Lemma 55.2.4. Let $L$ be a finite free $\mathbf{Z}$-module endowed with an integral symmetric bilinear positive definite form $\langle \ ,\ \rangle : L \times L \to \mathbf{Z}$. Let $A \subset L$ be a submodule with $L/A$ torsion free. Set $B = \{ b \in L \mid \langle a, b\rangle = 0,\ \forall a \in A\} $. Then we have injective maps
\[ A^\# /A \leftarrow L/(A \oplus B) \rightarrow B^\# /B \]
whose cokernels are quotients of $L^\# /L$. Here $A^\# = \{ a' \in A \otimes \mathbf{Q} \mid \langle a, a'\rangle \in \mathbf{Z},\ \forall a \in A\} $ and similarly for $B$ and $L$.
Proof.
Observe that $L \otimes \mathbf{Q} = A \otimes \mathbf{Q} \oplus B \otimes \mathbf{Q}$ because the form is nondegenerate on $A$ (by positivity). We denote $\pi _ B : L \otimes \mathbf{Q} \to B \otimes \mathbf{Q}$ the projection. Observe that $\pi _ B(x) \in B^\# $ for $x \in L$ because the form is integral. This gives an exact sequence
\[ 0 \to A \to L \xrightarrow {\pi _ B} B^\# \to Q \to 0 \]
where $Q$ is the cokernel of $L \to B^\# $. Observe that $Q$ is a quotient of $L^\# /L$ as the map $L^\# \to B^\# $ is surjective since it is the $\mathbf{Z}$-linear dual to $B \to L$ which is split as a map of $\mathbf{Z}$-modules. Dividing by $A \oplus B$ we get a short exact sequence
\[ 0 \to L/(A \oplus B) \to B^\# /B \to Q \to 0 \]
This proves the lemma.
$\square$
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