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Is $\mathbb{Z}[x]/(x^2+1)$ isomorphic to $\mathbb{Z}[i]$?

My attempt is that try to define a mapping $g$ from $\mathbb{Z}[x]$ to $\mathbb{Z}[i]$ by $g(f(x))= f(i)$, for $f(x)\in\mathbb{Z}[x]$. If it is possible then $\ker g$ is $(x^2+1)$? Am I on the right track? Please Help.

egreg
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EuReka
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    They are isomorphic, and a mapping along the lines you suggest will prove it. – André Nicolas Dec 17 '13 at 08:16
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    Yes, you're on the right track: proving that $\ker g = (x^2 + 1)$ is the way to go. Note that it is easy to see that $x^2 + 1 \in \ker g$. For the other way around, take an element $f$ in $\ker g$ and use the fact that you already know that $x^2 + 1 \in \ker g$ to show that $f$ is equivalent, modulo $\ker g$, to a polynomial of degree at most 1. – Magdiragdag Dec 17 '13 at 09:05

1 Answers1

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You are certainly on the right track. Let's define a homomorphism $\phi:\mathbb{Z}[x] \rightarrow \mathbb{Z}[i]$ as follows:

$$\phi(f(x)) = f(i)$$

Is $\phi$ a surjective homomorphism? Certainly: consider any $a+bi \in \mathbb{Z}[i]$. Note that $(bx + a) \mapsto (a+bi)$. Next, you should convince yourself that $\ker(\phi) = \langle x^2+1 \rangle$. At this point, we simply apply the isomorphism theorem:

$$\mathbb{Z}[x]/\langle x^2+1 \rangle \cong \mathbb{Z}[i]$$

Edward Evans
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Kaj Hansen
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    I can conclude that $<x^2+1> \subset ker(\phi)$. What about the converse? Since in $\mathbb{R}[X]$ division algorithm is valid, one can tell that since $i$ and $-i$ are roots of the polynomial from Kernel, we have $x^2+1$ is a factor of the polynomial. But in $\mathbb{Z}[X]$ this argument is not valid. right? How to conclude that $Ker(\phi) \subset <x^2+1>$ – Saikat Dec 05 '20 at 15:27