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Let $X$ be a smooth projective surface not contained in $\mathbb{P}^3$. Is there any known condition on $X$ under which I can embed it into $\mathbb{P}^3$ such that the its image contains at most simple surface singularities. Any idea or references on this topic will be very helpful.

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I am not an expert on this, but at least over C, cant you always embed into P^3? Just by taking lines through a point which is not on any chord or tangent reduces dimension by by one. Dimension the variety of chords and tangents shows you can keep finding such a point until you get down to P^3. –  Steve May 10 '13 at 17:45
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@Steve, definitely not every surface embeds in $\mathbb{P}^3$ (for example, Abelian surfaces). With regards to your argument, you can't find lines which intersect your variety only once (without tangency) in general. –  Karl Schwede May 10 '13 at 18:52
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What is your definition of simple surface singularity? –  rita May 10 '13 at 20:52
    
oops. I was thinking geometric surface, but algebraic curve. –  Steve May 10 '13 at 21:49
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1 Answer

Let $S' \subset \mathbb{P}^3$ be the birational projection of a smooth surface $S \subset \mathbb{P}^4$. The generic projection theorem of Gruson-Peskine (http://arxiv.org/abs/1010.2399v2) tells you that either $S'$ is smooth or has a curve of double points.

For instance, if $S$ is the Veronese surface in $\mathbb{P}^4$, then its projection in $\mathbb{P}^3$ (the Steiner surface) has a curve of double points.

So the answer to your question is "never", unless your surface embedds in $\mathbb{P}^3$ (which you don't want).

Edit : If by "embed into $\mathbb{P}^3$", you mean project down to $\mathbb{P}^3$ from another embedding, my answer is correct, but there is a much simpler proof : you can always embed your surface in $\mathbb{P}^5$ because the secant variety of a surface in any projective space has at most dimension $5$. Then a projection to $\mathbb{P}^3$ factors as a projection to $\mathbb{P}^4$ and a projection to $\mathbb{P}^3$. If you choose your projection generically from $\mathbb{P}^5$ to $\mathbb{P}^4$ then it has only simple singularities in $\mathbb{P}^4$. But a simple count of dimension shows that for any point $p \in \mathbb{P}^4$, there is a $1$-dimensional family of bisecants to $S$ passing through $p$. This gives you a curve of double points in $\mathbb{P}^3$.

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