the Gaussian integers are isomorphic to $\mathbb{Z}[x]/(x^2+1)$

I am trying to prove that $\mathbb{Z}[i]\cong \mathbb{Z}[x]/(x^2+1)$.

My initial plan was to use the first isomorphism theorem. I showed that there is a map $\phi: \mathbb{Z}[x] \rightarrow \mathbb{Z}[i]$, given by $\phi(f)=f(i).$ This map is onto and homorphic. The part I have a question on is showing that the $ker(\phi) = (x^{2}+1)$.

One containment is trivial, $(x^2+1)\subset ker(\phi)$. To show $ker(\phi)\subset (x^2+1)$, let $f \in ker(\phi)$, then f has either $i$ or $-i$ as a root. Sot $f=g(x-i)(x+i)=g(x^2+1).$ How can I prove that $f \in \mathbb{Z}[x]\rightarrow g \in \mathbb{Z}[x]$?

Solution 1:

Let me elaborate on Daniel Fischers comment. You have a ring homomorphism $\phi: \mathbb Z[x]\to\mathbb Z[i]$ given by $x\mapsto i$. Take $f \in \ker \phi$. By the division algorithm, $$ f = q\cdot(x^2+1) + r,$$ where $\deg r < 2$ and $q,r\in\mathbb Z[x]$, since $x^2+1$ is monic. Applying $\phi$ to this equation yields $$ 0 = \underbrace{\phi(q)\cdot(i^2+1)}_0 + \phi(r).$$ Since $r$ is of degree $<2$, we can write $r = ax+b$ for some $a,b\in\mathbb Z$. Then $\phi(r)=0$ gives $$ai+b=0.$$ This equation in $\mathbb Z[i]$ implies $a=b=0$, so we have $r=ax+b=0\in\mathbb Z[x]$ and therefore $$ f = q\cdot (x^2+1) \in \langle x^2+1\rangle. $$ We conclude that $\ker \phi \subseteq \langle x^2+1\rangle$.

Solution 2:

When quotient out an ideal, we consider what happens to the ring when all the elements in the ideal are considered as identity elements.

Now if $x^2+1=0\Rightarrow x=\pm i$ let us take the "+" root.

$\mathbb{Z}[x]/(x^2+1)=\{f\in\mathbb{Z}[x]\,|x^2+1=0\}=\{f\in\mathbb{Z}[x]\,|x=i\}=\{a+bi|a,b\in\mathbb{Z}\}=\mathbb{Z}[i]$

I.e they are isomorphic.

I have answered a similar question here Is this quotient Ring Isomorphic to the Complex Numbers