How to find ideals? Let $R=\mathcal{C}([0,1],\mathbb{R})$ be the ring (standard one) of continuous functions. For each $\gamma\in[0,1]$, let $I_\gamma=\{f\in R; f(\gamma)=0\}$. It is easy to prove that $I_\gamma$ is an ideal, in fact, a maximal one.
My question is: how to find other ideals (not necessarily maximal), that is, different of the type of $I_\gamma$?
 A: An intersection of two different maximal ideals cannot, of course, be maximal. After a little thought you can see how this ideal is described by using the same idea as the maximal ideals, except the zero sets have been merged, and now you're looking at the functions zero on two specific points.
From here you can go whole hog and use any subset of the interval as a zero set. Be aware though that these zero sets don't correspond to the ideals in a one-to-one way. Suppose for example, you use the subset of rationals of that interval as a zero set. Can you compute the corresponding ideal for me?
Another interesting fact about the ring you described (I hope my memory isn't wrong!) is that all the maximal ideals appear that way. I think this fact is false if the interval is unbounded, though! 
A: For $f\in R$ let $Z(f)=\{x\in[0,1]:f(x)=0\}$, the zero-set of $f$. A family $\mathscr{F}$ of zero-sets of $[0,1]$ is a zero-set filter if 


*

*$\varnothing\notin\mathscr{F}$,  

*$\mathscr{F}$ is closed under finite intersections, and  

*$Z\in\mathscr{F}$ whenever $Z$ is a zero-set and $Z\supseteq Z_0$ for some $Z_0\in\mathscr{F}$.


Suppose that $I$ is an ideal in $R$. If $F$ is a finite subset of $I$ such that $\bigcap_{f\in F}Z(f)=\varnothing$, let $g=\sum_{f\in F}f^2$; then $Z(g)=\bigcap_{f\in F}Z(f)=\varnothing$, so $1/g\in R$, and $1_R\in I=R$. Thus, for a proper ideal $I$ the family $\mathscr{F}=\{Z(f):f\in I\}$ is closed under finite intersections and does not contain $\varnothing$. Now suppose that $Z$ is a zero-set of $[0,1]$ such that $Z\supseteq Z(f)$ for some $f\in I$. Let $g\in R$ be such that $Z(g)=Z$; then $fg\in I$ and $Z(fg)=Z$, so $Z\in\mathscr{F}$. That is, $\mathscr{F}$ is a zero-set filter on $[0,1]$.
Conversely, suppose that $\mathscr{F}$ is a zero-set filter on $[0,1]$ and let $I=\{f\in R:Z(f)\in\mathscr{F}\}$; then $I$ is an ideal of $R$.


*

*If $f,g\in I$, then $Z(f-g)\subseteq Z(f)\cap Z(g)\in\mathscr{F}$, so $f-g\in I$.

*If $f\in I$ and $g\in R$, then $Z(fg)=Z(f)\cup Z(g)\supseteq Z(f)$, so $g\in I$, and $RI\subseteq I$. (In fact $RI=I$, since $R$ has a multiplicative identity.) Let $Z=\bigcap\mathscr{F}$; $[0,1]$ is compact, so $Z\ne\varnothing$, and $I\subseteq I_Z=\{f\in R:Z(f)\supseteq Z\}$. However, $I$ may be a proper subset of $I_Z$. For example, fix a zero-set $Z\subsetneqq[0,1]$, let $\mathscr{Z}=\{Z_n:n\in\omega\}$ be a family of zero-sets such that $$Z_0\supseteq\operatorname{int}Z_0\supseteq Z_1\operatorname{int}Z_1\supseteq Z_1\supseteq\ldots\supseteq Z\;,$$ and let $\mathscr{F}$ be the zero-set filter generated by $\mathscr{Z}$, and let $I$ be the associated ideal. Let $f$ be any member of $R$ such that $Z(f)=Z$; then $f\in I_Z\setminus I$.
A: As rschwieb describes, you can take any $A\subseteq [0,1]$ and consider the ideal $I_A=\bigl\{f\in R\mid f(A)=\{0\}\bigr\}$. There may be several different $A$s that give rise to the same ideal in this way, but unless I'm mistaken, every such ideal is produced by exactly one closed $A$.
On the other hand, there are also ideals that are not produced in this way. One example is the set of all functions that are zero on some neighborhood of $0$. The only common zero of all these functions is $0$, but they form a proper subset of $I_0$ -- for example $f(x)=x$ is not there.
There are also ideals that are not defined solely by their zero sets. For example, consider the principal ideal generated by $f(x)=x^2$. This is again a proper subset of $I_0$, but the possible zero sets of functions in it are exactly the closed subsets of $[0,1]$ that contain $0$, just as for $I_0$.
