Finding all the maximal ideals of $C([0,1])$ and also prime ideals in it which are not maximal

To supplement egreg's answer, here is a proof of the result they mentioned:

If $X$ is a completely regular space, then there is a natural bijection between points of the Stone-Cech compactification $\beta X$ and maximal ideals in $C(X)$.

For $f\in C(X)$, we will write $V(f)$ for $f^{-1}(\{0\})$. Given a point $x\in \beta X$, define $$M_x=\{f\in C(X):x\in\overline{V(f)}\}.$$ (Here and below, the overline denotes closure in $\beta X$, not closure in $X$.) I claim that $x\mapsto M_x$ is our desired bijection.

First, $M_x$ is a maximal ideal. It is clear that it is closed under multiplication by elements of $C(X)$. If $f,g\in M_x$, observe $V(f)\cap V(g)\subseteq V(f+g)$. We will show $x\in\overline{V(f)\cap V(g)}$ and hence $f+g\in M_x$.

So, suppose $x\not\in\overline{V(f)\cap V(g)}$. We can then choose a closed neighborhood $A$ of $x$ which is disjoint from $\overline{V(f)\cap V(g)}$ and a function $h:\beta X\to [0,1]$ which vanishes on $A$ and is $1$ on $\overline{V(f)\cap V(g)}$. Now consider $F=(f^2+h,g^2+h)$ as a function $X\to\mathbb{R}^2$. Observe that $F$ never vanishes, since when $f$ and $g$ both vanish, $h$ is $1$. Choose a continuous function $s:\mathbb{R}^2\setminus\{(0,0)\}\to [0,1]$ such that $s(a,0)=0$ for all $a$ and $s(0,b)=1$ for all $b$. Then $s\circ F$ extends continuously to a map $k:\beta X\to [0,1]$. Observe that $k=1$ on $V(f)\cap A$ and $k=0$ on $V(g)\cap A$. But $A$ is a neighborhood of $x$, so $x\in\overline{V(f)\cap A}$ and also $x\in\overline{V(g)\cap A}$. Thus by continuity, $k(x)=1$ and $k(x)=0$, which is a contradiction.

We have therefore shown that $M_x$ is an ideal. For maximality, suppose $f\in C(X)\setminus M_x$. Then $x\not\in\overline{V(f)}$, so as above we can choose $h:\beta X\to [0,1]$ which vanishes on a neighborhood of $x$ and is $1$ on $\overline{V(f)}$. Then $h|_X\in M_x$ since it vanishes on an entire neighborhood of $x$, and $h|_X+f^2$ vanishes nowhere on $X$. Thus $h|_X+f^2$ is a unit in $C(X)$ and thus the ideal generated by $M_x$ and $f$ is not proper. Since $f\not\in M_x$ was arbitrary, this shows $M_x$ is maximal.

Now I claim that for $x\neq y$, $M_x\neq M_y$. To prove this, choose $h:\beta X\to [0,1]$ vanishing on a closed neighborhood of $x$ but not vanishing at $y$. Then $h|_X\in M_x$ and $h|_X\not\in M_y$.

Finally, I claim that any maximal ideal $M\subset C(X)$ is equal to $M_x$ for some $x\in \beta X$. To prove this, let $Z=\{\overline{V(f)}:f\in M\}$. Note that $V(f)$ is nonempty for all $f\in M$ since $M$ cannot contain a unit, and also that $V(f)\cap V(g)\supseteq V(f^2+g^2)$. It follows that $Z$ has the finite intersection property, and so by compactness of $\beta X$ there is some $x\in \beta X$ which is in every element of $Z$. But this just means that $M\subseteq M_x$, and so by maximality of $M$, $M=M_x$.

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As for prime ideals that are not maximal, there are a lot more possibilities and it is much harder to describe all of them. As mentioned in Max's answer, you can construct some using ultrafilters. In fact, instead of considering functions which are $0$ along an ultrafilter as he does, you can instead consider functions which converge to $0$ at a certain rate, and in this way you can obtain large nested collections of prime ideals. You can find the details of such a construction at this answer of mine, which constructs a chain of prime ideals in $C(X)$ which is order-isomorphic to $[0,\infty)$ from any element of $C(X)$ that is not locally constant.

As pointed out by Eric Wofsey, my previous answer was incorrect (and so I deleted it) but it can be corrected in the following way : pick $a\in (0,1)$ and a sequence $(a_n)$ of distinct elements converging to $a$.

Let $\mathcal{U}$ be an nonprincipal ultrafilter on $\mathbb{N}$ and consider the set of $f\in C(0,1)$ that are $0$ on a $\{a_n, n\in U\}$, $U\in \mathcal{U}$ (for the rest of the answer I will simply say "$f$ is $0$ on $U$").

This implies that $f(a)=0$ in particular (because the subsequence $(a_n)_{n\in U}$ converges to $a$, as $U$ is not finite, and consists of zeroes of $f$).

But obviously this set is not equal to $\{f\in C(0,1)\mid f(a)=0\}$.

Now this set is closed under multiplication by any $g\in C(0,1)$ because $\mathcal{U}$ is upward closed; it contains $0$ because $\mathbb{N}\in \mathcal{U}$; and finally it is closed under addition because if $f=0$ on $U$, $g=0$ on $V$, then $f+g =0 $ on $U\cap V$.

So this set $I$ is an ideal.

It is prime because if $fg=0$ on $U$ then $U \subset \{n, f(a_n)=0 \}\cup\{n, g(a_n) =0\}$ so this union is in $\mathcal{U}$, and since $\mathcal{U}$ is an ultrafilter, this implies that one of them is in $\mathcal{U}$, so $f$ or $g$ is in $I$.

As $I$ is properly contained in $\{f, f(a)=0\}$, this is an example of a prime ideal that's not maximal.

A way of seeing it, perhaps more clearly, and to confirm that I haven't made a mistake this time (I hope) is that $I=\mathrm{Ker}\varphi$ where $\varphi : C(0,1) \to \mathbb{R}^* := \mathbb{R}^\mathbb{N}/\mathcal{U}$ is defined by $f\mapsto [(f(a_n))_n]$ which is a composition of ring morphisms so a ring morphism; now $\mathbb{R}^*$ is a field so this implies $C(0,1)/\mathrm{Ker}f$ is isomorphic to a subring of a field, hence an integral domain, hence $I=\mathrm{Ker}f$ is prime. And it's not maximal for the reason stated above

I edited to go to $C(0,1)$ instead of $C[0,1]$ but it actually works for both; actually this is an example that works for $C(X)$ for any $X$ as long as $X$ has a limit point that satisfies that there is $f: X\to \mathbb{R}$ that vanishes only on that point locally (it yields a prime ideal that is not maximal). I don't know much about maximal ideals of this ring but egreg gave some ideas about that.

You can even modify it by going from a sequence to a net