I have the following problem:
Let $R:=\mathbb{R}[X,Y]/(Y-X^2,Y+X)$
i) Show that $R\cong_\text{Ring} \mathbb{R}\times \mathbb{R}$
ii) Conclude that $|\operatorname{Spec}R|=2$. How do these two ideals look like?
So for the first one I got the hint to eliminate one variable at a time and see what happens. Doing that I got:
$1.) \mathbb{R}[X]/(-X^2,X)\cong \mathbb{R}[X]/(-X^2)\times\mathbb{R}[X]/(X)\cong\mathbb{R}[X](-X^2)\times \mathbb{R}$
$2.) \mathbb{R}[Y]/(Y,Y)=\mathbb{R}[Y]/(Y)\cong\mathbb{R}$
using the chinese remainder theorem in $1.)$. But since $\mathbb{R}[X]/(-X^2,X)\times\mathbb{R}[Y]/(Y)\ncong\mathbb{R}[X,Y]/(Y-X^2,Y+X)$ and the fact that there is still $\mathbb{R}[X]/(-X^2)$ left in the first equation, I don't see how this leads me to the desired isomorphism. For the second one I don't see how an isomorphism can lead me to the number of ideals, because I'm pretty sure $A\cong B$ doesn't imply $\operatorname{Spec}A \cong \operatorname{Spec}B$ for rings $A,B$ (I've seen it in a book).
Can someone help me with this problem? Thanks in advance.