is there some connection between a curve in the algebraic geometry sense, e.g.
Separated scheme of finite type over spec($k$)
for a field $k$
and a curve in the sense of a smooth map from an interval in $\mathbb R$ to $\mathbb R^n$?
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1$\begingroup$ In addition to the answer of Simon Rose there are a couple more distinctions one could note. First, you could take a constant map $\mathbf R \rightarrow \mathbf R^n$. The image of that is defined by polynomials, but it's definitely not a curve. Less nitpickily, the set of real points of an irreduclble 1-dimensional scheme over $\mathbf R$ might have different topology from the image of an interval: it could be empty, or disconnected. $\endgroup$– Lazzaro CampeottiAug 26, 2016 at 10:12
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2 Answers
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Well, first of all, a separated scheme of finite type over $Spec(k)$ is not necessarily a curve. A one dimensional separated scheme of finite type etc. etc. may be a curve, but this is also not quite a curve in the sense that you describe.
A curve in the algebro-geometric sense is a one dimensional variety. However, depending on your base field $k$ this may look very different than a real curve. For example, we could consider the scheme $$ Spec\ \mathbb{C}[x, y]/(y^2 - x^3 - 1) $$ which produces an affine elliptic curve. Topologically though[1], this is a real 2-dimensional topological space. It is however, one complex dimensional, and so we call it a curve.
However, if you were to look at the real points of $$ Spec\ \mathbb{R}[x, y](y^2 - x^3 - 1) $$ (i.e. corresponding to maps $Spec\ \mathbb{R} \to Spec\ \mathbb{R}[x, y](y^2 - x^3 - 1)$ then you do obtain something that is a one (real) dimensional topological space, whose curve is exactly what you think it is, at least once you choose the correct topology again.
As for curves viewed as maps $\gamma : [0, 1] \to \mathbb{R}^N$: these are not necessarily (in fact, not likely!) algebro-geometric curves, mostly because the image of such a curve may not be the zero locus of a collection of polynomial (or even analytic) functions.
Really, I think that viewing a curve as a "one dimensional thing", whatever that means in your context, is how you should think of it. So in the algebraic setting, it's going to be something that is locally the spectrum of a ring of Krull-dimension one. In the topological setting, it may be a topological space of Hausdorff dimension one.
[1] There is maybe a point that one should make about the Zariski versus the analytic topology here...
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I think when you mention a smooth map, the condition seems appear in the category of diffential geometry, which is quite different. The concept of 1 dimensional separated scheme (since we consider curves) is in fact a generalization of algebraic curves(not necessarily smooth). More precisely, an algebraic curve is a geometric object plus some algebraic conditions for example the sheaf of polynomials on it, scheme in some sense generalize the concept of sheaves of regular function on it!
By the way, even a smooth algebraic curve may not be just a single map as you mentioned. This answer is somehow conceptual ,hope it useful
