The following question is a part of this problem.
Let $R = \{ f: [0,1] \to \mathbb{R} \; | \; f \text{ is a continuous map} \}$ be the ring of continuous functions. We define $M_c = \{ f \in R \; | \; f(c)=0 \}$. It is very easy to prove that $M_c$ is a maximal ideal. The final result we want to prove is:
$M_c$ is not a finitely generated ideal.
One of the claims which I saw while reading from here is the following:
If $M_c$ is finitely generated, then it is a principal ideal.
(Later they use this result that since $M_c$ is not a principal ideal, and hence $M_c$ is not finitely generated.)
Proof. Assume that $M_c = \left<f_1,\dots,f_n\right>$. The function $f_{1}^{1/3}$ is also in $M_c$ so write it as $\sum_{i=1}^n g_if_i$ for some $g_i \in R$. Cubing both side, we have: $f_1 = \sum h_if_i$ where, in particular, $h_1$ vanishes at $c$ (gathering up coefficients of $f_1$ in $(\sum g_if_i)^3$; we see the coefficient $h_i$ is in the ideal $<f_1, \dots, f_n>^2$).
So, $(1 − h_1)f_1 \in <f_1, \dots, f_n>$. Since $1 − h_1$ does not vanish at $c$, we can find an interval $B = [a, b]$ containing $c$ where it also doesn’t vanish.
Here is what I don't understand:
Doubt: In the ring $S$ of continuous functions on $B, \; 1 − h_1$ is a unit, hence on this ring $S$, the ideal $M_c$ is generated by $<f_2|_{B}, \dots, f_n|_{B}>$. How does this follow?
