I want to understand the definition of an $K$-rational point of an affine scheme with this example.
Let $f \in R[x,y]$ be a polynomial such that $f=0$ has no solutions in $\mathbb{R}$
$X=\operatorname{Spec}(\mathbb{R}[x,y]/(f))$ is an affine curve or more generally an affine scheme over $\mathbb{R}$. The closed points of are the points in $X$, maximal ideals in $\mathbb{R}[x,y]/(f)$. Under the base change morphism $X_C=X \otimes_\mathbb{R}\mathbb{C} \to X$ there are two closed points in $X_\mathbb{C}$ lying above each point in $X$.
Now a $\mathbb{C}$-rational point is a section of the morphism $X \to \operatorname{Spec}(\mathbb{C})$. I guess I either don't understand the definition correctly or it is wrong, since $\operatorname{Spec}(C)=\operatorname{Spec}(R)=\{(0)\}$, so the rational points would be the same? I read that the closed and rational points of a scheme have correspondence. Where does it come from?
I have almost no previous knowledge of algebraic geometry, so working with schemes is not very natural to me.
Thanks for your answers!
$\DeclareMathOperator{\Spec}{Spec}$
There is no morphism $X \rightarrow \Spec \mathbb C$.
$$\begin{matrix} X_{\mathbb C} & \rightarrow & \Spec \mathbb C \\ p\downarrow & & \downarrow \\ X & \rightarrow & \Spec \mathbb R\end{matrix}$$
Instead, a $\mathbb C$-rational point is (1) a section of the morphism $X_{\mathbb C} \rightarrow \Spec \mathbb C$, or equivalently can be thought of as (2) a morphism $\Spec \mathbb C \rightarrow X$ such that the diagram
$$\begin{matrix} \Spec \mathbb C \\ \downarrow & \searrow \\ X & \rightarrow & \Spec \mathbb R\end{matrix}$$
is commutative. The universal property of fiber products shows that $f \mapsto p \circ f$ is a bijection between morphisms of the first description and morphisms of the second description.
For $A = \mathcal O_X(X)$, the ring-theoretic characterization of this is that a $\mathbb C$-rational point of $X$ is (1) a $\mathbb C$-algebra homomorphism $A \otimes_{\mathbb R} \mathbb{C} \rightarrow \mathbb C$ or (2) an $\mathbb R$-algebra homomorphism $A \rightarrow \mathbb C$.
As for the distinction between closed and rational points: for the $\mathbb R$-scheme $X$, the closed points of $X$ correspond to the maximal ideals of $A$. The $\mathbb R$-rational points can be identified with those closed points (maximal ideals) $\mathfrak m$ for which the residue field $A/\mathfrak m$ is equal to $\mathbb R$. The $\mathbb C$-rational points of $X$ do not correspond to points of the scheme $X$, but rather to points of the scheme $X_{\mathbb C}$
For the $\mathbb C$-scheme $X_{\mathbb C}$, the closed points of $X_{\mathbb C}$ again correspond to the maximal ideals of the global section $A \otimes_{\mathbb R} \mathbb C$. Again, the $\mathbb C$-rational points of $X_{\mathbb C}$ correspond to those maximal ideals of $A \otimes_{\mathbb R} \mathbb C$ for which the residue field is $\mathbb C$. Since $\mathbb C$ is algebraically closed, $\mathbb C$-rational points and closed points of $X_{\mathbb C}$ are the same.
Finally, what about $\mathbb R$-rational points of $X_{\mathbb C}$? This is nonstandard terminology in the modern language of schemes, and does not make sense unless we "remember" the $\mathbb R$-scheme $X$ from which $X_{\mathbb C}$ was obtained by base change. If so, then the $\mathbb R$-rational points of $X_{\mathbb C}$ correspond to those maximal ideals of $A \otimes_{\mathbb R} \mathbb C$ whose contraction to $A$ has residue field $\mathbb R$, i.e. is a rational point of $X$. In terms of algebra homomorphisms, the $\mathbb R$-rational points of $X_{\mathbb C}$ consist of those $\mathbb C$-algebra homomorphisms $A \otimes_{\mathbb R} \mathbb C$ whose restriction to $A$ has its image contained in $\mathbb R$.
Although $p: X_{\mathbb C} \rightarrow X$ is surjective and not injective, it is easy to see that it is bijective on the $\mathbb R$-rational points of $X_{\mathbb C}$ and $X$.