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Let $k$ be a field, and $X$ a scheme locally of finite type over $k$. Let $\overline{k}$ be the algebraic closure of $k$. Is it true that the set of closed points of $X$ is in bijection with $$\frac{X(\overline{k})}{Gal(\frac{\overline{k}}{ k})}\quad ?$$ I think so, but I don't recall seeing this anywhere before, so either it's not true or I wasn't paying attention. The way the (alleged) correspondence goes is that given a closed point $x \in X$, Zariski Lemma says that $\frac{\kappa(k)}{k}$ is a finite extension, and hence $\kappa(x)$ can be embedded in $\overline{k}$ in $Aut(\frac{\kappa(x)}{k})$ ways, and so that's why I am quotienting out by the Galois action. Then the composite $O_{X, x} \to \kappa(k) \to \overline{k}$ gives rise to a $\overline{k}$-valued point of $X$.

usr0192
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1 Answers1

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This might be too late, but here is a reference for your result. This is Proposition 5.4. in Görtz, Wedhorn, Algebraic Geometry I: Schemes With Examples and Exercises.

Let X be a $k$-scheme locally of finite type. Let $s$ be the unique point in $\mathrm{Spec}(\bar k)$. The map $$\alpha : X(\bar k) \to X, \bar x \mapsto \bar x(s)$$ induces a bijection between the set of $\mathrm{Aut}(\bar k/k)$-orbits in $X(\bar k)$ and the set of closed points of $X$.

Watson
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