Artin's Algebra exercise special case of some theorem/problem? The following exercise is from Artin's Algebra Text:

Show that there is a one to one correspondence between maximal ideals of $ \bf R$$[x]$ and complex upper half plane.

Solution: Follows from the fact that $ \bf R$$[x]$ is a PID,and any irreducible polynomial of  $ \bf R$$[x]$ is either of degree $1$ or of degree $2$.
Is the above problem a special case of some theorem/Problem? In the case of algebraically closed field $k$ It's well known that maximal ideals of polynomial ring in $n$ variable over $k$ corresponds to points of affine space $ \bf A_k^n$.
 A: Let $K$ be a field and $\bar K$ its algebraic closure. Since every element of $\bar K$ has an unique  monic minimal polynomial over $K$, we can define an equivalence relation on $\bar K$ by calling two elements of $\bar K$ equivalent iff they have the same minimal polynomial. If you know some Galois theory, you will recognise these equivalence classes as the orbits of $\bar K$ under the action of the Galois group of $\bar K/K$ in the case that $\bar K$ is separable over $K$ (perhaps also in the non-separable case, but I'm not sure).
There is a bijective correspondence between these equivalence classes and the maximal ideals of $K[x]$. The reason of course is that on one hand every minimal polynomial is irreducible, so it generates a non-zero prime ideal, which is maximal since $K[x]$ is a PID. On the other hand, again using that $K[x]$ is a PID, every ideal is generated by a unique monic polynomial, which is the minimal polynomial for some element of $\bar K$. 
Your statement is the special case $K = \mathbb R$, $\bar K=\mathbb C$. The only non-trivial Galois action is the complex conjugation, so the equivalence above just identifies every element of $\mathbb C$ with its complex conjugate, which allows you to choose a representative of each equivalence class in the upper half plane. 
A: Most probably, this fact is also true for a real closed field, because its algebraic closure has dimension $2$. Choosing one of the square roots of $-1$ will give you an analogue of the upper half plane.
