Rational points in the field extension I am reading Mumford's The Red Book of Varieties and Schemes
In Section 4 of Chapter 2,

Let $X_0$ be a prescheme over a field $k_0$, and $k$ is a field extension of $k_0$. The prescheme $X$ over $k$ is defined to be $X_0 \times_{\mathrm{Spec}k_0} \mathrm{Spec}k$. 

So if $X_0 = \mathrm{Spec}R$ is affine, then $X =\mathrm{Spec}R \otimes_{k_0}k$. The following example is:

Here are questions I have and points I am in need of help in order to understand


*

*Is the set of prime ideals of $\mathbb C[X,Y]$ $$\{(0)\} \cup \{(X-a,Y-b)|a,b \in \mathbb C\} \cup \{(f(X,Y))| f(X,Y) \text{ is irreducible over } \mathbb C[X,Y]\}?$$ 

*Let $R$ be $\mathbb R[X,Y]/(X^2+Y^2-1)$. So $X = \mathrm{Spec}R_0$ where $R_0 = \mathbb C[X,Y]/(X^2+Y^2-1)$? The prime ideals of $R_0$ are the image of prime ideals of $\mathbb C[X,Y]$ under the canonical map $\mathbb C[X,Y] \rightarrow \mathbb C[X,Y]/(X^2+Y^2-1)$? 

*How can I get the correspondence between the set of prime ideals and points on the surface in the first figure?


Moreover, how is the correspondence between the rational points, other points and the prime ideals obtained? I don't understand the last sentence in bracket.
Thanks for everyone.
 A: *

*Your list of prime ideals in $\mathbb C[x,y]$ is correct! More precisely, the maximal ideals are exactly those of the form $(x-a,y-b)$, with $a,b\in \mathbb C$. The other ones are primes but not maximal. Moreover, $[(0)]$ is the generic point of $\textrm{Spec }\mathbb C[x,y]=\mathbb A^2$, and $[(f)]$ is the generic point of the irreducible affine curve $\textrm{Spec }\mathbb C[x,y]/f\subset \mathbb A^2$.

*Prime ideals of $R_0$ correspond to prime ideals of $\mathbb C[x,y]$ that contain the ideal $(x^2+y^2-1)$. Algebraically, this is just a fundamental isomorphism theorem; geometrically, it reminds us that a closed point $\mathfrak p=[(x-a,y-b)]\in \mathbb A^2$ is in $\textrm{Spec }\mathbb C[x,y]/f$ if and only if $f(a,b)=0$. In other words, if you are acquainted with the interpretation of a scheme as "functions on that scheme": $\mathfrak p\in \textrm{Spec }\mathbb C[x,y]/f \iff f(\mathfrak p)=0$. 

*Let $X=\textrm{Spec }\mathbb C[x,y]/(x^2+y^2-1)$. You can look at its $\mathbb R$-rational points, and at its $\mathbb C$-rational points. You find $X(\mathbb C)=X$, and $$X(\mathbb R)=\{[(x-a,y-b)]\in X\,|\,a,b\in \mathbb R\}=\textrm{the real circle}.$$ In other words, once we know that for a couple $(a,b)\in\mathbb C^2$ the condition $a^2+b^2-1=0$ holds, i.e. we are on the surface (or, further translation: $[(x-a,y-b)]\in X$), we only have two possibilities: either $a,b$ are real (so we are on the circle in the picture), or they are not both real (then we are still on the sphere, but outside that circle).


Added. I find it hard to visualize complex points on a sheet of paper, as they have 4 real coordinates. By the way, here is a small insight: take a real point $P=(\alpha,\beta)\in\mathbb A^2_\mathbb C$. If it is outside the real circle (i.e. $\alpha^2+\beta^2>1$), there are two tangents at the circle, passing through $P$. If it lies on the circle, there is one tangent through $P$, which is given by $\alpha x+\beta y=1$. What if $P$ is inside the circle? you can still consider the equation $\alpha x+\beta y=1$, but the points of tangency are, in this case, (strictly) complex. Now, the sum of the ideals is the ideal of the intersection, thus an ideal of the shape $(x^2+y^2-1,\alpha x+\beta y-1)$ represents the intersection between the real line $\alpha x+\beta y=1$ and the complex circle (you may want to try to do the explicit computation of this intersection). 
It is necessarily rational over $\mathbb C$ and not over $\mathbb R$ by what we said (the tangents passing through an interior point are complex).
