I've just read the definition of a $T$-valued point on an $S$-scheme from Ravi Vakil's notes (i.e. the morphisms $T \to S$), and something about the definition doesn't seem to make sense intuitively.
To make everything as "nice" as possible, I'm going to phrase my question with $\overline{k}$-varieties. Let's say I have the $\mathbb{C}$-variety $V := \text{Spec}(\mathbb{C}[x] / ((x-i)(x-\pi))$. In classical algebraic geometry, one could easily talk about the algebraic points on $V$, i.e. the point $i$. However, using the definition of a $\overline{\mathbb{Q}}$-valued point on $V$, we get $V(\overline{\mathbb{Q}}) = \emptyset$ since there exist no morphisms $\mathbb{C}[x] / ((x-i)(x-\pi)) \to \mathbb{Q}$. Am I missing something here, or is this just an unfortunate side-effect of using schemes?
Since $\overline{\mathbb{Q}}$ is no $\mathbb{C}$-algebra, it doesn't make sense to talk about the $\overline{\mathbb{Q}}$-valued points of a $\mathbb{C}$-variety. Of course you could look at the underlying variety over $\mathbb{Q}$, say, but then there are again no points, as you have observed correctly. Your $\mathbb{C}$-variety has exactly two $\mathbb{C}$-valued points.