Show that $\frac{\mathbb{R}[x]}{\langle x^{2}+1 \rangle}$ is isomorphic to $\mathbb{C}$ without using fundamental theorem.

Question

The task is to prove that $\dfrac{\mathbb{R}[x]}{\langle x^2+1\rangle}$ is isomorphic to $\mathbb{C}$ without using the fundamental theorem, and insteadusing equivalence classes, with the relation $p(x)\sim q(x)$ if $p-q \in N$, where $N$ is the set of polynomials divisible by $x^2+1$.

The equivalence classes looks like $[ax+b]$. The function $f$ is defined from equivalence classes to $\mathbb{C}$ as $f(a+bx)=a+ib$.

I'm stuck at prove the condition $f([p] \cdot [q])=f([p])\cdot f([q])$. Suppose $p(x)=ax+b$ and $q(x)=cx+d$, then we have to prove $f(acx^{2}+(ad+bc)x+bd)=(a+ib)\cdot (c+id)$. But I stuck here. Can anyone help me?

Answer 1

Note that, in $\Bbb R[x]/\langle x^2+1\rangle$, $x^2=-1$. So,$$acx^2+(ad+bc)x+bd=(ad+bc)x+bd-ac,$$and therefore\begin{align}f\bigl((ax+b)(cx+d)\bigr)&=f\bigl(acx^2+(ad+bc)x+bd\bigr)\\&=f\bigl((ad+bc)x+bd-ac\bigr)\\&=(ad+bc)i+bd-ac\\&=(ai+b)(ci+d).\end{align}