About the ideal of the product of rings If $R_1, \cdots,R_n$ are rings with identities and $I$  is an ideal of $R_1\times\cdots\times R_n$, then $I=A_1\times\cdots\times A_n$ where each $A_i$ is an ideal of $R_i$ ($1\le i\le n$).
What I want to know is that how does the condition "$R_1,\cdots,R_n$ have identities" works in the conclusion? Is the condition necessary?
Is the conclusion necessarily true if the rings don't have identities?
 A: The assumption is needed (it could be relaxed, though). Let $R$ be a nontrivial (additive) abelian group and consider the trivial ring structure on $R$, where $rs=0$ for every $r,s\in R$; then every subgroup of $R\times R$ (which is a trivial ring as well) is an ideal; the subgroup $\{(x,x):x\in R\}$ is an ideal that's not the product of two ideals.

If we want to plunge more deeply in the question, it is sufficient to analyze the case of two rings. Suppose $I$ is an ideal of $R\times S$. Then $A=\{x\in R:(x,0)\in I\}$ is an ideal of $R$ and $B=\{y\in S:(0,y)\in I\{$ is an ideal of $S$. The verification is elementary.
It is also clear that $A\times B\subseteq I$, because $(x,y)=(x,0)+(0,y)$.
If $R$ has an identity, then $A\times\{0\}=I(1,0)=(1,0)I$; indeed, if $x\in A$, then $(x,0)\in I$ and $(x,0)=(x,0)(1,0)$. Similarly if $S$ has an identity. As a consequence, $I=A\times B$.
However, if the rings are not supposed to have an identity, the statement can be false and the counterexample above proves this.
A sufficient condition is that $I(R\times S)=I$.
A: Suppose $I$ is an ideal of $R_1×\cdots ×R_n$ (assume commutative rings for convenience). Define $A_i:=\{a_i|(0,...,0,a_i,0,...,0)\in I\}$. Then $I=A_1×\cdots× A_n$. This is because if $(a_1,...,a_n)\in I$ then $(1,0,...,0)(a_1,...,a_n)=(a_1,0,...,0)\in I$ (this is where identity is required) so $a_1\in A_1$ and similarly $a_i\in A_i$ for each $i$. Conversely if $(a_1,..,a_n)\in A_1\times\cdots \times A_n$ then by definition of $A_i$'s $(0,...,0,a_i,0,...,0)\in I$, adding these we get $(a_1,...,a_n)\in I$. Now $I$ is a group with the additive structure hence each of the $A_i$ 's are closed under addition and similarly closed under taking inverses and contains the zero of each ring, $R_i$. Take an $A_i$ and consider the element $(0,...,0,a_i,0,...,0)\in I=A_1×\cdots ×A_n$ with $a_i\in A_i$. Then $(0,...,0,r_i,0,..,0)(0,...,0,a_i,0,...,0)=(0,...,0,r_ia_i,0,...,0)\in I$ hence $r_ia_i\in A_i$ for all $r_i\in R_i$, showing $A_i$ is an ideal.
