1) Let $R$ be a ring with a maximal ideal $\mathcal{I}$. Show that the quotient ring $R/\mathcal{I}$ is a field. Previously, I proved that
$*$ Let $R$ be a ring. $R$ is a field if and only if the only ideals of $R$ are $\{0\}$ and $R$.
I saw online that the prove of (1) that states
there is an obvious correspondence between the ideals of $R/\mathcal{I}$ and ideals of $R$ that contain $\mathcal{I}$. The result therefore follows immediately from $*$.
How is this possible?
Let $\pi \colon R \to R/I$ be the canonical projection map. Here are some facts you can use to prove the quoted result.
1) For any ideal $J$ of $R/I$, $\pi^{-1}(J)$ is an ideal of $R$.
2) Every ideal of a ring contains $0$, so in particular, every $0+I \in J$ for every ideal $J$ of $R/I$.
3) If $J, K \subseteq R/I$ are ideals such that $J$ is a proper subset of $K$, then $\pi^{-1}(J)$ is a proper subset of $\pi^{-1}(K)$.
Note that (1)+(2) imply that for any ideal $J$ in $R/I$, $I = \pi^{-1}(0+I) \subseteq \pi^{-1}(J)$. Certainly, $\langle 0 + I \rangle$ and $R/I$ are two ideals of $R/I$. Suppose there were any ideal $J$ of $R/I$ such that $\langle 0 + I \rangle \subset J \subset R/I$ with both inclusions proper. Then $I = \pi^{-1}(\langle 0 + I \rangle) \subset \pi^{-1}(J) \subset \pi^{-1}(R/I) = R$, which contradicts the fact that $I$ is a maximal ideal of $R$. Hence, the only two ideals of $R/I$ are $\{0\}$ and $R/I$.
More generally, one can in fact very slightly modify the above to show that there is a bijective correspondence between ideals of $R/I$ and ideals of $R$ which contain $I$. (1)+(2)+(3) show that the correspondence $J \mapsto \pi^{-1}(J)$ as a map between ideals of $R/I$ and ideals of $R$ which contain $I$ is injective and makes sense, and surjectivity is straightforward to show as well.