$R＝\Bbb{R}[x,y]/(x＋y-a,x^2-ax＋b)$ is integral domain when $a^2-4b＜0$.

Question

Let $a,b∈\Bbb{R}$ and let $R＝\Bbb{R}[x,y]/(x＋y-a,x^2-ax＋b)$.

I want to prove that if $a^2-4b＜0$ , then $R$ is integral domain.

My try and thougt;

This condition means $x^2-ax＋b$ is irreducible, I think this is condition which says $\Bbb{R}[x]/(x^2-ax＋b)$ is isomorphic to $\Bbb{C}$. I'm struggling how to relate $R$ and $\Bbb{R}[x]/(x^2-ax＋b)$ .Thank you for your help. Just hint is also appreciated.

Answer 1

Once you have that $\mathbb{R}[x]/(x^2-ax+b)\simeq\mathbb{C}$, let $\zeta\in\mathbb{C}$ correspond to $x$ under this identification.

Concretely $\zeta$ is a complex root of $x^2-ax+b=0$.

Then $$ \frac{\mathbb{R}[x,y]}{(x+y-a,x^2-ax+b)}\simeq \frac{\mathbb{C}[y]}{(y+\zeta-a)}\simeq\mathbb{C}. $$