Classifying prime ideals of $\mathbb{Q}[x]$ I have taken an undergraduate course in algebraic geometry and after that wanted to do some self-study with use of the "Foundations of Algebraic Geometry" textbook by Ravi Vakil. The course was very basic, we've only started talking about sheaves and schemes towards the end of it. I decided to start with the chapter about schemes, since I have already taken an introductory course in sheaf theory, but got stuck very early, namely on the exercise 3.2.C, which reads: Describe the set $\mathbb{A}^{1}_{\mathbb{Q}}$. It does not ask about the structure sheaf, so in practice it comes down to classifying prime ideals of $\mathbb{Q}[x]$. My attempt went as follows:
We know that every ideal in $\mathbb{Q}[x]$ is a principal one, so we can associate an element of $\mathbb{A}^{1}_{\mathbb{Q}}$ with the zeroes of the polynomial generating it and the zeroes of polynomials over $\mathbb{Q}$ are algebraic numbers, so I need to get all the algebraic numbers up to some equivalence. So we obviously have the rational numbers associated with the ideals of the form $(x-q)$, where $q$ is rational and the $0$ ideal.
Unfortunately that was the only thing I managed to classify (which wasn't even really done by me, since it was shown before the exercise). I am not even entirely sure what would it mean to classify them, it was very easy for the case of complex and real polynomials, but here, there seems to be much more variation here. For the numbers expressible with radicals, I believe I can show that two such numbers are zeroes of the same irreducible polynomial when they only differ by the sign of roots of even degree (so, for example, $2+\sqrt{5}+\sqrt[7]{2}$ would be equivalent to $2-\sqrt{5}+\sqrt[7]{2}$), but doing only that would not help me classify the polynomials with such roots and even if I did manage to do that, there are still all the algebraic numbers not expressible with radicals, with which I have no idea how to begin looking for a possible equivalence describing which of them must be zeroes of the same polynomials and how to describe those polynomials.
To formulate a question, how would one approach this? Is it a good first step to separate them into cases (those expressible with radicals and those that are not)? I don't necessarily look for a full solution, but maybe some clue or a key fact that should be used.
 A: As mentioned in the comments, the correct description of (the closed points of) this space is indeed the space of simple orbits of $\overline{\Bbb Q}$ under the action of $\operatorname{Gal}(\overline{\Bbb Q},\Bbb Q)$. This connects to the previous example where it's shown that closed points of $\operatorname{Spec} \Bbb R[x]$ are exactly a copy of $\Bbb C$ with complex-conjugate elements identified: we're exactly looking at $\overline{\Bbb R}$ under the action of $\operatorname{Gal}(\overline{\Bbb R},\Bbb R)$ in that question.
To prove this, we can use the fact that $\Bbb Q[x]$ is a PID: this means every non-zero prime ideal is of the form $(f)$ for $f$ an irreducible polynomial of degree $n>0$. Such a polynomial has distinct roots $\overline{\Bbb Q}$ because it is coprime to its derivative, and these roots $r_1,\cdots,r_n$ exactly form an orbit of $\overline{\Bbb Q}$ under the $G=\operatorname{Gal}(\overline{\Bbb Q},\Bbb Q)$-action: if any nontrivial subset was $G$-stable, then $\prod_{i\in S} (x-r_i)$ would be an element of $\Bbb Q[x]$ dividing $f$. Conversely, given any $G$-invariant subset $S$ of $\overline{\Bbb Q}$ which has no nontrivial invariant subsets, the polynomial $\prod_{s\in S} (x-s)$ is in $\Bbb Q$ and irreducible by the same logic. It is direct to see that these procedures are mutually inverse, so we have a bijection between $\overline{\Bbb Q}/G$ and the nonzero prime ideals of $\Bbb Q[x]$, aka the closed points of $\operatorname{Spec} \Bbb Q[x]$.
The moral of this story is that if $F$ is a perfect field, you can/should think about $\operatorname{Spec} F[x]$ as $\operatorname{Spec} \overline{F}[x]$ modulo the Galois action. So on the level of closed points, you're folding up $\overline{F}$ in an interesting way.
