Sublattice of index 2

Question

Let $M \subset L$ be two lattice of $\mathbb{R}^2$ and $|L:M|=2$. Let $v_1,v_2 $ be basis of $M$ and linearly independent in $L$. Assume further that $||v_1||\leq ||v_2||$.

If we fix $v_1$, can we replace $v_2$ by $v_2'$ such that $(v_1,v_2')$ is a basis of $L$?

Edited: I have counterexample in general and I have edited with further assumption that length of $v_1$ is shorter than $v_2$.

Answer 1

The answer is no. Here is a counterexample:

Let $L=\mathbb Z \times \mathbb Z$ and $M=2\mathbb Z \times \mathbb Z$. Then $M$ is a sublattice of $L$ of index $2$.

Take $v_1=(2,0)$ and $v_2=(0,1)$. Then $v_1,v_2$ is a basis for $M$ but $v_1$ is never part of a basis for $L$.

Even with the length restriction, this is still false: take $v_1=(2,0)$ and $v_2=(2,1)$.