## Characterization of ideals in a direct product of unital rings

Let $R$ and $S$ be rings with 1. Prove that every ideal of $R \times S$ is of the form $I \times J$ for some ideals $I \subseteq R$ and $J \subseteq S$.

First let $I \subseteq R$ and $J \subseteq S$ be ideals. We know that $I \times J \subseteq R \times S$ is an additive subgroup. Now let $(a,b) \in I \times J$ and $(r,s) \in R \times S$. Then $(a,b)(r,s) = (ar,bs) \in I \times J$ and $(r,s)(a,b) = (ra,sb) \in I \times J$, so that $I \times J$ is a two-sided ideal of $R \times S$.

Now let $K \subseteq R \times S$ be a two-sided ideal. Define $I = \{ a \in R \ |\ (a,b) \in K\ \mathrm{for\ some}\ b \in S \}$ and $J = \{ b \in S \ |\ (a,b) \in K\ \mathrm{for\ some}\ a \in R \}$. We certainly have $K \subseteq I \times J$. Now let $(a,b) \in I \times J$. Then $(a,b^\prime), (a^\prime,b) \in K$ for some $a^\prime$ and $b^\prime$. Then $(a,b) = (1,0)(a,b^\prime) + (0,1)(a^\prime,b) \in K$. Thus $K = I \times J$. It remains to be seen that $I$ and $J$ are ideals of $R$ and $S$, respectively. Certainly both are nonempty. Now suppose $i_1,i_2 \in I$, with $(i_1,s_1), (i_2,s_2) \in K$ for some $s_1,s_2 \in S$. Now if $r \in R$, we have $(r,0)(i_1,s_1) + (i_2,s_2) = (ri_1 + i_2, s_2) \in K$, so that $ri_1+i_2 \in I$. Hence $I$ is an ideal. Similarly, $J$ is an ideal.

Also you should show that $I, J$ are ideals.