Consider the $2$-dimensional $\mathbb Q$-vector space $\mathbb Q^2$. Let $\{v,w \}$ be a basis of this vector space. Consider the following lattices \begin{align} &\mathcal{L}_1=\{iv+jw\},~i,j \in \mathbb Z \\ & \mathcal{L}_2=\{u+iv+jw\},~\text{for some vector}~u \in \mathbb Q^2. \end{align}
What is the similarity or difference between $\mathcal{L}_1$ and $\mathcal{L}_2$ ?
I think one similarity is $\mathcal{L}_1 \supset \mathbb Z^2$ and $\mathcal{L}_2 \supset \mathbb Z^2$.
Further $\mathcal{L}_2$ is a translation of $\mathcal{L}_1$.
What is the difference ?
A correction to something that you mentioned in the question: it may not necessarily be the case that $\mathcal L_2\supset \mathbb Z^2$, depending on the value of the vector $u$ and the values of $v,w$. For instance, if $v=(1,0)$ and $w=(0,1)$ (the standard basis) and $u=(\tfrac{1}{2}, \tfrac{1}{2})$, then the elements of $\mathcal L_2$ will take the form $$\big(i+\tfrac{1}{2},j+\tfrac{1}{2}\big)$$ where $i,j$ range over $\mathbb Z$, meaning that in fact none of its points belong to $\mathbb Z^2$. So this is a possible difference between $\mathcal L_1$ and $\mathcal L_2$.