$\newcommand{\Frac}{\operatorname{Frac}}\newcommand{\spec}{\operatorname{Spec}}\newcommand{\trdeg}{\operatorname{trdeg}_k}\newcommand{\p}{\mathfrak{p}}\newcommand{\q}{\mathfrak{q}}$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.
Best Answer
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.