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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.

Thanks in advanced.

user111636
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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 04 '14 at 08:04
  • $R/I$ is a field. – user111636 Jan 04 '14 at 08:06
  • $I$ 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 04 '14 at 08:08
  • 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 04 '14 at 08:11
  • @Magdiragdag Can you give me more details? – user111636 Jan 04 '14 at 08:14
  • $I$ is not maximal. The assumption on $I$ says that $I$ is proper and not prime; and that no proper, non-prime ideal $J$ strictly contains $I$. – Magdiragdag Jan 04 '14 at 08:24
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    @TheNumber23: Not maximal with respect to all ideals, maximal with respect to a certain class of ideals. – Jim Jan 04 '14 at 08:24
  • Can anyone think of an example? Going to the quotient ring, $R/I$ must have exactly two maximal ideals and no other proper ideals besides those two maximal ideals and $(0)$. Furthermore, $R/I$ must have zero-divisors. – Magdiragdag Jan 04 '14 at 08:30

1 Answers1

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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 $I$.

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