Difficulty understanding the claim: base change according to a purely inseparable field extension is a homeomorphism

Question

$\newcommand{\Frac}{\operatorname{Frac}}\newcommand{\spec}{\operatorname{Spec}}\newcommand{\trdeg}{\operatorname{trdeg}_k}\newcommand{\p}{\mathfrak{p}}\newcommand{\q}{\mathfrak{q}}$In Liu's algebraic geometry proposition $2.7$ we have the following scenario:

Let $k$ be any field, $K/k$ a purely inseparable algebraic field extension. Let $X$ be an algebraic variety over $k$, meaning a scheme covered by finitely many (not necessarily reduced) affine schemes associated to finitely generated $k$-algebras.

Consider the base change $X_K=X\times_{\spec(k)}\spec(K)$. The map $X_K\to X$ is a homeomorphism.

The proof for this was miserably terse, and there are two points I don't really understand. Firstly, he claims it is immediate from the case of $K/k$ being finite to deduce the general case, by appealing to the following lemma:

Lemma: for any (reduced) closed subvariety $W$ of $X_K$ there exists a finite subextension $K'$ of $K$ and a (reduced) closed subvariety $Z$ of $X_{K'}$ such that $W=Z_K$

I would then know that $X_{K'}\to X$ is a homeomorphism, and that the restriction of $X_K\to X$ to $W$ would look like $W\to Z\hookrightarrow X_{K'}\cong X$. But then I'm stuck in the same rut, because $W\to Z$ need not be a homeomorphism; I have no idea, since $K/K'$ is not a finite extension! The best I can say is that $W$ has closed image, namely the image of $Z$ under this homeomorphism. So, $X_K\to X$ is certainly a continuous, closed surjection. Why should it be injective though? That, that I don't understand.

The second point, which is more important. We may assume $X=\spec A$ for some f.g. $k$-algebra $A$. In the simplest possible case, say $K=k(\gamma^{1/p^k})$ is a simple purely inseparable extension, $p=\mathrm{char}(k)$. $X_K$ is just the affine scheme associated to $A_K:=A\otimes_k K$. According to Liu, to see that the map $X_K\to X$ is injective it suffices to show:

For every $\p$ a prime of $A$, the radical ideal $\sqrt{\p A_K}$ is prime

This deeply baffled me. With a lot of help from a friend I got some kind of argument but I'm not sure about the details. Let $\q:=\sqrt{\p A_K}\subset A_K$; assume primality. We can see that $\q\cap A=\p$ and want to show it is unique in that respect. Every other such prime $\q'$ must contain $\q$, so really we want to show that $\q\subsetneq\q'$ with $\q'$ lying over $\p$ cannot possibly happen. Let the dimension of a point $x$ of a scheme refer to the topological dimension of $\overline{\{x\}}$.

The friend suggest $\dim\q\ge\dim q'+1$ follows immediately from $\q\subsetneq\q'$, which I agree with, but moreover that $\dim_{X_K}\q=\dim_X\p=\dim_{X_K}\q'$ must follow if both $\q,\q'$ lie over $\p$, creating a contradiction. So now the goal is to check the base change $X_K\to X$ preserves dimensions of a point.

We have $\dim\p=\trdeg\Frac(A/\p)$. The claim I don't understand is that the primes of $A_K$ lying over $\p$ are the same thing as the primes of $\Frac(A/\p)\otimes_k K$. Assuming this for the moment, since $K$ is algebraic over $k$ it should be true that $(\Frac(A/\p)\otimes_k K)/\q$ is an algebraic extension of $\Frac(A/\p)$ hence has the same transcendence degree over $k$, and we need to justify $(\Frac(A/\p)\otimes_k K)/\q\cong\Frac((A\otimes_k K)/\q)$ as $k$-field extensions, from which $\dim\p=\dim\q$ would follow by looking at transcendence degrees and we'd have a contradiction, $\dim\q=\dim\p=\dim\q'\le\dim\q-1$. But there are a lot of manipulations there which I'm not sure of, and I also wonder if there is an easier way to see Liu's claim since this feels like too much to justify not alluding to. I remain a bit confused by this argument, for example $\Frac(A/\p)\otimes_k K$ is not a field and I see no reason for prime ideals of it to be maximal, so why is $(\Frac(A/\p)\otimes_k K)/\q$ a field?

I would appreciate help clarifying these two "immediate" claims of Liu.

Answer 1

For your first question, if you follow this post, you can find an argument that fits the lemma proposed by Qing Liu. Here is a different approach. So far, you have realized that everything boils down to proving that $X_K \longrightarrow X$ is injective if $K/k$ is algebraic + purely inseparable. In fact, we can show that this morphism is universally injective (remains being injective under any base change) but here I just stick to the injectivity of $f \colon X_K \longrightarrow X$ itself. This morphism is injective if and only if for any point $x \colon \operatorname{Spec}(\kappa(x)) \longrightarrow X$, the fiber product $X_K \times_X \operatorname{Spec}(\kappa(x))$ has at most one point. Assume that it is not empty so its underlying topological space coincides with $\operatorname{Spec}(\kappa(x) \otimes_k K)$ and we need to show that this is one point space. That being said, from beginning you can assume that $X = \operatorname{Spec}(k)$. and show that $\operatorname{Spec}(L \otimes_k K)$ is one point for any $k \hookrightarrow L$. However, this can be seen as a characterization of purely inseperable extensions.

For your second question, this comes from a much more general result: if $f \colon A \longrightarrow B$ is an integral extension of rings, and if $\mathfrak{p}$ be a prime ideal of $A$, then there is at most one prime ideals $\mathfrak{q}_1,\mathfrak{q}_2$ of $B$ so that $f^{-1}(\mathfrak{q}_i) = \mathfrak{p}$ and $\mathfrak{q}_1 \subset \mathfrak{q}_2$, then $\mathfrak{q}_1=\mathfrak{q}_2$. This is standard commutative algebra, look at chapter 5 of Atiyah & McDonald for instance. Your extesion $f \colon A \longrightarrow A \otimes_k K$ is clearly integral because $K/k$ is algebraic. Assume this, you may want to show that any $\mathfrak{q}$ with $f^{-1}(\mathfrak{q}) = \mathfrak{p}$ then $ \mathfrak{q} \subset \sqrt{\mathfrak{p}A_K}$; however, this is already written in the proof of proposition 2.7, what remains is to prove $\sqrt{\mathfrak{p}A_K}$ is a prime ideal.