(5)

(b) Let E be a right A-module, F an (A, C)-bimodule and G a right C-module. For every mapping $h \in \mathrm{Hom}_C(E \otimes_A F, G)$, let $h'$ be the mapping of E into $\mathrm{Hom}_C(F, G)$ defined by $(h'(x))(y) = h(x \otimes y)$ for $x \in E, y \in F$. The mapping $h \mapsto h'$ is an isomorphism

(6)

In particular B and C may be taken to be a subring $\Gamma$ of the centre of the ring A; then for every $\Gamma$-module G, the three $\Gamma$-modules

are canonically isomorphic to the $\Gamma$-module of $\Gamma$-*bilinear* mappings of $E \times F$ into G which satisfy (1). More particularly:

(ii) When $E$ is a finitely generated projective $A$-module, the canonical homomorphism (7) is bijective.

(i) Fixing $E$ and $G$, for every left $B$-module $F$, we write

for every left $B$-module homomorphism $u : F \to F'$, we write $T(u) = 1 \otimes u$ (1 here denoting the identity mapping of $\mathrm{Hom}_A(E, G)$),

on the other hand we write $\nu_F$ instead of $\nu$. Then we have the following lemmas:

is commutative.

The verification is immediate.

The commutativity follows from Lemma 1, the other assertions from § 1, no. 6, Corollary 2 to Proposition 6 and § 3, no. 7, Proposition 7.

This follows from Lemma 2 and § 1, no. 6, Corollary 1 to Proposition 6.

Then Lemma 3, together with § 2, no. 2, Proposition 4, shows that it suffices to consider the case where $F$ is a free module. But, if $(b_\mu)$ is a basis of $F$, every element of $\mathrm{Hom}_A(E, G) \otimes_B F$ may then be written uniquely as $\sum_\mu u_\mu \otimes b_\mu$, where $u_\mu \in \mathrm{Hom}_A(E, G)$ (§ 3, no. 7, Corollary 1 to Proposition 7); the image of this element under $\nu$ is the $A$-linear mapping $x \mapsto \sum_\mu u_\mu(x) \otimes b_\mu$; it cannot be zero for all $x \in E$ unless $u_\mu(x) = 0$ for all $x \in E$ and all $\mu$, which is equivalent to saying that $u_\mu = 0$ for all $\mu$; hence $\nu$ is injective. When also $F$ admits a finite basis, Lemma 3 shows (by induction on the number of elements in the basis of $F$) that to prove that $\nu$ is surjective, it suffices to do so when $F = B_s$; but in this case the two sides of (7) are canonically identified with $\mathrm{Hom}_A(E, G)$ (§ 3, no. 4, Proposition 4) and $\nu$ becomes the identity.

(ii) To show the proposition when $E$ is projective and finitely generated, this time we fix $F$ and $G$ and write, for every left $A$-module $E$,

and, for every left $A$-module homomorphism $v : E \to E'$,

on the other hand, we write $\nu_E$ instead of $\nu$. Then we have the two lemmas:

They are proved as Lemmas 1 and 2, taking account of § 1, no. 6, Corollary 2 to Proposition 6, § 2, no. 1, Proposition 1 and § 3, no. 7, Proposition 7.

The remainder of the proof then proceeds as in (i) and is reduced to the case where $E = A_s$; the two sides of (7) are then canonically identified with $G \otimes_B F$ and $\nu$ becomes the identity.

In particular take $B = A$ and $G$ the $(A, A)$-bimodule ${}_sA_d$ ($§ 3$, no. 4), so that the right $A$-module $\mathrm{Hom}_A(E, {}_sA_d)$ is just the dual $E^*$ of $E$ and $({}_sA_d) \otimes_A F$ is canonically identified with $F$ ($§ 3$, no. 4, Proposition 4). Homomorphism (7) then becomes a canonical $\mathbf{Z}$-homomorphism

and $\theta(x^* \otimes y)$ is the linear mapping of $E$ into $F$

Remark (1). The characterization of projective $A$-modules given in $§ 2$, no. 6, Proposition 12, can also be expressed as follows: for a left $A$-module $E$ to be projective, it is necessary and sufficient that the canonical homomorphism

be such that $1_E$ belongs to the image of $\theta_E$.

Even when $E$ and $F$ are both finitely generated, $\theta$ is not necessarily surjective, as is shown by the example $A = \mathbf{Z}, E = F = \mathbf{Z}/2\mathbf{Z}$; the right hand side of (11) is non-zero but $E^* = 0$. On the other hand, examples can be given where $E$ is free, but (11) is neither injective nor surjective (Exercise 3(b)).

When E admits a finite basis $(e_i)$, the inverse isomorphism $\theta^{-1}$ of $\theta$ can be found explicitly as follows. Let $(e_i^*)$ be the dual basis of $(e_i)$ (§ 2, no. 6); for all $u \in \mathrm{Hom}(E, F)$ and all $x = \sum_i \xi_i e_i$ with $\xi_i \in A$,

and therefore $u = \sum_i \theta(e_i^* \otimes u(e_i))$, in other words

In particular, if further $F = E$, it is seen that the image under $\theta_E^{-1}$ of the identity mapping $1_E$ is the element $\sum_i e_i^* \otimes e_i$, which is therefore independent of the basis $(e_i)$ considered in E.

Note on the other hand that when E is a finitely generated projective module the ring structure on $\mathrm{End}_A(E)$ can be transported by $\theta_E^{-1}$ to $E^* \otimes_A E$; it is immediately verified that, for $x, y$ in E, $x^*, y^*$ in $E^*$, in the ring $\mathrm{End}_A(E)$,

Remark (2). Let E be a right A-module; replacing E by $E^*$ in (11), we obtain a canonical $\mathbf{Z}$-homomorphism

On the other hand, there is a canonical A-homomorphism $c_E : E \to E^{**}$, whence there is a $\mathbf{Z}$-homomorphism $c_E \otimes 1_F : E \otimes_A F \to E^{**} \otimes_A F$; composing the latter with the homomorphism (14), we hence obtain a canonical $\mathbf{Z}$-homomorphism

such that $\theta'(x \otimes y)$ is the linear mapping

If E and F are projective modules, the mapping (15) is injective. For $c_E$ is then injective (§ 2, no. 7, Corollary 4 to Proposition 13) and as F is projective, the $\mathbf{Z}$-homomorphism $c_E \otimes 1_F : E \otimes_A F \to E^{**} \otimes_A F$ is also injective (§ 3, no. 7, Corollary 6 to Proposition 7); finally, it has been seen (Proposition 2) that the homomorphism (14) is injective, whence the conclusion.

If E is projective and finitely generated, the mapping (15) is bijective for the two mappings of which it is composed are then bijective (§ 2, no. 7, Corollary 4 to Proposition 13 and Proposition 2 above).

The two mappings $(u, v) \mapsto \mathrm{Tr}(u \circ v), (u, v) \mapsto \mathrm{Tr}(v \circ u)$ of

into $C$ are $C$*-bilinear*; it therefore suffices to verify (20) when $u$ is of the form $x \mapsto \langle x, a^* \rangle b$ and $v$ of the form $y \mapsto \langle y, b^* \rangle a$, with $a \in E, a^* \in E^*, b \in F, b^* \in F^*$. But then $v \circ u$ is the mapping $x \mapsto \langle x, a^* \rangle \langle b, b^* \rangle a$ and $u \circ v$ the mapping $y \mapsto \langle y, b^* \rangle \langle a, a^* \rangle b$. Formula (17) shows that the values of the two sides of (20) are equal to $\langle a, a^* \rangle \langle b, b^* \rangle$.

for $1 \leq i \leq p$ ("invariance of the trace under cyclic permutation").

It suffices to apply (20) to the product

Note that on the other hand it is not necessarily true that

for three endomorphisms $u, v, w$ of E.

It is obviously sufficient to perform the proof for the ordered pairs $(E_1, F_1)$ and $(E_1, E_2)$.

We consider first the case of the ordered pair $(E_1, F_1)$; we fix $E_2, F_1, F_2$ and write for every C-module $T(E) = \operatorname{Hom}(E, F_1) \otimes_c \operatorname{Hom}(E_2, F_2)$ and $T'(E) = \operatorname{Hom}(E \otimes E_2, F_1 \otimes F_2)$ and, for every C-homomorphism $v : E \to E'$,

and

Then Lemmas 4 and 5 (no. 2) (where v is replaced by $\lambda$) are valid and are proved by completely analogous methods.

We next fix $E_2$ and $F_2$ and this time write, for every C-module F,

and $T'(F) = \operatorname{Hom}(C \otimes E_2, F \otimes F_2)$ and, for every C-homomorphism $u : F \to F'$,

and

This time it is immediately verified that Lemmas 1 and 2 (no. 2) (where $\lambda$ always replaces v) are valid.

This being so, we show the proposition first when $E_1 = C$ and $F_1$ is projective and finitely generated. The argument of no. 2 (which rests on Lemmas 1 and 2), together with the above remarks, reduces this to proving the proposition when also $F_1 = C$; then $\operatorname{Hom}(E_1, F_1), E_1 \otimes E_2$ and $F_1 \otimes F_2$ are identified with C, $E_2$ and $F_2$ respectively (§ 3, no. 4, Proposition 4); the two sides of (21)

THE HOMOMORPHISM $\operatorname{Hom}_c(E_1, F_1) \otimes_c \operatorname{Hom}(E_2, F_2)$

are then both canonically identified with $\operatorname{Hom}(E_2, F_2)$ and, after these identifications, it is verified that $\lambda$ becomes the identity.

We now suppose that $F_1$ is projective and finitely generated; the argument of no. 2 (depending this time on Lemmas 4 and 5) reduces the proof for $E_1$ any finitely generated projective module to the case where $E_1 = C$, that is to the first case dealt with.

For the ordered pair $(E_1, E_2)$, the procedure is similar, this time applying Lemmas 4 and 5 twice; we leave the details to the reader.

Note that when $E_1 = C^{(I)}$, $E_2 = C^{(J)}$ are free (finitely generated or not), then $\operatorname{Hom}(E_1, F_1) = F_1^I$, $\operatorname{Hom}(E_2, F_2 = F_2^J$ and

to within canonical isomorphisms and (21) is then identical with a special case of the canonical homomorphism (22) of § 3, no. 7.

When $E_2 = C$, the canonical homomorphism (21) gives, after identifying $\operatorname{Hom}(E_2, F_2)$ with $F_2$ and $E_1 \otimes E_2$ with $E_1$, a canonical homomorphism

for any three $C$-modules $E, F, G$ which is just the homomorphism (7) of no. 2 for $A = B = C$.

Note that when $F = C$ the canonical homomorphism (22) again gives (11) (no. 2) for the case of a commutative ring.

Suppose now that $F_1 = F_2 = C$; as $F_1 \otimes F_2$ is identified with $C$, there is this time a canonical homomorphism

for two $C$-modules $E, F$; for $x^* \in E^*, y^* \in F^*$, the image of $x^* \otimes y^*$ under the canonical homomorphism (23) is the linear form $u$ on $E \otimes F$ such that Moreover, if $E_1, E_2, F_1, F_2$ are four $C$-modules, $f : E_1 \to E_2, g : F_1 \to F_2$ two linear mappings, it follows immediately from (24) that the diagram is commutative.

By linearity, it suffices to consider the case where u₁ is of the form $x_1 \mapsto \langle x_1, x_1^* \rangle y_1$ and u₂ of the form $x_2 \mapsto \langle x_2, x_2^* \rangle y_2$; then the image of $x_1 \otimes x_2$ under $u_1 \otimes u_2$ is by definition

$x_1^* \otimes x_2^*$ being canonically identified under $\mu$ with an element of $(E_1 \otimes E_2)^*$. As $\langle y_1 \otimes y_2, x_1^* \otimes x_2^* \rangle = \langle y_1, x_1^* \rangle \langle y_2, x_2^* \rangle$, formula (26) follows in this case from (17).