Tangent space of schemes and manifolds

Suppose $X$ is the set of $\mathbb R$-points of a scheme $\mathfrak X$. Let $x\in \mathfrak X$ be a simple point with a smooth neighbourhood $\mathfrak U$. I presume the (scheme-theoretic) tangent space at $x\in\mathfrak U$ coincides with the (differential-geometric) tangent space at $x\in U$, where $U$ denotes the set of $\mathbb R$-points of $\mathfrak U$ with a manifold structure.

Do you know of a reference for this? (If the statement is true, that is, and I haven't failed to assume algebraic closure or something like that.)

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It is not clear to me what you mean by smooth neighborhood in general. What is the smooth structure on $X$ if the scheme is, say, $\text{Spec } \mathbb{R}[x_1, x_2, ... ]$? – Qiaochu Yuan Jul 20 '12 at 4:59
Does it help to assume that $\mathfrak X$ is reduced? I am thinking of the situation when $x\in\mathfrak X(\mathbb R)$ has a neighbourhood of a smooth algebraic variety and I would like to consider this neighbourhood as a manifold... – user1205935 Jul 20 '12 at 5:14

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The example I gave is reduced. Since this is a local question, you are free to restrict to affine schemes, in which case I think you want to restrict to a finitely-generated reduced ring and a non-singular point (maybe this is what you mean by simple). In that case you have an embedding into $\mathbb{R}^n$ ($n$ the number of generators) so you can talk about the induced smooth structure away from the singular locus, and you should now be able to write down explicitly what the two tangent spaces are and verify that they're the same. – Qiaochu Yuan Jul 20 '12 at 5:20
– user1205935 Jul 20 '12 at 5:30
It's not a duality (the correspondence, to the degree that it exists, is covariant). Most people prefer to talk about the complex points in which case you want to look up the analytification functor (en.wikipedia.org/wiki/Algebraic_geometry_and_analytic_geometry). The study of real points is more subtle and, as I understand it, has a very different flavor (en.wikipedia.org/wiki/Real_algebraic_geometry), but in this case you can get away with a relatively straightforward computation. – Qiaochu Yuan Jul 20 '12 at 14:08
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