I'm reading the Gortz's Algebraic Geometry, proof of Theorem 6.28 (p.159) and some question arises.
Let $k$ be a field, $X$ a $k$-scheme locally of finite type, and let $x \in X$ be a closed point.
And let $K$ be an algebraically closed extension of $k$. And write $X_K = X \otimes_k K$.
Let $y$ be a point of $X_K$ which is lying over $x$.
Then the residue field $\kappa(y)$ is $K$?
My first strategy is using that : https://stacks.math.columbia.edu/tag/0C4Y
There, the Lemma 33.5.1. states that
Lemma 33.5.1. Let $K/k$ be an extension of fields. Let $X$ be scheme over $k$ and set $Y=X_K$. If $y∈Y$ with image $x∈X$, then
(1) $\mathcal{O}_{X,x} → \mathcal{O}_{Y,y}$ is a faithfully flat local ring homomorphism,
(2) with $\mathfrak{p}_0:=\operatorname{Ker}(κ(x)⊗_kK→κ(y))$ we have $κ(y)=κ(\mathfrak{p}_0)$,
(3)$\mathcal{O}_{Y,y}=(\mathcal{O}_{X,x}⊗_k K)_{\mathfrak{p}}$ where $\mathfrak{p}⊂\mathcal{O}_{X,x}⊗_k K$ is the inverse image of $\mathfrak{p}_0$.
(4)we have $\mathcal{O}_{Y,y}/\mathfrak{m}_x\mathcal{O}_{Y,y}=(κ(x)⊗_kK)_{\mathfrak{p}_0}$
We will use the Lemma 33.5.1.-(2).
Here, $\psi : \kappa(x) \otimes_{k} K \to \kappa(y)$ maybe $(a \otimes b) \mapsto \varphi_1(a) \varphi_2(b)$ , where $\varphi_1 : \kappa(x) \to \kappa(y)$, $\varphi_2 : K \to \kappa(y)$ are induced field homomorphisms from $\pi_{1,y}^{\sharp} : \mathcal{O}_{X,x} \to \mathcal{O}_{Y,y}$ and $\pi_{2,y}^{\sharp} : \mathcal{O}_{\operatorname{Spec}K, \pi_{2}(y)} \to \mathcal{O}_{Y,y}$.
( $\pi_1 : Y=X_K \to X, \pi_2 : Y=X_K \to \operatorname{Spec}(K)$)
Now, from the Lemma 33.5.1., we want to show that $\kappa(\mathfrak{p}_0)= K$. Let $A:=\kappa(x) \otimes _k K$. Then to show $\kappa(\mathfrak{p}_0)= K$, is suffices to show that $Q(A/\mathfrak{p}_0) \cong K$, where $Q(A/\mathfrak{p}_0)$ is the quotient field.
Since $x\in X$ is a closed point(assumption), the residue field $\kappa(x)$ is a finite extension of $k$ (his book prop.3.33). In particular, there exists a $k$-embedding $i: \kappa(x) \hookrightarrow K$
Now, I'm trying to construct an isomorphism $\eta : Q(A/\mathfrak{p}_0) \to K$ as
$$\frac{(f_1 \otimes f_2) + \mathfrak{p}_0}{(g_1 \otimes g_2)+\mathfrak{p}_0} \mapsto \
\frac{i(f_1)f_2}{i(g_1)g_2} $$
And I'm stuck at showing that $\eta$ is well-defined homomorphism.
It seems possible to show the injectivity and surjectivity (true?)
Should we try to use that $X$ is locally of finite type over $k$ and $K$ is an algebraically closed? Then, how?
And.. there is also a possibility for the proof that first we construct a ring homomorphism $K \to Q(A/\mathfrak{p}_0)$ and second, showing that the field extension $K \to Q(A/\mathfrak{p}_0)$ is algebraic, and use the algebraically closedness of $K$.
Further progress : My question is, can we choose such a $k$-embedding (above)
$i: \kappa(x) \to K$ such that $\varphi_1 = \varphi_2 \circ i$? If so, it seems possible to construct an isomorphism $K \to Q(A/\mathfrak{p}_0)$ ; i.e., let's define
$$ \eta : K \to Q(A/\mathfrak{p}_0)$$
$$ \frac{f}{g} \mapsto \frac{(1 \otimes f) + \mathfrak{p}_0}{(1 \otimes g) + \mathfrak{p}_0}$$
Note that if our question ($\varphi_1 = \varphi_2 \circ i$) is true, then for $f_1 \in \kappa(x)$ and $f_2 \in K$,
$ (1\otimes i(f_1)f_2) + \mathfrak{p}_0 = (f_1 \otimes f_2) + \mathfrak{p}_0 $, by the definition of $\mathfrak{p}_0 = \operatorname{ker} \psi$.
Then, we maybe show that the $\eta$ is surjective since for
$u := \frac{(f_{1,1} \otimes f_{1,2} + \cdots + f_{n,1}\otimes f_{n,2}) + \mathfrak{p}_0}{(g_{1,1} \otimes g_{1,2} + \cdots + g_{m,1} \otimes g_{m,2} ) + \mathfrak{p}_0} \in Q(A/\mathfrak{p}_0) $ ($A:=\kappa(x) \otimes _k K$),
$$\eta( \frac{i(f_{1,1})f_{1,2} + \cdots + i(f_{n,1})f_{n,2} }{ i(g_{1,1})g_{1,2} + \cdots + i(g_{m,1})g_{m,2} } ) = u$$
(True?)
So, the question $\varphi_1 = \varphi_2 \circ i$ is true?
Or..is there any other simpler method?
Anyway, can anyone help?
This question originates from next proof of the theorem 6.28 in Gortz's book :
Q. Why the underlined statement is true?
Best Answer
Let $y \in X_K$ lying above $x$. In particular, $\underline{y}:\mathrm{Spec}\,\kappa(y) \rightarrow X_K$ is the composition of $i: \mathrm{Spec}\,\kappa(y) \to \mathrm{Spec}\,\kappa(x) \times_k \mathrm{Spec}\,K$ and $\underline{x}_K: \mathrm{Spec}\,\kappa(x) \times_k K \rightarrow X_K$.
Since $x$ is a closed point, the natural morphism $\underline{x}: \mathrm{Spec}\,\kappa(x) \rightarrow X$ is a closed immersion, thus $\underline{x}_K$ is a closed immersion.
Since $x$ is a closed point, $\kappa(x)$ is a finite $k$-algebra, hence $\kappa(x) \otimes_k K$ is a finite $K$-algebra and its spectrum is finite discrete. Thus the set-theoretic image of $\underline{x}_K$ is a finite discrete set of closed points, and in particular $y$ is a closed point.
Therefore $\kappa(y)$ is a finite extension of $K$. Since $K$ is algebraically closed, $\kappa(y)=K$.