I have to study the scheme theoretic inverse image of points under the morphism $f: \operatorname{Spec}\mathbb{R}[x,y]/<x^2+y^2-1> \mapsto \operatorname{Spec}\mathbb{R}[x]$ which projects the circle $x^2+y^2-1$ onto the $x-$axis.
Now, we have geometrically that $f^{-1}(y)$ is empty if $|y| > 1$; it has a point if $|y|=1$ and it has two points otherwise. I want to show that this situation is the same for schemes. I'll write $X'$ to denote the scheme theoretic inverse image at any level $y$. The morphism $f$ induces $f^{*}: \mathbb{R}[x] \mapsto \mathbb{R}[x,y]/<x^2+y^2-1>=A$. According to Shafarevich, the closed embedding $Y' \in Y$ is defined by a quotient homomorphism $\mathbb{R}[x] \mapsto \mathbb{R}[x]/I$ (with $Y'$ is a closed subscheme of $Y$ in general; in our case it's a point);
if $f^*(I)A=A$, then $X'=\emptyset$;
otherwise, the scheme $X'=\operatorname{Spec}[A/f^*(I)A]$ is the desidered scheme theoretic inverse image.
Who guarantees that $f^*(I)A$ is an ideal? And how can I finish the exercise?