# Sums of Strongly z-ideals

In the rings of continuous functions,i.e.$(C(X))$ an ideal $I$ is called strongly $z$-ideal if it is an intersection of some maximal ideals of $C(X)$. i.e. $$I=\cap_{\alpha \in A} \mathcal{M_{\alpha}}$$ Where each $\mathcal{M_{\alpha}}$ is a maximal ideal of $C(X)$.

Question: If $I$ and $J$ are two strongly $z$-ideals of $C(X)$, Is the ideal $I+J$ strongly $z$-ideal or all of $C(X)$?

PS: If $X$ is a $P$- space, then each ideal of $C(X)$ is strongly $z$- ideal, And conversly if each ideal of $C(X)$ is strongly $z$- ideal than $X$ is a $P$-space.

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Dear Asaf. You didn't notice that $M_{x_1}+M_{x_2}=C(X)$ and in this case $C(X)$ is the intersection of empty set of maximal ideals. Maybe its better for you to consider this Question In this form: If $I$ and $J$ are two strongly $z$-ideals, is it true that $I+J$ is strongly $z$-ideal or all of $C(X)$? – Ali Reza Aug 16 '12 at 7:01
When I added this comment, Asaf had given an answer to my Question. But Know he deleted it. the above comment is for that answer. – Ali Reza Aug 16 '12 at 7:04
Dear Asaf. I do not think that every z-ideal in $C[0,1]$ is of the form of a principal ideal $(f)$, because in this case this ideal must be generate by an idempotent and this shows that $[0,1]$ has a nontrivial clopen set and this is false. On the other hand I think this case is different from my case. because you are thinking about z-ideals but its not true that every z-ideal is a strongly z-ideal. Your proof reminds me that the sum of two z-ideal, is a z-ideal and it's a bit similar to my Question. – Ali Reza Aug 16 '12 at 8:26