About the rank of submodules over PID

Question

If $M$ is a free $R$-module of finite rank $n$ and $R$ is a PID, do proper submodules of $M$ have strictly less rank than $M$?

I know that in this case, every submodule of $M$ is free and has finite rank $m\leq n$, but can we guarantee that if the submodule is proper, its rank will be strictly less than the rank of $M$? In other words,

does the rank of a submodule $N$ of $M$ equals $n$ if and only if $N=M$?

I think this is not necessarily true, but I can't find a counter-example. Some help would be very appreciated. Thanks!

Answer 1

Both $2\mathbb{Z}$ and $\mathbb{Z}$ are free rank $1$ $\mathbb{Z}$-modules, with $2\mathbb{Z}\subsetneq \mathbb{Z}$.

Answer 2

take $Z+Z+Z=M$ and $2Z+3Z+4Z=N$. Then $N$ is proper in the free $M$ but has same rank $3$ of $M$.

Answer 3

The rank of a submodule $N$ of $M$ will be equal to the rank of $M$ if and only if the quotient module $M/N$ is a torsion module, by the fundamental theorem of finitely generated modules over P. I. D.s. Hence its rank will be strictly less than the rank of $M$ if and only if $M/N$ contains non-torsion elements.