# Projective varieties - basics

I am taking an introductive course in (real) algebraic geometry and I got stuck at some basic exercises.

They regard affine and (real) projective varieties, as follows:

1. Prove that the punctured projective space, $\mathbb{P}^n - \{x\}$ is neither projective, nor quasi-affine, when $n \geq 2$.

2. Prove that $\mathbb{P}^1 \times \mathbb{P}^1$ and $\mathbb{P}^2$ are birationally equivalent, but not isomorphic.

Now, to clarify some things. We study more or less based on I. R. Shafarevich - Basic Algebraic Geometry (or something like that).

Our definitions of the notions involved are:

• X is quasi-affine if it is a Zariski open set in an affine variety

• $\mathbb{P}^n$ is supposed to mean $\mathbb{P}^n(k)$, for an algebraically closed field $k$ and is the space of "directions" in $\mathbb{A}^{n+1}-\{0\}$. Specifically, $\mathbb{P}^n=\mathbb{A}^n/\sim$, where $x\sim y \Leftrightarrow \exists \lambda \in k, \ s.t. x=\lambda\cdot y$.

• birational equivalence means that there exist rational functions from either to the other, whose composite is the identity (either way), but that these need not be defined everywhere

• isomorphism is usually treated in terms of isomorphic fields under the isomorphism induced by the initial morphism.

Note that the course is absolutely basic, without (co)homology, schemes, sheaves etc. Just the basics that I listed, along with Krull dimension.

Thank you.

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1) a) The variety $V=\mathbb{P}^n - \{x\}$ is not projective because it is not compact (an argument valid for $k=\mathbb C$ ).
b) It is not quasi-affine because the global functions $\Gamma(V,\mathcal O_V)=k \;$ do not generate the sheaf $\mathcal O_V$.
[This argument may be a bit premature with respect to your present knowledge; if that is the case, come back to this answer a little later. Anyway the concept of quasi-affine variety is essentially useless in an introductory course. You have plenty of vital notions to absorb before.]

2) a) Represent $\mathbb{P}^1 \times \mathbb{P}^1$ as a quadric $Q\subset \mathbb{P}^3$ and project $Q$ on a plane $P \subset \mathbb{P}^3$ from a point $p \in \mathbb{P}^3$ not on $Q$. This will yield a birational equivalence.
b) Two curves in $\mathbb{P}^2$ always intersect (weak Bézout) , whereas for $a\neq b$ the curves $\lbrace a\rbrace \times \mathbb{P}^1\subset \mathbb{P}^1 \times \mathbb{P}^1$ and $\lbrace b\rbrace \times \mathbb{P}^1\subset \mathbb{P}^1 \times \mathbb{P}^1$ are disjoint.

Edit Here are two alternative proofs addressing Adrian's request in his comment.

1) a) Over an arbitrary algebraically closed field $k$, the variety $V=\mathbb{P}^n_k \setminus \lbrace x\rbrace$ is not projective.
Suppose $x=[1:0:...:0]$ and consider the curve $\;\mathbb A^1_k\setminus \lbrace 0\rbrace\to V:t\mapsto [1:t:...:0]$.It cannot be extended across $t=0$ but if $V$ were projective it could.

1) b) The variety $V$ is not quasi-affine either. If it were an open subset of the affine variety $W$, its $k$-algebra of global regular functions $k[V]$ would suffice to separate its points: just take the restrictions $f|V$ of the global functions $f$ on $W$.
This is not the case at all since the global regular functions on $V$ extend to $\mathbb{P}^n$ and are thus constant : $k[V]=k$

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Thank you very much, but: 1.a) We generally work over a generic algebraically closed field, so I suppose a particular argument for $k=\mathbb{C}$ would not do... b) I have zero knowledge about sheafs... I am quite sure that there is an elementary argument, since our prof usually gives easy questions :) 2. a&b) Great, thank you! –  AdrianM Jan 22 '12 at 14:18
Dear @Adrian, I have written an edit giving a more general or more elementary proof of the statements you mention. –  Georges Elencwajg Jan 22 '12 at 16:46
Thanks again. It's not entirely clear, to be honest, but surely it will become once I study a bit more. I find it quite hard many times when I study something which I don't understand or like from a textbook to accept some explanations which don't resemble the textbook. It is the case here my knowledge regarding algebraic geometry is veeeery fragile :) Anyway, sincerely thank you! –  AdrianM Jan 22 '12 at 17:43
Dear @Adrian: algebraic geometry is a hard subject to learn for everybody because of the numerous techniques involved. It can also be frustrating because intuitively obvious results like "$\mathbb A^n$ has dimension $n$ " are difficult to prove rigorously. So don't be discouraged by your difficulties: you will overcome them! –  Georges Elencwajg Jan 22 '12 at 18:46
Ha-ha! Thank you, sir! The problem is that I have a final examination tommorow morning and questions like these are far from my reach. It bugs me, because they seem basic, like stuff you obtain as a direct consequence of the definitions. However, thankfully I did grasp most of the other subjects proposed. :) –  AdrianM Jan 22 '12 at 19:26