Restriction of a scheme to an open subset.

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If $(X,mathscr O_X)$ is a scheme and $U$ an open subset of $X$, how does it follow that $(U,mathscr O_X_U)$ is a scheme? I found this as a remark in Bosch's book, 'Algebraic Geometry and Commutative Algebra' just after the definition of a scheme.



I don't get this even when the scheme is affine(except when the open set is basic).




algebraic-geometry schemes




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    In an affine scheme, every open set is the union of basic open sets.
    – Lord Shark the Unknown
    Jul 22 at 18:59














up vote
1
down vote

favorite












If $(X,mathscr O_X)$ is a scheme and $U$ an open subset of $X$, how does it follow that $(U,mathscr O_X_U)$ is a scheme? I found this as a remark in Bosch's book, 'Algebraic Geometry and Commutative Algebra' just after the definition of a scheme.



I don't get this even when the scheme is affine(except when the open set is basic).




algebraic-geometry schemes




share|cite|improve this question















  • 1




    In an affine scheme, every open set is the union of basic open sets.
    – Lord Shark the Unknown
    Jul 22 at 18:59












up vote
1
down vote

favorite









up vote
1
down vote

favorite











If $(X,mathscr O_X)$ is a scheme and $U$ an open subset of $X$, how does it follow that $(U,mathscr O_X_U)$ is a scheme? I found this as a remark in Bosch's book, 'Algebraic Geometry and Commutative Algebra' just after the definition of a scheme.



I don't get this even when the scheme is affine(except when the open set is basic).




algebraic-geometry schemes




share|cite|improve this question











If $(X,mathscr O_X)$ is a scheme and $U$ an open subset of $X$, how does it follow that $(U,mathscr O_X_U)$ is a scheme? I found this as a remark in Bosch's book, 'Algebraic Geometry and Commutative Algebra' just after the definition of a scheme.



I don't get this even when the scheme is affine(except when the open set is basic).





algebraic-geometry schemes





share|cite|improve this question










share|cite|improve this question




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asked Jul 22 at 18:54









Jehu314

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457







  • 1




    In an affine scheme, every open set is the union of basic open sets.
    – Lord Shark the Unknown
    Jul 22 at 18:59












  • 1




    In an affine scheme, every open set is the union of basic open sets.
    – Lord Shark the Unknown
    Jul 22 at 18:59







1




1




In an affine scheme, every open set is the union of basic open sets.
– Lord Shark the Unknown
Jul 22 at 18:59




In an affine scheme, every open set is the union of basic open sets.
– Lord Shark the Unknown
Jul 22 at 18:59










1 Answer
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Pick an affine cover $left mathcalU_i right$ of $X$ and a cover by distinguished open sets $left operatornameD left( f_ij right) right$ for each $mathcalU_i$.



Then for any open subset $mathcalU$ of X, $mathcalU cap mathcalU_i$ is an open subset of the affine $mathcalU_i$. Thus $mathcalU cap mathcalU_i$ is covered by distinguished affines $operatornameD left( f_ij right)$ for some j's.



And so $mathcalU$ has an affine cover by $operatornameD left( f_ij right)$ for some i,j's.






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  • So the restriction of an affine scheme is not necessarily affine?
    – Jehu314
    Jul 22 at 19:14






  • 2




    @Jehu312 Yes, open subschemes of affine schemes are not necessarily affine. Consider the punctured plane as an example.
    – Jake
    Jul 22 at 19:16










  • Thank you very much, I hadn't considered that possibility.
    – Jehu314
    Jul 22 at 19:18










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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










Pick an affine cover $left mathcalU_i right$ of $X$ and a cover by distinguished open sets $left operatornameD left( f_ij right) right$ for each $mathcalU_i$.



Then for any open subset $mathcalU$ of X, $mathcalU cap mathcalU_i$ is an open subset of the affine $mathcalU_i$. Thus $mathcalU cap mathcalU_i$ is covered by distinguished affines $operatornameD left( f_ij right)$ for some j's.



And so $mathcalU$ has an affine cover by $operatornameD left( f_ij right)$ for some i,j's.






share|cite|improve this answer























  • So the restriction of an affine scheme is not necessarily affine?
    – Jehu314
    Jul 22 at 19:14






  • 2




    @Jehu312 Yes, open subschemes of affine schemes are not necessarily affine. Consider the punctured plane as an example.
    – Jake
    Jul 22 at 19:16










  • Thank you very much, I hadn't considered that possibility.
    – Jehu314
    Jul 22 at 19:18














up vote
3
down vote



accepted










Pick an affine cover $left mathcalU_i right$ of $X$ and a cover by distinguished open sets $left operatornameD left( f_ij right) right$ for each $mathcalU_i$.



Then for any open subset $mathcalU$ of X, $mathcalU cap mathcalU_i$ is an open subset of the affine $mathcalU_i$. Thus $mathcalU cap mathcalU_i$ is covered by distinguished affines $operatornameD left( f_ij right)$ for some j's.



And so $mathcalU$ has an affine cover by $operatornameD left( f_ij right)$ for some i,j's.






share|cite|improve this answer























  • So the restriction of an affine scheme is not necessarily affine?
    – Jehu314
    Jul 22 at 19:14






  • 2




    @Jehu312 Yes, open subschemes of affine schemes are not necessarily affine. Consider the punctured plane as an example.
    – Jake
    Jul 22 at 19:16










  • Thank you very much, I hadn't considered that possibility.
    – Jehu314
    Jul 22 at 19:18












up vote
3
down vote



accepted







up vote
3
down vote



accepted






Pick an affine cover $left mathcalU_i right$ of $X$ and a cover by distinguished open sets $left operatornameD left( f_ij right) right$ for each $mathcalU_i$.



Then for any open subset $mathcalU$ of X, $mathcalU cap mathcalU_i$ is an open subset of the affine $mathcalU_i$. Thus $mathcalU cap mathcalU_i$ is covered by distinguished affines $operatornameD left( f_ij right)$ for some j's.



And so $mathcalU$ has an affine cover by $operatornameD left( f_ij right)$ for some i,j's.






share|cite|improve this answer















Pick an affine cover $left mathcalU_i right$ of $X$ and a cover by distinguished open sets $left operatornameD left( f_ij right) right$ for each $mathcalU_i$.



Then for any open subset $mathcalU$ of X, $mathcalU cap mathcalU_i$ is an open subset of the affine $mathcalU_i$. Thus $mathcalU cap mathcalU_i$ is covered by distinguished affines $operatornameD left( f_ij right)$ for some j's.



And so $mathcalU$ has an affine cover by $operatornameD left( f_ij right)$ for some i,j's.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 22 at 19:12


























answered Jul 22 at 19:06









Jake

9241621




9241621











  • So the restriction of an affine scheme is not necessarily affine?
    – Jehu314
    Jul 22 at 19:14






  • 2




    @Jehu312 Yes, open subschemes of affine schemes are not necessarily affine. Consider the punctured plane as an example.
    – Jake
    Jul 22 at 19:16










  • Thank you very much, I hadn't considered that possibility.
    – Jehu314
    Jul 22 at 19:18
















  • So the restriction of an affine scheme is not necessarily affine?
    – Jehu314
    Jul 22 at 19:14






  • 2




    @Jehu312 Yes, open subschemes of affine schemes are not necessarily affine. Consider the punctured plane as an example.
    – Jake
    Jul 22 at 19:16










  • Thank you very much, I hadn't considered that possibility.
    – Jehu314
    Jul 22 at 19:18















So the restriction of an affine scheme is not necessarily affine?
– Jehu314
Jul 22 at 19:14




So the restriction of an affine scheme is not necessarily affine?
– Jehu314
Jul 22 at 19:14




2




2




@Jehu312 Yes, open subschemes of affine schemes are not necessarily affine. Consider the punctured plane as an example.
– Jake
Jul 22 at 19:16




@Jehu312 Yes, open subschemes of affine schemes are not necessarily affine. Consider the punctured plane as an example.
– Jake
Jul 22 at 19:16












Thank you very much, I hadn't considered that possibility.
– Jehu314
Jul 22 at 19:18




Thank you very much, I hadn't considered that possibility.
– Jehu314
Jul 22 at 19:18












 

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