Lattice in $\mathbb{R}^3$ 
A lattice in $\mathbb{R^3}$ is a finite $\mathbb{Z}$-Submodul $F\in\mathbb{R^3}$. Give some examples and describe $\mathbb{R^3}/F$. Moreover, what can we say about the rang?

I am a beginner in Algebra so I would really need some help and insight. I read that that for every basis of $\mathbb{R^3}$ the subgroup of all the linear combinations of the basis vectors with integer coefficients creates a lattice and that there also will be an isomorphism with $\mathbb{Z^3}$. I also read about the Bravias lattices, I don't know if they can help me in this case.
Thanks in advance for the help
 A: Let $B$ be any basis of $\mathbb{R}^3$. Then, it is easy to see that $$\mathcal{L}(B) = \{Bz | \forall z \in \mathbb{Z}^3\}$$ is a finitely generated $\mathbb{Z}$-submodule because $\forall v, w \in \mathcal{L}(B), z_1, z_2 \in \mathbb{Z}$ we have $z_1v + z_2w \in \mathcal{L}(B)$ and it is finitely generated due to $B$. To show that this is isomorphic to $\mathbb{Z}^3$, consider the mapping $\phi : \mathcal{L}(B) \rightarrow \mathbb{Z}^3$ defined as $f(v) = B^{-1}v$. This is well-defined because we assumed $B$ to be a basis of $\mathbb{R}^3$ and hence it is invertible. Also, any $v \in \mathcal{L}(B)$ can be written as $v = Bz$ where $z \in \mathbb{Z}^n$, therefore, $f(v) = B^{-1}v = B^{-1}Bz = z \in \mathbb{Z}^n$. Since $B$ is invertible, $f$ is one-one and onto (i.e., bijective). Also, $$f(v + w) = B^{-1}(v+w) = B^{-1}v + B^{-1}w = f(v) + f(w)$$$\forall v, w \in \mathcal{L}(B)$, so $f$ is homomorphic too. This implies that $f$ is an isomorphism between $\mathcal{L}(B)$ and $\mathbb{Z}^3$. Usually, the rank of the lattice is defined as the dimension of subspace spanned by the vectors of the lattice, then rank of $\mathcal{L}(B)$ is $3$.
Finally, we show that $\mathbb{R}^3/\mathcal{L}(B) = \{Br + \mathcal{L}(B) | \forall r \in [0,1)^{n} \}$, i.e., the entries of $r$ are reals in the range of $0$ to $1$. Any element of $\mathbb{R}^3/\mathcal{L}(B)$ will look like $x + \mathcal{L}(B)$ where $x \in \mathbb{R}^3$. Since $B$ is a basis of $\mathbb{R}^3$, we can write $x = Bc = Bz + Br$ where $z = \lfloor{c} \rfloor$ (i.e., each entry is rounded below to the nearest integer) and $r = c - z$. Observe that $r \in [0,1)^n$ and $Bz + \mathcal{L}(B) = \mathcal{L}(B)$, therefore, $x + \mathcal{L}(B) = Br + \mathcal{L}(B)$.
