Suppose $M$ is a finitely generated free $\mathbb{Z}$-module of rank $n$, so $M=u_1\mathbb{Z} \oplus \cdots \oplus u_n\mathbb{Z}$. Suppose that $M'$ is a submodule of full rank, so it is also rank $n$, and let $\widetilde{M}=u_1\mathbb{Z} \oplus \cdots \oplus u_{n-1}\mathbb{Z}$.
Now, I'm pretty sure that $M'\cap\widetilde{M}$ is a submodule of $\tilde{M}$ that is full rank, i.e., of rank $n-1$. I'm not sure why that's true, though. Does this follow from some standard result, and/or is the proof real easy?
Thanks in advance.
There is a short exact sequence $$0\to M'\cap\tilde{M}\to M'\to M'/(M'\cap\tilde{M})\to0.$$ Note also that $M'/(M'\cap\tilde{M})$ is isomorphic to the image of $M'$ in $M/\tilde{M}\cong\mathbb{Z}$, so it free of rank at most $1$. It follows that the exact sequence splits, so the rank of $M'\cong M'\cap\tilde{M}\oplus M'/(M'\cap\tilde{M})$ is the sum of the ranks of $M'\cap\tilde{M}$ and $M'/(M'\cap\tilde{M})$. Since $M'$ has rank $n$ and $M'\cap\tilde{M}$ has rank at most $n-1$, the only way this can happen is if $M'\cap\tilde{M}$ has rank $n-1$ and $M'/(M'\cap\tilde{M})$ has rank $1$.