Does there always exist finitely presented submodules?

Question

Suppose $M\neq 0$ is an $A$-module where $A$ is a commutative ring with $1$. Is it always possible to find a finitely presented submodule $N\neq 0$ of $M$?

It is not interesting when the ring $A$ is Noetherian.

Just for fun.

Thank you very much!

Answer 1

No. Let $A = k [x_0, x_1, x_2, \ldots]$ be the ring of polynomials in (countably) infinitely many variables over a field $k$. Let $M$ be $k$ regarded as an $A$-module where each $x_i$ acts as zero. Clearly, $M$ is finitely generated as an $A$-module, but $M$ is simple (i.e. the only submodules of $M$ are $0$ and $M$ itself) and $M$ is not finitely presented.

To see that $M$ is not finitely presented, simply note that $M$ is the colimit of the diagram $$k [x_0, x_1, x_2, \ldots] \to k [x_1, x_2, \ldots] \to k [x_2, \ldots] \to \cdots$$ where at each step we annihilate one of the variables; if $M$ were a finitely presented $A$-module, then one of the quotient maps $k [x_n, \ldots] \to M$ would have to split as an $A$-module homomorphism, and that is impossible.