# Suppose that $I$ is an ideal of $R$ which is maximal with respect to the property that it is proper and not prime.

Let $R$ be a commutative ring with $1$. Suppose that $I$ is an ideal of $R$ which is maximal with respect to the property that it is proper and not prime. Deduce that $I$ is contained in at most two other proper ideals of $R$.

I don't have idea. Help me.

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Hint: look at $R/I$. Ideals of $R/I$ correspond to ideals of $R$ containing $I$, so what does the condition on $I$ say about $R/I$? – Magdiragdag Jan 4 '14 at 8:04
$R/I$ is a field. – user111636 Jan 4 '14 at 8:06
$I$ is not a maximal ideal; it is only maximal with respect to some funny property; $I$ is, by assumption, not even prime. Can you think of an example, actually? – Magdiragdag Jan 4 '14 at 8:08
All maximal ideals are prime. I don't understand the wording here. Also maximal ideals are contained in exactly 1, proper ideal of $R$ namely itself. – TheNumber23 Jan 4 '14 at 8:11
@TheNumber23: Not maximal with respect to all ideals, maximal with respect to a certain class of ideals. – Jim Jan 4 '14 at 8:24

Let $I$ be such an ideal, and let $P$ be a minimal prime ideal over $I$. Then in $R/P$, every proper ideal is prime. Hence for any nonzero $a \in R/P$, $(a^2)$ is prime, so it contains $a$, so there is some $b$ such that $a^2b = a$; that is, $a(ab-1)=0$ and since $R/P$ is an integral domain, it follows that $ab=1$ and $a$ has an inverse. So that means that $R/P$ is a field; hence $P$ is maximal.
Now consider $R/I$. The nonzero proper ideals of $R/I$ correspond exactly to the proper ideals of $R$ strictly containing $I$, and they are all maximal. Use the result of your previous question to deduce that there are most two maximal ideals in $R/I$, and hence at most two proper ideals of $R$ strictly containing $I$.