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

1. $\overline{T}$ is a closed subset of $X_{\overline{k}}$,

2. 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 $k \subset k'$ 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 $P(T) = T^ d + a_1T^{d - 1} + \ldots + a_0$ the we see that $a_ i \in I := A \cap \overline{I}$. By Algebra, Lemma 10.37.5 we see that $x$ 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. $\square$

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