Corresponding between prime ideals in $C(X)$ and $C^*(X)$

Question

we know that every maximal ideal in $C(X)$ is in this form:

$$M^p=\left\{\,f \in C^*(x):\ p\in cl_{\beta X} Z\left(f\right)\,\right\}$$

and every maximal ideal in $C^*(X)$ is

$$M^{*p}=\left\{\,f\in C^*(X):\ f^{\beta}\left(p\right)=0\,\right\}$$

and it is not necessary that

$$ M^p \cap C^*(X) = M^{*p}$$

My question is:

If are prime ideals of $C(X)$ contained in $M^p$, in a one to one corresponding to that of $C^*(X)$ contained in $M^{*p}$ for $p\in \beta X$ ?

$M^p$ and $M^{*p}$ are maximal ideals respectively in $C(X)$ and $C^*(X)$ correspond to $p$. $\beta X$ is Ston-cech compactification of the space $X$ for terminology and notions you can refer to here.

Edite

you can find $\beta X$ at the beginning of Ch.6 and if you want more you should continue. also in Ch.7 section 7.11, it deal whit corresponding between maximal ideals specially.

Answer 1

In general, the prime ideals of $C(X)$ and $C^*(X)$ is not in a one to one corresponding. In fact, The prime ideals of $C(X)$ contained in $M^p$ is in one to one corresponding with that of $C^*(X)$ contained in $M^{*p}$ if and only if $p\in \upsilon X$, where $\upsilon X$ is the Hewitt-Nachbin space (real compactification of X). For more information one can see here Chapter 8 specially p. 119, Problem 7. J and Problem 7. K.

Answer 2

Yes - the correspondence $M^p \mapsto M^{*p}$ (and vice versa) is exactly what you are looking for.