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From a Masters Qual. Practice Exam:

Let $R$ be a commutative ring with identity, and let $I$ be an ideal of $R$. Prove there is a bijection between the intermediate ideals $J$ such that $I \subseteq J \subseteq R$ and the ideals of the quotient ring $R/I$. Thus prove that if $I$ is maximal ideal, then $R/I$ is a field.

I've read other proofs that if $I$ is a maximal ideal, $R/I$ is a Field, but I'm having a hard time understanding them, I can't even tell if they use this same technique or not.

Edit: I think now I have the bijection, we let $\phi(I) = J/I$, and this gives us a correspondence between intermediate ideals and ideals of $R/I$.

I still don't know how to get that $R/I$ is a field after this.

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    $\begingroup$ This is just advice: When someone says "commutative ring with 1", think "the integers" first. Work out what this says if $R = \mathbb{Z}$ and understand why, in this special case. This often generates much light. $\endgroup$
    – B. Goddard
    Commented Jan 6, 2017 at 3:23
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    $\begingroup$ One way to think about this, which might help recall sitting for the exam, is to consider what ideals are found in a field. The general relationship of ideals of $R$ to ideals of a quotient $R/I$ then solves the problem. $\endgroup$
    – hardmath
    Commented Jan 6, 2017 at 3:24
  • $\begingroup$ @hardmath Brilliant. I think I get it now. $\endgroup$ Commented Jan 6, 2017 at 3:25
  • $\begingroup$ @B.Goddard Thanks for the advice, I'll remember that. $\endgroup$ Commented Jan 6, 2017 at 3:25
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    $\begingroup$ @setholopolus If you have a hard time understanding a proof here, you should first use a comment to prod the poster. If you were talking about proofs you found IRL, then you should be able to find the duplicates on this site and take a look. $\endgroup$
    – rschwieb
    Commented Jan 6, 2017 at 3:43

2 Answers 2

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If $I$ is amaximal ideal, then there are only two intermediate ideals, namely, $I$ and $R$. Thus, $R/I$ has only two ideals.

What kind of rings have only two ideals?

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from the theorem: let $R$ a ring and $I$ an ideal in $R$ then we have a Corresponding $1-1$ between ideals in $R$ contain $I$ and ideals in $R/I$

then $I$ a maximal ideal in $R$ if and only if $I=< \overset- 0 >$ a maximal ideal in $R/I$ if and only if $R/I$ field.

A ring $R$ is a field $\Leftrightarrow$ the only ideals are $(0),R$ . Where $(0)$ is maximal ideal and $R$ minimum ideal .

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