Generalization of Weak Nullstellensatz?

Question

I believe the following is standard, namely when $k = \bar{k}$ is algebraically closed there is a bijection between points and maximal ideals

\begin{eqnarray*} k^n &\longrightarrow& \operatorname{Specm}(k[X_1, \ldots, X_n]) \\ x &\longrightarrow& \ker(\operatorname{ev}_x) \end{eqnarray*}

where surjectivity follows from Zariski's Lemma. It seems like the following should also be true, by essentially the same argument, but I couldn't find a reference. For $k$ not algebraically closed there is a bijection

\begin{eqnarray*} \bar{k}^n / \operatorname{Aut}(\bar{k} / k) &\longrightarrow \operatorname{Specm}(k[X_1, \ldots, X_n]) \end{eqnarray*}

Is there a canonical reference for this ?

Answer 1

The closest reference in literature I have encountered is in Mumford's Red book of varieties and schemes, II.4 Theorem 1. More precisely, a direct citation is as follows:

Let $X_0$ be a prescheme over $k_0$, let $X = X_0 \times_{k_0} k$, and let $p: X \rightarrow X_0$ be the projection. Assume that $k$ is an algebraic closure of $k_0$. Then

1. $p$ is surjective and both open and closed (i.e., maps open/closed sets to open/closed sets).
2. For all $x,y \in X$, $p(x) = p(y)$ if and only if $x = \sigma_X(y)$, some $\sigma \in Gal(k/k_0 )$. In other words, for all $x \in X_0$ $p^{-1}(x)$ is an orbit of $Gal(k/k_0)$. Moreover, $p^{-1}(x)$ is a finite set.

The theorem applied to $X_0=\mathrm{Spec}\,k_0[x_1, \dots, x_n]$ together with the standard weak Nullstellensatz for $X=\mathrm{Spec}\,k[x_1, \dots, x_n]$ should give your statement.

Answer 2

Just discovered, it's demonstrated in Bourbaki Commutative Algebra Chapter V Section 3.3 Proposition 2.

Answer 3

Or see Proposition 2.4.6 in Bjorn Poonen's book Rational Points on Varieties (link). This is almost exactly the result you conjectured, just a bit more general:

Let $X$ be a $k$-variety. Then the map $$\left\{\text{$\operatorname{Gal}_k$-orbits in $X(\overline{k})$}\right\}\rightarrow \left\{ \text{closed points of $X$} \right\}$$ given by mapping the orbit of $f \colon \operatorname{Spec} \overline{k} \to X$ to $f(\operatorname{Spec} \overline{k} )$ is a bijection.