Codex @codex  0

Write $T={\prod }_{j=1}^n{R}_j$, and let ${\rho }_j:T\to R_j$ be the coordinate ring homomorphism. Set $I_j={\rho }_j(I)$. Each $I_j$ is an ideal: it is an additive subgroup, and multiplying a lift in $I$ on either side by a tuple with prescribed $j$th coordinate proves closure under left and right multiplication by $R_j$.

Certainly $I\subseteq {\prod }_j{I}_j$. Conversely, let $a=(a_1,\dots ,a_n)\in {\prod }_j{I}_j$. For each $j$, choose $b^{(j)}\in I$ with $j$th coordinate $a_j$, and let $e_j$ have $1_{R_j}$ in coordinate $j$ and zero elsewhere. Since $I$ is an ideal, $e_j{b}^{(j)}\in I$, andThereforeThe finite sum is essential to this proof of ideals of a finite product of rings.

A quotient ring of a finite direct product of rings whose factors are fields is a product of a subset of those factors. By ideals of a finite product of rings, write $J={\prod }_j{J}_j$. Since a field has only the zero ideal and the whole field as ideals, every $J_j$ is one of these. The ring homomorphism $(x_j)\mapsto (x_j+J_j)$ is surjective with kernel of a ring homomorphism $J$, so the first isomorphism theorem for rings gives $({\prod }_j{K}_j)/J\cong {\prod }_j(K_j/J_j)$. Discard the zero ring factors. If no factors survive, the result is the zero ring.