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Let R and S be rings . Consider K an ideal in RS. Show that there exist an ideal I in R and J in S such that K=IJ . How can I proceed this ?

Moumita Ghosh
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  • What is $RS$? Is it suppose to be the direct product of the rings $R$ and $S$? – Mike Sep 11 '19 at 14:57
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    it is the cartesian product – Moumita Ghosh Sep 11 '19 at 15:00
  • I am not sure this claim is true. For instance, consider $\mathbb{Z}$ as a group. Define multplication on $\mathbb{Z}$ by $ab = 0$ for any $a,b \in \mathbb{Z}$. Then, this group can be thought of as a ring. Moreover, the ideals are precisely the subgroups of $\mathbb{Z}$ which have the form $n\mathbb{Z}$. Now, in the direct product ring $\mathbb{Z} \times \mathbb{Z}$, the diagonal subgroup ${ (a,a): a \in \mathbb{Z} }$. This is a subgroup (and hence an ideal of $\mathbb{Z} \times \mathbb{Z}$), but cannot be realized as a direct product of subgroups of $\mathbb{Z}$. – Mike Sep 11 '19 at 15:20
  • Hello: next time, please do a search before asking. I found the duplicate by searching "ideals product of rings". And when you ask, please choose a better title. "question of rings and ideal" is just inane when you could have basically just used the body of your question as the title, and it would have been much better. The things you used for your title should rather be added to the tags. – rschwieb Sep 11 '19 at 15:33

1 Answers1

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I am going to assume that both $R$ and $S$ are unary rings. The first thing to notice is that if $ (k_1,k_2) \in K $, then $ (k_1,0),(0,k_2) \in K $. Thus $ K $ is of the form $ K = \{ (a,b) : a\in I, \; b\in J \} $, where $ I \subseteq R $ and $ J \subseteq S $. All that remains is to show that $ I $ and $ J $ are indeed ideals.

Erin
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  • what about the ring I proposed in the comments above. I believe that $K \subset I \times J$ but not conversely. – Mike Sep 11 '19 at 15:24
  • You are correct. I think the problem disappears if we assume that R and S are unary. – Erin Sep 11 '19 at 15:31