How to see $\operatorname{Spec} k[x]$ for non necessarily algebraic closed field $k$? 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 $\operatorname{Spec} k[x]$? 
 A: 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.
A: You note correctly in a comment that the points of $\operatorname{Spec} k[x]$ correspond to (monic) irreducible polynomials, and you are correct that this can be identified with the set $k\cup\{*\}$ when $k$ is algebraically closed.
Suppose $\bar k$ is an algebraic closure of $k$. Then the inclusion $k[x]\to\bar k[x]$ induces a morphism $\pi:\operatorname{Spec} \bar k[x]\to\operatorname{Spec} k[x]$ which is surjective as a map of sets.
If we take a point of $\operatorname{Spec} k[x]$ and think of it as a monic irreducible polynomial $p(x)$, the fibre of $\pi$ over that point is exactly the set of roots of $p$. For simplicity, let's assume that $k$ is a perfect field (e.g. it has characteristic zero or it's a finite field), so that the extension $\bar k/k$ is Galois. The Galois group $G:=\operatorname{Gal}(\bar k/k)$ acts on $\operatorname{Spec} \bar k[x]\simeq \bar k\cup\{*\}$ in the obvious way (fixing the point $*$), and the fibres of $\pi$ are exactly the orbits of this action, so we can think of $\operatorname{Spec} k[x]$ as the quotient $(\bar k\cup\{*\})/G$.
For example, if $k=\mathbb{R}$, we can think of $\operatorname{Spec}\mathbb{R}[x]$ as something like what we get by taking the complex plane $\mathbb{C}$ and identifying complex conjugates, i.e. folding it over on itself across the real line. (That, along with the generic point $*$.)
