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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 )

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    What you are asking is whether every ideal is contained in a maximal ideal - which is true. However, not every ideal is maximal. For instance $I = {f\in C[0,1] : f(0) = f(1) = 0}$ – Prahlad Vaidyanathan Mar 03 '15 at 15:37
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    @PrahladVaidyanathan : Oh yes , your observation is true , and then it just boils down to zorn's lemma as $C[0,1]$ is a commutative ring with unity , thank you very much , you should make your comment an answer –  Mar 03 '15 at 15:43

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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]$.

Davide Giraudo
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