Lemma 33.7.10 (038B)—The Stacks project

Lemma 33.7.10. Let $k$ be a field. Let $\overline{k}/k$ be a (possibly infinite) Galois extension. Let $X$ be a scheme over $k$. Let $\overline{T} \subset X_{\overline{k}}$ have the following properties

$\overline{T}$ is a closed subset of $X_{\overline{k}}$,
for every $\sigma \in \text{Gal}(\overline{k}/k)$ we have $\sigma (\overline{T}) = \overline{T}$.

Then there exists a closed subset $T \subset X$ whose inverse image in $X_{\overline{k}}$ is $\overline{T}$.

Proof. This lemma immediately reduces to the case where $X = \mathop{\mathrm{Spec}}(A)$ is affine. In this case, let $\overline{I} \subset A \otimes _ k \overline{k}$ be the radical ideal corresponding to $\overline{T}$. Assumption (2) implies that $\sigma (\overline{I}) = \overline{I}$ for all $\sigma \in \text{Gal}(\overline{k}/k)$. Pick $x \in \overline{I}$. There exists a finite Galois extension $k'/k$ contained in $\overline{k}$ such that $x \in A \otimes _ k k'$. Set $G = \text{Gal}(k'/k)$. Set

\[ P(T) = \prod \nolimits _{\sigma \in G} (T - \sigma (x)) \in (A \otimes _ k k')[T] \]

It is clear that $P(T)$ is monic and is actually an element of $(A \otimes _ k k')^ G[T] = A[T]$ (by basic Galois theory). Moreover, if we write $P(T) = T^ d + a_1T^{d - 1} + \ldots + a_ d$ the we see that $a_ i \in I := A \cap \overline{I}$. Combining $P(x) = 0$ and $a_ i \in I$ we find $x^ d = - a_1 x^{d - 1} - \ldots - a_ d \in I(A \otimes _ k \overline{k})$. Thus $x$ is contained in the radical of $I(A \otimes _ k \overline{k})$. Hence $\overline{I}$ is the radical of $I(A \otimes _ k \overline{k})$ and setting $T = V(I)$ is a solution. $\square$

Comments (5)

Comment #7100 by WhatJiaranEatsTonight on March 07, 2022 at 18:55

I cannot get why $x$ is contained in the radical of $I(A\otimes \overline{k})$ . In the lemma 00H6, we need $A\otimes_k \overline{k}$ to be a subring of some field.

Comment #7101 by WhatJiaranEatsTonight on March 07, 2022 at 19:02

I think that $P(x)=0$ and $a_i\in I$ implies that $x^d =0 -a_1x^{d-1}-\ldots-a_d\in I(A\otimes_k \overline{k})$ . Hence $x$ is obviously contained in the radical of $I(A\otimes_k\overline k)$ . And it does not have relation to the lemma 00H6.

By the way, the last $a_0$ in the expression of $P(T) = T+d+a_1T^{d-1}+\ldots+a_0$ should be $a_d$ .

Comment #7265 by Johan on May 02, 2022 at 14:50

Much better! Thanks and fixed here.

Comment #8280 by Et on March 24, 2023 at 07:29

I think the reduction to the affine case is a bit confusing as stated, specifically the fact T is closed. Mainly, if X is not q.c you may end up taking an union of infinitely many closed sets in X and then it's not clear why you would get a closed set.

A better way to phrase the reduction to the non affine case is to note the construction given agrees on passing to principal opens in X, and so it glues together tp give the desired closed set T.

Comment #8914 by Stacks project on April 10, 2024 at 10:19

What you say works, but that is not how I think about it. If you have a $T$ as in the lemma, then $T$ is the image of $\overline{T}$ . So really one is just checking that the image of $\overline{T}$ is closed in $X$ which may be done locally.

I am going to leave as is for now.

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