# $X_K$ is Reduced iff $X_F$ is reduced for every finite extension $k \subset F \subset K$

I have a couple of questions about a proof given in the answer for this question: Base change and irreducibility/reducedness/connectedness in Qing Liu's book (3.2.7 and 3.2.11 using 3.2.6)

Let $$X$$ be a scheme of finite type over a field $$k$$ and $$k \subset K$$ is an algebraic field extension. Remark: in the linked topic a variety is by definition a scheme of finite type over $$k$$. The CLAIM is:

$$X_K$$ (abbrev. for fiber product $$=X \times_k \operatorname{Spec} K$$) is reduced iff $$X_F$$ is reduced for every finite extension $$k \subset F \subset K$$.

The proof uses frequently following LEMMA which we assume as known:

LEMMA: Let $$X$$ be be a scheme of finite type over $$k$$, and let $$K$$ be an algebraic extension of $$k$$. Then for any reduced closed subscheme $$W$$ of $$X_K$$, there exist a finite subextension $$K'$$ of $$K$$, and a unique (for fixed $$K'$$) reduced closed subscheme $$Z$$ of $$X_{k'}$$ such that $$W = Z_K$$.

The proof of the CLAIM works as follows:

"$$\Leftarrow$$": Let $$k\subset K$$ be an algebraic field extension. Suppose $$X_K$$ is not reduced. Then $$X_K^{red}$$, the reduction, is a closed reduced subscheme. Applying LEMMA, we can find an intermediate field $$k\subset K'\subset K$$ and a reduced closed subscheme $$Z\subset X_{K'}$$ so that $$X_K^{red}=Z_K$$. We note that such a $$Z$$ cannot be equal to $$X_{K'}$$, as $$(X_{K'})_K=X_K\neq X_K^{red}$$.

"$$\Rightarrow$$": On the other hand, if there is a finite subextension $$k\subset F\subset K$$ so that $$X_F$$ is non-reduced, then $$(X_F^{red})_K$$ gives closed subscheme of $$X_K$$ which contains all the points of $$X_K$$ but is not equal to $$X_K$$, and thus $$X_K$$ is not reduced.

There is a couple of steps I not really understand:

On "$$\Leftarrow$$": I not see why the observations that $$X_K^{red}=Z_K$$ and $$Z \neq X_{K'}$$ imply that $$X_F$$ is not reduced. The LEMMA says that such $$Z \subset X_F$$ that is unique with property $$X_K^{red}=Z_K$$ exist. But there is no hint why this $$Z$$ should be the "maximal" reduced closed subbscheme of $$X_F$$. So the conclusion isn't clear.

Now on "$$\Rightarrow$$": What do we know about $$(X_F^{red})_K$$? Pure topologically it coinsides with $$X_K$$. But how we obtain the consequence $$(X_F^{red})_K \neq X_K$$ and if we assume we know the later, why this imply that $$X_K$$ is not reduced?

Since we can check reducedness on affine opens, let us work with an affine scheme $$X=\operatorname{Spec} R$$ with $$R$$ a finitely generated $$k$$-algebra. Let $$N \subset R$$ denote the nilradical of $$R$$ and let $$N_K \subset R_K$$ denote the nilradical of $$R_K=R \otimes_k K$$. [The nilradical is the set of all nilpotent elements, and a ring is reduced if and only if it is the zero ideal].
If $$k \subset F \subset K$$ then $$N_k \subset N_F \subset N_K$$ so it follows that if $$N_K=0$$ then $$N_F$$ is zero for all fields $$k \subset F \subset K$$.
For the reverse direction, we remark that $$N_K$$ is a finitely generated ideal, because $$R_K$$ is finitely generated over $$K$$, hence Noetherian. So $$N_K=(x_1, \cdots, x_n)$$, which means that it suffices to show that $$N_F$$ is zero, with $$F=k(x_1, \cdots, x_n)$$. Note that $$F/k$$ is a finite extension, because it is algebraic (integral) and finitely generated.