Intuition behind primitive sublattices Let $\Gamma$ be an integral lattice in $\mathbb{R}^n$. A $\mathbb{Z}$-submodule of $\Gamma$ is called sublattice of $\Gamma$. A sublattice $\Lambda$ of $\Gamma$ is called primitive if $\Gamma/\Lambda$ is a free $\mathbb{Z}$ module.
Is there some way to think about primitive lattices? And why are they named "primitive"?
 A: My idea is that a primitive sublattice $ \Lambda\subseteq \Gamma\subset\mathbb R^n $ is such that $ \Lambda=\mathrm{span}(\Lambda)\cap\Gamma $ so they're basically the equivalent for lattices of subspaces for a vector space.
The following should be a valid proof but I would like to have a feedback before calling it conclusive.
The inclusion $ \Lambda\subseteq\mathrm{span}(\Lambda)\cap\Gamma $ is obvious, let's prove the opposite inclusion assuming $ \Lambda $ primitive.
Suppose $ \mathrm{rk}(\Gamma)=n $ and $ \mathrm{rk}(\Lambda)=r\leq n $, then we can write
$ \Gamma=\bigoplus_{i=1}^n w_i\mathbb Z, \;\Lambda= \bigoplus_{j=1}^r v_j\mathbb Z $
with $ v_j=\sum_{i=1}^na_{j,i}w_i $ and $ A:= [a_{j,i}] $ such that $ \mathrm{rk}(A)=r $.
Now let $ v\in \mathrm{span}(\Lambda)\cap\Gamma $ therefore there are $ b_i\in\mathbb Z $ and $ x_j\in\mathbb R $ such that
$ v=\sum_{i=1}^n b_iw_i=\sum_{j=1}^rx_jv_j=\sum_{j=1}^rx_j\sum_{i=1}^na_{j,i}w_i= \sum_{i=1}^n\left(\sum_{j=1}^ra_{j,i}x_j\right)w_i $
from which $ \sum_{j=1}^ra_{j,i}x_j = b_i $ or $ A^Tx=b $.
Suppose $ A' $ is a $ r\times r $ submatrix of $ A^T $ with non-zero determinant, by reordering the basis we can consider $ b = [b'\,|\, b''] $ such that $ A'x=b' $. Now $ A' $ and $ b' $ have integer coefficients and the matrix is invertible, then $ x=A'^{-1}b' $ and has rational coefficients, more precisely $ \det(A') x\in \mathbb Z^r $.
Now consider $ \Gamma/\Lambda $ free. If we suppose $ 0\neq v\notin\Lambda $ we have that the only integer multiple of $ v $ that can lie in $ \Lambda $ is $ 0 $ but $ \det(A')v $ has integer components on $ {v_j} $ and is therefore in $ \Lambda $ despite $ \det(A') \neq 0 $, absurd, and therefore $ v\in\Lambda $.
This proves primitive $\Rightarrow \Lambda=\mathrm{span}(\Lambda)\cap\Gamma $.
For the opposite inclusion we suppose that $ [v]\in \Gamma/\Lambda $ such that it exists $ k\in\mathbb Z $ for which $ k[v]=0 $ or equivalently $ kv\in \Lambda=\mathrm{span}(\Lambda)\cap\Gamma $ and dividing by $ k $ also $ v\in\mathbb{span}(\Lambda) $. But $ v $ was also in $ \Gamma $ by hypothesis and therefore $ v\in\mathbb{span}(\Lambda)\cap\Gamma=\Lambda $ and $ [v]=0 $, which proves $ \Gamma/\Lambda $ free.
