Real Points of resolution of singularities of $\mathrm{Spec} \mathbb{R}[x,y]/(x^2+y^2)$

Consider the scheme $X = \mathrm{Spec} \mathbb{R}[x,y]/(x^2+y^2)$. Scheme-theoretically, it’s a one dimensional scheme with one real, singular point (there are, of course, other complex points). I’m interested in finding the structure of the real points of the resolution of $X$, but I’m having trouble visualizing exactly what I get.

I think I want to do something like the following: embedd $X\hookrightarrow \mathbb{A}^2_\mathbb{R}$ and then blow up at the origin and see what I get. Here’s the part where I start to get confused. Over the complex numbers, I know if I were to do this, since $(x^2+y^2)=(x+iy)(x-iy)$, I would have my scheme be two lines intersecting transversally, which under the blow up would turn in to two disjoint lines, and the resolution of singularities would be from two disjoint copies of $\mathbb{A}^1_\mathbb{C}$ to two transversally intersecting lines.

When I try and think about doing this same sort of business over $\mathbb{R}$, I get confused about where the lines live. I know I ought to have something that looks like two disjoint lines embedded in $\mathbb{A}^2_\mathbb{R}\times\mathbb{P}^1_\mathbb{R}$, but I don’t really understand how much of these lines are real points and how much of them are complex points.

I’d like it if someone could give me some advice on how to approach the problem or perhaps some sort of solution which makes explicit where the real points are in a (the?) resolution of singularities of this scheme.

Answer

Let $A=\mathbb{R}[x,y]/(x^2+y^2)$. If you blow up, the resolution is covered by two open sets of the form $A[x/y],A[y/x]$ and both are isomorphic to a ring of the form $\mathbb{R}[x,y]/(x^2+1)$ and thus the resolution has no real points.

Attribution
Source : Link , Question Author : KReiser , Answer Author : Mohan

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