TITLE: The set of continuous functions vanishing at $c$ is not the principal ideal generated by $x-c$
QUESTION [5 upvotes]: Let $R = \mathcal C([0,1], \mathbb R)$ and $M_c = \{f \in R \mid  f(c) = 0\}$.
I would like to show that:
a. $M_c \neq M_b$ if $b \neq c$ ;
b. $M_c$ is not the principal ideal generated by $x-c$;
c. $M_c$ is not finitely generated.
For a. I think its clear that if $f(x)=x-c$ then $ f \in M_c $ and $f \in M_b$ iff $b=c$.
b. is trickier. I'm thinking about something like $f=|x-c|$ because then if $f=g(x-c)$ then $g=|x-c|/(x-c)$ and then  $g \notin R$.
This doesn't feel good enough. How can I make this precise or am I even on the right track?
EDIT; Suppose $c \neq 1$. Let  $f(x)=(x-c)^{1/2}\chi_{x > c}$. Clearly $f \in M_c$. Suppose $(x-c)$ divides $f$ Then $f= g(x-c)$ for some $g \in \mathcal{R}$. I want to go for $g \notin \mathcal{R}$ but I'm pretty sure I cannot say $g=\frac{f}{x-c}$ because $x-c$ is not a unit...so I'm stuck here.
EDIT2; Ok I think I have it.  Suppose $x>c$. If we calculate:
$$|g(x)-g(c)|=|\frac{f(x)}{x-c}-g(c)|\geq|x-c|^{-1/2}-|g(c)| \geq 1 - |g(c)|$$  So $g$ has no right limit as $x \rightarrow c$ and therefore $g \notin R$

REPLY [7 votes]: Suppose that $M_c$ is finitely generated by $\{f_1, \dots, f_n\}$.
$$f(x) = \sum_{i=1}^n \sqrt{\vert f_i(x) \vert}$$ is an element of $M_c$. Therefore, $f= \sum_{i=1}^n g_if_i$ with $g_1, \dots, g_n \in R$ and
$$\vert f(x) \vert \le M \sum_{i=1}^n \vert f_i(x) \vert$$ where $M = \sup\limits_{1 \le i \le n} \Vert g_i \Vert_\infty$. By continuity, it exists an open $U$ containing $c$ such that for all $x \in U$ and $1 \le i \le n$:
$$\sqrt{\vert f_i(x) \vert} \le \frac{1}{2M}.$$
Then
$$\vert f(x) \vert \le M \sum_{i=1}^n \vert f_i(x) \vert =M \sum_{i=1}^n \sqrt{\vert f_i(x) \vert}\sqrt{\vert f_i(x) \vert} \le \frac{1}{2}\sum_{i=1}^n \sqrt{\vert f_i(x) \vert} = \frac{1}{2} \vert f(x) \vert.$$
Consequently $f, f_1, \dots, f_n$ vanish on $U$. A contradiction as $x \mapsto \vert x - c \vert$ belongs to $M_c$ and only vanishes at $c$.