Proving $|X_{(\mathbb{F}_{q^n})}|=\sum_{\deg x\mid n}\deg x$

Question

Given a scheme $X$ of finite type over a finite field $k=\mathbb{F}_q$ and a finite field extension $k\subset K=\mathbb{F}_{q^n}$. For any point $x\in X$, denote by $k_x$ the residue field of $X$ at $x$ and $\deg x=[k_x:k]$.
How do we prove that the number of closed points of $X_{(K)}=X\times_k\operatorname{Spec}K$ is equal to $\sum_{\deg x\mid n} \deg x$ where the sum is over all closed points $x$ of $X$ such that $\deg x$ divides $n$ ?
Thank you in advance for your answers!

Answer 1

It suffices to treat one point $x\in X$ at a time. As any closed point of $X_K$ must live over a closed point of $X$, every closed point of $X_K$ is in the fiber $(X_K)_x$ for some $x$. But $(X_K)_x$ fits in the following diagram, where every rectangle is a fiber product:

$$\require{AMScd} \begin{CD} (X_K)_x @>>> X_K @>>> \operatorname{Spec} K\\ @VVV @VVV @VVV \\ \operatorname{Spec} k(x) @>>> X @>>> \operatorname{Spec} k \end{CD}$$

Thus $(X_K)_x$ is the spectrum of $K\otimes_k k(x)$ and we can just figure out what happens with this tensor product. Because $k$ is perfect, every finite extension is separable, so we can write $k(x)\cong k[t]/(f(t))$ for some separable irreducible polynomial $f$. Then $K\otimes_k k(x)\cong K\otimes_k k[t]/(f(t))\cong K[t]/(f(t))$.

This splits as a product of field extensions of $K$. We get a $K$-point for each trivial extension among these field extensions, each of which corresponds to a root of $f$ over $K$. But if any root of $f$ is in $K$, all roots of $f$ are, so any time we have a $K$-point we have $\deg f$ of them, and we get a $K$-point iff $f$ has a root over $K$. Such a root generates a field extension of $k$ of degree $n$ - since there's only one of these up to isomorphism, $f$ has a root in $K$ iff $K$ contains a subextension of degree $\deg f$, which happens iff $\deg x=\deg f\mid n$. Therefore we get $\deg x$ $K$-points over $x$ when $\deg x\mid n$ and no $K$-points otherwise.