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1. • CommentRowNumber1.
   • CommentAuthorTodd_Trimble
   • CommentTimeApr 18th 2013
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexIn the Idea section of [[Zariski site]], I included a little patch which includes the [[little site]] notion, as well as the big site.

   In the Idea section of Zariski site, I included a little patch which includes the little site notion, as well as the big site.

2. • CommentRowNumber2.
   • CommentAuthorIngoBlechschmidt
   • CommentTimeNov 20th 2014
   • (edited Nov 20th 2014)
   • PermaLink
   Author: IngoBlechschmidt
   Format: MarkdownItexAdded to _[[Zariski site]]_ an explicit presentation of the Kripke–Joyal semantics of the big and little Zariski toposes.

   Added to Zariski site an explicit presentation of the Kripke–Joyal semantics of the big and little Zariski toposes.

3. • CommentRowNumber3.
   • CommentAuthorIngoBlechschmidt
   • CommentTimeMay 8th 2015
   • (edited May 8th 2015)
   • PermaLink
   Author: IngoBlechschmidt
   Format: MarkdownItexAt _[[Zariski site]]_, there was a remark that the canonical morphisms $Spec A[S^{-1}] \to Spec A$ are always open immersions. This is false, as can be deduced from a [MathOverflow discussion](http://mathoverflow.net/questions/20782/ring-theoretic-characterization-of-open-affines). I corrected the article; luckily, the rest of the article didn't depend on the wrong statement. Furthermore, I believe that such a morphism is an open immersion if and only if localizing away from $S$ is the same as localizing away from a certain single element. I might add this observation somewhere as well. *Edit:* The purported proof goes like this. The second criterion stated at the MathOverflow page says that $Spec A[S^{-1}] \to Spec A$ is an open immersion if and only if the extension of the ideal $T = \{ f \in A | A[f^{-1}] \to A[S^{-1},f^{-1}] \text{ is an isomorphism} \} \subseteq A$ in $A[S^{-1}]$ is the unit ideal. A short calculation shows that $T = \bigcap_{s \in S} \sqrt{(s)}$. Therefore $(T) = (1)$ as ideals of $A[S^{-1}]$ if and only if $T \cap S \neq \emptyset$, i.e. if and only if there exists an element $s_0 \in S$ such that localizing away from $s_0$ is the same as localizing away from $S$.

   At Zariski site, there was a remark that the canonical morphisms $Spec A[S^{-1}] \to Spec A$ are always open immersions. This is false, as can be deduced from a MathOverflow discussion. I corrected the article; luckily, the rest of the article didn’t depend on the wrong statement.

   Furthermore, I believe that such a morphism is an open immersion if and only if localizing away from $S$ is the same as localizing away from a certain single element. I might add this observation somewhere as well.

   Edit: The purported proof goes like this. The second criterion stated at the MathOverflow page says that $Spec A[S^{-1}] \to Spec A$ is an open immersion if and only if the extension of the ideal $T = \{ f \in A | A[f^{-1}] \to A[S^{-1},f^{-1}] \text{ is an isomorphism} \} \subseteq A$ in $A[S^{-1}]$ is the unit ideal. A short calculation shows that $T = \bigcap_{s \in S} \sqrt{(s)}$. Therefore $(T) = (1)$ as ideals of $A[S^{-1}]$ if and only if $T \cap S \neq \emptyset$, i.e. if and only if there exists an element $s_0 \in S$ such that localizing away from $s_0$ is the same as localizing away from $S$.

4. • CommentRowNumber4.
   • CommentAuthorDavidRoberts
   • CommentTimeMay 8th 2015
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   Author: DavidRoberts
   Format: MarkdownItexPresumably also the same if $S$ is finite.

   Presumably also the same if $S$ is finite.

5. • CommentRowNumber5.
   • CommentAuthorIngoBlechschmidt
   • CommentTimeMay 8th 2015
   • PermaLink
   Author: IngoBlechschmidt
   Format: MarkdownItexAdded a short remark that sheafification does not change the set of sections over local rings.

   Added a short remark that sheafification does not change the set of sections over local rings.

6. • CommentRowNumber6.
   • CommentAuthorIngoBlechschmidt
   • CommentTimeMay 8th 2015
   • (edited May 8th 2015)
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   Author: IngoBlechschmidt
   Format: MarkdownItexDavid: Indeed; but localizing away from a finite set of elements is the same as localizing away from their product, so this is not yet a counterexample.

   David: Indeed; but localizing away from a finite set of elements is the same as localizing away from their product, so this is not yet a counterexample.

7. • CommentRowNumber7.
   • CommentAuthorDavidRoberts
   • CommentTimeMay 8th 2015
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   Author: DavidRoberts
   Format: MarkdownItexIngo - I was thinking more in terms of finite intersections of opens are open, rather than algebraic properties of localisations. And conversely, there's little reason to assume in general that infinite intersections of opens (one for each element of $S$) are open.

   Ingo - I was thinking more in terms of finite intersections of opens are open, rather than algebraic properties of localisations. And conversely, there’s little reason to assume in general that infinite intersections of opens (one for each element of $S$) are open.