Polynomial ring isomorphisms Let $f = aX^2 + bX + c \in \mathbb{R}[X]$. I want to show that: 
$\mathbb{R}[X]/(f) \cong \mathbb{C}$ if $b^2 - 4ac < 0$,
$\mathbb{R}[X]/(f) \cong \mathbb{R}[\epsilon]$ if $b^2 - 4ac = 0$,
$\mathbb{R}[X]/(f) \cong \mathbb{R} \times \mathbb{R}$ if $b^2 - 4ac > 0$.
My idea so far is to prove that in the first case, $(f) = (X^2 + 1)$, in the second case $(f) = (X^2)$ , in the third case $(f) = (X^2 - 1)$, which would lead to the desired result. However, I don't know how to prove this or if I'm on the correct road at all. Does anyone have any ideas?  
 A: You're on the right track, but it is not generally the case that the ideal $f$ itself is of one of those three. Luckily, we can fix this by using the right changes of variables. In the following, I will repeatedly use the following fact: If $R$ is a ring, $I$ is an ideal and $\sigma:R\to R$ is an automorphism, then $R/I \cong R/\sigma(I)$. I will also use the terms "change of variables" and "automorphism of $\mathbb R[x]$" interchangably. Note that all isomorphisms are actually $\mathbb R$-algebra isomorphisms, not just ring isomorphisms.
I'll assume $a\neq 0$.
$f=aX^2+bX+c=\frac{1}{4a}(4a^2X^2+4abX+b^2+4ac-b^2)=\frac{1}{4a}((2aX+b)^2+b^2-4ac)$.
We have that $f(X)\mapsto f(2aX+b)$ is an automorphism of $\mathbb R[X]$. After we apply the inverse of this automorphism, we get the ideal $g=(\frac{1}{4a}(X^2+D))$ where $D=b^2-4ac$. Now $\frac{1}{4a}$ is just a unit, so we are actually are dealing with the ideal $g=(X^2+D)$. If $D=0$, then obviously $\mathbb R[x]/(g) \cong \mathbb R[\epsilon]$. If $D\neq 0$, then we make another change of variables via $X \mapsto \sqrt{|D|}X$. After this change of variables, we get the ideal $(X^2-1)$ or $(X^2+1)$, depending on the sign of $D$. The rest should be obvious.
A: You need $a\ne0$, which I'll assume at the outset.
Your task is to find, in each case, a suitable surjective ring homomorphism from $\mathbb{R}[X]$ to the corresponding ring and having $I$ as kernel.
For the case $b^2-4ac>0$, the idea is to consider the two distinct (real) roots $\alpha$ and $\beta$ of the polynomial $aX^2+bX+c$ and the homomorphism
$$
\varphi\colon\mathbb{R}[X]\to\mathbb{R}\times\mathbb{R}
$$
defined, for $p(X)\in\mathbb{R}[X]$, by
$$
\varphi(p)=(p(\alpha),p(\beta))
$$
Then $p\in\ker\varphi$ if and only if it is divisible by $X-\alpha$ and $X-\beta$, which is the same as saying it is divisible by $a(X-\alpha)(X-\beta)$. Complete the argument.
For the case $b^2-4ac<0$ it is essentially the same: let $\alpha$ be a (complex) root of $aX^2+bX+c$; then the required homomorphism is
$$
\psi\colon\mathbb{R}[X]\to\mathbb{C}
\qquad
\psi(p)=p(\alpha)
$$
Why is it surjective? Why is its kernel $I$?
Can you find a suitable homomorphism for the case $b^2-4ac=0$?
