$I$ is maximal ideal $\implies$ $R/I$ has no proper ideals

Question

I'm reading through a proof in a book on commutative algebra and in the proof it uses the fact that $I$ is a maximal ideal $\implies$ $R/I$ has no proper ideals, by using the correspondence theorem. I'm fairly new to the theory of ideals, so I don't see how the correspondence theorem can be applied to show this. The book doesn't go too much into this, so I'm sorry if this is a very basic concept. Also just to confirm, here the book is referring to the quotient group $R/I=\{a+I|a\in R\}$ and not something else I'm unaware of.

Answer 1

Noting $\pi : R \to R/$ the quotient map, the ideals $J$ of $R/I$ are in increasing bijection with the ideals of $R$ containing $I$ via the the inverse image by $\pi$, namely, the map $J\mapsto \pi^{-1}(J)$ whose inverse is $K\mapsto \pi(K)$. As $I$ is maximal, you see that the only ideals of $A$ containing it are $I$ and $R$, which give the zero ideal and $R/I$.

Answer 2

Suppose $a\in R\setminus I$. Since $I$ is maximal, $(a,I)=R$, therefore the ideal $(a,I)+I$ in $R/I$ must be all of $R/I$.