In a PID if the rank (finite) of a free module and its submodule are the same, then are they necessary equal?

Question

let $R$ be a PID and $M$ a free (unital) R-module of finite rank $a$. If $N$ is a free $R$-submodule of $M$ with $\operatorname{rank}(N) = a$, then is $N = M$?

I think it doesn't holds if rank is not finite for eg if $F=\Bbb Z$ and $M=\langle 2\rangle$ then $\operatorname{rank}F= \operatorname{rank} M$ but $F$ is not equal to $M$.

Any help will be appreciated. Thanks in advance.

Answer 1

Given a commutative ring with unit $R$ and $x\in R$ which is not a zero divisor, the $R$-modules $R$ and $xR$ are isomorphic and free of rank 1. If $x\notin R^*$ as well, then $xR\subsetneq R$.