Trying to solve a problem, I got stuck on the following question.
If $P\in\mathbb R[X]$ is an irreducible polynomial and $\operatorname {deg} P=2$, then is it true that $\mathbb{R}[X]/(P)\cong\mathbb C?$
I believe it must be true from the bits I've been hearing throughout my studies, but of which I haven't seen proofs. I've been unable to find anything on the internet other than this single page. It tells me that there is something called Weierstrass' theorem which should be relevant, but can't find even the precise statement of the theorem, let alone the proof.
Here's what I have tried.
First of all, I can assume that $P$ is monic, because for any polynomial of degree $2$ there's a monic polynomial generating the same ideal in $\mathbb R[X].$ So I have
$P(x)=x^2+bx+c$
for some real numbers $b,c$ such that $b^2<4c$.
I want to assume there is an isomorphism $\phi:\mathbb R[X]/(P)\to\mathbb C$ and work out what it should look like. I will denote by $[S]$ the coset of $S\in \mathbb R[X]$ in $\mathbb R[X]/(P).$ I'd like to find $S$ such that $\phi([S])=i.$
I have
$ [x^2+bx+c-1]\mapsto -1, $
So the coset that is mapped to $-1$ has to be a square root of $[x^2+bx+c-1]$ in $\mathbb R[X]/(P).$ Such a coset must be represented by a linear polynomial $kx+l$. Thus $(kx+l)^2-(x^2+bx+c-1)\in (P),$
that is $ (kx+l)^2-(x^2+bx+c-1)\equiv a(x^2+bx+c) $
for some real number $a$, and further
$ (k^2-1-a)x^2+(2kl-b-ab)x+l^2-c-ac+1\equiv 0. $
This gives
$ \begin{cases} k^2=1+a\\ l^2=(1+a)c+1\\ 2kl=(1+a)b \end{cases} $
The problem is that solving this gives formulas that are very difficult to handle. And I need to handle them if I want to prove that what I'm constructing is indeed an isomorphism. Is there a better way of doing it?