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There are a few posts about proving that $\mathbb{Z}[x]/(x^2+1)\simeq \mathbb{Z}[i]$. For example, in this post, the answer requires that we convice ourselves that kernel of the map is $(x^2+1)$. I use this notation for the ideal generated by $x^2+1$. So I am trying to make a complete proof that this.

As in the title, we define the homomorphism as follows. It is surjective since $bx+a \mapsto a+bi$ for any $a,b\in \mathbb{Z}$.

$$\phi: \mathbb{Z}[x] \to \mathbb{Z}[i],\ p(x) \mapsto p(i)$$

We can see that if $p(x) \in (x^2+1)$, then $p(x) = q(x)(x^2+1)$ for some $q(x) \in \mathbb{Z}[x]$. $$\phi(p(x))=\phi(q(x))\phi(x^2+1)=\phi(q(x))(i^2+1)=0$$

The harder part is to show that $\phi(p(x)) \ne 0$ when $p(x) \in \mathbb{Z}[x] \setminus (x^2+1)$, i.e., when $p(x)$ is not a multiple of $x^2+1$.

This is where I am having trouble. Since $\mathbb{Z}[x]$ is not a Euclidean domain, I don't have an explicit form for representing $p(x)=q(x)(x^2+1)+r(x)$, which I thought would be nice, but can't use here.

Updated Attempted Answer

After looking back through my notes, $\mathbb{Z}[x]$ is a unique factorization domain because $\mathbb{Z}$ is as well. This means that any polynomial, $p(x)$, in $\mathbb{Z}[x]$ can be factored into a product of irreducible elements, $q_i(x)$. $$p(x)=q_1(x)q_2(x)...q_n(x)$$

I can show by brute force that $x^2+1$ is irreducible in $\mathbb{Z}[x]$. This means that $$p(x)\in(x^2+1) \iff q_k(x)=x^2+1$$ for some $k \in {1,...,n}$.

I am still having trouble pulling the last bit together, that when $(x^2+1)$ is not a factor of $p(x)$, then this needs to imply that $\phi(p(x))\ne 0$.

jeffery_the_wind
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    You correctly identified the obstacle that the normal Euclidean algorithm won't work smoothly in $\Bbb{Z}[x]$. However, the remedy is also near. You don't need the full Euclidean algorithm to find $q(x)$ and $r(x)$. After all, you are only carrying out the polynomial division once (in other words, you won't need to continue with $r(x)$ as the next divisor). And that's where salvation comes from. Because $x^2+1$ is a monic polynomial, the usual long division by $x^2+1$ still produces the wanted $q(x)$ and $r(x)$. Do check it out! – Jyrki Lahtonen May 03 '20 at 17:10
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    (cont'd) The fact that $\Bbb{Z}[x]$ is not a Euclidean domain manifests itself in that you cannot "continue". You have no control of the leading coefficient of $r(x)$, and hence the next step might take you outside of $\Bbb{Z}[x]$ (into $\Bbb{Q}[x]$). – Jyrki Lahtonen May 03 '20 at 17:10
  • This is a great comment, thanks. I remember reading a lot about special properties of monic polynomials. That must be the key. – jeffery_the_wind May 03 '20 at 18:53
  • @JyrkiLahtonen I went back through my notes, and I updated my answer, drawing on the fact that $\mathbb{Z}[x]$ is a ufd and $x^2+1$ is irreducible. Although it seems to bring me to the same impasse. So I definitely need to use that fact that you are alluding to, but I am not quite seeing it. How can I conclude that a factorization of $p(x)$ exists in the form $p(x)=q(x)(x^2+1)+r(x)$ with $deg(r(x)) < deg(x^2+1)$, like in the euclidean algorithm? – jeffery_the_wind May 04 '20 at 12:20

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By the fundamental theorem of algebra every polynomial $p$ can be written as $p(x)=c \prod (x-a_l)$. So if $p(x) \in \mathbb{R}[x] \supset \mathbb{Z}[x]$, then $p(i)=0$ if and only if $a_j=i$ for some $j$. Now since $0=\overline{p(i)} = p(\overline{i})$, where $\overline{a+ib} = a-ib$ denotes the complex conjugation in $\mathbb{C}$, we also have $a_k=-i$ for $k \neq j$ (because $\deg(p) \geq 2$). Thus $x^2+1$ divides $p$ and hence $p(i)=0$ if and only if $x^2+1$ is a divisor of $p(x)$.

$\textbf{Update:}$ I think I have figured out what your real problem is now. I will explain what is meant by @Jyrki Lahtonen in the comments. You can also write $p(x) \in \mathbb{Z}[x]$ as $p(x) = \sum\limits_{k =0}^n p_k x^k$ with $p_k \in \mathbb{Z}$. Now from what we have seen before $p(i) = 0$ if and only if $x^2+1$ divides $p(x)$ in $\mathbb{R}[x]$. But then the euclidean algorithm in $\mathbb{R}[x]$ is valid for $p(x)$, that is $p(x) = (x^2+1)q(x)$ for $q(x) \in \mathbb{R}[x]$, right? Now note, that \begin{eqnarray} \sum\limits_{k=0}^n p_k x^k \colon (x^2+1) = p_n x^{n-2} + p_{n-1}x^{n-3} + (p_{n-2}-p_{n})x^{n-4} + ... = q(x) \end{eqnarray} So we see, if we carry on with that division, that all coefficients of $q(x)$ are generated by subtracting or summing up some coefficients $p_j$ of $p(x)$. Since all $p_j \in \mathbb{Z}$ by assumption, we deduce, that $q(x) \in \mathbb{Z}[x]$. Thus we have proven \begin{eqnarray} p(x) \in \ker(\phi) \iff p(i) = 0 \iff p(x) \in (x^2+1) \end{eqnarray} and therefore $\ker(\phi) = (x^2+1)$.

h3fr43nd
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    You forgot the leading coefficient: $p(x)= (something) \prod(x-a_l)$. – Batominovski May 03 '20 at 18:07
  • Thanks, I corrected it! :) – h3fr43nd May 03 '20 at 20:05
  • This is a great answer. I need to catch up on my fundamental theorem of algebra. I understand that $p(i)=0 \implies a_j=i$ for some j. However I am not understanding the next part, $0=\bar{p(i)}=p(\bar{i})$. I assume you use the bar for equivalence class, but which equivalence class are you referring to? This is still the piece I'm not quite seeing. Why if $p(x) \notin (x^2+1) \implies \phi(p(x))=p(i)=0$? – jeffery_the_wind May 04 '20 at 12:37
  • Sorry I should have made that more clear! It is not meant to denote some kind of equivalence class, but rather the complex conjugation, i.e. $\overline{a+ib}=a-ib$ for $a,b \in\mathbb{R}$. Do you see why $$0=p(i)=\overline{p(i)}=p(\overline{i}) =p(-i)$$ Is the rest clear then? – h3fr43nd May 04 '20 at 13:34
  • I have updated my answer once more! Let me know, if it is clear now. – h3fr43nd May 05 '20 at 09:34
  • I see that. Well the problem I am seeing is that you are using the euclidean algorithm, but we are not in $\mathbb{R}[x]$, we are in $\mathbb{Z}[x]$ which is not a euclidean domain. I think the point I am getting getting hung up on is: why, when we assume $p(i) \ne 0 \implies p(x) \notin (x^2+1)$. Just wanted to add, thanks for you help. It seems, in the end, if could be proven using polynomial long division? – jeffery_the_wind May 05 '20 at 12:29
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    My point was, that we can use the euclidean algorithm in $\mathbb{R}[x]$ to deduce there is a factorization $p=(x^2+1)q$ and then we can show that $q \in \mathbb{Z}[x]$. The message is, that if $p \in \mathbb{Z}[x]$ with $p(i)=0$, then $p=(x^2+1)q$ with $q \in \mathbb{Z}[x]$, which is as i said what you wanted to have. You do not need the euclidean algorithm in $\mathbb{Z}[x]$, since as you yourself said, that doesn't work. However my solution circumvents that problem, do you see that? – h3fr43nd May 05 '20 at 13:14
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    It is totally irrelevant if you are in $\mathbb{Z}[x]$ at first. We can still go bigger in the beginning and consider the factorization $p = (x^2+1)q$ in $\mathbb{R}[x]$. If $q \in \mathbb{Z}[x]$, which is what we verified, then you know that all this also works out in $\mathbb{Z}[x]$ i.e. we have proven the existence of $q \in \mathbb{Z}[x]$ such that $p = (x^2+1)q$ in $\mathbb{Z}[x]$, so we weren't really ever going bigger than $\mathbb{Z}[x]$. – h3fr43nd May 05 '20 at 13:26