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$R$ and $S$ are two rings. Let $J$ be an ideal in $R\times S$. Then there are $I_{1}$, ideal of $R$, and $I_{2}$, ideal of $S$ such that $J=I_{1}\times I_{2}$.

For me is obvious why $\left\{ r\in R\mid \left(r,s\right)\in J\text{ for some } s\in S\right\}$ is an ideal of $R$ (and the same for $S$) so I can prove $J$ is a subset of the product of two ideals and also that the product of these two ideals is also an ideal.

But I can't see how to prove equality without assuming existence of unity or commutativity. Trying basically to show that if $\left(r,s\right)\in J$ then also $\left(r,0\right),\left(0,s\right)\in J$.

Thank you for your help!

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1 Answer 1

If you have an identity, multiplying by $(1_R,0)$ and $(0,1_S)$ will help. If you don't, then the result need not be true. E.g. if you put the zero product on the abelian group $\mathbb Z_2\oplus \mathbb Z_2$, then the ideal ($=$subgroup) generated by $(\overline 1,\overline 1)$ is a counterexample.

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I dont exactly understand what "put the zero product on the abelian group" mean. can you explain please? and Thank you for your quick answer =) –  IIJ Jan 22 '12 at 2:29
@IIJ: If $(G,+)$ is an abelian group and you define a multiplication by $a\cdot b =0$ for all $a,b\in G$, then $G$ becomes a ring with the zero product. E.g. $\mathbb Z_2$ with the zero product has the usual addition, but $\overline 0\cdot \overline 0 =\overline 1\cdot \overline 0 =\overline 0\cdot \overline 1=\overline 1\cdot\overline 1 = \overline 0$. –  Jonas Meyer Jan 22 '12 at 2:32
Oh yes! Thank you!. have a great day! –  IIJ Jan 22 '12 at 2:44
Hello Sir, I have a question, what if J in this case is prime ideal? how do we prove that its a product of one prime ideal of R and S itself or product of R and prime ideal of S? I came up with the R*S/J being integral domain.. But I'm stuck. Thankyou –  d13 Apr 2 '14 at 10:54
@d13: Sorry, I don't understand your comment. –  Jonas Meyer Jun 27 '14 at 7:21

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