# Ring Isomorphism :$\Bbb{Z}[i]/\langle p\rangle\simeq \Bbb{Z}[x]/\langle x^2+1,p\rangle$

I would like, given a prime number $p$ in $\Bbb{N}$, to prove that $$\Bbb{Z}[i]/\langle p\rangle\simeq \Bbb{Z}[x]/\langle X^2+1,p\rangle$$

Given this, I can conclud with the third isomorphism theorem because $\Bbb{Z}[i]\simeq \Bbb{Z}[x]/\langle x^2+1\rangle$ which is a consequence of the euclidean algorithm.

But I would like to use the first one (just for "training"), Here we go :

Let $\phi:\Bbb{Z}[x]\to \Bbb{Z}[i]/\langle p\rangle$ such that $\phi(f(x)=f(i)+(p).$

1. $\phi$ is surjective

Let $\overline{x}\in\Bbb{Z}[i]/\langle p\rangle$, I have to find an element of $g\in\Bbb{Z}[x]$ such that $\phi{(p(x))}=\overline{x}.$ Such a polynom is $(a+bx)+p(c+dx)$ because $\phi\bigl((a+bx)+p(c+dx)\bigr)=(a+bi)+p(c+di).$

1. $\ker\phi=\{f(x)\in\Bbb{Z}[x]: \phi(f(x))=\overline{0} \}.$

Now I am not sure how can continue, I now that if $f(i)=0$ then $f(-i)=0$; Plus, $x^2+1$ can be view as polynomial in $\Bbb{Z}[x]$ and monic, I can perform the euclidean algorithm. But I am stuck proving that ther kernel is include $\langle x^2+1\rangle+\langle p\rangle:=\langle x^2+1,p\rangle.$

• Your analysis of the situation in #2 is not correct. After all, $X^2+1$ is certainly in the kernel, isn’t it? So the kernel is definitely larger than $\Bbb Z[X]\cap (p)$. – Lubin Nov 25 '15 at 15:26
• @Lubin clearly Yep, I don't know why I wrote that. – JeSuis Nov 25 '15 at 15:29

• thansk, I like the idea, but I cannot justify that the conditions implies that $h(x^2)-wp$ is a multiple of $x^2+1$. We have $h(-1)=0$ ok but next ? Can you give more details? Please. – JeSuis Nov 25 '15 at 16:26