Let $F$ be a non-archimedian local field with valuation $\nu$. Then $\mathcal{O}=\{x\in F: \nu(x)\geq 0\}$ is the ring of integers of $F$. $\mathfrak{m}=\{x\in F: \nu(x)> 0\}$ is the maximal ideal of $\mathcal{O}$. Let $\pi \in \mathcal{O}$ such that $\nu(\pi)=1$.
A lattice of $F^d$ is a finitely generated $\mathcal{O}$-module $L$ which contains a basis of $F^d$. Let $v_1, \ldots, v_d$ be a basis of $F^d$ and $L=v_1\mathcal{O}+\cdots v_d \mathcal{O}$. It is said that $L/\pi L \cong (\mathcal{O}/\pi \mathcal{O})^d \cong k^d$, $k=\mathcal{O}/\pi \mathcal{O}$. How to show that $L/\pi L \cong (\mathcal{O}/\pi \mathcal{O})^d$? If we have two vector spaces $A, B$, and $C, D$ are sub-vector spaces of $A, B$ respectively. It seems that we do not have $(A+B)/(C+D) = A/B + C/D$.
Thank you very much.
You know more than $L = v_1 \mathcal{O}+\dots + v_d\mathcal{O}$, because $v_1,..., v_d$ are linearly independent over $F$ and thus also over $\mathcal{O}$. So we know $L \cong \mathcal{O} \times \dots \times \mathcal{O}=\mathcal{O}\,^d$. This isomorphism defines a map $$\phi: L \rightarrow \mathcal{O}\,^d \rightarrow (\mathcal{O}/\pi \mathcal{O})^d.$$ The $\mathcal{O}$-module homomorphism $\phi$ is onto. If we quotient out the kernel, we get an isomorphism of $\mathcal{O}$-modules. Now let $v=a_1v_1+\dots+a_dv_d \in \ker \phi$, then $a_i \in \pi \mathcal{O}$ for every $i$. So this shows that $v \in \pi L$. It is clear that $\pi L \subset \ker \phi$. All in all we get the desired isomorphism $$L/\pi L \cong (\mathcal{O}/\pi \mathcal{O})\,^d.$$