Example 1

Let us apply Proposition 5 of VIII, p. 104, to the specific case $V = B_s$.

a) The mapping $J\mapsto$ PJ is an isomorphism from the ordered set $D(B_s)$ of left ideals of B to the ordered set D(P) of A-submodules of P. The inverse mapping sends an A-submodule M of P to the left ideal J(M) of B consisting of the elements $b$ of B such that M contains $Pb$.

b) The mapping $K\mapsto$ KP is an isomorphism from the ordered set $D(A_d)$ of right ideals of A to the ordered set D(P) of B-submodules of P. The inverse mapping sends a B-submodule V of P to the right ideal K(V) of A consisting of the elements $a$ of A such that V contains $aP$.

Indeed, the A-module $P\otimes_BB_s$ can be canonically identified with P. If J is a left ideal of B, then the canonical image of $P\otimes_BJ$ in $P\otimes_BB_s$ corresponds to PJ through this identification. Consequently, the mapping $J\mapsto$ PJ is an isomorphism of ordered sets from $D(B_s)$ to D(P). Let $J\in D(B_s)$. Denote the set of elements $b$ of B such that PJ contains $Pb$ by $J'$. It is a left ideal of B that contains J, and we have PJ$'\subset$ PJ. Since the mapping $J\mapsto$ PJ is an isomorphism of ordered sets, we necessarily have PJ$'=$ PJ and $J = J'$. This proves a).

Assertion b) follows from assertion a) applied to the invertible $(B^o,A^o)_k$-bimodule P.

Note that the ring A is a field if and only if the B-module P is simple.