Proposition 1

Let $(A_i)_{i\in I}$ be a family of rings, and for every $i\in I$, let $M_i$ be an $A_i$-module. Set $A =\prod A_i,M =\prod M_i$, and $N =\bigoplus M_i$. Endow M with the structure of an A-module with law of action $((a_i),(x_i))\mapsto (a_ix_i)$. The set N is an A-submodule of M.

a) The mapping $(u_i)\mapsto \prod u_i$ from $\prod$ End$_{\mathbf{Z}}(M_i)$ to End$_{\mathbf{Z}}(M)$ (II, §1, No. 5, p. 200) restricts to ring isomorphisms from $\prod(A_i)_{M_i},\prod(A_i)'_{M_i}$, and $\prod(A_i)''_{M_i}$ to $A_M,A'_M$, and $A''_M$, respectively.

b) The mapping $(u_i)\mapsto \bigoplus u_i$ from $\prod$ End$_{\mathbf{Z}}(M_i)$ to End$_{\mathbf{Z}}(N)$ (II, §1, No. 6, p. 203) restricts to ring isomorphisms from $\prod(A_i)_{M_i},\prod(A_i)'_{M_i}$, and $\prod(A_i)''_{M_i}$ to $A_N,A'_N$, and $A''_N$, respectively.

The ring isomorphisms defined in Proposition 1 are called canonical. The product $\prod(A_i)_{M_i}$ is often identified with $A_M$, and $\prod(A_i)'_{M_i}$ with $A'_M$, etc., through these isomorphisms.

The mapping $\varphi : (u_i)\mapsto \prod u_i$ from $\prod$ End$_{\mathbf{Z}}(M_i)$ to End$_{\mathbf{Z}}(M)$ is an injective ring homomorphism. By the definition of the A-module structure on M, we have $\varphi (\prod(A_i)_{M_i}) = A_M$. Let $u\in A'_M$. For every $i\in I$, denote by $h_i$ the element of A with all components equal to 1 except that of index $i$, which is equal to 0 . If $x$ is an element of M with component of index $i$ equal to 0, then we have $x=h_ix$, and therefore pr$_i(u(x)) =$ pr$_i(u(h_ix)) =$ pr$_i(h_iu(x)) = 0$. Consequently, there exists a unique group homomorphism $u_i: M_i\mapsto M_i$ such that pr$_i(u(y)) =u_i$(pr$_i(y)$) for every $y\in M$. We have $u=\prod u_i$. Since the mapping $u$ is A-linear, the mapping $u_i$ is $A_i$-linear for every $i\in I$. This proves that the image of $\prod(A_i)'_{M_i}$ under $\varphi$ contains $A'_M$; the reverse inclusion is obvious. By applying this to the countermodule of M, we deduce that $\varphi$ induces an isomorphism from $A''_M$ to $\prod_i(A_i)''_{M_i}$. This proves assertion a).

The proof of b) is the same as that of a) mutatis mutandis.