Maximal and prime ideals in $C$ The set $C(X) = C$ consists of all continuous, real valued functions on topological space $X$ and 
$C^*=C^* (X) = \{ f \in C(X) \mid f\text{ is bounded} \}$.
Can anyone help me to solve the following problems?
An ideal $P $ in $A$  is prime if $ab \in P $   implies  $a \in P$ or $b \in P$, i.e.,  $A/ P $ is an integral domain.

1: An ideal $ P $ in $C$ is prime if and only if $ P \cap
 C^{*} $ is a prime ideal in $C^{*}$.

(By definition, the product $ I J$ is the smallest ideal containing  all products $fg$, where $ f \in I$ and $g \in J $.)
[If $ f \in P $, then $f^{1/3} \in P $]

2: If $P$ and $Q$ are prime ideals in $C$ or in $C^{*}$, then $ P Q = P \cap Q $. In particular $P ^{2} = P $, hence $M^{2}= M $ for every  maximal ideal $M$ in $C$ or $C^{*}$.

And the last one:

3: An ideal $I$ in a commutative ring is an intersection of prime
  ideals if and only if $ a^{2} \in I $ implies $a \in I $

 A: Ad 1:
Suppose $P$ is prime in $C$. Let $a,b\in C^*$ with $ab\in P\cap C^*$. Then $ab\in P$, hence $a\in P$ or $b\in P$. Then automatically $a\in P\cap C^*$ or $b\in P\cap C^*$. We conclude that $P\cap C^*$ is prime.
On the other hand, suppose $P\cap C^*$ is prime. Let $a,b\in C$ with $ab\in P$. For $f\in C$, write $\bar f:= \min\{\max\{f,-1\},1\}$ for the function $f$ "clamped" into $[-1,1]$. Note that $\bar f=c_f\cdot f$ where 
$$c_f(x)=\begin{cases}1/f(x)&f(x)>1\\-1/f(x)&f(x)<-1\\1&\text{otherwise}\end{cases}$$
As $c_f$ is continuous with continuous reciprocal, we conclude that

an ideal of $C$ contains $f$ if and only it contains $\bar f$.

Hence $ab\in P$ implies $\bar a\bar b=c_ac_b\cdot ab\in P$
But then $\bar a\bar b\in P\cap C^*$, hence $\bar a\in P\cap C^*\subseteq P$ or $\bar b\in P\cap C^*\subseteq P$.
Henec $a=\frac1{c_a}\bar a\in P$ or $b=\frac1{c_b}\bar b\in P$. We conclude that  $P$ is prime.
[$a^3=a\cdot a^2\in P$ implies $a\in P$ or $a^2=a\cdot a\in P$, hence $a\in P$ or $a\in P$ or a\in P$]
Ad 2.
Clearly, $PQ$ is the set of functions $f$ of the form $f=\sum_{k=1}^np_kq_k$ with $p_k\in P$, $q_k\in Q$. For such a function, every summand $p_kq_k$ is in $P$ and in $Q$, hence in $P\cap Q$, and so is the sum. We conclude $PQ\subseteq P\cap Q$. Use $f^{1/3}\in P,Q$ to show $P\cap Q\subseteq PQ$
