Prove that \(\mathbb{R}[x]/(x^2+bx+c)\) is isomorphic to \(\mathbb{R}^2\) or \(\mathbb{C}\) - First try

A few minutes ago I stumbled upon this meme on some math and science forum:

Prove that Rx/(x^2+bx+c) is isomorphic to R^2 or C

Here is my thought process:

\(b^2-4c\) is just the discriminant of the polynomial \(f(x) = x^2+bx+c\), and I will refer to it from now on as \(\Delta(f)\).

We know that if \(\Delta(f)>0\) then the formula \(\frac{-b\pm\sqrt{b^2-4c}}{2} = -\frac{b}{2} \pm \sqrt{\left(\frac{b}{2}\right)^2 - c}\) gives us the zeros \(x_1, x_2 \in \mathbb{R}\) of \(f\). Let’s look at \(\mathbb{R}[x] / (f)\). And identify each part now.

• The principal ideal \((f) \overbrace{=}^{\mathbb{R}[x] \text{ commutative}} \{ p(x)\cdot \underbrace{f(x)}_{x^2+bx+c \\ = (x-x_1)(x-x_2) } \mid p(x) \in \mathbb{R}[x]\}\)

• The quotient ring \(\mathbb{R} [x]/(f)\), which is the same as \(\mathbb{R}[x]/\sim\) where \(\sim\) is an equivalence relation: \(a\sim b \Leftrightarrow a-b \in (f)\)

  When performing polynomial division \(\frac{p(x)}{f(x)}, \> p(x)\in\mathbb{R}[x]\) if \(\deg(p(x))>2\) we can always perform at least one step and the remainder must be of the form \(ax+b \in \mathbb{R}[x]/(f)\)

The solution to this problem can be found in this blog post