Discuss $\mathbb R[X]/(aX^2 +bX + c)$ in terms of $\Delta = b^2-4ac$

Discuss $\mathbb R[X]/(aX^2 +bX + c)$ in terms of $\Delta = b^2-4ac$.

I've already found that $\mathbb R[X]/(aX^2 +bX + c) \simeq \mathbb{R} \times \mathbb{R}$ if $\Delta > 0$ and $\mathbb R[X]/(aX^2 +bX + c) \simeq \mathbb{C}$ if $\Delta < 0$.

For the case $\Delta = 0$ I don't really find an suitable morphism. The polynomial has only one root of multiplicity $2$ and is reducible but I can't really see how I have to proceed from here.

I've already read this question but it didn't really help me: Polynomial ring isomorphisms

EDIT: MY ATTEMPT (after reading you suggestions)

Consider the morphism $\psi: \mathbb{R}[X] \to \frac{\mathbb{R}[X]}{(X^2)}$ defined by $\psi(P) = P(X + r)$ where $r$ is the double root of the polynomial $aX^2 + bX + c$.

It is easy to show that the kernel of $\psi$ is the ideal $(X-r)^2$. An element of $\frac{\mathbb{R}[X]}{(X^2)}$ can be written as $P(X) + (X^2)$ where $P(X)$ is a polynomial of $\mathbb{R}[X]$. If we take $Q \in \mathbb{R}[X]$ such that $Q = P(X-r)$, then $\psi(Q) = Q(X+r) + (X^2) = P(X - r + r) + (X^2) = P(X) + (X^2) $. So $\psi$ is surjective and we conclude by using the first isomorphism theorem to find that

$$ R[X]/(aX^2 +bX + c) \eqsim R[X]/ (X^2) $$

Is this enough? or should I add the proof that $\psi$ is an morphism?

Solution 1:

I suspect that this question is essentially a duplicate, but I couldn't find an earlier question addressing it.

Hint If the discriminant $\Delta = b^2 - 4 a c$ is $0$, then the quadratic polynomial $aX^2 + bX + c$ has a double root and so can be written as $a (X - r)^2$ for some $r$. Assuming that $a \neq 0$, our ring is $$\Bbb R[X] / (X - r)^2 ,$$ and by writing $Y := X - r$, we regard our ring as $$\Bbb R[Y] / (Y^2) .$$ It remains to describe this ring and in particular determine whether it is isomorphic to $\Bbb C$, to $\Bbb R \times \Bbb R$, or neither. (See also dual numbers and well as this answer to a related question.)