Proof.
By Lemma 55.2.3 the rank of $A$ is $n - 1$. The composition
\[ \mathbf{Z}^{\oplus n} \xrightarrow {\text{diag}(m_1, \ldots , m_ n)} \mathbf{Z}^{\oplus n} \xrightarrow {(a_{ij})} \mathbf{Z}^{\oplus n} \xrightarrow {\text{diag}(m_1, \ldots , m_ n)} \mathbf{Z}^{\oplus n} \]
has matrix $a_{ij}m_ im_ j$. Since the cokernel of the first and last maps are torsion of order prime to $\ell $ by our restriction on $\ell $ we see that it suffices to prove the lemma for the matrix with entries $a_{ij}m_ im_ j$. Thus we may assume $m = (1, \ldots , 1)$.
Assume $m = (1, \ldots , 1)$. Set $V = \{ 1, \ldots , n\} $ and $E = \{ (i, j) \mid i < j\text{ and }a_{ij} > 0\} $. For $e = (i, j) \in E$ set $a_ e = a_{ij}$. Define maps $s, t : E \to V$ by setting $s(i, j) = i$ and $t(i, j) = j$. Set $\mathbf{Z}(V) = \bigoplus _{i \in V} \mathbf{Z}i$ and $\mathbf{Z}(E) = \bigoplus _{e \in E} \mathbf{Z}e$. We define symmetric positive definite integer valued pairings on $\mathbf{Z}(V)$ and $\mathbf{Z}(E)$ by setting
\[ \langle i, i \rangle = 1\text{ for }i \in V, \quad \langle e, e \rangle = a_ e\text{ for }e \in E \]
and all other pairings zero. Consider the maps
\[ \text{d} : \mathbf{Z}(V) \to \mathbf{Z}(E), \quad i \longmapsto \sum \nolimits _{e \in E,\ s(e) = i} e - \sum \nolimits _{e \in E,\ t(e) = i} e \]
and
\[ \text{d}^*(e) = a_ e(s(e) - t(e)) \]
A computation shows that
\[ \langle d(x), y\rangle = \langle x, \text{d}^*(y) \rangle \]
in other words, $\text{d}$ and $\text{d}^*$ are adjoint. Next we compute
\begin{align*} \text{d}^*\text{d}(i) & = \text{d}^*( \sum \nolimits _{e \in E,\ s(e) = i} e - \sum \nolimits _{e \in E,\ t(e) = i} e) \\ & = \sum \nolimits _{e \in E,\ s(e) = i} a_ e(s(e) - t(e)) - \sum \nolimits _{e \in E,\ t(e) = i} a_ e(s(e) - t(e)) \end{align*}
The coefficient of $i$ in $\text{d}^*\text{d}(i)$ is
\[ \sum \nolimits _{e \in E,\ s(e) = i} a_ e + \sum \nolimits _{e \in E,\ t(e) = i} a_ e = - a_{ii} \]
because $\sum _ j a_{ij} = 0$ and the coefficient of $j \not= i$ in $\text{d}^*\text{d}(i)$ is $-a_{ij}$. Hence $\mathop{\mathrm{Coker}}(A) = \mathop{\mathrm{Coker}}(\text{d}^*\text{d})$.
Consider the inclusion
\[ \mathop{\mathrm{Im}}(\text{d}) \oplus \mathop{\mathrm{Ker}}(\text{d}^*) \subset \mathbf{Z}(E) \]
The left hand side is an orthogonal direct sum. Clearly $\mathbf{Z}(E)/\mathop{\mathrm{Ker}}(\text{d}^*)$ is torsion free. We claim $\mathbf{Z}(E)/\mathop{\mathrm{Im}}(\text{d})$ is torsion free as well. Namely, say $x = \sum x_ e e \in \mathbf{Z}(E)$ and $a > 1$ are such that $ax = \text{d}y$ for some $y = \sum y_ i i \in \mathbf{Z}(V)$. Then $a x_ e = y_{s(e)} - y_{t(e)}$. By property (2) we conclude that all $y_ i$ have the same congruence class modulo $a$. Hence we can write $y = a y' + (y_1, y_1, \ldots , y_1)$. Since $\text{d}(y_1, y_1, \ldots , y_1) = 0$ we conclude that $x = \text{d}(y')$ which is what we had to show.
Hence we may apply Lemma 55.2.4 to get injective maps
\[ \mathop{\mathrm{Im}}(\text{d})^\# /\mathop{\mathrm{Im}}(\text{d}) \leftarrow \mathbf{Z}(E)/(\mathop{\mathrm{Im}}(\text{d}) \oplus \mathop{\mathrm{Ker}}(\text{d}^*)) \rightarrow \mathop{\mathrm{Ker}}(\text{d}^*)^\# /\mathop{\mathrm{Ker}}(\text{d}^*) \]
whose cokernels are annihilated by the product of the $a_ e$ (which is prime to $\ell $). Since $\mathop{\mathrm{Ker}}(\text{d}^*)$ is a lattice of rank $1 - n + e$ we see that the proof is complete if we prove that there exists an isomorphism
\[ \Phi : M_{torsion} \longrightarrow \mathop{\mathrm{Im}}(\text{d})^\# /\mathop{\mathrm{Im}}(\text{d}) \]
This is proved in Lemma 55.2.5.
$\square$
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