Primes of $\Bbb R[x]$ and Galois theory in terms of affine schemes I want to find all prime ideals of $\Bbb R[x]$. These are generated by the monic irreducible polynomials. Since $\Bbb C/\Bbb R$ is a degree two extension, and is algebraically closed, any other algebraic extension of $\Bbb R$ must be of degree $2$ or $1$, so is either $\Bbb R/\Bbb R$ or is isomorphic to $\Bbb C$. Then any monic irreducible polynomial in $\Bbb R[x]$ has degree at most two (otherwise we would obtain $\Bbb R[x]/(f)$ an algebraic extension of degree greater than two which is impossible). Then since $\text{Gal}(\Bbb C/\Bbb R)$ is generated by conjugation, and this degree two monic polynomial cannot have any $\Bbb R$-roots, the roots must be complex conjugate.
Then $\text{Spec}(\Bbb R[x])=\{(0),(x-a),(x^2-2ax+a^2+b^2)\mid a\in \Bbb R, b\in \Bbb R\backslash\{0\}\}$. Is this correct?
Then $\Bbb A^1_{\Bbb R}$ looks like $\Bbb C$, with all the real points $(x-a)$ on the real line, but we identify points that are complex conjugate - so in some sense we are left with the upper half plane $\mathcal{H}=\{a+bi\mid a\in\Bbb R, b\in\Bbb R_{\geq 0}\}\subset \Bbb C$. 
Additionally, in general, for $K$ a field which is not algebraically closed, is there a way to treat the theory of Galois extensions $L/K$ in terms of automorphisms of $\text{Spec}(K[x])$? It seems that the closed points which aren't codimension one encode a Galois action at that point? Like these $(x^2-2a+a^2+b^2)$ points can already see the $\Bbb C$-conjugate points in $\Bbb C[x]$.
 A: Suppose $K$ is a field and $L/K$ is a Galois extension. Take any $L$-rational point of $K$-schemes in $\Bbb{A}^1_K=\mathrm{Spec}(K[x])$, i.e. a $K$-algebra morphism $\mathrm{Spec}(L) \to \mathrm{Spec}(K[x])$ which corresponds to a morphism $\phi:K[x] \to L$. Then $\ker(\phi)$ is a maximal ideal in $K[x]$ of the form $\ker(\phi)=(f)$ where $f$ is irreducible over $K$, but splits over $L$. Let $\alpha_1, \dots, \alpha_n \in L$ be the roots of $f$. Then we can take the morphism $\mathrm{Spec}(L[x]) \to \mathrm{Spec}(K[x])$ and look at the fiber over $(f)$. We get $$\mathrm{Spec}(\kappa(f)) \otimes_{K[x]} \mathrm{Spec}(L[x]) \cong \mathrm{Spec}(L[x]/(f))=\coprod_{i=1}^n\mathrm{Spec}(L[x]/(x-\alpha_i))$$ Thus the projection $\mathrm{Spec}(\kappa(f)) \otimes_{K[x]} \mathrm{Spec}(L[x]) \to \Bbb{A}^1_L$ corresponds to the inclusion of the roots of $f$. (Which are given by the Galois orbit of $\alpha_1$)
If we have any $L$-rational point of $\Bbb{A}^1_L$ (which is necessarily of the form $(x-a)$ for a unique $a \in L$), then we can compose with $\mathrm{Spec}(L[x]) \to \mathrm{Spec}(K[x])$ to obtain a $L$-rational point of $\Bbb{A}^1_K$ (which will be $(f)$ for $f$ the minimal polynomial of $a$ over $K$) and then apply the above construction to obtain the Galois orbit of $a$.
To summarize this, there's a commutative diagram $$\require{AMScd} \begin{CD}
L @>{\cong}>> \Bbb{A}^1_L(L)\\ @VVV @VVV\\
L/G @>>{\cong}> \Bbb{A}^1_K(L)
\end{CD}$$
Where $G=\mathrm{Gal}(L/K)$ and $L \to L/G$ is the map sending an element of $L$ to its Galois orbit and the map $\Bbb{A}^1_L(L) \to \Bbb{A}^1_K(L)$ is given by composing with $\Bbb{A}^1_L \to \Bbb{A}^1_K$
If $L$ is algebraically closed, then $\Bbb{A}^1_K(L)$ corresponds to all closed points in $\Bbb{A}^1_K$. (In general, it only corresponds to those closed points whose residue field embeds into $L$) Even if $L/K$ is not separable, the above construction works, but it will give rise to non-reduced fibers. (Non-separable extensions correspond geometrically to ramification)
