Over $\mathbb{R}$, is it still the case that the singular locus of a nonempty variety is a proper subvariety?

Question

I thought through the argument in Hartshorne for algebraically closed fields and I don't see any issues, but I would appreciate it if someone checked their intuition against this, since I don't know anything about real algebraic geometry. I imagine that something weird could happen because points are missing.

Roughly the argument would go like this:

Given a real affine variety $X$...

1. Because $R$ is a perfect field, there is a separable trascendence basis for the function field. Hence $X$ is birational to a hypersurface.
2. The singular locus of a nonempty irreducible hypersurface $Z(f)$ is a proper closed subvariety, since otherwise $f$ divides all of its derivatives, which implies that all of its derivatives are zero, hence it is the constant function.

(I just want to conclude that the Grassmanian is a manifold by identifying it with the locally closed space of orthogonal projections inside $M_n(\mathbb{R})$, and observing that it is homogeneous because $GLn(R)$ acts transitively on it, hence the fact that it is generically smooth implies that it is smooth. (I know how to describe the Grassmanian as a manifold without this argument.))

Answer 1

If you care about the real points of an affine scheme over $\mathbb R$, then it might happen that all real points are in the singular locus of $X$. For example, if we consider the curve in $\mathbb A^2$ given by $\mathbb R[x,y]/(x^2 + y^2)$, then the singular locus is the origin. This also happens to be the only real point!

However, in this case one might say that the manifold we get is still smooth, because it is just a point. It is also enlightening to consider examples like $$\mathbb R[x,y,z]/((x^3-y^2)^2 + z^2).$$ For this variety the real points are the curve $x^3 = y^2$, which of course has a singularity. However, the singular locus is only the origin, so we're happy.

It seems to me that this behaviour is typical (i.e. even if the real points are fewer than we expect, it is still smooth almost everywhere), although I do not know how to prove this. In both examples above, we do however get a manifold of dimension one less than we would (algebraically) expect.