Let $R$ be a ring and $M$ be a nonzero left $R$-module. If we take a nonzero $m\in M$, then the map $R\to M$ given by $r\mapsto rm$ has some kernel $I$, which is a left ideal of $R$ and thus $M$ admits a left $R$-submodule isomorphic to $R/I$. So, arbitrary modules contain submodules isomorphic to quotients of $R$ by left ideals.

In the commutative world, a remarkable fact is that sometimes you can ensure that an $R$-module contains a submodule isomorphic to $R/P$ for some prime $P$! This happens when $R$ is Noetherian and $M$ is finitely generated. Then the zero divisors of $R$ on $M$ (thos elements $r\in R$ such that $rm = 0$ for some nonzero $m\in M$) by the theory of associated primes is a set that is the union of primes, each of which is the annihilator of a nonzero element of $M$.

Therefore, if $R$ is Noetherian and $M$ is finitely generated, you can always find an $m\in M$ such that $R\to M$ given by $r\mapsto rm$ has kernel $P$ where $P$ is a prime ideal! Comes in handy on occasion.

*Can you find a counterexample of a Noetherian ring $R$ and an infinitely generated module $M$ where $M$ does not contain any submodule isomorphic to one of the form $R/P$ where $P$ ranges over the primes?*