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Let $I$ and $J$ be two ideals of a commutative ring $R$ with $1.$ Give a necessary and sufficient condition so that $R/IJ\cong R/I\times R/J.$ Prove your claim. Then decide whether the following ring isomorphisms are true or not: $\mathbb{Q}[x]/\langle x^2-1\rangle\cong \mathbb{Q}[x]/\langle x-1\rangle\times \mathbb{Q}[x]/\langle x+1\rangle,$ $\mathbb{Z}[x]/\langle x^2-1\rangle\cong \mathbb{Z}[x]/\langle x-1\rangle\times \mathbb{Z}[x]/\langle x+1\rangle.$

It has been a while since I took Abstract Algebra and I am preparing for the prelims. I am not sure how to tackle this one. Any help/suggestion/hint will be much obliged. This is what I have so far:

This is a simplified version of the chinese remainder theorem. The map $R\to R/I\times R/J$ defined by $r\mapsto (r+I,r+J)$ is a ring homomorphism with kernel $I\cap J.$ If the ideals $I$ and $J$ are comaximal, then this map is surjective and $I\cap J=I\cdot J$, so $R/(I\cdot J)=R/(I\cap J)\cong R/I\times R/J.$

My question is about the second part of the question, do I have to check whether: $\langle x-1\rangle$ and $\langle x+1\rangle$ are comaximal and $\mathbb{Q}[x]$ and/or $\mathbb{Z}[x]$ are ring homomorphisms? I believe that these ideals are comaximal in both cases but forget how to prove it.

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    @SriPot: Yes, that's exactly what I mea$n$t.2012-06-25

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Note that $I$ and $J$ are comaximal if and only if neither ideal is all of $R$, and $I+J=R$, if and only if $1\in I+J$, if and only if there exist $a\in I$ and $b\in J$ such that $a+b=1$.

In the case of $R=\mathbb{Q}[x]$, what can you say about $\langle x+1\rangle+\langle x-1\rangle = \langle x+1,x-1\rangle$? Does it necessarily contain $1$? (Answer should be easy to spot)

In the case of $\mathbb{R}=\mathbb{Z}[x]$, it is not so easy to spot the answer... If they are comaximal, then any homomorphism that maps, say, $x-1$ to $0$ must map $x+1$ to a unit (can you see why?); conversely, if every homomorphism that maps $x-1$ to $0$ also maps $x+1$ to a unit, then they are comaximal (can you see why?). Now, consider the map $\mathbb{Z}[x]\to\mathbb{Z}$ which is given by "evaluation at $1$". The map will send $x-1$ to $0$...

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    To answer your first question: Yes, $x+1\in \langle x+1 \rangle $ and $x-1\in \langle x-1\rangle $ , so over $\mathbb{Q}$ we have $1/2(x+1)-1/2(x-1)=1\in \langle x+1,x-1\rangle.$2012-06-25