= Solution
Write $T=\prod_{j=1}^nR_j$, 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 $I\subseteq\prod_jI_j$. Conversely, let $a=(a_1,\ldots,a_n)\in\prod_jI_j$. 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>, $e_jb^{(j)}\in I$, and
$$
a=\sum_{j=1}^ne_jb^{(j)}\in I.
$$
Therefore
$$
\boxed{I=I_1\times\cdots\times I_n}.
$$
The finite sum is essential to this proof of <ideals of a finite product of rings>.
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