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If I am not mistaken there is a theorem that says any curve $C$ can be embedded in $\mathbf{P}^3$. What can be said about surfaces? Do we have a theorem like:

All surfaces can be embedded in $\mathbf{P}^{N}$ for some fixed $N$.

And if not what is the simplest counterexample? A counterexample would be an infinite sequence of surfaces: $\{S_1, S_2, \cdots \}$ and infinite sequence of strictly increasing positive integers: $\{n_1, n_2, \cdots \}$ such that $S_i$ can not be embedded in $\mathbf{P}^{n_i}$.

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A related question:… – Georges Elencwajg Jan 20 '11 at 13:25
up vote 17 down vote accepted

All smooth projective surfaces can be embedded in $\mathbb{P}^5$ using a (linear) projection. This is a classical theorem in algebraic geometry. In general, it is known that any smooth projective variety of dimension $n$ can be embedded into $\mathbb{P}^{2n+1}$.

The proof runs as follows: The secant variety $Sec(X)$ (the closure of points on secants on $X$) and the tangent variety $Tan(X)$ (the closure of the union of tangent spaces) both have dimension $\le 2n+1$, so if $X$ lies in a big projective space $\mathbb{P}^N$, successive projections from points outside $Sec(X)\cup Tan(X)$, gives you a morphism $\mathbb{P}^{2n+1}$ which is injective, finite, unramified hence a closed embedding.

This result makes surfaces in $\mathbb{P}^4$ an interesting study. A remarkable theorem of Ellingsrud-Peskine says that surfaces in $\mathbb{P}^4$ are very special: those not of general type belong to finitely many families, hence one has some hope of classifying them all. For example, Hartshorne conjectured that the degree of rational surfaces in $\mathbb{P}^4$ is bounded (it is now belived that this bound is equal to 12).

There are however non-projective surfaces, which cannot be embedded in any projective space, see e.g., this question.

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Nonprojective surfaces are singular. If you allow singularities, there is no bound on the embedding dimension, even for projective surfaces (or one-point schemes, for that matter). – Laurent Moret-Bailly Jan 20 '11 at 15:07
"Nonprojective surfaces are singular". Why? A non-algebraic complex torus is certainly smooth as a complex Kaehler manifold. Probably you meant "non-projective, proper algebraic" or something like this. – Francesco Polizzi Jan 20 '11 at 16:31
@Francesco: Indeed, I was assuming from the start that we were talking about proper algebraic surfaces. Smooth proper algebraic surfaces are known to be projective; together with compactification and resolution of singularities, this implies that smooth algebraic surfaces are quasiprojective. – Laurent Moret-Bailly Jan 20 '11 at 16:40

The explanation above can be expanded with the following classical fact. Consider the Veronese surface $V$ in $\mathbb{P}^{5}$ which is the image of $\mathbb{P}^{2}$ via the embedding induced by $\mathcal{O}_{\mathbb{P}^{2}}(2)$. Note that $\mathbb{P}^{5}$ parametrizes conics and $V$ parametrizes conics of rank $1$ i.e. squares of linear forms. Since the general conic can not be written as sum of two squares the variety of secant lines $Sec(V)$ has dimension $4$ instead of the expected dimension $5$. So $Sec(V)$ is a cubic hypersurface in $\mathbb{P}^{5}$ singular along $V$. This means that if $p\in \mathbb{P}^{5}$ is a general point the projection

$$\pi_{p}:V\rightarrow \mathbb{P}^{4}$$

defines an isomorphism between $V$ and a smooth surface of degree $4$ in $\mathbb{P}^{4}$. Indeed Severi proved that the Veronese surface $V$ is the only surface in $\mathbb{P}^{5}$ which can be isomorphically projected in $\mathbb{P}^{4}$.

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