Ideals of a finite product of rings ($I⊴{\prod }_{j=1}^n{R}_j⟹I={\prod }_{j=1}^n{I}_j$)

Every two-sided ideal $I$ of a finite direct product of rings $T={\prod }_{j=1}^n{R}_j$ is ${\prod }_j{I}_j$, where $I_j$ is its image under the $j$th coordinate ring homomorphism. These images are ideals because the coordinate maps are surjective. To prove the reverse inclusion, lift $a_j\in I_j$ to $b_j\in I$ and multiply by the coordinate idempotent $e_j=(0,\dots ,1,\dots ,0)$. Then ${\sum }_j{e}_j{b}_j=(a_1,\dots ,a_n)\in I$. Finiteness matters: in ${\prod }_{j\ge 0}\mathbb{Z}$, the finitely supported sequences form an ideal whose every coordinate image is $\mathbb{Z}$, but which is not the whole product.