Proposition 12

The mapping $cl : \mathfrak{g} \to P(A)$ defines, by taking quotients, an isomorphism from $\mathfrak{c} = \mathfrak{g}/\theta(B^*)$ onto the kernel of the canonical homomorphism

(no. 4).

In other words, there is an exact sequence

It follows from Proposition 11 of no. 6 and the definition of addition in $P(A)$ that $cl(M.N) = cl(M) + cl(N)$ for $M, N$ in $\mathbf{3}$, which shows that $cl$ is a homomorphism. If $M \in \mathfrak{J}$ is isomorphic to $A$, there exists $b \in B$ such that $M = Ab$ and, as $M$ is invertible, there exists $b' \in B$ such that $b'b = 1$, in other words $b$ is invertible in $B$; the converse is immediate. Hence the kernel of $cl$ in $\mathfrak{J}$ is $\theta(B^*)$.

Let us now determine the image of $cl$. If $M \in \mathfrak{J}$, then $M \otimes_A B = S^{-1}M = B$ (no. 5, Proposition 8 (c)), whence $cl(M) \in \mathrm{Ker}(\phi)$. Conversely, let $P$ be a projective $A$-module of rank 1 such that $P_{(B)} = P \otimes_A B$ is $B$-isomorphic to $B$. As $P$ is a flat $A$-module, the injection $u : A \to B$ defines an injection $u \otimes 1 : P \to P_{(B)} = B$ and $P$ is thus identified with a sub-$A$-module of $B$; by virtue of Proposition 8 (c) of no. 5, $P$ is non-degenerate and Theorem 4 of no. 6 shows that $P$ is invertible. The kernel of $\phi$ is therefore equal to the image of $cl : \mathfrak{J} \to P(A)$.