If $I$ is a proper ideal of $C[0,1]$ , then should there exist $a \in [0,1]$ such that $f(a)=0 , \forall f \in I$?

Question

Let $C[0,1]$ be the ring of all real valued continuous functions under point-wise addition and multiplication . We know that for every $a \in [0,1]$ , $\{f \in C[0,1] : f(a)=0\}$ is a proper ideal of $C[0,1]$ , I would like to ask ; if $I$ is a proper ideal of $C[0,1]$ , then should there exist $a \in [0,1]$ such that $f(a)=0 , \forall f \in I$ ? ( I know that if $I$ is maximal , then such an $a \in [0,1]$ must exist , so in other words I am asking , whether every proper ideal of $C[0,1]$ is maximal or not )

Answer 1

Suppose not: then the family $\left(f^{-1}(\mathbf R\setminus \{0\}\right)_{f\in I}$ is an open over of the unit interval. Extract a finite subcover $\left(f_j^{-1}(\mathbf R\setminus \{0\}\right)_{j=1}^n$: the map $f\colon x\mapsto\sum_{j=1}^nf_j(x)^2$ belongs to $I$ and does not vanish anywhere on $[0,1]$. Hence $1/f\cdot f\in I$ and we conclude that $ I=C[0,1]$.