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

  • 1
    $\begingroup$ It is enough to have $f=g(x^2+1)$, because this says $f\in (x^2+1)$, and we are done. $\endgroup$ – Dietrich Burde Apr 22 '14 at 13:34
  • 1
    $\begingroup$ 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. $\endgroup$ – Daniel Fischer Apr 22 '14 at 13:35
  • $\begingroup$ @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 $\mathbb{Z}$? $\endgroup$ – kslote1 Apr 22 '14 at 13:57
  • $\begingroup$ 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. $\endgroup$ – Daniel Fischer Apr 22 '14 at 14:06
  • $\begingroup$ 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. $\endgroup$ – Hurkyl Aug 9 '17 at 16:40

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


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.


I.e they are isomorphic.

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


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,

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


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

Which is exactly what you want.

The map is obviously bijective and a ring homomorphism.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.