= Ideals of a finite product of rings
{title2=$I\trianglelefteq\prod_{j=1}^nR_j\Longrightarrow I=\prod_{j=1}^nI_j$}
Every two-sided <ideal> $I$ of a finite <direct product of rings> $T=\prod_{j=1}^nR_j$ is $\prod_jI_j$, 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> $e_j=(0,\ldots,1,\ldots,0)$. Then $\sum_je_jb_j=(a_1,\ldots,a_n)\in I$. Finiteness matters: in $\prod_{j\geq0}\mathbb Z$, the finitely supported sequences form an <ideal> whose every coordinate image is $\mathbb Z$, but which is not the whole product.
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