# Verification of a proof that the ring of continuous functions on $[0,1]$ is not noetherian

My problem is to show that the ring of continuous real-valued functions on the interval $[0, 1]$ is non-noetherian.

I'd like to know if this argument holds:

Let $I$ be the set of functions such that for every $f \in I$, there exists an open neighborhood $[0, t)$ of $0$ such that $f(x) = 0$ whenever $x \lt t$. Let's call the largest such neighborhood the zero neighborhood of $f$.

This set is an abelian group, since the zero function obviously has such a neighborhood, and given any pair of function $f$ and $g$, we can intersect their zero neighborhoods to get the zero neighborhood of the sum. Lastly, this set is closed under multiplication in $\mathbb{R}$: scaling by $0$ gives you the zero function, and scaling by anything else preserves the zero neighborhood.

Then, to prove our ring non-noetherian, we must show this ideal cannot be finitely generated. Suppose it was, $I = (f_1, f_2, ... ,f_k)$ with each $f_i$ having a zero neighborhood of $[0, x_i)$. We will find a function $f$ which cannot be written as a finite linear combination of these $f_i$.

We noted above that closure under addition and under scaling may only affect the zero neighborhood in certain ways. Addition of two nonzero functions with zero neighborhoods $[0, x_1)$ and $[0, x_2)$ will leave you with a function whose zero neighborhood is $[0, \min_{ x_1, x_2} )$. Scaling by a nonzero real will preserve the zero neighborhood. This means that the linear combination of $f_i$ must necessarily either have $[0, 1]$ as its zero neighborhood, or else it must have $[0, x_i)$ for some $i$.

To construct a counterexample, we define $\widetilde{x}$ to be the midpoint between $\max {x_i}$ and $1$. Then we let $f(x)$ be uniformly zero until $x_i$, then let it grow linearly from there on out.

Wordy, but hopefully clear. There are some subtle points I'm missing. For one, I know that I would need to exclude the zero function from the final $\max {x_i}$ calculation. I also need to show that if a continuous function is zero on $[0, 1)$, it is the zero function. These two guarantee the choice at the end is legitimate.

Does this seem like a legitimate approach? Or did I miss anything?

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For the last point, about such a function being the zero function or not, the inverse image of a closed set is closed. If $[0, 1) \subseteq f^{-1}(\{0\})$, then the closure of $[0, 1)$ must also be contained in $f^{-1}(\{0\})$. Also, I think it's easier to show that there exists an infinite, strictly increasing chain of ideals, for instance $$(f_1) \subseteq (f_1, f_2) \subseteq (f_1, f_2, f_3) \subseteq \cdots$$ where $f_n$ is some function which is zero on $[0, \frac{1}{n}]$ and non-zero otherwise. – Arthur Aug 23 '13 at 6:46
You likely want to let $\tilde x$ to be the midpoint of min $x_i$ and 0. You can certainly have large neighborhoods that vanish in your ideal, the trick is constructing a small one. Also note that personally I think it is MUCH MUCH easier to just create a non-terminating chain of ideals though I don't see any reason why your proof wouldn't work. – PVAL Aug 23 '13 at 6:48
Addition of two non-zero functions with zero neighborhoods $[0,x_1)$ and $[0, x_2)$ won't necessarily give a function with zero neighborhood $[0, \min(x_1,x_2))$. The zero neighborhood of the sum may be larger: just think of the case where you're adding a function $f$ and its negation $-f$. – Magdiragdag Aug 23 '13 at 8:34
You're making the problem much harder than it is. Why don't you consider the chain $$(\sqrt{x}) \subseteq (x^{\frac{1}{4}}) \subseteq (x^{\frac{1}{8}})\ldots?$$ – user38268 Aug 23 '13 at 9:32
@Peter Oops. Yes. That is an issue, isn't it? – Tac-Tics Aug 26 '13 at 16:06

Denote your ring by $R$. You actually started out being pretty close to a very simple approach.
Hint: Show that $I_{ab}=\{f\in R\mid f([a,b])=\{0\}\}$ is an ideal of $R$ for any subinterval $[a,b]$ of $[0,1]$. Notice that if $[c,d]\subseteq [a,b]$, then $I_{cd}\supseteq I_{ab}$. Show that if $[c,d]$ is properly inside $[a,b]$ then the associated ideals are properly contained too.