Proof of Gelfand Kolmogorov theorem about the Stone-Cech compactification I'm not able to understand the proof of the G-K theorem stating that given $X$ completely regular the subsets of $C(X)$ of the form $$M^p=\{f\in C(X)\ |\ cl_{\beta X}Z(f)\ni p\}$$ where $p$ is a point of the compactification $\beta X$ are maximal ideals and are in one-to-one corrispondence with the points of $\beta X$. Every proof I found begins with "Let's show that $M^p$ is a (maximal) ideal showing that $Z[M^p]$ is a z-(ultra)filter, where $$Z[M^p]=\{Z(f)\ |\ f\in M^p\}.$$ So far so good, the empty set is not in $Z[M^p]$ and it is closed under supersets. The problem is when they show that $Z[M^p]$ has the finite intersection property instead of showing it is closed under intersection. Are those equivalent in this case? Am I missing something? [NOTE: The proof would be correct if gizen two zero sets $Z_{1,2}$ in $Z[M^p]$ whose intersection is nonempty, $clZ_1\cap clZ_2=cl(Z_1\cap Z_2)$, but thats not the case right?]
 A: I will just show how some properties of intersections could be derived from the definitions. I'll start from the characterization of $\beta X$ as a space containing $X$ and satisfying the following two properties.


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*Compactification: $\beta X$ is compact and Hausdorff, and $X$ is dense in $\beta X$

*$C^*$-embedding: every bounded function $X\to\mathbb R$ extends to a continuous $\beta X\to\mathbb R$


I will show:


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*If $Z(f),Z(g)\in Z[M^p]$ then $Z(f)\cap Z(g)\neq\emptyset$

*If $Z(f),Z(g)\in Z[M^p]$ then $Z(f)\cap Z(g)\in Z[M^p]$
For (1.), suppose for contradiction that $Z(f)\cap Z(g)=\emptyset$. Then $\frac{f^2}{f^2+g^2}$ is a continuous function $X\to[0,1]$ so has an extension $h:\beta X\to[0,1]$ whose level sets $h^{-1}(\{0\})$ and $h^{-1}(\{1\})$ are disjoint closed sets in $\beta X$ separating $Z(f)$ from $Z(g)$. This shows $cl_{\beta X}(Z(f))\cap cl_{\beta X}(Z(g))$ is empty so doesn't contain $p,$ contradicting the assumption that $p$ is in both $cl_{\beta X}(Z(f))$ and $cl_{\beta X}(Z(g))$.
For (2.) suppose for contradiction that the closure of $Z(f)\cap Z(g)$ in $\beta X$ does not contain $p$. Since $\beta X$ is normal there is a function $h:\beta X\to [0,1]$ with $h(p)=0$ and $h(Z(f)\cap Z(g))=\{1\}$. Note $Z(f)$ can be covered by two sets $Z(f)_1=Z(f)\cap h^{-1}[0,\tfrac 1 2]$, and $Z(f)_2=Z(f)\cap h^{-1}[\tfrac 1 2,1]$, and $cl_{\beta X}(Z(f)_2)$ is contained in $h^{-1}[\tfrac 1 2,1]$ so doesn't contain $p$. So $p\in cl_{\beta X}(Z(f)_1)$. Since $Z(f)_1$ can be expressed as a zero set $Z(f)_1=Z(f^2+\max(0,h-\tfrac 1 2))$ we have $Z(f)_1\in Z[M^p]$. The similarly-defined function $Z(g)_1$ is also in $Z[M^p]$. But $Z(f)_1\cap Z(g)_1=\emptyset$ by construction, contradicting (1.).
