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Let $C[0,1]$ denote the ring of continuous functions on $[0,1]$. Consider the maximal ideal $M_a =\{ f\in C[0,1] | f(a)=0 \}$ for $a$ $\in$ $[0,1]$. Is $M_a$ finitely generated?

Note: It may seem that $M_a = \langle x-a\rangle$ which is certainly not true because $ \vert x-a \vert $ is in $M_a$ but not in $\langle x-a\rangle$. Any ideas for solving this question?

user26857
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Arpit Kansal
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    $(x - a)(x - b) \in M_a$ for each real $b \neq a \in \mathbb{R}$ – r9m Sep 23 '15 at 05:24
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    @r9m;if we keep $x-a$ in the set of generators of $M_a$ and then choose $g(x)\in C[0,1]$ accordingly then we can write $(x-a)(x-b)..(x-z)=(x-a)g(x) ;g(x)=(x-b)..(x-z)$ then where is the contradiction – Learnmore Sep 23 '15 at 06:06
  • Given any $n$ elements in $M_a$, say ${f_1,f_2,\cdots,f_n}$, consider ideal generated by them - $J=\langle f_1,f_2,\cdots,f_n\rangle$. Consider function $e^{r_1f_1+r_2f_2+\cdots + r_nf_n}-1$. Here, $r_i$'s are any elements from ring $C[0,1]$. Then, for every choice of $r_i$'s does this function always belong to $J$? (I don't know answer; but I was proceeding in this way for answer). – Groups Sep 23 '15 at 07:09
  • I have a question. Please see http://www.math.lsa.umich.edu/~kesmith/593hmwk2-2014-solutions.pdf. Can anyone explain why in Step 1 $M_c$ restricted on $S$ is finitely generated, so is $M_c$ on $[0,,1]$? – Knt Aug 01 '19 at 04:42

2 Answers2

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The following proof is found here. One can actually show that $M_a$ is not generated by countably many elements.

Assume the contrary that $M_a$ is generated by $(f_1, f_2, \dots, f_n, \dots)$. By normalizing we assume $|f_i(x)| \le 1$ for all $i, x$. Then define

$$f(x) = \sum_{i=1}^n \frac{\sqrt{|f_i(x)|}}{2^i}.$$

Then $f \in M_a$ and there exist $g_1, \dots, g_r \in C[0,1]$ so that

$$f = f_1 g_1 +\cdots +f_r g_r \Rightarrow |f(x)| \le M \sum_{i=1}^r |f_i(x)|,$$

where $M$ is an upper bound for $g_1, \dots, g_r$.

Now by continuity there is an open $U$ containing $a$, so that

$$\sqrt{|f_i(x) |} \le \frac{1}{2^{i+1}M} \ \ \ \ \ \forall x\in U,\ i=1, \dots, r.$$

This implies

$$|f(x) \vert \le M \sum_{i=1}^r |f_i(x)| \le M \sum_{i=1}^r \frac{\sqrt{|f_i(x)|}}{2^{i+1}M} \le \frac{1}{2} |f(x)| <|f(x)|$$

for all $x\in U$. This is impossible. Thus $M_a$ is not generated by countably many elements.

user26857
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  • Nice solution,Thank you! – Arpit Kansal Sep 23 '15 at 10:14
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    "Then define $f(x) = \sum_{i=1}^n \frac{\sqrt{|f_i(x)|}}{2^i}.$" I can't understand this. Where does $n$ come from? Or is this meant to say "Then define $f(x) = \sum_{i=1}^\infty \frac{\sqrt{|f_i(x)|}}{2^i}.$"? – Smiley1000 Dec 14 '24 at 10:16
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If we assume $M_a$ is finitely generated, say $M_a = \langle f_1,\dots, f_n \rangle$, such that $|f_k(x)| \le 1$ for $x \in [0,1]$.

Consider, $f(x) = \sum\limits_{k=1}^{n} \sqrt{|f_k(x)|} \in M_a$.

Then, there are functions $\{g_k\}_{1 \le k \le n} \in C[0,1]$ such that $f(x) = \sum\limits_{k=1}^{n} g_k(x)f_k(x)$, and set $M = \underset{\substack{1 \leq k \leq n \\ x \in [0, 1]}}{\operatorname{max}} |g_k(x)|$.

Then, choose an open interval $U \subset [0,1]$ near $a$, such that $|f_k(x)| < \dfrac{1}{M^2}$, for each $k$ (which should exist by virtue of continuity of $f_k$'s and $f_k(a) = 0$).

So for $x \in U \setminus \{ a \} $, we have, $|f(x)| \le M\sum\limits_{k=1}^{n} |f_k(x)| < \sum\limits_{k=1}^{n} \sqrt{|f_k(x)|} = f(x)$ (contradiction!).

Note that the strict inequality in the last line is due to $$|f_k(x)| < M^2 \implies \sqrt{|f_k(x)|} < \frac{1}{M} \implies \frac{1}{\sqrt{|f_k(x)|}} > M \implies \frac{|f_k(x)|}{\sqrt{|f_k(x)|}} > M |f_k(x)| \implies \sqrt{|f_k(x)|} > M |f_k(x)|.$$

Smiley1000
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r9m
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