Maximal ideals in $C(X)$ and Axiom of Choice The following result are true if we assume full axiom of choice:


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*A. If $X$ is a compact Hausdorff space, then every maximal ideal of the ring $C(X)$ has the form $A_p=\{f\in C(X); f(p)=0\}$.

*B. If $X$ is a compact Hausdorff space, then every ideal of the ring $C(X)$ is contained in an ideal of the form $A_p=\{f\in C(X); f(p)=0\}$.
I wonder how much choice is needed. To be precise, do we get a statement equivalent to some known form of AC if we assume validity of A/B for every compact space, for every compact metric space, for every complete totally bounded metric space or for the case $X=[0,1]$?
I've tried to answer A at least partially in my answer here.
If I did not make a mistake there, I've shown that in ZF the claim A holds for complete totally bounded metric space, B holds for compact spaces. I've also gathered a few relevant references in that answer. According to those references validity of A of every compact regular space is equivalent to ultrafilter theorem. According to the same book, in ZF it can be shown that A holds if and only if $X$ the form $[0,1]^I$ and B holds if and only of $X$ is compact.
As I am not experienced with working in ZF (without AC), I'll be glad if you check my work there and point out any mistakes and add any additional references/proofs/insights.
This question is also related, but not identical: https://math.stackexchange.com/questions/97603/realizing-a-homomorphism-mathcalcx-to-mathbbr-as-an-evaluation
 A: EDIT : The following does not answer the question, because it uses Urysohn's lemma, as pointed out in the comments. However, the only choice needed is, I think, countable dependent choice. It also gives a proof in ZF of both A and B simultaneously, in the case of compact metric space.

First, let's show that $A_p$ is a maximal ideal. Let $J$ by an ideal of $C(X)$ such that $A_p \subsetneq J$. Let's show that necessarily $J=C(X)$. Let $g \in J$, $g \notin A_p$. So $g(p) \neq 0$, and we can write
$$1 = \frac{g(p)-g}{g(p)} + \frac{g}{g(p)}$$
The first term vanishes at $p$, so is in $A_p$, and the second term belongs to $J$. It follows that the function $1 \in J$, and thus $J=C(X)$.
Now, let $M$ be a maximal ideal of $C(X)$. Suppose that for all $p \in X$, there exists $g_p \in M$ such that $g_p(p) \neq 0$. By continuity, $g_p \neq 0$ on some neighborhood $V_p$ of $p$. By compacity of $X$, there exists $p_1, p_2, \dots, p_n$ such that
$$X=V_{p_1} \cup V_{p_2} \cup \dots \cup V_{p_n}.$$
Let $h_1, h_2, \dots, h_n$ be a partition of unity subordinate to $V_{p_1}, \dots, V_{p_n}$ (This follows from Urysohn's lemma). Now write
$$1= h_1 + \cdots + h_n = \frac{h_1}{g_{p_1}}g_{p_1} + \frac{h_2}{g_{p_2}}g_{p_2} + \cdots +  \frac{h_n}{g_{p_n}}g_{p_n}.$$
The functions $h_j/g_{p_j}$ are continuous in $X$, since the support of each $h_j$ is contained in $V_{p_j}$, and $g_{p_j} \neq 0$ in $V_{p_j}$. It follows that $1 \in M$, a contradiction.
We have shown that there exists a $p \in X$ such that for all $g \in M$, $g(p)=0$, i.e. $M \subseteq A_p$. By maximality of $M$, $M=A_p$.
The second part of the proof also shows that $B$ holds.
A: Let me mention here at least some case, which I was able to solve. (Hopefully, I have not missed some hidden use of AC.)

$\newcommand{\mc}[1]{\mathcal{#1}}\newcommand{\inv}[1]{{#1}^{-1}}\newcommand{\emps}{\emptyset}\newcommand{\Invobr}[2]{\inv{#1}(#2)}$
We say that an ideal $A\subseteq C(X)$ is fixed, if there is a point $p$ such that $f(p)=0$ for each $f\in A$, i.e., there is a point $p$ such that $A\subseteq A_p$.
Claim 1 (ZF) If $X$ is compact, then every proper ideal in $C(X)$ is fixed.
Proof. Let $A$ be a proper ideal in $C(X)$. Then $\{\Invobr f0; f\in A\}$ is a system, which has finite intersection property. (Suppose that $\Invobr f0\cap\Invobr g0=\emps$ for some $f,g\in A$. This implies that $h=f^2+g^2$ is a non-zero function such that $h\in A$. Consequently $1\in A$ and $A=C(X)$, a contradiction.)
By compactness, the above system has a non-empty intersection, i.e., there exists a point $p\in\bigcap\limits_{f\in A} \Invobr f0$. This means that $A\subseteq A_p$. $\square$
The basic idea used in the following proof is that in a complete regular space every closed set is an intersection of zero-sets.
Claim 2 (ZF) If $X$ is completely regular, and every proper ideal in $C(X)$ is fixed, then $X$ is compact.
Proof.
Let $\mc K$ be a system of closed subsets of $X$ which has finite intersection property.
Let $$A=\{f\in C(X); \text{ there exists a finite set }\mc F\subseteq\mc K\text{ such that }\Invobr f0\supseteq \bigcap\mc F\}.$$
Then $A$ is an ideal in $C(X)$. (It is easy to see that $f\in A$ implies $gf\in A$ for any $g\in C(X)$. If $f,g\in A$ then $\Invobr f0\supseteq\bigcap\mc F_1$ and $\Invobr g0\supseteq\bigcap\mc F_2$ for some finite sets $\mc F_1$ and $\mc F_2$. Thus for $h=f+g$ we get $\Invobr h0\supseteq\bigcap (\mc F_1\cup\mc F_2)$.) Clearly $1\notin A$ and thus $A$ is proper.
Using complete regularity we can show that $\bigcap\limits_{f\in A} \Invobr f0 = \bigcap \mc K$. (It is clear from the definition of $A$ that $\bigcap\mc K\subseteq\bigcap\limits_{f\in A} \Invobr f0$. On the other hand if  $p\notin\bigcap\mc K$, then we have a closed set $K\in\mc K$ such that $p\notin K$. Since $X$ is complete regular, there exists a function $f\in C(X)$ such that $f(p)=1$ and $f|_K=0$. This function belongs to $A$. Thus we get that $p\notin \bigcap\limits_{f\in A} \Invobr f0 $.)
By the assumption, the intersection $\bigcap\limits_{f\in A} \Invobr f0$ is non-empty. So we see that $\bigcap\mc K$ is non-empty. $\square$
