Maximal ideal of ring of continuous real-valued functions on $[0, 1]$ is not finitely generated.

Question

Consider the ring $R = C([0,1])$. Let $c\in [0,1]$ and consider the ideal $I_{c} = \{f \in R : f(c) = 0 \}$. I want to show that this ideal is not finitely generated. I was thinking that if $I_{c} = \langle \{f_{1},...,f_{n} \} \rangle$, then I could consider $R/I$, as it is a field and maybe derive some contradiction, but I am otherwise stuck.

Answer 1

Suppose $I_c = \langle f_1, \ldots, f_n\rangle$. Define $f = \sum_{i} |f_i|$. Clearly, $\sqrt f \in I_c$. Thus, $\sqrt f = \sum_i h_i f_i$ for some $h_i \in R$. Define $h = \sum_i |h_i|$. We have \begin{align*} \sqrt f &= \sum_i h_i f_i \\ &\le \sum_i |h_i| \cdot |f_i| \\ &\le \sum_i |h_i| \sum_i |f_i| \\ &= hf. \end{align*}

The last inequality follows from Cauchy-Schwarz.

Note that for $x \ne c$, we must have at least one $i$ with $f_i(x) \ne 0$ since $f_i$ generate $I_c$. Thus, $f(x) \gt 0$ for $x \ne c$. It follows that $h \ge 1/\sqrt{f}$ for all $x \ne c$. Since $f(c) = 0$, this implies that $h$ is unbounded, but this contradicts the compactness of $[0, 1]$.

(I take no credit for this proof. I found it in my notes. I think I first saw it as an exercise in Atiyah-Macdonald or Dummit & Foote. I don't remember.)