Is it possible that $\mathbb{P}^n$ is an algebraic vector bundle over some algebraic variety? This is an interesting question that my friend asked in a student seminar. I believe that the answer is NOT. Because the only global sections of $\mathcal{O}\_{\mathbb{P}^n}$ are constants. However as a total space of vector bundles $E$ over $X$, the global sections are all in $\Gamma(X, \oplus Sym^n E)=\oplus \Gamma(X,Sym^n E)$ which may not be $\mathbb{C}$ in general. The trouble case is when $\Gamma(X, E)=0$? Am I correct? Is there any other explanation?


Edit: Thanks to everyone for your comments and answer. It seems that compactness is the correct way to follow. 


However, I still wondering that why the argument on sections doesn't work. In other words, is there an example of a non-proper algebraic variety whose structure sheaf only has constant global sections.



Can $\mathbb{P}^n$ be regarded as an algebraic vector bundle over some algebraic variety?
Stephen Griffeth's argument works over any field. The total space of a vector bundle is never proper (follows by, e.g., valuative criterion for properness). On the other hand, $P^n$ is always proper.


Here is an argument that the total space of a vector bundle is not proper: A fiber of a vector bundle is isomorphic to $A^n$ . Moreover, the fiber considered as a subscheme is a closed subscheme. You can map $(A^1 \setminus 0)$ into the fiber by a map like $f(x)=1/x$. This map can't extend over zero, because if it did, then zero would be sent to something in the closure of the fiber. But the fiber is already closed, so zero would be sent to the fiber, contradiction.




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Edit: Of course, amaanush's answer is a much better answer than mine!


