Theorem 4

Let $M$ be a non-degenerate sub-A*-module* of $B$. The following properties are equivalent: (a) $M$ is invertible. (b) $M$ is projective. (c) $M$ is projective of rank 1. (d) $M$ is a finitely generated $A$-module and, for every maximal ideal $m$ of $A$, the $A_m$-module $M_m$ is monogenous.

Let us show first the equivalence of properties (a), (b) and (c). If (a) holds and N is sub-A-module of B such that M . N = A, then there is a relation

For all $x \in M$, set $v_i(x) = n_i x$; the $v_i$ are linear forms on M and by (9) $x = \sum_{i=1}^{n} m_i v_i(x)$ for all $x \in M$; this proves (Algebra, Chapter 11, §2, no. 6, Proposition 12) that M is projective and generated by the $m_i$; hence M is a finitely generated projective module.

Let m be a maximal ideal of A; we show that the integer $r = \mathrm{rg}_m(M)$ is equal to 1. Let S' be the image of S in $\mathbf{A}_{m'}$; as the elements of S are not divisors of 0 in A, those of S' are not divisors of 0 in $\mathbf{A}_{m'}$ since $\mathbf{A}_{m'}$ is a flat A-module (§ 2, no. 4, Theorem 1 and Chapter I, §2, no. 4, Proposition 3); then ${S'}^{-1}A_m \neq 0$ and, as $M_m$ is a free $A_m$-module of rank $r$, ${S'}^{-1}M_m$ is a free ${S'}^{-1}A_m$-module of rank $r$. But if T' is the image of A $-m$ in $S^{-1}A$, ${S'}^{-1}A$, (resp. ${S'}^{-1}M_m$) is canonically identified with ${T'}^{-1}(S^{-1}A)$ (resp. ${T'}^{-1}(S^{-1}M)$) (§ 2, no. 3, Proposition 7). Now $S^{-1}M = B$ (Proposition 8 (c)) and hence ${T'}^{-1}(S^{-1}M)$ is a free A-module of rank 1 over ${T'}^{-1}(S^{-1}A)$, which proves that $r = 1$ and shows the implication (a) $\Rightarrow$ (c).

The implication (c) $\Rightarrow$ (b) is trivial. Let us show that (b) $\Rightarrow$ (a). There exists by hypothesis a family (not necessarily finite) $(f_\lambda)_{\lambda \in L}$ of linear forms on M and a family $(m_\lambda)_{\lambda \in L}$ of elements of M such that, for all $x \in M$, the family $(f_\lambda(x))$ has finite support and $x = \sum_{\lambda \in L} m_\lambda f_\lambda(x)$ (Algebra, Chapter II, §2, no. 6, Proposition 12). Since M is non-degenerate, $f_\lambda(x) = n_\lambda x$ for some $n_\lambda \in A$: M by virtue of Proposition 9 of no. 5. Taking x as an element of M $\cap S$ (no. 5, Proposition 8), it is seen that of necessity $n_\lambda = 0$ except for a finite number of indices and $\sum_{\lambda \in L} m_\lambda n_\lambda = 1$. This obviously implies M . (A: M) = A, whence (a).

By virtue of Definition 2 of no. 3, (c) implies (d). Let us show the converse. As M is non-degenerate, its annihilator is zero (Proposition 8 (b)), then so is the annihilator of $M_m$ (§ 2, no. 4, formula (9)). As M, is assumed to be a monogenous &-module, it is therefore free of rank 1 and it then follows from no. 3, Theorem 2 that M is projective of rank 1.