prime ideals in C([0,1])

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 $c \in [0,1]$.

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.

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:

1. $Z(I)$ is closed in $[0,1]$ for any ideal $I$
2. $Z(I) = Z(I')$
3. $Z(I(A))=A'$
4. $I(A)$ is closed in $C([0,1])$ for any $A \subset [0,1]$
5. $I(A) = I(A')$
6. $I(Z(I))=I'$ for any ideal $I$.