Finding a basis for a complex lattice given a nondivisible vector in the lattice

Question

If I am given some lattice defined as, say

$$L=\{Az_1+Bz_2\ |\ A,B \in\mathbb{Z}\}$$

and a vector $v=az_1+bz_2$ , where $\gcd(a,b)=1$, I would like to find another vector $\,w\in L\,$ such that $v$ and $w$ form a basis for $L$.

I'm a bit stuck, but I can see how this would be accomplished in a lattice where $z_1=1$ and $z_2=i$ if I was given $v=1+i$: inspecting the lattice I could choose $w=i$ (or $w=1$) and still cover all the lattice points (diagonal lines of slope $1$ along all the lattice points).

If $v=2+i$, I can see how $w=1+i$ would work, where along each row of lattice "squares" you could could go across $2$, up $1$ ($v=2+i$), then up $1$, across $1$ ($w=1+i$), then subtract a $v$, so you are now at

$$(2+i)+(1+i)-(2+i)=1+i=w$$

then add a $w$ so you are at $2+2i$, then subtract a $v$ so you are at

$$(2+2i)-(2+i)=i$$

and continue ad nauseum.

My questions are:

1. In general, are we only guaranteed a single choice of $w$?
2. How can I use the fact that there are always $s,t$ where $sa+tb=1$ to help find a a $w$?

Thanks in advance.

Answer 1

A lattice in $\,\Bbb C_{\Bbb R}\,$ is a free abelian group of the form $\,w_1\Bbb Z+w_2\Bbb Z\,$ , with $\,\{w_1,w_2\}\,$ a basis for (the real vector space) $\,\Bbb C_{\Bbb R}\,$.

This means that $\,w_1\Bbb Z+w_2\Bbb Z\,$ is a lattice iff

$$\{w_1,w_2\}\,\,\,\text{are linearly independent over }\,\,\Bbb R\Longleftrightarrow \frac{w_1}{w_2}\notin\Bbb R\Longleftrightarrow \operatorname{Im}\frac{w_i}{w_2}\neq 0$$

The above is the reason why we can always do the following:

$$\tau:=\frac{w_1}{w_2}\Longrightarrow w_1\Bbb Z+w_2\Bbb Z=Z+\tau\Bbb Z$$

and taking the basis $\,\{1,\tau\}\,$ for our lattice.

As usually done with elliptic curves, we can even choose $\,\tau\,\,\,s.t.\,\,\,\operatorname{Im}(\tau)>0\,$

Thus, if you already have some given, fix $\,z_1\,$ in the lattice (I used above $\,w_1\,$), then any element

$\,z_2\,$ (I used $\,w_2\,$) in the lattice s.t. $\,\displaystyle{\frac{z_1}{z_2}\notin\Bbb R}\,$ will do the work!

Answer 2

In this case $z_1$ and $z_2$ seem to be linear independent vectors. Hence $L$ is isomorphic to $\mathbb Z^2$ via $e^i\mapsto z_i$. The preimage of $v$ in $\mathbb Z^2$ is $(a,b)$ which I will just call $v$. Now you need a vector $w=(w_1,w_2)$ such that the equation $r\cdot v+ s\cdot w=u$ has an integral solution for all $u\in\mathbb Z^2$. This is equivalent to the matrix $(v,w)$ having determinant $\pm 1$, since for having an integral solution the matrix has to be invertible over $\mathbb Z$.

Now you use the second statement and just insert $s$ and $t$ from your equation into $w$ such that the determinant becomes the equation. I think $w=(-t,s)$ is correct. The image of this vector then is $t\cdot z_1+s\cdot z_2$ the second basis vector in your lattice $L$.

Since there are infinitely many vectors orthogonal to $v$ the choice is not unique. You may add or subtract $v$ to $w$ and the result will stay a basis.

Hope this helps.