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

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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.


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 $Z(f) \subseteq Z(g)$.

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thanks for your answer. –  sh gh Oct 21 '12 at 11:23

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