Finite generation of ideal in function ring

Question

Let $R$ be the ring of continuous functions from $[0,1]$ to the real numbers. Fix $c \in [0,1]$ and let $M_c$ = ker $E_c$ where $E_c$ denotes evaluation at $c$, a ring homomorphism from $R$ to the real numbers. That is, $E_c(f) = f(c)$ for $f \in R$. What is a nice clean way to show $M_c$ is not finitely generated? I figure that one way is to suppose that \begin{equation} A = \{ f_1, f_2,\dots, f_n \} \end{equation} is a minimal generating set and try to come up with $f_{n+1}$ which cannot possibly be an R-linear combination of the $f_k$, but that feels like a lot of guesswork. (And I didn't succeed.) Thanks for the help.

Answer 1

Your idea does work, and unfortunately there does not seem to be a way around the guesswork. Suppose $\{ f_1, \ldots, f_n \}$ is a set of continuous functions that generate the ideal of functions vanishing at $c$. Then there is a continuous function $$f = \left| f_1 \right| + \cdots + \left| f_n \right|$$ and it is clear that $f (x) = 0$ if and only if $x = c$. (Indeed, if $f (x) = 0$, then non-negativity means $f_1 (x) = \cdots = f_n (x) = 0$, so if $x \ne c$ that is tantamount to saying that any continuous function that vanishes at $c$ also vanishes at $x$, which is absurd for the closed interval $[0, 1]$.) Let $g = \sqrt{f}$. This is also a continuous function vanishing at $c$, so $$g = h_1 f_1 + \cdots + h_n f_n$$ for some continuous functions $h_1, \ldots, h_n$. Let $$h = \left| h_1 \right| + \cdots + \left| h_n \right|$$ and observe that $g (x) \le h (x) f (x)$ for all $x$. But $f (x) = g (x)^2$, so that means $h (x) \ge 1 / g(x)$ for all $x \ne c$, and since $g (c) = 0$, $h$ cannot be continuous at $c$, contradicting the fact that $h_1, \ldots, h_n$ are continuous.

Answer 2

It can be proved that $M_c$ is not even countable generated. In fact $M_c$ is countable generated iff $c$ is an isolated point of the space.