Let $[u_1,u_2]$ denote the free $\Bbb{Z}$-module of rank $2$ with generators $u_1$ and $u_2$, i.e., $$[u_1,u_2]:=u_1\Bbb{Z}\oplus u_2\Bbb{Z}.$$ Let $M=[1,\theta]$. Except for $\theta$, which is irrational, every variable in sight here is an integer.
It is clear that any rank-$2$ submodule of $M$ can be written in the form $[a,b+c\theta]$, where we can assume w.l.o.g. that $a$ and $c$ are positive and that $0\le b\le a-1$. Subject to these restrictions, the representation is unique, and can be taken as a canonical representation.
For submodules with canonical representations:
$M_1=[a,b+c\theta]\subseteq M \\ M_2=[d,e+f\theta]\subseteq M,$
I want to write down the canonical representation of the submodule $$M_1+M_2=\{m_1+m_2\mid m_1\in M_1, m_2\in M_2\},$$ but I'm stuck on a detail. I have this:
$M_1+M_2=[g_1,h+g_2\theta]$
where $g_1=\gcd(a,d,bf-ce)$ and $g_2=\gcd(c,f)$. For $h$, I have the system of linear congruences:
$\begin{cases}\dfrac{c}{g_2}h\equiv b\pmod{g_1} \\ \dfrac{f}{g_2}h\equiv e\pmod{g_1}\end{cases}$.
I need to know that such an $h$ exists, and that we can determine it uniquely up to congruence modulo $g_1$, and that's where I'm stuck. For example, the first congruence has a solution iff $\gcd\left(\frac{c}{g_2}, g_1\right)| b$. It's clear to me that, if $t|\frac{c}{g_2}$ and $t|g_1$, then we have $t|bf$, but we need $t|b$. From looking at examples, the system always seems to have a unique solution, even though both congruences might individually have multiple ones.
I don't know whether I'm missing something obvious, or whether my approach is all wrong, or what. Any help, hints or insight anyone can offer would be very much appreciated.
I figured it out.
I have the wrong expression there for $g_1$. The other one, $g_2=\gcd(c,f)$ is perfect, and we should define $c'=\dfrac{c}{g_2}, f'=\dfrac{f}{g_2}$, two relatively prime integers. Then the formula we really need is:
$g_1=\gcd(a,d,bf'- c'e)$
Then the congruences we need are:
$\begin{cases}c'h\equiv b\pmod{g_1} \\ f'h\equiv e\pmod{g_1}\end{cases}$
Now, any divisor of $c'$ and $g_1$ is relatively prime to $f'$. By the definition of $g_1$, it must divide $bf'$, but because of relative primeness to $f'$, it must also divide $b$. Similarly with the second congruence. Uniqueness is also not hard to show, once we have the correct definition for $g_1$.
I figured this out by just trying different examples to poke at different parts of it. The one that finally made it clear was adding $M_1=[9,4+18\theta]$ and $M_2=[3,1+6\theta]$. Their sum contains $1=(4+18\theta) -3(1+6\theta)$, but my previous definition of $g_1$ was not capturing that.