x^{-2}}=g e^{-x^{-2}}$ over all $\Bbb R$. By the Krull intersection theorem, $C(\Bbb R)$ cannot be Noetherian. Recall that the Krull intersection theorem says that if you have a noetherian ring $R$ and an ideal $\mathfrak a$, then whenever $x\in\bigcap\mathfrak a^n$ then $x\in x\mathfrak a$.

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