Show that $\mathbb{Z}[x]/(x^2+1)\cong\mathbb{Z}[i]$. Is there an easy way of checking multiplication?

Question

I'm trying to show that $\mathbb{Z}[x]/(x^2+1)\cong\mathbb{Z}[i]$ (to be clear, I'm really taking $\mathbb{Z}[x]/I$, where $I$ is the ideal generated by $x^2+1$). First, $p,q\in\mathbb{Z}[x]/(x^2+1)$ are considered equivalent iff $x^2+1|p(x)-q(x)$. Let $p(x)-q(x)=\sum_{a=0}^{2n+1}c_ax^a$ where $c_a\in\mathbb{Z}$ (if the order of $p-q$ is even, just take $c_{2n+1}=0$). Then, $p\sim q$ iff $(p-q)(\pm i)=0$, which occurs iff $$\sum\limits_{a=0}^n(-1)^ac_{2a}+\left(\sum\limits_{a=0}^n(-1)^ac_{2a+1}\right)i=0\,\iff\,\sum\limits_{a=0}^n(-1)^ac_{2a}=\sum\limits_{a=0}^n(-1)^ac_{2a+1}=0.$$ Therefore, if we consider the map $\phi:\mathbb{Z}[x]/(x^2+1)\to\mathbb{Z[i]}$ given by $$\phi(r)=\sum\limits_{a=0}^n(-1)^ar_{2a}+\left(\sum\limits_{a=0}^n(-1)^ar_{2a+1}\right)i$$ where $r_a$ are the coefficients of $r$, then $\phi(r)=\phi(j)$ iff $r\sim j$. Moreover, each Gaussian integer has some element of $\mathbb{Z}[x]/(x^2+1)$ which maps to it under $\phi$, so $\phi$ is a bijection. Now, we need to check that $\phi$ is a homomorphism, which would imply that it is an isomorphism. It's easy to check that $\phi(1)=1$, and that $\phi(r+j)=\phi(r)+\phi(j)$. However, checking that $\phi(rj)=\phi(r)\phi(j)$ is a massive pain, and I really would rather not do the computation. I'm wondering if there's an easy way of checking that doesn't involve actually multiplying polynomials by hand and doing some manipulation.

Answer 1

Any element of $\ \mathbb{Z}[x]/I\ $, where $\ I=\left\langle x^2+1\right\rangle\ $, can be written in the form $\ a+bx+I\ $, where $\ a+bx\in \mathbb{Z}[x]\ $ is the common remainder mod $\ x^2+1\ $ of all of the polynomials belonging to the equivalence class $\ a+bx+I\ $. You don't need to write $\ a\ $ and $\ b\ $ out as a sum of coefficients and negative coefficients of some higher degree polynomial in the equivalence class, because you can simply show that \begin{align} (a+bx+I)(c+dx+I)&=(ac-bd)+(ad+bc)x+bd(x^2+1)+I\\ &=(ac-bd)+(ad+bc)x+I\ . \end{align}

Answer 2

Well, the quotient ring $\Bbb Z[x]/\langle x^2+1\rangle$ contains a zero of the polynomial $x^2+1$, namely the residue class of $x$. Call it $[x]$. Then $[x]^2+1 = [x^2+1] = [0]$, i.e., $[x]^2 = [-1]$.

This shows that the arithmetic in $\Bbb Z[x]/\langle x^2+1\rangle$ is the same as in $\Bbb Z[i]$, where $i^2=-1$.