Showing that if $I$ is a maximal ideal, $R/I$ is a Field

Question

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.

Answer 1

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?

Answer 2

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 .