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continuous functions that have a root 0

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hbghlyj Posted at 2023-3-1 04:58:52 |Read mode
Last edited by hbghlyj at 2023-3-1 09:20:00The ring of continuous functions $C(\Bbb R)$ has an ideal $S=\set{f\in C(\Bbb R):f(0)=0}$. Is $S$ finitely generated?

I proved $S$ is acyclic:
For any $f\in S$, $\langle f\rangle\ne S$, since $\sqrt{\abs f}\in S$ and $\sqrt{\abs f}\over f$ is not continuous at 0, the function $\sqrt{\abs f}$ is not in $\langle f\rangle$.

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Czhang271828 Posted at 2023-3-2 13:17:45
hbghlyj 发表于 2023-3-1 16:24
"cyclic" corrected, thanks!

For any $f_1,\cdots,f_n\in S$, suppose $\langle f_1,\cdots,f_n\rangle=S ...
First, set $O:C(\mathbb R)\to(P,\leqq)$, where $O(f)\leqq O(g)$ whenever $\lim_{x\to 0}\dfrac{f(x)}{g(x)}=0$. One can verify such partial order is well-defined.

Consider the ideal $S'$ generated by $\{f_i\}_{i=1}^n\subseteq S$. Then for arbitrary $g\in S'$, the proposition
$$
(O(g)\leqq O(f_1))\lor(O(g)\leqq O(f_2))\lor \cdots \lor (O(g)\leqq O(f_n))
$$
holds ture. Now we shall prove the existence of $g\in S$ such that
$$
(O(f_i)\leqq O(g))\land (O(f_i)\neq O(g))
$$
holds true for every $i\in \{1,\ldots ,n\}$.

The pivot is to understand the non-existence of function in $S$ with the lowest damping rate. The construction is easy. Consider
$$
g(x):=\sqrt{\max_{1\leq i\leq n}(|f_i(x)|)}.
$$

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