$\mathbb R[x]/(f)\cong\mathbb R\times\mathbb R$

Question

I’m trying to use The First Isomorphism Theorem for rings to answer some questions.

I need to show that $$\mathbb R[x]/(f)\cong\mathbb R\times\mathbb R$$ where $f=ax^2+bx+c$ whose discriminant is greater than $0$. I know that all I need to do is define a surjective homomorphism from $\mathbb R[x]\rightarrow \mathbb R\times\mathbb R$ whose kernel is $(f)$. But I have no intuition with this and just need some guidance

Answer 1

Hint: One approach might involve the fact that evaluation maps are ring homomorphisms. To get $\mathbb{R}^2$ you clearly need to evaluate at two points $x \in \mathbb{R}$. What points could these possibly be?

Answer 2

$f(x)=a(x-c_1)(x-c_2)$ for $c_1\neq c_2 \in \Bbb R$ (two real distinct roots).

Send $p(x) \in \Bbb R[x]$ to $(p(c_1),p(c_2)) \in \Bbb R^2$.

The kernel is $(f)$ as $(x-c_1)|p(x)$ and $(x-c_2)|p(x)$ implies $(x-c_1)(x-c_2) | p(x)$ and so $f(x)|p(x)$ etc.