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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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Are you sure that some of those $(x - 1)$'s shouldn't be $x + 1$? Not that the quotients are any different. Also, note that over $\mathbb Q$ you have that $(x + 1) - (x - 1)$ is a unit. –  Dylan Moreland Jun 25 '12 at 4:20
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You only gave a sufficient condition for the map to be an isomorphism. You haven't shown that it is also necessary, so you aren't done with the first part of the question, either. –  Arturo Magidin Jun 25 '12 at 4:23
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@SriPot: But the point is: your condition is, so far, only sufficient. If you don't know whether it is necessary or not, then you don't know whether you need to check it to establish the isomorphisms in question. If the condition were to turn out to be only sufficient but not necessary, and you check the examples in the second part and find they fail to satisfy that sufficient condition, what does that tell you? Nothing. –  Arturo Magidin Jun 25 '12 at 4:32
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I'm not asking a question (except retorically). I'm trying to point out that you should not be approaching the second question until you know the answer to the first question. If all you have is a sufficient condition for your conclusion, and you test a given example for that condition, and the examples fails to satisfy the sufficient condition, then that does not tell you whether the example has the desired property or not. For instance: it is sufficient for a number to end in $0$ for that number to be even. But it is not necessary.(cont) –  Arturo Magidin Jun 25 '12 at 5:02
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(cont) So say I run into the number $4$. I notice that it fails my sufficient condition: it does not end in $0$. Now, this failure does not tell me whether $4$ is even or not. It only tells me my test didn't give me any information. You have what has, so far, only been established to be a sufficient condition. Unless and until you show it is necessary as well, it will not necessarily help you with the second part. So the answer to your question, "Do I have to check if... they are comaximal?" Well, unless your condition is necessary, no, you don't "have to". It may tell you nothing. –  Arturo Magidin Jun 25 '12 at 5:04

1 Answer 1

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 $1/2(x+1)-1/2(x-1)=1\in \langle x+1,x-1\rangle.$ –  Lyapunov Jun 25 '12 at 4:57

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