Algebraic Geometry – Picture of Spec(R[x])

Question

Follow-up to Comparing the prime spectra of $\mathbb{Q}[x],\mathbb{R}[x]$ and $\mathbb{C}[x]$. I'm trying to draw a picture of $\mathrm{Spec}(\mathbb{R}[x])$.

I've already drawn a picture of $\mathrm{Spec}(\mathbb{C}[x]) = \{(0)\} \cup \{(x – \alpha) \mid \alpha \in \mathbb{C}\}$ as a complex plane plus the generic point $(0)$:

Now, $\mathrm{Spec}(\mathbb{R}[x]) = \{(0)\} \cup \{(p) \mid p \text{ irreducible in }\mathbb{R}\}$.
At first, I thought of identifying each irreducible polynomial with a complex root, but this isn't well-defined as some irreducible polynomials will have more than one root.
But the irreducibles in $\mathbb{R}[x]$ are linear polynomials and quadratics with no real root, and the latter have two conjugate complex roots.
So I thought we can represent this spectrum as the upper half of the complex plane, identifying each quadratic irreducible with its complex root with positive imaginary part, as follows:

Is this an accurate picture?

Answer 1

Yes, this is correct. More generally, for a Galois extension $k\subset K$, one can think of the natural map $\Bbb A^n_K\to\Bbb A^n_k$ coming from the natural inclusion $k[t_1,\cdots,t_n]\to K[t_1,\cdots,t_n]$ as the quotient by the Galois action, which is what you've done here: folded the complex plane over the real axis.