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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 $f \in \mathbb{Z}[x]\rightarrow g \in \mathbb{Z}[x]$?

kslote1
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    It is enough to have $f=g(x^2+1)$, because this says $f\in (x^2+1)$, and we are done. – Dietrich Burde Apr 22 '14 at 13:34
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    By the division algorithm, $f = q\cdot (x^2+1) + r$ with $\deg r < 2$. Then you just need to check that $r(i) = 0$ implies $r = 0$ if $\deg r < 2$. (Note: $\deg 0 = -\infty$) You could also appeal to Gauß' lemma. – Daniel Fischer Apr 22 '14 at 13:35
  • @DanielFischer, I think that is the answer I am looking for since $g \in \mathbb{R}[x]$ which is a euclidean domain and I am allowed to make that claim. Do you care to expand on that? As in, how can I be certain that the coefficients of $q$ are in $\mathbb{Z}$? – kslote1 Apr 22 '14 at 13:57
  • Don't use $\mathbb{R}[x]$ here, use $\mathbb{Q}[x]$. But you need never leave $\mathbb{Z}[x]$ even potentially. The point is that $x^2+1$ is monic, i.e. has lead coefficient $1$. Thus if you do polynomial division, you always get an integer coefficient, and the existence of $q,r\in\mathbb{Z}[x]$ with $f = q\cdot (x^2+1) + r$ and $\deg r < 2$ follows. – Daniel Fischer Apr 22 '14 at 14:06
  • As an aside, this is one of those problems where it's easier to show the isomorphism directly, by writing down a homomorphism and its inverse, rather than the indirect route of finding a surjective homomorphism with zero kernel. –  Aug 09 '17 at 16:40

3 Answers3

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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 $\ker \phi \subseteq \langle x^2+1\rangle$.

Christoph
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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

Community
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Ellya
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You can define your isomorphism as follows

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

$\phi(a+bi)=(a+bx)+\langle x^2+1 \rangle$

If you compute $[a+bx+\langle x^2+1 \rangle]\times [c+dx+\langle x^2+1 \rangle]$

We get $ac+(ad+bc)x+bdx^2+\langle x^2+1 \rangle$. We reduce by long division to get,

$(ac-bd)+(ad+bc)x+\langle x^2+1 \rangle$.

Therefore,

$\phi([a+bi][c+di])=(ac-bd)+(ad+bc)x+\langle x^2+1 \rangle$.

Which is exactly what you want.

The map is obviously bijective and a ring homomorphism.