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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?

gerardlouw
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1 Answers1

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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.

Martin Brandenburg
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  • Right, but since $i$ is indeed an element of $\overline{\mathbb{Q}}$, my intuition tells me that it should correspond to a $\overline{\mathbb{Q}}$-valued point of $V$ (if we define $T$-valued points "correctly"). Why is my intuition failing here, or is there perhaps some more general definition of $T$-valued points? – gerardlouw May 23 '15 at 13:15
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    You have to think of $\mathbb{C}[x]/((x-i)(x-\pi))$ as a whole, not just of one piece of it. Put differently, you cannot write down the equation $(x-i) \cdot (x-\pi)=0$ over $\overline{\mathbb{Q}}$. – Martin Brandenburg May 23 '15 at 16:07