Let $R$ be a ring and $P$ be an ideal containing an ideal $I$ of $R.$ Show that $P$ is a maximal ideal of $R$ iff $P/I$ is a maximal ideal of $R/I.$
My attempt so far is as follows:
Let $P/I$ be a maximal ideal of $R/I$ then, let $U$ be a ideal of $R$ such that $P\subseteq U\subseteq R.$
This means, $P/I\subseteq U/I\subseteq R/I.$ We have two cases:
If $P/I=U/I$ then, let $u\in U.$ This means $\exists p\in P$ such that $$p+I=u+I\implies u-p\in I\subseteq P\implies u\in P\implies U\subseteq P\implies U=P.$$
If $U/I=R/I$ then, let $r\in R$. Thus, $\exists u\in U$ such that $$u+I=r+I\implies r-u\in I\implies r-u\in P\subseteq U\implies r-u\in U\implies r\in U\implies R\subseteq U\implies U=R.$$
Thus, $P$ is a maximal ideal.
Next, we assume that $P$ is a maximal ideal of $R$ and try to prove that, $P/I$ is a maximal ideal of $R/I.$
This is the portion precisely, where I am stuck. Any help regarding this will be greatly appreciated.
Useful for this question is the ideal correspondence theorem, which states that there's an order-preserving bijection between ideals of $R/I$ and ideals of $R$ containing $I$. The order here is set containment. Let $\pi: R \to R/I$ be the natural projection, the map \begin{align*} \Phi:\{U \subseteq R\ :\ \text{$U$ an ideal such that $I \subseteq U \subseteq R$}\} &\longrightarrow \{J \subseteq R/I\ :\ \text{$J$ an ideal}\}\\ U &\longmapsto U/I \end{align*} is a bijection with inverse \begin{align*} \Psi: \{J \subseteq R/I\ :\ \text{$J$ an ideal}\} &\longrightarrow \{U \subseteq R\ :\ \text{$U$ an ideal such that $I \subseteq U \subseteq R$}\}\\ J &\longmapsto \pi^{-1}(J) \end{align*} In particular, $U \subseteq V \subseteq R$ if and only if $\Phi(U) \subseteq \Phi(V) \subseteq R/I$.