Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $Rng$ be the category of commutative rings. Let $Loc$ be the category of locally ringed spaces. Let $(X, \mathcal{O}_X)$ be an locally ringed space. Then $\Gamma(X) = \Gamma(X, \mathcal{O}_X)$ is an commutative ring. Hence $\Gamma(X)$ induces an functor $\Gamma\colon Loc \rightarrow Rng^o$, where $Rng^o$ is the oposite category of $Rng$. Does $\Gamma$ have an adjoint functor?

share|improve this question
    
See my answer to this question: math.stackexchange.com/questions/56854/… –  Keenan Kidwell Nov 23 '12 at 23:51
    
@KeenanKidwell: I was just about to provide the link to your very nice answer :) –  Rankeya Nov 24 '12 at 0:05
    
Dear @Rankeya, Thanks for the kind words. –  Keenan Kidwell Nov 24 '12 at 0:13
1  
I just want to mention that when I learned this result, particularly the fact that, when you have a morphism $X\rightarrow\mathrm{Spec}(A)$ with $X$ an LRS, the map on global sections determines the underlying map of topological spaces in the only possible way it could, it completely changed my understanding of schemes, and how to work with affine opens of general (non-affine) schemes. I use to be bothered by the definition of an affine open of an abstract scheme as "an open which is isomorphic as an LRS to the spectrum of some ring." Does the isomorphism matter? Are there lots of them? –  Keenan Kidwell Nov 24 '12 at 0:30
1  
With this result you can prove that a LRS (not just a scheme!) $X$ admits a unique morphism of LRS $can:X\rightarrow\mathrm{Spec}(\mathscr{O}_X(X))$, and it is an isomorphism if and only $X$ is an affine scheme in the sense that it admits some isomorphism to the spectrum of some ring. I actually greatly prefer to think of the definition of an affine scheme as "a locally ringed space for which the canonical morphism to the spectrum of its global sections is an isomorphism." I felt like I broke through a major barrier in understanding of schemes when I learned this result in the context of LRS. –  Keenan Kidwell Nov 24 '12 at 0:32

1 Answer 1

up vote 5 down vote accepted

The $Spec$ functor is the desired adjoint functor to the category of locally ringed spaces (it is right adjoint to $\Gamma$). I think Hartshorne has an exercise where he asks us to prove this when $Loc$ is replaced by the category of schemes.

I like Anton Geraschenko's answer here: http://mathoverflow.net/questions/731/points-in-algebraic-geometry-why-shift-from-m-spec-to-spec/756#756

share|improve this answer
    
I need a proof of the stated fact. –  Makoto Kato Nov 23 '12 at 23:44
    
Look at the second answer here: mathoverflow.net/questions/57777/… –  Rankeya Nov 23 '12 at 23:47
    
+1. But I still would like to know a complete proof. I don't have EGA I second edition, by the way. –  Makoto Kato Nov 24 '12 at 0:04
    
Apparently the Stacks Project has a proof of this. But, I am not being able to locate it at the moment. In any case, always try the Stacks Project. It has a great new search feature which is pretty useful, and all its exposition is pretty detailed as well. –  Rankeya Nov 24 '12 at 0:07
1  
It's Lemma 01I1 in Stacks. –  Zhen Lin Nov 24 '12 at 8:39

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.