What is finitely generated module?

Example 3.10 (A Smith reduction, in full)

Reduce $M = \begin{pmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{pmatrix}$ over $\Z$. The corner $1$ divides everything: clear its row and column ($L_2 \leftarrow L_2 - 4L_1$, $L_3 \leftarrow L_3 - 7L_1$, then $C_2 \leftarrow C_2 - 2C_1$, $C_3 \leftarrow C_3 - 3C_1$):

In the inner block, the corner $-3$ divides all entries: $L_3 \leftarrow L_3 - 2L_2$ and $C_3 \leftarrow C_3 - 2C_2$ clear it to $\operatorname{diag}(-3, 0)$. Adjusting signs (multiply a row by $-1$, a legal operation):

Cross-check by determinantal divisors: $D_1 = \gcd(\text{entries}) = 1$; every $2\times2$ minor of $M$ is a multiple of $3$ (e.g. $\det\bigl(\begin{smallmatrix}1 & 2\\ 4 & 5\end{smallmatrix}\bigr) = -3$) and one equals $-3$: $D_2 = 3$; $D_3 = \det M = 0$. Hence $d_1 = 1$, $d_2 = 3$, $d_3 = 0$: same answer. Two lessons: a zero invariant factor records the rank drop (the cokernel picks up a free $\Z$ summand), and the divisibility chain $1 \mid 3 \mid 0$ is the Smith certificate — a diagonal reduction that violates the chain (say $\operatorname{diag}(2, 3)$, which the careless can produce from $\bigl(\begin{smallmatrix}2 & 0\\ 0 & 3\end{smallmatrix}\bigr)$ by stopping too early: correct Smith form $\operatorname{diag}(1, 6)$, as $D_1 = 1$ here!) is not finished.