Corollary

Let $(A_i)_{i\in I}$ be a finite family of simple rings and $f$ an isomorphism from A to $\prod_{i\in I}A_i$. For every $i\in I$, there exists a unique element $\varphi (i)$ of $\mathscr{S}$ such that the kernel of pr$_i\circ f$ is $\mathfrak{b}_{\varphi(i)}$. The mapping $\varphi$ is a bijection from I to $\mathscr{S}$; the mapping pr$_i\circ f$ induces an isomorphism $f_i$ from $A/\mathfrak{b}_{\varphi(i)}$ to $A_i$ for every $i\in I$.

Thus, $f$ is the composition of the canonical isomorphism from A to $\prod_{\lambda\in\mathscr{S}}A/\mathfrak{b}_{\lambda}$ and the isomorphism from $\prod_{\lambda\in\mathscr{S}}A/\mathfrak{b}_{\lambda}$ to $\prod_{i\in I}A_i$ deduced from the $f_i($“uniqueness of the decomposition of a semisimple ring into a product of simple rings”).

Let us prove the corollary. Let $i\in I$; denote the kernel of pr$_i\circ f$ by $\mathfrak{b}'_i$. Since the simple ring $A_i$ is isomorphic to $A/\mathfrak{b}'_i$, the two-sided ideal $\mathfrak{b}'_i$ of A is maximal. By Proposition 8, c), there consequently exists a unique element $\varphi (i)$ of $\mathscr{S}$ such that we have $\mathfrak{b}'_i=\mathfrak{b}_{\varphi(i)}$. When passing to the quotient, pr$_i\circ f$ defines an isomorphism $f_i$ from $A/\mathfrak{b}_{\varphi(i)}$ to $A_i$. Moreover, we have $\mathfrak{b}'_i+\mathfrak{b}'_j= A$ if $i\not=j$ and $\cap_{i\in I}\mathfrak{b}'_i= 0$ (cf. I, §8, No. 11, p. 110, Proposition 10). It follows from this and Proposition 8 that $\varphi$ is a bijection from I to $\mathscr{S}$.