Are two submodules (where one is contained in the other) isomorphic if their quotientmodules are isomorphic?

Question

Let $M$ be an $R$ module and $N_1 \subset N_2$ be submodules of $M$ such that $M / N_1 \cong M / N_2$. Can I know conclude $N_1 \cong N_2$ or even $N_1 = N_2$? I know that a proper submodule can be isomorphic to the original module (I know some examples when considering vectorspaces) which is why I am almost certain that $N_1 \neq N_2$ in the general case, but $N_1 \cong N_2$ seems intuitive enough.
I would appreciate any help, since I am relatively new to modules.

Answer 1

Nice question. No, this is not true in general, even if $R$ is a field.

Say the ring is a field, and let $V$ be a vector space over it with uncountable dimension. We can pick two subspaces $W_f$ and $W_\infty$ of finite, respectively countable, dimension. Then the quotients will both have the same dimension as $V$ and thus be isomorphic to each other (we are using the axiom of choice here), but the subspaces are not isomorphic to each other.

Answer 2

Let $R$ be a nontrivial (necessarily noncommutative) ring with the property that $R\cong R\oplus R$ as a left $R$-module. (This is the case if $R$ is the endomorphism ring of an infinite-dimensional vector space.) Then $R$ has a proper submodule $N$ with $N\cong R$ and $R/N \cong R$ as $R$-modules. Take $M=R$, $N_1=0$ and $N_2=N$. Clearly $N_1\not\cong N_2$.