TITLE: How to see $\operatorname{Spec} k[x]$ for non necessarily algebraic closed field $k$?
QUESTION [10 upvotes]: I know that $\operatorname{Spec} \mathbb{C}[x]$ can be identified with the set $\mathbb{C}\cup *$, where $*$ is a generic point via the correspondence 
$$
\prod_{i}(x-a_i) \leftrightarrow \{a_i\}_i
, \ \ \ \ (0)\leftrightarrow *.
$$
This correspondence holds for any $\operatorname{Spec} k[x]$ with algebraic closed field $k$. 
When $k$ is not algebraic closed, how should one understand $\operatorname{Spec} k[x]$?

REPLY [2 votes]: Let $\bar k$ be an algebraic closure of $k$.
Let $G = Aut(\bar k/k)$.
$\bar k/G$ be the set of $G$-orbits.
We will show that there exists a canonical bijection $Spec$ $k[x] \rightarrow \bar k/G\cup \{*\}$.
Let $Y =$ $Spec$ $k[x] - \{(0)\}$.
It suffices to prove that there exists a canonical bijection $\psi\colon Y \rightarrow \bar k/G$.
Let $p \in Y$.
There exists a unique monic irreducible polynomial $f(x) \in k[x]$ such that $p = (f(x))$.
Let $S$ be the set of roots of $f(x)$ in $\bar k$.
It is clear that $S$ is $G$-stable.
Let $\alpha, \beta \in S$.
There exists a $k$-isomorphism $k(\alpha) \rightarrow k(\beta)$ transforming $\alpha$ to $\beta$.
It can be extended to a $k$-automorphism of $\bar k$.
Hence there exists $\sigma \in G$ such that $\sigma(\alpha) = \beta$.
Hence $S \in \bar k/G$.
Hence we get a map $\psi\colon Y \rightarrow \bar k/G$ such that $\psi(p) = S$.
Clearly $\psi$ is injective.
Let $T \in \bar k/G$.
Let $\alpha \in T$.
Let $f(x)$ be the minimal polynomial of $\alpha$ over $k$.
Clearly $\psi((f(x)) = T$.
Hence $\psi$ is surjective.