Proposition 4

Let $A, B$ be two commutative rings, $\rho : A \to B$ a ring homomorphism, $E$ an $A$-algebra and $F$ a $B$-algebra. Then the canonical $A$-module isomorphism

(II, § 5, no. 2, Proposition 6) is an $A$-algebra isomorphism.

The verifications are trivial on account of § 1, no. 5.

In particular, the $A$-algebra structure on $B \otimes_A E$, obtained by restricting the ring $B$ of scalars to $A$, is identical with the structure of the algebra $B \otimes_A E$, the tensor product of the $A$-algebras $B$ and $E$.

Finally, if $(A_i, \phi_{ji})$ is a direct system of commutative rings, $(E_i, f_{ji})$ and $(F_i, g_{ji})$ two direct systems of $A_i$-algebras ($§ 1$, no. 6) and $A = \lim \rightarrow A_i$, the canonical $A$-module isomorphism

(II, § 6, no. 3, Proposition 7) is also an $A$-algebra isomorphism, as follows from the definitions.

Examples of tensor products of algebras. (1) Let $A$ be a commutative ring and $M, N$ two $A$-modules; the canonical mapping

(II, § 4, no. 4) is an $A$-algebra homomorphism, as follows from II, § 3, no. 2, formula (5). When $M$ or $N$ is a finitely generated projective $A$-module, we know that this homomorphism is bijective (II, § 4, no. 4, Proposition 4). In particular we recover the definition of the tensor product of two square matrices.

(2) Let $S, T$ be two monoids and $A^{(S)}$ and $A^{(T)}$ the algebras of the monoids $S$ and $T$ over the ring $A$ (III, § 2, no. 6); then there is a canonical $A$-algebra isomorphism

The elements $e_s \otimes e_t$ (resp. $e_{(s,t)}$), where $s$ runs through $S$ and $t$ runs through $T$, form a basis of $A^{(S)} \otimes_A A^{(T)}$ by virtue of II, § 3, no. 7, Corollary 2 to Proposition 7 (resp. of $A^{(S \times T)}$; the desired isomorphism is obtained by mapping $e_s \otimes e_t$ to $e_{(s,t)}$ and it follows from the definitions that this is indeed an algebra isomorphism.