Rational points in the field extension

Question

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:

enter image description here

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.

Brenin told you that your classificaiton of ideals is correct. Let me just mention why. You know by the Nullstellensatz that the maximals are just $(x-a,y-b)$. Now, suppose that $\mathfrak{p}$ is non-zero non-maximal. Let $f(x,y)\in\mathfrak{p}$ and factor $f(x,y)$ into irreducibles $f_1(x,y)^{e_1}\cdots f_m(x,y)^{e_m}$. Since $\mathfrak{p}$ is prime, some irreducible factor must lie in $\mathfrak{p}$, let's say that $(f_1(x,y))\subseteq\mathfrak{p}$. — Alex Youcis, Oct 02 '13 at 22:53
Now, if $\mathfrak{p}\ne (f_1(x,y))$ then you have the chain of prime ideals $(0)\subseteq (f_1(x,y))\subseteq \mathfrak{p}$, and since $\dim \mathbb{C}[x,y]=2$ this would imply that $\mathfrak{p}$ is maximal, which is a contradiction. Thus, $\mathfrak{p}=(f_1(x,y))$. Geometrically, hypersurfaces (codimension $1$ closed subvarieties) always have principally generated prime (in fact, this is equivalent). So, the $0$-dimensional primes are just the points $(x-a,y-b)$, the $2$-dimensional subvariety is just $(0)$, and what's left are the one-dimensional subvarieties which, by what I said — Alex Youcis, Oct 02 '13 at 22:56
correspond to principal prime ideals $(f(x,y))$. Of course, the above generalizes to any algebraically closed field. — Alex Youcis, Oct 02 '13 at 22:56

Brenin · Accepted Answer · 2013-10-06T09:22:02.210

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).

This is a very illuminating answer. Thank you very much. But do you mind if I ask some further questions? I am wondering how I can write the ideal $(X^2+Y^2-1,\alpha X+ \beta Y -1)$ of $\mathbb C[X,Y]$ for $\alpha,\beta \in \mathbb C$, $\alpha^2+\beta^2 <1$ into the form $(f(X,Y))$ or $(X-a,Y-b)$. Moreover, why is $(X^2+Y^2-1,\alpha X+ \beta Y -1)$ not rational over $\mathbb R$ by definition (or what is its residue field, why is this residue field not isomorphic to $\mathbb R$)? Thanks again. :) — sunkist, Oct 04 '13 at 09:20
@sunkist: I am sorry for the late reply; I was notified late about your comment. I added just some small insight on how to think about the sentence in brakets. — Brenin, Oct 06 '13 at 09:23
I may still need a little more effort to understand everything, but your answer really helps great. Thank you very much. — sunkist, Oct 10 '13 at 08:15

Rational points in the field extension

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