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Given an affine variety, we can forget it into a variety. The other way round, given a variety $X$ we can look at $$\mathrm{spec}(\mathcal O_X(X)$$

Is this necessarily an affine variety (a priori its just an affien scheme)? If so, am I right that this implies that taking spec of the global functions is left adjoint to the forgetful functor on the level of varieties?

Also, what does this functor intuitively? If you feed in an affine variety, it spits out the same variety again. If you apply it to a complete irreducible variety (for example $\mathbb P^n$) it gets transformed into a point.

Simon Fraser
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Jan
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1 Answers1

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The functor $Sch \to Ring^{op}, X \mapsto \mathcal{O}_X(X)$ is left adjoint to the functor $\text{Spec} : Ring^{op}\to Sch$.

By a variety you mean a separated integral scheme $X$ of finite type over a field $k$? Then $O_X(X)$ does not have to be of finite type over $k$. Namely, there are integral $k$-subalgebras $A,B$ of a $k$-algebra $C$ such that $A,B$ are of finite type, but $A \cap B$ is not. Then glue $Spec(A)$ and $Spec(B)$ along $Spec(C)$ to get a counterexample.

Martin Brandenburg
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  • Argh okay, thanks Martin. However the forgetful functor from affine varieties to varieties still preserves finite limits? – Jan Oct 05 '10 at 10:54
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    Yes if you replace integral by geometrical-integral. Otherwise it is not clear at all why affine varieties have finite limits (and thus, it would be hard to check whether the forgetful functor preserves them). – Martin Brandenburg Oct 05 '10 at 11:08
  • Okay, thanks I am working over $\mathbb C$ anyway :) – Jan Oct 05 '10 at 11:22