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?


1 Answer 1


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.

  • $\begingroup$ Your answer proves the CLAIM, but not answers precisely my question. My question concerns the argumentation way in the proof of the CLAIM I quoted above. The two crucial parts which I not understand there are summarized in the last two paragraphs. It seems that KReiser either used some additional imput implicitely or oversees some gaps in his proof or ... I just too fool to understand the arguments $\endgroup$
    – user784905
    May 11, 2020 at 15:07

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