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Is it true that if $n_1, \dots, n_m \in \mathbf{Z}$ are distinct, then $\mathbf{Z}[x]/((x-n_1)\dots(x-n_m)) \cong \mathbf{Z}^m$? I think the answer is yes, but I'm not sure my argument is correct.

Let $f(x) = (x-n_1)\dots(x-n_m)$ and let $R = \mathbf{Z}[x]/(f(x))$. Since $f(x)$ is monic and the degree of $f(x)$ is $m \geq 1$, $R$ is free over $\mathbf{Z}$ of rank $m$ with basis $\left\{\bar{1}, \bar{x}, \dots, \bar{x}^{m-1}\right\}$. Thus $R \cong \mathbf{Z}^n$.

Is this a correct solution? What if $n_1, \dots, n_m$ aren't necessarily distinct. Is it still true then? It seems so.

user26857
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user313349
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  • What you've shown is that $R$ is isomorphic to $\mathbb{Z}^m$ as an abelian group. But there's a stronger question you could ask (which, if this is homework, is probably the intended question), which is whether $R$ is isomorphic to $\mathbb{Z}^m$ as a ring. – Qiaochu Yuan Feb 14 '16 at 07:05
  • Just use Chinese Remaind Theorem, since $(x-n_i)$ are coprime you can write a product of $m$ copies of $\mathbb Z$. – Maffred Feb 14 '16 at 07:05
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    @Maffred: this does not work. In particular, I claim that $\mathbb{Z}[x]/(x^2 - 1)$ is not isomorphic to $\mathbb{Z}^2$ as a ring. (Reduce $\bmod 2$.) The problem is that $(x - 1)$ and $(x + 1)$ are not coprime in $\mathbb{Z}[x]$, in the sense that they do not generate the unit ideal. – Qiaochu Yuan Feb 14 '16 at 07:06
  • What do you mean? How can CRT doesn't work? – Maffred Feb 14 '16 at 07:09
  • You saying $a(x)(x-1)+b(x)(x+1)=1$ cannot hold since putting $x=2$ makes $2b(1)=1$ which is absurd? – Maffred Feb 14 '16 at 07:11
  • $x=1$ ofc!!!!!!! – Maffred Feb 14 '16 at 07:14
  • @QiaochuYuan I forgot to use the @. Btw it's like to say I can't pretend that two prime ideals are coprime if I'm not in a Dedekind Domain? – Maffred Feb 14 '16 at 07:24
  • @Qiaochu: How does the proof go that $\mathbf{Z}[x]/(x^2-1)$ is not isomorphic to $\mathbf{Z}^2$? – user313349 Feb 14 '16 at 07:35

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