prime ideals in C([0,1]) It is clear that each maximal ideal in ring of continuous functions over $[0,1]\subset \mathbb R$ corresponds to a point and vice-versa. 
So, for each ideal $I$ define $Z(I) =\{x\in [0,1]\,|\,f(x)=0, \forall f \in I\}$. But map $I\mapsto Z(I)$ from ideals to closed sets is not an injection! (Consider the ideal $J(x_0)=\{f\,|\,f(x)=0, \forall x\in\hbox{ some closed interval which contains }x_0\}$)
How can we describe ideals in $C([0,1])$ ? Is it true that prime ideals are maximal for this ring?
 A: In answer to your first question, you probably want to think about closed ideals (i.e. closed in the sup norm).  (Bars and overlines seem to render poorly, so I'll use a prime to denote closure.)  For $A \subset [0,1]$ we let $I(A) = \lbrace f : f = 0 \text{ on } A \rbrace$, which is clearly an ideal.
The following are all true and mostly easy to show:


*

*$Z(I)$ is closed in $[0,1]$ for any ideal $I$

*$Z(I) = Z(I')$

*$Z(I(A))=A'$

*$I(A)$ is closed in $C([0,1])$ for any $A \subset [0,1]$

*$I(A) = I(A')$

*$I(Z(I))=I'$ for any ideal $I$.  

A: I found this paper on Project Euclid, Have a look.

PRIME IDEALS IN RINGS OF CONTINUOUS FUNCTIONS  by CARL W.KOHLS

http://projecteuclid.org/download/pdf_1/euclid.ijm/1255454113
A: Here is a way to construct a non-maximal prime ideal: consider the multiplicative set $S$
of all non-zero polynomials in $C[0,1]$. Use Zorn lemma to get an ideal $P$ that is disjoint from $S$ and is maximal with this property. $P$ is clearly prime (for this you only need $S$ to be multiplicative.) On the other hand $P$ cannot be any one of the maximal ideals, since
it does not contain $x-c$ for every $c \in [0,1]$.
A: There are many prime ideals that are not maximal.  You can find some things by Googling "prime ideals in $C(X)$" (e.g. that every maximal ideal is the sum of two proper prime ideals). 
The problem is that prime ideals are not closed unless they are maximal. 
The closed ideals in $C(X)$ for $X$ compact Hausdorff are in 1-1 correspondence with the closed subsets of $X$.  This fact is in many books; e.g., it is an exercise in chapter 11 of Rudin's Functional Analysis.
EDIT 8/17: There is a lot of information about prime ideals in $C(X)$ in Chapters 7 and 14 of the book Rings of Continuous Functions by Gillman and Jerison, mentioned already by Yemon.
A: I think this paper may be of some help. Although this may not be the earliest source, it has a proposition there which says that given a point $p\in [0,1]$, one can find non-maximal prime ideals $\mathfrak{p}_1,\mathfrak{p}_2$ such that $\mathfrak{m}_p$, the maximal ideal at $p$, is the sum of $\mathfrak{p}_1$ and $\mathfrak{p}_2$. They take a sequence of points $x_n$ converging to $p$ and then use two different ultrafilters on $D= \{ x_n \}_{n\geq 1}$ to define the two prime ideals. 
A: Let $ \mathbf U\ $ be a non-trivial ultrafilter of sets in an infinite Hausdorff compact space $\ X\ $ (e.g. $\ X:=[0;1]).\ $ Then
$$ J_\mathbf U:=\{f\in C(X): f^{-1}(0)\in\mathbf U\} $$ 
is a prime ideal; this ideal is contained in exactly one maximal ideal, namely $\ J_\mathbf U\subseteq J_{(x)},\ $ where $\ x\ $ is the (unique) limit point of $\ \mathbf U,\ $ and $\ (x)\ $ is a filter (maximal) with the base $\ \{\{x\}\}$. When $\ \{x\}\ $ is a $\ G_\delta$-set (as for instance for $\ X:=[0;1])\ $ then the inclusion for the said ideals is sharp, $\ J_\mathbf U\subset J_{(x)}$.
