Proposition 2

Suppose that B and B' are provided with anti-automorphisms J and I such that

Let E be a left A-module and F a left B-module; put $E' = A' \otimes_A E$ and $F' = B' \otimes_B F$. For every sesquilinear mapping (for J) $\Phi$ of $E \times F$ into $G$, there exists one and only one sesquilinear mapping (for I) $\Phi'$ of $E' \times F'$ into $G'$ such that

for all $a' \in A'$, $b' \in B'$, $x \in E$, $y \in F$.

The uniqueness of $\Phi'$ follows from the fact that the tensor products $a' \otimes x$ and $b' \otimes y$ generate $E'$ and $F'$, respectively. To establish its existence, consider the mapping

from $A' \times E \times B' \times F$ into $G'$. It is obviously $\mathbf{Z}$-multilinear, and, in view of (11) and (15), satisfies $m(a', ax, b', y) = m(a'h(a), x, b', y)$ and $m(a', x, b', by) = a'.u(\Phi(x, y)).h'(b^J){b'}^I = m(a', x, b'h'(b), y)$ ($a \in A, b \in B, a' \in A', b' \in B', x \in E, y \in F$). There therefore exists a $\mathbf{Z}$-bilinear mapping $\Phi'$ of $E' \times F'$ into $G'$ satisfying (16) (chap. III, 2e éd., App. II, no 1, prop. 2). This relation, together with the definition of the module structures of E' and F' by (12), shows, in view of (15), that $\Phi'$ is sesquilinear for I, which completes the proof.

The most important examples of $(A', B')$-bimodules G', provided with $\mathbf{Z}$-linear mappings $u$ from G into G' satisfying (11), are the following:

1) We take for $G'$ the tensor product $A' \otimes_A G \otimes_B B'$ (chap. III, 2nd ed., App. II, No. 9) and for $u$ the mapping

from G into G'. The pair $(G', u)$ thus defined is clearly universal in the following sense: for every (A', B')-bimodule $G'_1$ and every $\mathbf{Z}$-linear mapping $u_1$ of G into $G'_1$ satisfying the analogue of (11), there exists one and only one $\mathbf{Z}$-linear mapping $f$ of G' into $G'_1$ such that $f(a'g'b') = a'f(g')b'$ ($a' \in A'$, $g' \in G'$, $b' \in B'$; in other words $f$ is a homomorphism for the bimodule structures of G' and $G'_1$) and such that $u_1 = f \circ u$.

2) When $A = B = G$ (the structure of (A, A)-bimodule of A being defined by the left and right homotheties), $A' = B'$, and $h = h'$ one can take for G' the ring $A'$ and for $u$ the homomorphism $h$ of A into $A'$.

3) Suppose that $A = B$, $A' = B'$, $h = h'$, that the rings A and A' are commutative, and that the structure of left A-module of G coincides with its structure of right A-module. One can then take for G' the tensor product $A' \otimes_A G$ (the structure of right A'-module of G' coinciding with its structure of left A'-module) and for $u$ the mapping $g \to 1 \otimes g$ ($g \in G$) from G into G'. We shall then say that the bilinear mapping (resp. sesquilinear mapping) $\Phi'$ defined by prop. 1 (resp. prop. 2) is obtained from $\Phi$ by extension of the base ring, or by extension of scalars.

The following is valid equally for bilinear mappings and for sesquilinear mappings; the hypotheses and notations are those of prop. 1 (resp. prop. 2). Given a submodule M of E or F, we shall denote by M' the submodule of E' or F' generated by the canonical image of M.