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Why is $\mathbb A^1$ not projective? i.e. why is it not isomorphic to a projective variety?

I'm drawing a complete blank here...

Any help would be appreciated. Thanks!

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1 Answer 1

The ring of regular functions of a projective variety is the underlying field, however the ring of regular functions of $\mathbb{A}^{1}$ is $k[t]$. In fact the only projective and affine varieties are the singletons. This can be shown using the bijection between:

$\operatorname{Hom}(X,Y) \leftrightarrow \operatorname{Hom}(A(Y),\mathcal{O}(X))$, where $X$ is any variety and $Y$ an affine variety.

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what are regular functions? why The ring of regular functions of a projective variety is the underlying field, however the ring of regular functions of $\mathbb{A}^1$ is $k[t]$? –  Une Femme Douce May 18 '12 at 8:37
    
This is not correct unless you require k to be algebraically closed. F.ex when $k=\mathbb{R}$, the function $y^2/(x^2+y^2)$ is regular on all of $\mathbb{P}^1$. –  Fredrik Meyer May 18 '12 at 13:46

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