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.

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