The 'Locally Ringed' condition in the definition of a scheme.

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Is the 'Locally Ringed' condition in the definition of a Scheme redundant? My question is, if it admits a cover by Affine Schemes, does it follow that the Ringed space is Locally Ringed?



More generally, if a Ringed Space $(X,mathscrO_X)$ admits a cover $U_i$ such that each $(U_i,mathscrO_X_U_i)$ is Locally Ringed, is $(X,mathscrO_X)$ itself Locally Ringed?




algebraic-geometry schemes ringed-spaces




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  • I think you're right.
    – Suzet
    Aug 26 at 8:37










  • Have you got a proof?
    – Jehu314
    Aug 26 at 8:44






  • 3




    Take any $xin X$ and fix $U_i$ such that $xin U_i$. Then the stalk of $mathcal O_X$ at $x$ is the same as the stalk of its restriction $mathscrO_X_U_i$ to $U_i$, hence it is a local ring.
    – Suzet
    Aug 26 at 8:48






  • 2




    I also agree. I think the literature phrases the definition as it does to highlight that a scheme is indeed a locally ringed space, which is often important. Introducing it as a ringed space for the sake of minimalism might be confusing in a larger number of cases compared to doing it this way =).
    – Jesko Hüttenhain
    Aug 26 at 8:49







  • 3




    It's important that a morphism of schemes is a morphism in the category of locally ringed spaces, not in the category of ringed spaces.
    – Lord Shark the Unknown
    Aug 26 at 9:26














up vote
3
down vote

favorite












Is the 'Locally Ringed' condition in the definition of a Scheme redundant? My question is, if it admits a cover by Affine Schemes, does it follow that the Ringed space is Locally Ringed?



More generally, if a Ringed Space $(X,mathscrO_X)$ admits a cover $U_i$ such that each $(U_i,mathscrO_X_U_i)$ is Locally Ringed, is $(X,mathscrO_X)$ itself Locally Ringed?




algebraic-geometry schemes ringed-spaces




share|cite|improve this question




















  • I think you're right.
    – Suzet
    Aug 26 at 8:37










  • Have you got a proof?
    – Jehu314
    Aug 26 at 8:44






  • 3




    Take any $xin X$ and fix $U_i$ such that $xin U_i$. Then the stalk of $mathcal O_X$ at $x$ is the same as the stalk of its restriction $mathscrO_X_U_i$ to $U_i$, hence it is a local ring.
    – Suzet
    Aug 26 at 8:48






  • 2




    I also agree. I think the literature phrases the definition as it does to highlight that a scheme is indeed a locally ringed space, which is often important. Introducing it as a ringed space for the sake of minimalism might be confusing in a larger number of cases compared to doing it this way =).
    – Jesko Hüttenhain
    Aug 26 at 8:49







  • 3




    It's important that a morphism of schemes is a morphism in the category of locally ringed spaces, not in the category of ringed spaces.
    – Lord Shark the Unknown
    Aug 26 at 9:26












up vote
3
down vote

favorite









up vote
3
down vote

favorite











Is the 'Locally Ringed' condition in the definition of a Scheme redundant? My question is, if it admits a cover by Affine Schemes, does it follow that the Ringed space is Locally Ringed?



More generally, if a Ringed Space $(X,mathscrO_X)$ admits a cover $U_i$ such that each $(U_i,mathscrO_X_U_i)$ is Locally Ringed, is $(X,mathscrO_X)$ itself Locally Ringed?




algebraic-geometry schemes ringed-spaces




share|cite|improve this question












Is the 'Locally Ringed' condition in the definition of a Scheme redundant? My question is, if it admits a cover by Affine Schemes, does it follow that the Ringed space is Locally Ringed?



More generally, if a Ringed Space $(X,mathscrO_X)$ admits a cover $U_i$ such that each $(U_i,mathscrO_X_U_i)$ is Locally Ringed, is $(X,mathscrO_X)$ itself Locally Ringed?





algebraic-geometry schemes ringed-spaces





share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 26 at 8:13









Jehu314

657




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  • I think you're right.
    – Suzet
    Aug 26 at 8:37










  • Have you got a proof?
    – Jehu314
    Aug 26 at 8:44






  • 3




    Take any $xin X$ and fix $U_i$ such that $xin U_i$. Then the stalk of $mathcal O_X$ at $x$ is the same as the stalk of its restriction $mathscrO_X_U_i$ to $U_i$, hence it is a local ring.
    – Suzet
    Aug 26 at 8:48






  • 2




    I also agree. I think the literature phrases the definition as it does to highlight that a scheme is indeed a locally ringed space, which is often important. Introducing it as a ringed space for the sake of minimalism might be confusing in a larger number of cases compared to doing it this way =).
    – Jesko Hüttenhain
    Aug 26 at 8:49







  • 3




    It's important that a morphism of schemes is a morphism in the category of locally ringed spaces, not in the category of ringed spaces.
    – Lord Shark the Unknown
    Aug 26 at 9:26
















  • I think you're right.
    – Suzet
    Aug 26 at 8:37










  • Have you got a proof?
    – Jehu314
    Aug 26 at 8:44






  • 3




    Take any $xin X$ and fix $U_i$ such that $xin U_i$. Then the stalk of $mathcal O_X$ at $x$ is the same as the stalk of its restriction $mathscrO_X_U_i$ to $U_i$, hence it is a local ring.
    – Suzet
    Aug 26 at 8:48






  • 2




    I also agree. I think the literature phrases the definition as it does to highlight that a scheme is indeed a locally ringed space, which is often important. Introducing it as a ringed space for the sake of minimalism might be confusing in a larger number of cases compared to doing it this way =).
    – Jesko Hüttenhain
    Aug 26 at 8:49







  • 3




    It's important that a morphism of schemes is a morphism in the category of locally ringed spaces, not in the category of ringed spaces.
    – Lord Shark the Unknown
    Aug 26 at 9:26















I think you're right.
– Suzet
Aug 26 at 8:37




I think you're right.
– Suzet
Aug 26 at 8:37












Have you got a proof?
– Jehu314
Aug 26 at 8:44




Have you got a proof?
– Jehu314
Aug 26 at 8:44




3




3




Take any $xin X$ and fix $U_i$ such that $xin U_i$. Then the stalk of $mathcal O_X$ at $x$ is the same as the stalk of its restriction $mathscrO_X_U_i$ to $U_i$, hence it is a local ring.
– Suzet
Aug 26 at 8:48




Take any $xin X$ and fix $U_i$ such that $xin U_i$. Then the stalk of $mathcal O_X$ at $x$ is the same as the stalk of its restriction $mathscrO_X_U_i$ to $U_i$, hence it is a local ring.
– Suzet
Aug 26 at 8:48




2




2




I also agree. I think the literature phrases the definition as it does to highlight that a scheme is indeed a locally ringed space, which is often important. Introducing it as a ringed space for the sake of minimalism might be confusing in a larger number of cases compared to doing it this way =).
– Jesko Hüttenhain
Aug 26 at 8:49





I also agree. I think the literature phrases the definition as it does to highlight that a scheme is indeed a locally ringed space, which is often important. Introducing it as a ringed space for the sake of minimalism might be confusing in a larger number of cases compared to doing it this way =).
– Jesko Hüttenhain
Aug 26 at 8:49





3




3




It's important that a morphism of schemes is a morphism in the category of locally ringed spaces, not in the category of ringed spaces.
– Lord Shark the Unknown
Aug 26 at 9:26




It's important that a morphism of schemes is a morphism in the category of locally ringed spaces, not in the category of ringed spaces.
– Lord Shark the Unknown
Aug 26 at 9:26















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