This is a reply to a question posed in Georges' answer. It started as a comment, but I was worried about space limitations. Upon reflection, it's better to have it as an answer, because -- unlike the comment I posted to Georges' answer -- I will be automatically notified of any responses to it.

Since Georges asked for it, the [rough!] lecture notes where I discuss the facts that any smooth curve over an infinite field can be embedded in $\mathbb{P}^3$ and "immersed" in $\mathbb{P}^2$ with only ordinary double point singularities are available here (see Section 5):

http://www.math.uga.edu/~pete/8320notes6.pdf

As you will see, I am merely repeating the argument in Hartshorne -- omitting the trickier details of the immersion result -- and explaining why the ground field need not be algebraically closed but does need to be infinite.

Concerning Horrocks-Mumford and Van de Ven: I was not familiar with these results until Georges' post. But all the non-embeddability statements carry over immediately: if you have an embedding into $\mathbb{P}^n$ over the ground field, then the base change to the algebraic closure is still an embedding, of course.

This leaves the question of the positive part of the Horrocks-Mumford result. In strongest form, the question is: is it true that for any field $k$, there is an abelian surface over $k$ that can be embedded in $\mathbb{P}^4$? [I can certainly do it with $\mathbb{P}^2 \times \mathbb{P}^2$ -- take a product of two elliptic curves -- and it is conceivable to me that one might be able to get from this an embedding into $\mathbb{P}^4$ by composing with a well-chosen birational isomorphism, but I haven't even tried to decide whether this would work.]

I would have to see the proof of H-M to see whether it can be adapted to answer this question. Can you post a link to the paper? Or, if you need to know ASAP, ask Bjorn Poonen -- he eats questions like this for breakfast.

Finally, let me remark that over a non-algebraically closed field, a principal homogeneous space under an abelian variety may have higher embedding dimension than the (Albanese) abelian variety itself. The easy example of this is that if a smooth curve of genus one can be emedded in $\mathbb{P}^2$, then for geometric reasons it must embed as a cubic and therefore has a rational point of degree at most $3$. [Actually, it is possible that this is the only example. By the same theorems Georges quoted above, the only other possibility is a phs which does not embed in $\mathbb{P}^4$ while its Albanese abelian surface does.]

This is a reply to a question posed in Georges' answer. It started as a comment, but I was worried about space limitations. Upon reflection, it's better to have it as an answer, because -- unlike the comment I posted to Georges' answer -- I will be automatically notified of any responses to it.

Since Georges asked for it, the [rough!] lecture notes where I discuss the facts that any smooth curve over an infinite field can be embedded in $\mathbb{P}^3$ and "immersed" in $\mathbb{P}^2$ with only ordinary double point singularities are available here (see Section 5):

http://alpha.math.uga.edu/~pete/8320notes6.pdf

As you will see, I am merely repeating the argument in Hartshorne -- omitting the trickier details of the immersion result -- and explaining why the ground field need not be algebraically closed but does need to be infinite.

Concerning Horrocks-Mumford and Van de Ven: I was not familiar with these results until Georges' post. But all the non-embeddability statements carry over immediately: if you have an embedding into $\mathbb{P}^n$ over the ground field, then the base change to the algebraic closure is still an embedding, of course.

This leaves the question of the positive part of the Horrocks-Mumford result. In strongest form, the question is: is it true that for any field $k$, there is an abelian surface over $k$ that can be embedded in $\mathbb{P}^4$? [I can certainly do it with $\mathbb{P}^2 \times \mathbb{P}^2$ -- take a product of two elliptic curves -- and it is conceivable to me that one might be able to get from this an embedding into $\mathbb{P}^4$ by composing with a well-chosen birational isomorphism, but I haven't even tried to decide whether this would work.]

I would have to see the proof of H-M to see whether it can be adapted to answer this question. Can you post a link to the paper? Or, if you need to know ASAP, ask Bjorn Poonen -- he eats questions like this for breakfast.

Finally, let me remark that over a non-algebraically closed field, a principal homogeneous space under an abelian variety may have higher embedding dimension than the (Albanese) abelian variety itself. The easy example of this is that if a smooth curve of genus one can be emedded in $\mathbb{P}^2$, then for geometric reasons it must embed as a cubic and therefore has a rational point of degree at most $3$. [Actually, it is possible that this is the only example. By the same theorems Georges quoted above, the only other possibility is a phs which does not embed in $\mathbb{P}^4$ while its Albanese abelian surface does.]