I know that maximal ideals of $C[0, 1]$ corresponds to singleton. Also, using Zorn's lemma one can construct a prime ideal in $C[0, 1]$ which is not maximal.



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> Is there any $\textbf{closed}$ prime ideal of $C[0, 1]$ which is not maximal?
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Any references or ideas?



Closed prime ideal in $C[0, 1]$
No. Note that an ideal in a commutative ring with identity is prime if and only if the quotient ring is an integral domain.


Now consider $C[0,1]$. It is known that the closed ideals in this Banach algebra are all of the form $J\_F = \{ f\in C[0,1] \colon {f\vert}\_F =0 \}$ for some closed subset $F\subseteq [0,1]$, and $C[0,1]/J\_F \cong C(F)$. If $F$ contains two distinct points then $C(F)$ is not an integral domain (just take a "spike" which peaks at $x$ and is supported on a small neighbourhood, and then do likewise for $y$).


The same should be true with $[0,1]$ replaced by any compact Hausdorff space (in the last part of the proof, Urysohn's lemma will produce functions with the desired properties).




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Update: following requests in the comments, here is a characterization of the closed ideals in $C(X)$, where $X$ is any compact Hausdorff space. Since I am doing this off the top of my head and writing for analysts, I'm going to use excluded middle freely; I'm sure the arguments could be reformulated to reduce or remove this.


Note that I do not need any ${\rm C}^\ast$-algebra theory or functional calculus, and several ingredients below will work for any regular Banach function algebra.$\newcommand{\hull}{\rm hull}$


Given an ideal $J\subseteq C(X)$ let $\ker(J)=\{ x\in X \colon f(x)=0 \,\forall\,f\in J\}$. Given a subset $S\subseteq X$ let $\hull(S)=\{ f\in C(X) \colon f(x)=0 \,\forall\,x\in S\}$. Note that $\ker(J)$ is always closed in $X$ and $\hull(S)$ is always norm-closed in $C(X)$


One can check that $\ker$ and $\hull$ form a Galois connection (adjunction between appropriate posets) but we only need the prosaic fact that $\ker(\hull(J))\supseteq J$. In fact, since $\ker$ of any subset is norm-closed, $\overline{J}\subseteq \ker(\hull(J))$. What I will now show is that this last inclusion is an equality.


So let $a\in \ker(\hull(J))$ and let $\varepsilon>0$. Let $U=\{ x\in X \colon |a(x)| <\varepsilon\}$; this is an open neighbourhood of $\hull(J)$.


Suppose we can produce $b\in J$ such that $b\geq 0$ and $b(x) \geq 1$ for all $x\in X\setminus U$.
Consider $a\cdot (\varepsilon +b)^{-1}b$, which belongs to $J$. Note that if $x\in U$ then
$$
\left\lvert a(x) - \dfrac{a(x)b(x)}{\varepsilon+b(x)} \right\rvert \leq |a(x)| < \varepsilon
$$
and if $x\in X\setminus U$ then
$$
\left\lvert a(x) - \dfrac{a(x)b(x)}{\varepsilon+b(x)} \right\rvert 
= \dfrac{\varepsilon \lvert a(x)\rvert}{\varepsilon+b(x)} \leq |a(x)| \dfrac{\varepsilon}{\varepsilon+1}
$$
Putting these together gives
$$
\lVert a - a\cdot(\varepsilon+b)^{-1}b \rVert\_\infty \leq \max\left(\varepsilon, \dfrac{\varepsilon}{\varepsilon+1}\lVert a\rVert\_\infty\right).
$$
Since $\varepsilon>0$ is arbitrary, we can conclude that $a\in \overline{J}$.


It remains to produce $b\in J$ with the desired properties, which we do using a routine compactness argument. For each $x\in X\setminus U$, since $x\notin \hull(J)$ we can pick $f\_x\in J$ such that $f\_x(x)=2$, and then let $U\_x$ be an open neighbourhood of $x$ such that $|f\_x(y)|^2 > 1$ for all $y\in X\setminus U\_x$. By compactness we can extract a finite subcover, relabelled as $U\_1,\dots, U\_m$ with corresponding $f\_1,\dots, f\_n \in J$ that satisfy $|f\_i(y)|^2 > 1$ for all $y\in U\_i$. Now put $b= f\_1 \overline{f\_1} + \dots + f\_n\overline{f\_n} \in J$. Clearly $b\geq 0$ and $b(y) \geq 1$ for all $y\in X\setminus U$, as required.


**Remark:** pursuing these ideas further one can show that the Jacobson topology = hull-kernel topology on $X$ (analogue of Zariski topology) coincides with the Gelfand topology. This is definitely not true for general commutative semisimple Banach algebras.


