Welcome to mathimatikoi.org forum; Enjoy your visit here.

Localization

Algebraic Geometry
Post Reply
PJPu17
Articles: 0
Posts: 10
Joined: Mon Oct 17, 2016 9:33 pm

Localization

#1

Post by PJPu17 » Mon Oct 17, 2016 9:51 pm

Let $R$ a ring , and $ X =Spec(R)$, we define the presheaf of localization by open subsets $ U\subset X$, as $R′(U)=R_{SU}$ wbere $SU=\{ f\in R : (f)_{0}\cap U=\emptyset \}$, Is in a general case a sheaf or do exists some examples of rings where this prehseaf is not a sheaf?,


When we work with basic open subsets i have proved that the sheaf condition is satisfied, but for the general case i´m not pretty sure. For Dedekind´s rings where all open set is basic it is obvius that the condition is trivally satisfied, but the important question is: Do exists some topological or algebraic conditions in the space to claim that the sheaf condition is satisfied?
Observation:
$(f)_{0}=\{ \mathbb{p} \in Spec(R) : f \subset \mathbb{p} \}$
Tsakanikas Nickos
Community Team
Community Team
Articles: 0
Posts: 314
Joined: Tue Nov 10, 2015 8:25 pm

Re: Localization

#2

Post by Tsakanikas Nickos » Mon Oct 17, 2016 11:09 pm

Hello!

Allow me to say the following: Note that
  • $ (f)_{0} = \left\{ \mathfrak{p} \in \text{Spec}(R) \ \big| \ f \in \mathfrak{p} \right\} = \left\{ \mathfrak{p} \in \text{Spec}(R) \ \big| \ (f) = fR \subset \mathfrak{p} \right\} = V(fR) $
  • $ SU = \left\{ f \in R \ \big| \ (f)_{0} \cap U = \emptyset \right\} = \left\{ f \in R \ \big| \ U \subset D(f) \right\} $
where $ D(f) = \text{Spec}(R) \smallsetminus V(fR) $. Now, by using the above observations, it is easy to see that $ U \mapsto R_{SU} $ defines a presheaf on $X$, say $ \mathscr{R} $.


Since you showed that the sheaf conditions are verified on the basis $ \mathcal{B} = \left\{ D(f) \right\}_{f \in R} $ of the Zariski topology on $X$, you conclude that the $ \mathcal{B} $-presheaf $ \mathscr{R} $ extends uniquely (up to isomorphism) to a sheaf on $X$. [For details, see "Liu - Algebraic Geometry and Arithmetic Curves - Remark 2.2.6"]


Does this answer your question?
PJPu17
Articles: 0
Posts: 10
Joined: Mon Oct 17, 2016 9:33 pm

Re: Localization

#3

Post by PJPu17 » Tue Oct 18, 2016 1:52 pm

I understand you but i´m asking for a particular example of topological space where this presheaf is not a sheaf, or an explanation about why this presheaf is not always a sheaf.
Tsakanikas Nickos
Community Team
Community Team
Articles: 0
Posts: 314
Joined: Tue Nov 10, 2015 8:25 pm

Re: Localization

#4

Post by Tsakanikas Nickos » Tue Oct 18, 2016 5:03 pm

I do not have an answer to this question! Maybe the lemma on how to glue sheaves [See for example Hartshorne / Algebraic Geometry / Chapter II / Exercise 1.22] helps you obtain the answer you want.

You could also try to see if this gluing construction works in "simple" examples of non-affine schemes that you know, e.g. you could try to see if it works for $ \mathbb{A}^{2}_{k} \smallsetminus \{ (0,0 \} $, the affine plane without the origin. Maybe thinking thoroughly on this example lead you to a conclusion whether you always get a sheaf or what is the problem and you do not get one on your topological space.

Personally, I thought that $ \mathbb{A}^{2}_{k} \smallsetminus \{ (0,0 \} $ might provide us with a counterexample, but now I feel that one can actually glue these sheaves (you defined) on $ D(x) \cong \text{Spec}(k[x,y]_{x}) $ and $ D(y) \cong \text{Spec}(k[x,y]_{y}) $ and obtain a sheaf on the whole $ \mathbb{A}^{2}_{k} \smallsetminus \{ (0,0 \} = D(x) \cup D(y) $. What do you think? On $ D(xy) = D(x) \cap D(y) $ the localisations of the corresponding rings are isomorphic (to $ k[x,y,\frac{1}{x},\frac{1}{y}] = k[x,y]_{xy} $). But if this does not hold in a general case, then you cannot get an isomorphism of the restriction of the sheaves to such an intersection and thus cannot glue the sheaves on the spectra to a sheaf on the whole space. (Maybe this is where the problem lies.)
Post Reply