why the z-ideals in C(X) are basically the sums of intersections of maximal ideals?



a questions about the sums of intersections of maximal ideals
Let $\mathcal{M}\_f$ be the intersection of all maximal ideals in $C(X)$ which contains $f$. it is easy to show that $$\mathcal{M} \_f=\Big(g \in C(X): Z(f) \subseteq Z(g)\Big)$$


Now Let $I$ be a $z-$ideal of $C(X)$. It means if $f \in I$ and $Z(f)\subseteq Z(g)$, then $g \in I$.


It suffices to show that for all $f \in I$, $\mathcal{M}\_f \subseteq I$. 


let $g \in \mathcal{M}\_f$, by the above characterization of $\mathcal{M}\_f$ we have $Z(f) \subseteq Z(g)$ and this implies that $g \in I$.


So we have shown that for all $f \in I$, $\mathcal{M} \_f \subseteq I$. this yields: 
$$I=\sum \_{f \in I} \mathcal{M} \_f$$.
And this is the case which you wanted.




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At the end let me show the sketch of the characterization of $\mathcal{M} \_f$ .


Suppose $g$ is in all maximal ideals which contains $f$. Fix $x \in Z(f)$, then $f \in M\_x$. but $M\_x$ is a maximal ideal which contains $f$ .so $g \in M\_x$ and this implies that $x \in Z(g)$. then we show that $Z(f) \subseteq Z(g)$.


For the converse let $Z(f) \subseteq Z(g)$. suppose on the cotrary that there exists a maximal ideal $M$ such that $f \in M$ but $g \notin M$. 


So we have $M+(g)=C(X)$. this shows that there exists $k \in M$ and $h \in C(X)$ so that $$k+h.g=1$$


This implies that $Z(k) \cap Z(g)= \varnothing$. But we Know that $f, k \in M$. this obviously shows that $Z(f) \cap Z(g) \neq \varnothing$. We came into a contradiction, because we supposed that $Z(f) \subseteq Z(g)$.


