Rational Points, classical versus modern notion

Question

In classical algebraic geometry, a $\mathbb Q$-rational point on a, say, complex affine variety $V\subseteq\mathbb C^n$ is a point $p=(p_1,\ldots,p_n)$ with $\forall i: p_i\in\mathbb Q$. Now, in modern language, a $\mathbb Q$-rational point is a morphism $\operatorname{Spec}(\mathbb Q)\to V$. Clearly, if $V$ is defined as $\operatorname{Spec}(\mathbb C[X_1,\ldots,X_n]/I)$, then $V$ has no $\mathbb Q$-rational points in this language. Of course, if $I$ is generated by polynomials with coefficients in $\mathbb Q$, we could look at $V_{\mathbb Q}=\operatorname{Spec}(\mathbb Q[X_1,\ldots,X_n]/I)$ so $V=V_{\mathbb Q}\times_{\operatorname{Spec}(\mathbb Q)}\operatorname{Spec}(\mathbb C)$ and the classically $\mathbb Q$-rational points are the $\mathbb Q$-rational points of $V_{\mathbb Q}$. However, the object $V_{\mathbb Q}$ now fails to capture all classically $\mathbb C$-rational points. For instance, the scheme $V_{\mathbb Q}=\operatorname{Spec}(\mathbb Q[x]/\langle x(x^2+1)\rangle)$ contains two points, one of which is $\mathbb Q$-rational. However, base-changing it to $\mathbb C$, we get $3$ points, none of which is $\mathbb Q$-rational.

This is somewhat unsatisfactory to me. The classical definition seems so straightforward and clean, and I have a single object $V$ containing all my $\mathbb C$-rational points while it is still possible to easily identify the $\mathbb Q$-rational points. Unfortunately, the usual textbooks don't really elaborate on the merits of modern language with regard to this particular aspect. I would be very grateful if someone could do so.

I guess my question should be phrased as follows: Given a scheme $V$ defined over $\mathbb C$, what is the correct way, in modern language, to identify what I would classically consider to be the $\mathbb Q$-rational points in $V$?

Answer 1

Note that if $V\subset \mathbb{C}^n$ is a complex subvariety, the fact that $V$ has a rational point really depends on the embedding (Take the line $X=0$ and the line $X=\pi$ in a plane).

Thus, the question is whether $V$ has a $\mathbb{C}$-point which happen to lie over a $\mathbb{Q}$-point of $\mathbb{C}^n$.

In other words, you are looking for commutative diagrams $$ \require{AMScd} \begin{CD}Spec\mathbb{C} @>>> Spec\mathbb{Q}\\ @VVV @VVV \\ V @>>> \mathbb{A}_{\mathbb{Q}}^n \end{CD} $$

Answer 2

Let $L$ be a Galois extension of $K$ and $G = \mathrm{Gal}(L/K)$. Let $X$ be a $K$-scheme. Then $G$ acts on the set $X(L)$ of $K$-morphisms $\mathrm{Spec}(L) \to X$, and $X(L)^G = X(K)$. So maybe one should take Galois invariants to define rational points?

I think it is also true that $X_L(L) = X(L)$, where $X_L = X \times_{\mathrm{Spec}(K)} \mathrm{Spec}(L)$ and $X_L(L)$ is the set of $L$-morphisms $\mathrm{Spec}(L) \to X_L$, but it seems somehow harder to see the Galois action on $X_L(L)$ since everything here is living over $L$ rather than over $K$, so it's hard to figure out which of its elements should be called $K$-rational.

For example, with $X = \mathrm{Spec}(\mathbb{Q}[x]/(x(x^2+1))$ like in your question, there are three $\mathbb{Q}$-morphisms $\mathrm{Spec}(\bar{\mathbb{Q}}) \to X$, which means three elements in $X(\bar{\mathbb{Q}})$. The absolute Galois group of $\mathbb{Q}$ fixes only one of them since conjugation interchanges the other two, so we're able to pick out the rational point by taking Galois invariants.

Edit. Two things. First thing is that in general, one doesn't really need $L$ to be Galois over $K$. An $L$-point of a $K$-scheme $X$ is just a point $x \in X$ and an embedding $\kappa(x) \hookrightarrow L$ of the residue field $\kappa(x)$ of $x$ into $L$. It's a $K$-point if this embedding factors through $K$. Equivalently, amongst the elements of $X(L)$, the $K$-points are the ones in the image of the map $X(K) \to X(L)$ induced by the inclusion $K \hookrightarrow L$. Then, when $L$ is Galois over $K$, one can recover this image by taking Galois invariants.

Second thing. I think it might be too much to hope for to identify the $K$-points of an $L$-scheme without somehow remembering that the $L$-scheme came from a $K$-scheme to begin with? Namely, I think if there were a way of doing it, it wouldn't be an isomorphism invariant on the category of $L$-schemes. For example, with $X = \mathrm{Spec}(\mathbb{Q}[x]/(x(x^2+1)))$, notice that $X_{\bar{\mathbb{Q}}}$ is isomorphic as a $\bar{\mathbb{Q}}$-scheme to $Y = \mathrm{Spec}(\bar{\mathbb{Q}}[x]/(x(x^2-1)))$, with the isomorphism given by $x \mapsto ix$ on the level of $\bar{\mathbb{Q}}$-algebras.