Lattice in a vector space of dim 2 over a valuated field. I'm reading "Arbres, amalgames et SL2" of J.P. Serre, and something is not clear to me, but is to him :)
Let $k$ be a field, with a discrete valuation $v$, ie a group epimorphism $v:k^\ast \to \mathbb{Z}$ such that $\forall x,y \in k, v(x+y) \leq min(v(x),v(y))$ and $v(0)=\infty.$
Let $A$ be the valuation ring $\lbrace x \in k, v(x) \geq0 \rbrace$, pick $\pi \in k$ such that $v(\pi)=1$.
Let $V$ a two dimensional vector space over $k$.
A lattice in $V$ is a sub-$A$-module finitely generated who spans $V$ as a $k$ vector space, he's then free of rank 1.
Serre says that if $L_1, L_2$ are two lattices, there exists a $A$-base $\lbrace e_1,e_2\rbrace$ of $L$, and integers $a,b$ such that $\lbrace\pi^a e_1,\pi^be_2\rbrace$ is a $A$-base of $L_2$ .
Comes from the theory of elementary divisors, he says. Can someone explain me something clear about that? Seems very intuitive, but I need somethinga bit formal.
Thanks a lot!
 A: First, a theorem about finitely generated free modules over a principal ideal domain $A$: if $L$ is a finitely generated free $A$-module, say of rank $n$, with a choice of basis $\{e_1,\ldots,e_n\}$,, and $L^\prime\subseteq L$ is an $A$-submodule, then there is an integer $m$, $0\leq m\leq n$, and non-zero elements $a_1,\ldots,a_m\in A$, such that $\{a_1e_1,\ldots,a_me_m\}$ is a basis for $L^\prime$. This is sometimes (usually?) proved along the way to proving the structure theorem for finitely generated $A$-modules.
Okay, now, in your setup, $A$ is a discrete valuation ring, in particular a PID, but especially nice, in that, up to units, it has a unique prime element $\pi$ (an element with normalized valuation equal to $1$). Let $\{e_1,e_2\}$ be an $A$-basis for $L_1$ and choose an $A$-basis $\{f_1,f_2\}$ for $L_2$. Writing each $f_i$ as an $k$-linear combination of the $e_i$ (which you can do, since $\{e_1,e_2\}$ is also a $k$-basis for $V$ by the definition of a lattice), since every element of $k$ is an element of $A^\times$ times an integer power of $\pi$, one sees that for some integer $n\geq 0$, $\pi^n f_1,\pi^n f_2\in L_1$, and hence $\pi^n L_2\subseteq L_1$. Now, by the theorem above, there are non-zero elements $\pi^au,\pi^b v\in A$ (with $u,v\in A^\times$) such that $\{\pi^a ue_1,\pi^b ve_2\}$ is an $A$-basis for $\pi^n L_2$.
Finally, if we let $e_1^\prime=ue_1,e_2^\prime=ve_2$, then $\{e_1^\prime,e_2^\prime\}$ is an $A$-basis for $L_1$, and $\{\pi^{a-n}e_1^\prime,\pi^{b-n}e_2^\prime\}$ is a $A$-basis for $L_2$, as desired.
