Corollary

Let L1, L_2 be twofree A-modules & the same rank n and let f: L_1 \to L_2 be a homomorphism. Let U be the matrix off with respect to bases of L_1 and L_2. For Coker(f) to be a torsion A-module, it is necessary and sufficient that \det(U) \neq 0 and then:*

L_1 and L_2 can be considered as lattices in $V_1 = L_1 \otimes_A K$ and $V_2 = L_2 \otimes_A K$ respectively, f extending to a K-homomorphism $f_{(K)}: V_1 \to V_2$. Then

and to say that $\operatorname{Coker}(f)$ is a torsion A-module means that $\operatorname{Coker}(f_{(K)}) = 0$; now, it amounts to the same to say that $f_{(K)}$ is surjective or that $\det(U) \neq 0$, whence the first assertion. On the other hand, we may write $f(L_1) = u(L_2)$, where $u$ is an endomorphism of $L_2$ of determinant $\det(U)$; as

formula (8) follows from (7) and (6).