Quotient lattice: definition, references

Question

I apologize in advance for this dumb question, but I don't know where to look and I'd like to understand.

Consider the lattice $\mathbb{Z}^n$ (for me a lattice is a free abelian group of finite rank), and consider an element $v=(v_1,\ldots,v_n)\in\mathbb{Z}^n$.

Consider the quotient $\mathbb{Z}^n/\mathbb{Z}v$: I want to prove this is a lattice of rank $n-1$.

Let us first understand what the quotient is: given $x,y\in\mathbb{Z}^n$, $x=(x_1,\ldots,x_n)$, $$x\sim y \iff x_1-y_1=tv_1,\hspace{0.3cm}\ldots,\hspace{0.3cm} x_n-y_n=tv_n,$$ where $t\in \mathbb{Z}$. At this point I get stucked: I can isolate $t=\frac{x_i-y_i}{v_1}$ and put it in the remaining $n-1$ equation, but I don't see how this is still a lattice.

I tried to do a small example, like $\mathbb{Z}^2$, $v=(2,3)$, then $\mathbb{Z}^2/\mathbb{Z}v$ has the relation

\begin{equation} \begin{cases} x_1-y_1=2t \\ x_2-y_2=3t, \\ \end{cases} \end{equation}

hence $x\sim y \iff 3x_1-2x_2=3y_1-2y_2$. But how to continue from this? I guess $$\mathbb{Z}^2/\mathbb{Z}v\simeq \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z},$$ but I don't see why this is a lattice, and of rank $1$.

I know my question is rather vague, but I really struggle with this construction and I need to understand it. Any help would be much appreciate.

Answer 1

In your example, let $w=(1,2)$. Then $\mathbb{Z}^2 = \mathbb{Z}v \oplus \mathbb{Z}w$ and so $$\mathbb{Z}^2 / \mathbb{Z}v = (\mathbb{Z}v \oplus \mathbb{Z}w)/(\mathbb{Z}v \oplus 0) \cong \mathbb{Z}w \cong \mathbb{Z}$$

This exact argument only works in the general case when the entries on $v$ are coprime.

Answer 2

Assuming that $\mathbb{Z}^n/v\mathbb{Z}$ is free and $v\neq 0$.

Why $\mathrm{rk}\;\mathbb{Z}^n/\mathbb{Z}v < n$? If the rank of $\mathbb{Z}^n/\mathbb{Z}v $ were $n$, then you can find a basis of $\mathbb{Z}^n/\mathbb{Z}v $ with $n$ elements, say $f_1,f_2,\ldots , f_n$. This basis would lift to a basis $e_1,e_2,\ldots,e_n$ of $\mathbb{Z}^n$. Express $v$ in this basis as $v=\sum_{i=1}^{n}\alpha_i e_i$. Then in the quotient you get $0=\sum_{i=1}^{n}\alpha_i f_i$ with not all of the $\alpha_i$'s are zero. This is a contradiction.

Why $\mathrm{rk}\;\mathbb{Z}^n/\mathbb{Z}v \geq n-1$? Suppose there are $\alpha_i\in\mathbb{Z}$, not all of them are zero, with $\sum_{i=1}^{n-1}\alpha_i(e_i+v\mathbb{Z})=0$. Then you get $$\sum_{i=1}^{n-1}\alpha_ie_i+\alpha v =0\quad \text{ for some } \alpha \in \mathbb{Z}.$$ If $\alpha =0$, you get a contradiction; otherwise you get $$v=\frac{1}{\alpha}\sum_{i=1}^{n-1}\alpha_ie_i \in \mathbb{Q}^n \quad (*).$$ WLOG, you might assume $\alpha_1\neq 0$, and consider the family $\beta=\{v,e_2,e_3,\ldots,e_{n-1}\}$. Then $\beta$ is free, because any nontrivial linear combination in $\beta$ would contradict $(*)$.