Proposition 16

(i) Let $0 \to M_1 \xrightarrow{\alpha} M_2 \xrightarrow{g} M_3 \to 0$ be an exact sequence of finitely generated A-modules. Then

(ii) If there exists apseudo-isomorphismfrom M1 to M_2, then $c(M_1) = c(M_2)$.

(iii) *If T is a torsion module, then $c(T) = -c(\chi(T))$.

(iv) If $a \neq 0$ is afractional ideal Of K, then

(v) *If L is a free A-module, then $c(L) = 0$.

To prove (i), consider a free sub-module $L_1$ (resp. $L_3$) of M, (resp. $M_3$) such that $M_1/L_1$ (resp. $M_3/L_3$) is a torsion module. Since $L_3$ is free and $g$ is surjective, there exists in $g^{-1}(L_3)$ a free complement $L_{23}$ of $\operatorname{Ker}(g)$ which is isomorphic to $L_3$ (Algebra, Chapter II, § 1, no. 11, Proposition 21); but $\operatorname{Ker}(g) = f(M_1)$ contains $f(L_1) = L_{12}$ which is free since $f$ is injective. The sum $L_2 = L_{12} + L_{23}$ is direct and $L_2$ is therefore a free submodule of $M$. There is moreover the commutative diagram:

where the rows are exact and the vertical arrows are injections. We therefore obtain from the snake diagram (Chapter I, § 1, no. 4, Proposition 2) the exact sequence:

As $M_1/L_1$ and $M_3/L_3$ are torsion modules, this exact sequence shows first that so is $M_2/L_2$ and then, by virtue of Proposition 10 of no. 5, that:

which proves (i).

Assertions (iii) and (v) are obvious from the definition. We prove (ii). Therefore let $f : M_1 \to M_2$ be a pseudo-isomorphism and let $L_1$ be a free submodule of $M_1$ such that $M_1/L_1$ is a torsion module. We set $L_2 = f(L_1)$; as $\mathrm{Ker}(f)$ is pseudo-zero, it is a torsion module, hence $\mathrm{Ker}(f) \cap L_1 = 0$ and therefore $L_2$ is free. Let $\tilde{f} : M_1/L_1 \to M_2/L_2$ be the homomorphism derived from $f$ by taking quotients; $\mathrm{Ker}(\tilde{f})$ is isomorphic to $\mathrm{Ker}(f)$ and $\mathrm{Coker}(\tilde{f})$ to $\mathrm{Coker}(f)$ and hence $\tilde{f}$ is a pseudo-isomorphism; moreover

is a torsion module and so is $f(M_1)/L_2 = \tilde{f}(M_1/L_1)$, hence $M_2/L_2$ is a torsion module and it follows from no. 5, Proposition 10 (ii) that

Finally it remains to prove (iv). Let $x \in K^*$ be such that $a \subset xA$. By considering the exact sequence $0 \to a \to xA \to xA/a \to 0$, we obtain

by (i) and (v). But $xA/a$ is isomorphic to $A/x^{-1}a$, whence, by virtue of (iii),

(no. 5, Proposition 12). This completes the proof.

When $M$ is a lattice of $V$ with respect to $A$, $\chi(M/L) = -\chi(M, L)$ (no. 6, Proposition 14); let $(e_i)_{1 \leq i \leq n}$ be a basis of $L$, $e = e_1 \wedge e_2 \wedge \ldots \wedge e_n$, and

$M_w = a \cdot e$ (notation of no. 6); then $\chi(M, L) = \operatorname{div}(a)$, whence $c(M) = c(\operatorname{div}(a))$, which generalizes Proposition 16 (v).