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Localization
Localization
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:
$(f)_{0}=\{ \mathbb{p} \in Spec(R) : f \subset \mathbb{p} \}$
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:
$(f)_{0}=\{ \mathbb{p} \in Spec(R) : f \subset \mathbb{p} \}$

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Re: Localization
Hello!
Allow me to say the following: Note that
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?
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\} $
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?
Re: Localization
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.

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Re: Localization
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 nonaffine 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.)
You could also try to see if this gluing construction works in "simple" examples of nonaffine 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.)