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 $V$?