Understand a quotient Suppose to work in $N=\mathbb{Z}^4$, with $e_1,\ldots,e_4$ its canonical basis. Suppose moreover I consider the element $$t=(1,1,-2,-1),$$ and I want to find the quotient lattice $N/\mathbb{Z}t$: I can consider the exact sequence
$$ 0 \rightarrow \mathbb{Z}t \rightarrow N \rightarrow A \rightarrow 0; $$
I suppose that the first map is given by the $4\times 1$-matrix $(1,1,-2,-1)^T$. I know I should end up with a $3$-dimensional lattice $A\simeq \mathbb{Z}^3$: what I don't understand is how to find $A$ in practice, and how the construct the explicit isomorphism $A\simeq \mathbb{Z}^3$. The problem also is that looking online I should use Smith normal form of (some) matrix, but I don't know which since my first matrix is not even squared.
In conclusion, I'd like to see, with this example, how to show explicitly $A\simeq \mathbb{Z}^3$ (with the explicit map).
I am very sorry, this is my first attempt in this topic and I really need to understand it, apologize for some mistakes.
 A: Hint: $\{t,e_2,e_3,e_4 \}$ is a $\mathbb Z$-basis for $\mathbb Z^4$.
A: First of all, Smith normal form does not require square matrices - it can work with any size of matrix.
So, you do indeed start with the matrix $A = \begin{bmatrix} 1 \\ 1 \\ -2 \\ -1 \end{bmatrix}$.  This is easy to put into Smith normal form by subtracting the appropriate multiples of the first row from the other rows.  If you apply this with the augmented matrix, you find that:
\begin{align*}
\begin{bmatrix} 1 & 0 & 0 & 0 \\
-1 & 1 & 0 & 0 \\
2 & 0 & 1 & 0 \\
1 & 0 & 0 & 1 \end{bmatrix}
A
\begin{bmatrix} 1 \end{bmatrix} =
\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}.
\end{align*}
Now, you just need to follow along with the general proof that the cokernel of a matrix is isomorphic to the cokernel of its Smith normal form: $\operatorname{cok}(A) \simeq \operatorname{cok}(\begin{bmatrix} 1 & 0 & 0 & 0 \end{bmatrix}^T)$ via the map $x + \operatorname{im}(A) \mapsto Px + \operatorname{im}(\begin{bmatrix} 1 & 0 & 0 & 0 \end{bmatrix}^T)$ where $P$ is the $4\times 4$ matrix in the expression above.
Then, you just need to combine this with the canonical isomorphism $\operatorname{cok}(\begin{bmatrix} 1 & 0 & 0 & 0 \end{bmatrix}^T) \simeq \mathbb{Z}^3$, $e_2 + \operatorname{im}(\cdots) \mapsto e_1$, $e_3 + \operatorname{im}(\cdots) \mapsto e_2$, $e_4 + \operatorname{im}(\cdots) \mapsto e_3$.  In this way, you get an exact sequence
$$ \mathbb{Z} \overset{A}{\longrightarrow} \mathbb{Z}^4 \overset{\begin{bmatrix} -1 & 1 & 0 & 0 \\ 2 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{bmatrix}}{\longrightarrow} \mathbb{Z}^3 \to 0. $$
