proposition II.5.9 in Hartshorne
$begingroup$
I have a question about the 1 to 1 correspondence between quasi-coherent sheaf of ideals and the closed subschemes.
suppose $(i,i^×):Ysubset X$ is a closed subscheme,where $i^×:O_Xrightarrow i_*O_Y$.denote $I$ the kernel of $i^×$ denote $Z=Supp(O_X/I)$,we need to check $Y=Z$.He says the unicity is clear.but I don't know how to prove $Y=Z$,it is easy to see $Z$ is closed and $Zsubset Y$.How to show is equals?
I also find that in QingLiu's book,proposition2.24.the proof is :using $0rightarrow Irightarrow O_Xrightarrow i_*O_Yrightarrow 0$ we deduce that $I_x=O_{X,x}$ iff $xnotin Y$,where we know $(i_*O_Y)_x=O_{Y,x}$ if $xin Y$,and 0,otherwise.
I can't understand this if $xin Y$,and $O_{Y,x}=0$,we can also get $I_x=O_{X,x}$.Of course,I know this can't be true,but how to prove this material.
Thanks for your help.
algebraic-geometry schemes affine-schemes
asked Dec 14 '18 at 15:43
SkySky
1,243312
$endgroup$
add a comment |
$begingroup$
I have a question about the 1 to 1 correspondence between quasi-coherent sheaf of ideals and the closed subschemes.
suppose $(i,i^×):Ysubset X$ is a closed subscheme,where $i^×:O_Xrightarrow i_*O_Y$.denote $I$ the kernel of $i^×$ denote $Z=Supp(O_X/I)$,we need to check $Y=Z$.He says the unicity is clear.but I don't know how to prove $Y=Z$,it is easy to see $Z$ is closed and $Zsubset Y$.How to show is equals?
I also find that in QingLiu's book,proposition2.24.the proof is :using $0rightarrow Irightarrow O_Xrightarrow i_*O_Yrightarrow 0$ we deduce that $I_x=O_{X,x}$ iff $xnotin Y$,where we know $(i_*O_Y)_x=O_{Y,x}$ if $xin Y$,and 0,otherwise.
I can't understand this if $xin Y$,and $O_{Y,x}=0$,we can also get $I_x=O_{X,x}$.Of course,I know this can't be true,but how to prove this material.
Thanks for your help.
algebraic-geometry schemes affine-schemes
asked Dec 14 '18 at 15:43
SkySky
1,243312
$endgroup$
$begingroup$
I don't understand your problem. If $xin Y$, why would $O_{Y,x}$ be zero ? Since $Y$ is a scheme, its structure sheaf $O_Y$ has non zero stalks everywhere.
$endgroup$
– Roland
Dec 14 '18 at 20:11
$begingroup$
@Roland thanks,this really helps.
$endgroup$
– Sky
Dec 15 '18 at 0:32
add a comment |
$begingroup$
I have a question about the 1 to 1 correspondence between quasi-coherent sheaf of ideals and the closed subschemes.
suppose $(i,i^×):Ysubset X$ is a closed subscheme,where $i^×:O_Xrightarrow i_*O_Y$.denote $I$ the kernel of $i^×$ denote $Z=Supp(O_X/I)$,we need to check $Y=Z$.He says the unicity is clear.but I don't know how to prove $Y=Z$,it is easy to see $Z$ is closed and $Zsubset Y$.How to show is equals?
I also find that in QingLiu's book,proposition2.24.the proof is :using $0rightarrow Irightarrow O_Xrightarrow i_*O_Yrightarrow 0$ we deduce that $I_x=O_{X,x}$ iff $xnotin Y$,where we know $(i_*O_Y)_x=O_{Y,x}$ if $xin Y$,and 0,otherwise.
I can't understand this if $xin Y$,and $O_{Y,x}=0$,we can also get $I_x=O_{X,x}$.Of course,I know this can't be true,but how to prove this material.
Thanks for your help.
algebraic-geometry schemes affine-schemes
asked Dec 14 '18 at 15:43
SkySky
1,243312
$endgroup$
I have a question about the 1 to 1 correspondence between quasi-coherent sheaf of ideals and the closed subschemes.
suppose $(i,i^×):Ysubset X$ is a closed subscheme,where $i^×:O_Xrightarrow i_*O_Y$.denote $I$ the kernel of $i^×$ denote $Z=Supp(O_X/I)$,we need to check $Y=Z$.He says the unicity is clear.but I don't know how to prove $Y=Z$,it is easy to see $Z$ is closed and $Zsubset Y$.How to show is equals?
I also find that in QingLiu's book,proposition2.24.the proof is :using $0rightarrow Irightarrow O_Xrightarrow i_*O_Yrightarrow 0$ we deduce that $I_x=O_{X,x}$ iff $xnotin Y$,where we know $(i_*O_Y)_x=O_{Y,x}$ if $xin Y$,and 0,otherwise.
I can't understand this if $xin Y$,and $O_{Y,x}=0$,we can also get $I_x=O_{X,x}$.Of course,I know this can't be true,but how to prove this material.
Thanks for your help.
algebraic-geometry schemes affine-schemes
algebraic-geometry schemes affine-schemes
asked Dec 14 '18 at 15:43
SkySky
1,243312
asked Dec 14 '18 at 15:43
SkySky
1,243312
asked Dec 14 '18 at 15:43
SkySky
1,243312
asked Dec 14 '18 at 15:43
SkySky
1,243312
asked Dec 14 '18 at 15:43
SkySky
1,243312
1,243312
$begingroup$
I don't understand your problem. If $xin Y$, why would $O_{Y,x}$ be zero ? Since $Y$ is a scheme, its structure sheaf $O_Y$ has non zero stalks everywhere.
$endgroup$
– Roland
Dec 14 '18 at 20:11
$begingroup$
@Roland thanks,this really helps.
$endgroup$
– Sky
Dec 15 '18 at 0:32
add a comment |
$begingroup$
I don't understand your problem. If $xin Y$, why would $O_{Y,x}$ be zero ? Since $Y$ is a scheme, its structure sheaf $O_Y$ has non zero stalks everywhere.
$endgroup$
– Roland
Dec 14 '18 at 20:11
$begingroup$
@Roland thanks,this really helps.
$endgroup$
– Sky
Dec 15 '18 at 0:32
$begingroup$
I don't understand your problem. If $xin Y$, why would $O_{Y,x}$ be zero ? Since $Y$ is a scheme, its structure sheaf $O_Y$ has non zero stalks everywhere.
$endgroup$
– Roland
Dec 14 '18 at 20:11
$begingroup$
I don't understand your problem. If $xin Y$, why would $O_{Y,x}$ be zero ? Since $Y$ is a scheme, its structure sheaf $O_Y$ has non zero stalks everywhere.
$endgroup$
– Roland
Dec 14 '18 at 20:11
$begingroup$
@Roland thanks,this really helps.
$endgroup$
– Sky
Dec 15 '18 at 0:32
$begingroup$
@Roland thanks,this really helps.
$endgroup$
– Sky
Dec 15 '18 at 0:32
add a comment |
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$begingroup$
I don't understand your problem. If $xin Y$, why would $O_{Y,x}$ be zero ? Since $Y$ is a scheme, its structure sheaf $O_Y$ has non zero stalks everywhere.
$endgroup$
– Roland
Dec 14 '18 at 20:11
$begingroup$
@Roland thanks,this really helps.
$endgroup$
– Sky
Dec 15 '18 at 0:32