If $R_1,..$ are rings with identity and $I$ is an ideal in $R_1\times.. R_n$, then $I=A_1\times...\times A_m$,where each $A_i$ is an ideal in $R_i$.

Question

If $R_1,...,R_n$ are rings with identity and $I$ is an ideal in $R_1\times...\times R_n$, then $I=A_1\times...\times A_m$,where each $A_i$ is an ideal in $R_i$.

My proof:

Given $I$ let $A_k=\pi_k(I)$,where $\pi_k$ is the canonical epimorphism. Since $\pi_k$ is an epimorphism and $I$ is an ideal in $R_1\times...\times R_n$,then $A_k=\pi_k(I)$ is an ideal in $R_k$. And I can get $I$ has the form $A_1\times...\times A_m$.

I don't know why I need the condition that rings $R_i$ have identities.

You've shown each projection of $I$ is an ideal, but how do you conclude from that that the ideal is a direct product? Doesn't work for groups, does it? — Gerry Myerson, Aug 20 '15 at 10:20
Until you write how you "can get $I$ has the form $A_1\times...\times A_m$" we can't know for sure if you have done it without using identities. It is certainly easy if you are given identities. — rschwieb, Aug 20 '15 at 10:20
@haiganghu might be worth answering your own question to explain what you realized then! — rschwieb, Aug 21 '15 at 09:58
@rschwieb I find a link which is exactly my question,http://math.stackexchange.com/a/734501/229474 ,and I think the crucial step is using identities to prove $A_1\times ,...,A_n\subset I$. — haigang hu, Aug 22 '15 at 04:36
For future reference, $R_i$ have identities is used in proving $I\supseteq A_1\times \cdots \times A_n$. I am also adding details: Let $(a_1,…,a_n)\in A_1\times \cdots \times A_n$. WLOG consider first coordinate. Since $a_1\in A=\pi_1(I)$, $\exists (x_1,…,x_n)\in I$ such that $x_1=a_1$. Let $e_1=(1,0,…,0)$. Since $I$ is an ideal, $(x_1,…,x_n)e_1=(a_1,0,…,0)\in I$. Similarly $(0,…,a_i,…,0)\in I$ for each $1\leq i\leq n$. Now add them all, $$\sum_{i=1}^n (0,…,a_i,…,0)=(a_1,…,a_n)\in I.$$ — user264745, Dec 19 '23 at 06:28
I made a typo in my previous comment. Correction: $e_1=(1_{R_1},0,…,0)$ and $e_i=(0,…,1_{R_i},…,0)$, where $1_{R_i}$ is in $i$th position and zero elsewhere. — user264745, Dec 19 '23 at 18:52

If $R_1,..$ are rings with identity and $I$ is an ideal in $R_1\times.. R_n$, then $I=A_1\times...\times A_m$,where each $A_i$ is an ideal in $R_i$.

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