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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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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

Lemma. Let $(A_i)_{i=1,\ldots,n}$ be a (finite) family of rings with $1$. Then ideals of $A = \prod_{i=1}^n A_i$ are of the form $I_1 \times \ldots\times I_n$, where $I_i$ is an ideal of $A_i$ for each $i\in\{1,\ldots,n\}$.

Proof. By induction on $n$, the case $n=0$ beeing trivial, one sees that it suffices to prove the assertion for $n=2$. Let $K$ be an ideal of a product $A\times B$ of two rings with $1$, and note $p : A\times B\to A$ and $q : A\times B\to B$ the two canonical projections. Then $I = p^{-1}(K)$ (resp. $J = q^{-1}(K)$) is an ideal of $A$ (resp. of $B$.) (This is general, the inverse image of an ideal by a ring morphism is always an ideal.) Obviously $K\subseteq I\times J$. To show the inverse inclusion, let $(a,b)\in I\times J$. Then $(a,b') \in K$ and $(a',b)\in K$ for some $(a',b')\in A\times B$. Then $(a,b) = (1,0)(a,b') + (0,1)(a',b) \in K$. $\square$

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Let us assume that $R$ has an identity element $1_R$. Let $I$ be a two-sided ideal of $R\times S$. Define $I_R = \left\{r \in R: \, \exists s \in S \, \, \, \text{s.t.} \, \, (r,s) \in I \right\}$ and similarly $I_S = \left\{s \in S: \, \exists r \in R \, \, \, \text{s.t.} \, \, (r,s) \in I \right\}$. Then $I_S, I_R$ are ideals of $R,S$ and $I \subset I_R \times I_S$. Now take $r \in I_R$ and $s \in I_S$. We will show that $(r,s) \in I$. By definition there is some $s' \in A_2$ such that $(r,s') \in I$. Similarly there is some $r' \in A_1$ such that $(r',s) \in I$. Since $I$ is a two-sided ideal, we have that $(1_R,s)(r,s') = (r,ss') \in I$. Similarly, $(r',s)(1_R,s') = (r',ss') \in I$. Hence $(r-r',0) \in I$ and so $(r,s) \in I$.

Remark: As we see, all we need is that only one of $R,S$ has an identity element.

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