Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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 at 8:04
    
$R/I$ is a field. –  user111636 Jan 4 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 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 at 8:11
1  
@TheNumber23: Not maximal with respect to all ideals, maximal with respect to a certain class of ideals. –  Jim Jan 4 at 8:24
show 3 more comments

1 Answer

up vote 0 down vote accepted

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 $I$.

share|improve this answer
    
Thank you very much –  user111636 Jan 6 at 0:24
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.