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.
Is there any $\textbf{closed}$ prime ideal of $C[0, 1]$ which is not maximal?
Any references or ideas?
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).
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.$\newcommand{\hull}{{\rm hull}}$
[UPDATE 2024-04-13] a commenter pointed out that in my definitions I had hull and kernel the wrong way round; this has now been fixed.
Given an ideal $J\subseteq C(X)$ let $\hull(J)=\{ x\in X \colon f(x)=0 \,\forall\,f\in J\}$. Given a subset $S\subseteq X$ let $\ker(S)=\{ f\in C(X) \colon f(x)=0 \,\forall\,x\in S\}$. Note that $\hull(J)$ is always closed in $X$ and $\ker(S)$ 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.
