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:
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Prove that the punctured projective space, $\mathbb{P}^n – \{x\}$ is neither projective, nor quasi-affine, when $n \geq 2$.
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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:
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X is quasi-affine if it is a Zariski open set in an affine variety
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$\mathbb{P}^n$ 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 $x\sim y \Leftrightarrow \exists \lambda \in k, \ s.t. x=\lambda\cdot y$.
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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
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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.
Best Answer
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 : $k[V]=k$