Let $S$ and $T$ be commutative rings each having a multiplicative identity and let $R=S\times T$. Prove that every ideal $I$ of $R$ is of the form $J\times K$ where $J$ is an ideal of $S$ and $K$ is an ideal of $T$.
This problem seems like it should be really straightforward, but I'm not sure I'm interpreting it right. Here are my thoughts:
Elements of $R$ are ordered pairs $(s,t)$ where $s\in S$ and $t\in T$. So if $I$ is an ideal of $R$, then given some $(a,b)\in I$, we have that $(a,b)\cdot(s,t)\in I$. So clearly $a\in S$ and $b\in T$. Then we need to show that, given some ideals $J$ and $K$ of $S$ and $T$ respectively, $a\in J$ and $b\in K$.
Is this the right idea? If so, how can I come to the desired conclusion from here?
