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There is an exercise in Hartshorne, II.2.10, which asks us to describe $\mathrm{Spec}(\mathbb{R}[x])$ and its topology. This ends up being

  • The generic point $(0)$
  • Closed points of the form $(x - a)$
  • Closed points of the form $(x - z)(x - \overline{z})$ for some $z \in \mathbb{C} \setminus \mathbb{R}$.

So the topological space ends up being the set of the above points with the closed sets being $\mathrm{Spec}(\mathbb{R}[x])$ and finite sets of closed points. I thought this was the end, but when I compared my answer to other solutions available, everyone seems to mention that the residue field at the points $(x - a)$ is $\mathbb{R}$ and at the points $(x - z)(x - \overline{z})$ is $\mathbb{C}$.

I guess that is interesting, but I don't see why this matters? And most solutions seem to include it. So in general, for a ring $A$ with a prime ideal $\mathfrak{p}$, what does $A/\mathfrak{p}$ say about $\mathrm{Spec}(A)$? The only real ring operation I know that relates to the spectrum is localisation at a prime ideal and the stalk at a point.

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    $\begingroup$ A scheme can encode solution sets (of polynomials) in various fields via the functor of points view. The residue field of a (closed) point kind of tells us which (extension, if the base field exists) field the solution belongs to. $\endgroup$ Commented Oct 25 at 7:24

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First, the residue field is $\operatorname{Frac}(A/\mathfrak{p}),$ not $A/\mathfrak{p}$ (I guess you meant to say $\mathfrak{p}$ is a maximal ideal). Now the question asks "how the topological space of $\operatorname{Spec}\mathbb{R}[x]$ compares to the sets $\mathbb{R}$ and $\mathbb{C}$". I think the main reason for the calculation of the residue fields is for parametrization and to establish the comparison it asks: If you do the same calculations with $\mathbb{Z}$ in place of $\mathbb{R}$ you end up with $\mathbb{A}^1_{\mathbb{Q}}\cup\mathbb{A}^1_{\mathbb{F}_p}.$ So I think the answer you are looking for is that the residue fields in this case establish a relationship with the complex numbers (in particular that as a set $\operatorname{Spec}\mathbb{R}[x]$ is naturally identified with $\require{enclose}\enclose{horizontalstrike}{\mathbb{C}}$). Edit: This is wrong, the identification is with $\mathbb{C}/(z\sim \bar{z})$ as you pointed out. The example I mentioned with the integers is example 2.1.8 in Qing Liu's book.

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  • $\begingroup$ Wait I'm a little confused at the end, I thought that given the description, we have $\mathrm{Spec}(\mathbb{R}[x]) \cong \mathbb{C}/(z \sim \overline{z})$ as a topological space, so it's not exactly $\mathbb{C}$? And some of the residue fields are $\mathbb{R}$ and others are $\mathbb{C}$, so I'm still a bit confused at this identification via the residue fields. I might be being silly though. $\endgroup$ Commented Oct 25 at 18:47
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    $\begingroup$ You're right, I apologize. I edited my answer :) $\endgroup$ Commented Oct 26 at 10:41

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