Let $D$ a PID, $F$ a free module rank $n$, $N$ a submodule of $F$.
I want to prove (or find a counterexample) of: $\text{rank}(F/N)=0 \Longleftrightarrow\text{rank}(F)=\text{rank}(N)$
- $\text{rank}(F/N)=0\Rightarrow F/N=0 \Rightarrow F=N \Rightarrow \text{rank}(F)=\text{rank}(N)$
- I know it to the left is false for a ring (I take $d$ a non-zero divisor and I prove that is $(d)$ is not free. I think taking $\mathbb{Z}$ and $2\mathbb{Z}$ can work, because $\{1\}$ is a basis of $\mathbb{Z}$ and $\{2\}$ appears to be a basis of $2\mathbb{Z}$, so their rank is 1 but I can't prove $\text{rank}(\mathbb{Z}/2\mathbb{Z})$ is non-zero (basically because I think it is not free).
Does this counterexample work?
Hint: $\text{rank}(F/N)=0\Rightarrow F/N$ is a torsion module, not $\{ 0\}.$ But the conclusion is true. Use Structure Theorem for Finitely Generated Modules over a PID
Since $N$ is submodule of $F,$ there is a basis $y_1, y_2, \cdots , y_n$ of $F$ and $d_1, d_2, \cdots , d_r \in D$ with $d_1| d_2| \cdots |d_r$ such that $d_1y_1, d_2y_2, \cdots , d_r y_r$ is a basis of $N$ (rank $N = r$) and $F/N \cong D^{n-r} \times D/d_1D \times \cdots \times D/d_rD.$
In this case, rank$(F/N) = 0 \Rightarrow r = n.$ On the other hand, let $r = n.$ Then $F/N$ is a torsion module and hence is or rank zero.