For a commutative ring $R$, the followings are equivalent.
(1) $R$ has exactly 1 prime ideal. (2) $\sqrt{(0)}$ is maximal.
For a $R$ module $M$, $N$ is the submodule of $M$ and the followings are equivalent.
(2) There is no submodule $P\subset M$, such that $N\subset P\subset M$ where $\subset$ is proper containment. (1) The only submodule of $M/N$ are $M/N$ and $\{0\}$.
ring's (1) and module's (1) is same if I replace $M$ by $R$. Similarly for (2)'s. Was there any fundamental reason?
I'm not sure how's it's the same if you replace $M$ by $R$. Could you elaborate on that?
Generally, I don't think there's going to be any deep analogy between these two theorems about modules and prime ideals. The reason is that the statement about modules is pretty trivial set-theoretical observation that can be easily derived from the definition of a quotient module, while the statement about prime ideals stems from a quite nontrivial fact that $\sqrt{(0)}$ is the intersection of all prime ideals.