Morphism from a scheme to the spectra of global section
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This is a small inquiry, but I just want to be sure about it.
Let $X$ be a scheme. I am trying to understand what it is meant by the cannonical morphism between $X rightarrow Spec (Gamma(X,mathcal{O}_{X})).$ Is it the morphism of schemes induced by the identity map of rings $Gamma(X,mathcal{O}_{X}) rightarrow Gamma(X,mathcal{O}_{X})$ and hence by bijection between morphisms of scheme to affine scheme and morphisms of rings of global sections, the aforementioned morphism of scheme is isomorphism?
algebraic-geometry schemes
asked Dec 7 '18 at 23:45
solgaleosolgaleo
6911
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add a comment |
$begingroup$
This is a small inquiry, but I just want to be sure about it.
Let $X$ be a scheme. I am trying to understand what it is meant by the cannonical morphism between $X rightarrow Spec (Gamma(X,mathcal{O}_{X})).$ Is it the morphism of schemes induced by the identity map of rings $Gamma(X,mathcal{O}_{X}) rightarrow Gamma(X,mathcal{O}_{X})$ and hence by bijection between morphisms of scheme to affine scheme and morphisms of rings of global sections, the aforementioned morphism of scheme is isomorphism?
algebraic-geometry schemes
asked Dec 7 '18 at 23:45
solgaleosolgaleo
6911
$endgroup$
1
$begingroup$
Yes, that's one way to think about it.
$endgroup$
– Qiaochu Yuan
Dec 8 '18 at 0:34
$begingroup$
Possibly related: math.stackexchange.com/questions/1523392
$endgroup$
– Watson
Dec 8 '18 at 10:00
add a comment |
$begingroup$
This is a small inquiry, but I just want to be sure about it.
Let $X$ be a scheme. I am trying to understand what it is meant by the cannonical morphism between $X rightarrow Spec (Gamma(X,mathcal{O}_{X})).$ Is it the morphism of schemes induced by the identity map of rings $Gamma(X,mathcal{O}_{X}) rightarrow Gamma(X,mathcal{O}_{X})$ and hence by bijection between morphisms of scheme to affine scheme and morphisms of rings of global sections, the aforementioned morphism of scheme is isomorphism?
algebraic-geometry schemes
asked Dec 7 '18 at 23:45
solgaleosolgaleo
6911
$endgroup$
This is a small inquiry, but I just want to be sure about it.
Let $X$ be a scheme. I am trying to understand what it is meant by the cannonical morphism between $X rightarrow Spec (Gamma(X,mathcal{O}_{X})).$ Is it the morphism of schemes induced by the identity map of rings $Gamma(X,mathcal{O}_{X}) rightarrow Gamma(X,mathcal{O}_{X})$ and hence by bijection between morphisms of scheme to affine scheme and morphisms of rings of global sections, the aforementioned morphism of scheme is isomorphism?
algebraic-geometry schemes
algebraic-geometry schemes
asked Dec 7 '18 at 23:45
solgaleosolgaleo
6911
asked Dec 7 '18 at 23:45
solgaleosolgaleo
6911
asked Dec 7 '18 at 23:45
solgaleosolgaleo
6911
asked Dec 7 '18 at 23:45
solgaleosolgaleo
6911
asked Dec 7 '18 at 23:45
solgaleosolgaleo
6911
6911
1
$begingroup$
Yes, that's one way to think about it.
$endgroup$
– Qiaochu Yuan
Dec 8 '18 at 0:34
$begingroup$
Possibly related: math.stackexchange.com/questions/1523392
$endgroup$
– Watson
Dec 8 '18 at 10:00
add a comment |
1
$begingroup$
Yes, that's one way to think about it.
$endgroup$
– Qiaochu Yuan
Dec 8 '18 at 0:34
$begingroup$
Possibly related: math.stackexchange.com/questions/1523392
$endgroup$
– Watson
Dec 8 '18 at 10:00
1
1
$begingroup$
Yes, that's one way to think about it.
$endgroup$
– Qiaochu Yuan
Dec 8 '18 at 0:34
$begingroup$
Yes, that's one way to think about it.
$endgroup$
– Qiaochu Yuan
Dec 8 '18 at 0:34
$begingroup$
Possibly related: math.stackexchange.com/questions/1523392
$endgroup$
– Watson
Dec 8 '18 at 10:00
$begingroup$
Possibly related: math.stackexchange.com/questions/1523392
$endgroup$
– Watson
Dec 8 '18 at 10:00
add a comment |
1 Answer
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$begingroup$
I want to respond to one particular statement you made, but first I'll make things a bit more precise.
We have the adjunction $$newcommandHom{operatorname{Hom}}newcommandSpec{operatorname{Spec}}newcommandcalO{mathcal{O}}Hom(X,Spec A)simeq Hom(A,Gamma(X,calO_X)),$$ and in particular this tells us that
$$Hom(X,Spec Gamma(X,calO_X))simeq Hom(Gamma(X,calO_X),Gamma(X,calO_X)).
$$
The identity map $Gamma(X,calO_X)toGamma(X,calO_X)$ therefore gives us a map of schemes $Xto Gamma(X,calO_X)$ as you noticed, and this is indeed the canonical map. However, you seem to be under the impression that this map is therefore an isomorphism.
In general this cannot possibly be true, since if the map were an isomorphism, $X$ would necessarily have to be affine. However, if $X$ is affine, this map is indeed an isomorphism.
Let's be a little more clear how this map works then.
In fact let's be a little more clear how it works in general. Let $phi : Ato Gamma(X,calO_X)$ be a ring morphism. Let's try to understand the induced map $f : Xto Spec A$.
Let $U$ be an affine open in $X$. Then we have the maps
$$newcommandtobyxrightarrow Atoby{phi}calO_X(X)toby{r_{XU}} calO_X(U).$$
Taking $Spec$ of this sequence gives
$$Utoby{Spec r_{XU}} Spec calO_X(X) toby{Spec phi} Spec A.$$
Gluing these maps together gives the desired map from $X$ to $Spec A$.
Observe then that if $phi=newcommandid{operatorname{id}}id$, that the map $Xto Spec Gamma(X,calO_X)$ is the result of gluing the maps obtained from applying the Spec functor to the restrictions $r_{XU}:calO_X(X)to calO_X(U)$.
If $X$ is affine, then we can take $U=X$, and there's no need to glue, the map $Xto SpecGamma(X,calO_X)$ is $Spec id=id$. On the other hand, if $X$ is not affine, for example, if $X$ is a projective $k$-scheme with $k$ algebraically closed, then $Gamma(X,calO_X)=k$, and $Xto SpecGamma(X,calO_X)$ is the $k$-scheme structure morphism $Xto Spec k$, which in general is clearly not an isomorphism.
answered Dec 8 '18 at 2:46
jgonjgon
14.5k22042
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
I want to respond to one particular statement you made, but first I'll make things a bit more precise.
We have the adjunction $$newcommandHom{operatorname{Hom}}newcommandSpec{operatorname{Spec}}newcommandcalO{mathcal{O}}Hom(X,Spec A)simeq Hom(A,Gamma(X,calO_X)),$$ and in particular this tells us that
$$Hom(X,Spec Gamma(X,calO_X))simeq Hom(Gamma(X,calO_X),Gamma(X,calO_X)).
$$
The identity map $Gamma(X,calO_X)toGamma(X,calO_X)$ therefore gives us a map of schemes $Xto Gamma(X,calO_X)$ as you noticed, and this is indeed the canonical map. However, you seem to be under the impression that this map is therefore an isomorphism.
In general this cannot possibly be true, since if the map were an isomorphism, $X$ would necessarily have to be affine. However, if $X$ is affine, this map is indeed an isomorphism.
Let's be a little more clear how this map works then.
In fact let's be a little more clear how it works in general. Let $phi : Ato Gamma(X,calO_X)$ be a ring morphism. Let's try to understand the induced map $f : Xto Spec A$.
Let $U$ be an affine open in $X$. Then we have the maps
$$newcommandtobyxrightarrow Atoby{phi}calO_X(X)toby{r_{XU}} calO_X(U).$$
Taking $Spec$ of this sequence gives
$$Utoby{Spec r_{XU}} Spec calO_X(X) toby{Spec phi} Spec A.$$
Gluing these maps together gives the desired map from $X$ to $Spec A$.
Observe then that if $phi=newcommandid{operatorname{id}}id$, that the map $Xto Spec Gamma(X,calO_X)$ is the result of gluing the maps obtained from applying the Spec functor to the restrictions $r_{XU}:calO_X(X)to calO_X(U)$.
If $X$ is affine, then we can take $U=X$, and there's no need to glue, the map $Xto SpecGamma(X,calO_X)$ is $Spec id=id$. On the other hand, if $X$ is not affine, for example, if $X$ is a projective $k$-scheme with $k$ algebraically closed, then $Gamma(X,calO_X)=k$, and $Xto SpecGamma(X,calO_X)$ is the $k$-scheme structure morphism $Xto Spec k$, which in general is clearly not an isomorphism.
answered Dec 8 '18 at 2:46
jgonjgon
14.5k22042
$endgroup$
add a comment |
$begingroup$
I want to respond to one particular statement you made, but first I'll make things a bit more precise.
We have the adjunction $$newcommandHom{operatorname{Hom}}newcommandSpec{operatorname{Spec}}newcommandcalO{mathcal{O}}Hom(X,Spec A)simeq Hom(A,Gamma(X,calO_X)),$$ and in particular this tells us that
$$Hom(X,Spec Gamma(X,calO_X))simeq Hom(Gamma(X,calO_X),Gamma(X,calO_X)).
$$
The identity map $Gamma(X,calO_X)toGamma(X,calO_X)$ therefore gives us a map of schemes $Xto Gamma(X,calO_X)$ as you noticed, and this is indeed the canonical map. However, you seem to be under the impression that this map is therefore an isomorphism.
In general this cannot possibly be true, since if the map were an isomorphism, $X$ would necessarily have to be affine. However, if $X$ is affine, this map is indeed an isomorphism.
Let's be a little more clear how this map works then.
In fact let's be a little more clear how it works in general. Let $phi : Ato Gamma(X,calO_X)$ be a ring morphism. Let's try to understand the induced map $f : Xto Spec A$.
Let $U$ be an affine open in $X$. Then we have the maps
$$newcommandtobyxrightarrow Atoby{phi}calO_X(X)toby{r_{XU}} calO_X(U).$$
Taking $Spec$ of this sequence gives
$$Utoby{Spec r_{XU}} Spec calO_X(X) toby{Spec phi} Spec A.$$
Gluing these maps together gives the desired map from $X$ to $Spec A$.
Observe then that if $phi=newcommandid{operatorname{id}}id$, that the map $Xto Spec Gamma(X,calO_X)$ is the result of gluing the maps obtained from applying the Spec functor to the restrictions $r_{XU}:calO_X(X)to calO_X(U)$.
If $X$ is affine, then we can take $U=X$, and there's no need to glue, the map $Xto SpecGamma(X,calO_X)$ is $Spec id=id$. On the other hand, if $X$ is not affine, for example, if $X$ is a projective $k$-scheme with $k$ algebraically closed, then $Gamma(X,calO_X)=k$, and $Xto SpecGamma(X,calO_X)$ is the $k$-scheme structure morphism $Xto Spec k$, which in general is clearly not an isomorphism.
answered Dec 8 '18 at 2:46
jgonjgon
14.5k22042
$endgroup$
add a comment |
$begingroup$
I want to respond to one particular statement you made, but first I'll make things a bit more precise.
We have the adjunction $$newcommandHom{operatorname{Hom}}newcommandSpec{operatorname{Spec}}newcommandcalO{mathcal{O}}Hom(X,Spec A)simeq Hom(A,Gamma(X,calO_X)),$$ and in particular this tells us that
$$Hom(X,Spec Gamma(X,calO_X))simeq Hom(Gamma(X,calO_X),Gamma(X,calO_X)).
$$
The identity map $Gamma(X,calO_X)toGamma(X,calO_X)$ therefore gives us a map of schemes $Xto Gamma(X,calO_X)$ as you noticed, and this is indeed the canonical map. However, you seem to be under the impression that this map is therefore an isomorphism.
In general this cannot possibly be true, since if the map were an isomorphism, $X$ would necessarily have to be affine. However, if $X$ is affine, this map is indeed an isomorphism.
Let's be a little more clear how this map works then.
In fact let's be a little more clear how it works in general. Let $phi : Ato Gamma(X,calO_X)$ be a ring morphism. Let's try to understand the induced map $f : Xto Spec A$.
Let $U$ be an affine open in $X$. Then we have the maps
$$newcommandtobyxrightarrow Atoby{phi}calO_X(X)toby{r_{XU}} calO_X(U).$$
Taking $Spec$ of this sequence gives
$$Utoby{Spec r_{XU}} Spec calO_X(X) toby{Spec phi} Spec A.$$
Gluing these maps together gives the desired map from $X$ to $Spec A$.
Observe then that if $phi=newcommandid{operatorname{id}}id$, that the map $Xto Spec Gamma(X,calO_X)$ is the result of gluing the maps obtained from applying the Spec functor to the restrictions $r_{XU}:calO_X(X)to calO_X(U)$.
If $X$ is affine, then we can take $U=X$, and there's no need to glue, the map $Xto SpecGamma(X,calO_X)$ is $Spec id=id$. On the other hand, if $X$ is not affine, for example, if $X$ is a projective $k$-scheme with $k$ algebraically closed, then $Gamma(X,calO_X)=k$, and $Xto SpecGamma(X,calO_X)$ is the $k$-scheme structure morphism $Xto Spec k$, which in general is clearly not an isomorphism.
answered Dec 8 '18 at 2:46
jgonjgon
14.5k22042
$endgroup$
I want to respond to one particular statement you made, but first I'll make things a bit more precise.
We have the adjunction $$newcommandHom{operatorname{Hom}}newcommandSpec{operatorname{Spec}}newcommandcalO{mathcal{O}}Hom(X,Spec A)simeq Hom(A,Gamma(X,calO_X)),$$ and in particular this tells us that
$$Hom(X,Spec Gamma(X,calO_X))simeq Hom(Gamma(X,calO_X),Gamma(X,calO_X)).
$$
The identity map $Gamma(X,calO_X)toGamma(X,calO_X)$ therefore gives us a map of schemes $Xto Gamma(X,calO_X)$ as you noticed, and this is indeed the canonical map. However, you seem to be under the impression that this map is therefore an isomorphism.
In general this cannot possibly be true, since if the map were an isomorphism, $X$ would necessarily have to be affine. However, if $X$ is affine, this map is indeed an isomorphism.
Let's be a little more clear how this map works then.
In fact let's be a little more clear how it works in general. Let $phi : Ato Gamma(X,calO_X)$ be a ring morphism. Let's try to understand the induced map $f : Xto Spec A$.
Let $U$ be an affine open in $X$. Then we have the maps
$$newcommandtobyxrightarrow Atoby{phi}calO_X(X)toby{r_{XU}} calO_X(U).$$
Taking $Spec$ of this sequence gives
$$Utoby{Spec r_{XU}} Spec calO_X(X) toby{Spec phi} Spec A.$$
Gluing these maps together gives the desired map from $X$ to $Spec A$.
Observe then that if $phi=newcommandid{operatorname{id}}id$, that the map $Xto Spec Gamma(X,calO_X)$ is the result of gluing the maps obtained from applying the Spec functor to the restrictions $r_{XU}:calO_X(X)to calO_X(U)$.
If $X$ is affine, then we can take $U=X$, and there's no need to glue, the map $Xto SpecGamma(X,calO_X)$ is $Spec id=id$. On the other hand, if $X$ is not affine, for example, if $X$ is a projective $k$-scheme with $k$ algebraically closed, then $Gamma(X,calO_X)=k$, and $Xto SpecGamma(X,calO_X)$ is the $k$-scheme structure morphism $Xto Spec k$, which in general is clearly not an isomorphism.
answered Dec 8 '18 at 2:46
jgonjgon
14.5k22042
answered Dec 8 '18 at 2:46
jgonjgon
14.5k22042
answered Dec 8 '18 at 2:46
jgonjgon
14.5k22042
answered Dec 8 '18 at 2:46
jgonjgon
14.5k22042
14.5k22042
add a comment |
add a comment |
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1
$begingroup$
Yes, that's one way to think about it.
$endgroup$
– Qiaochu Yuan
Dec 8 '18 at 0:34
$begingroup$
Possibly related: math.stackexchange.com/questions/1523392
$endgroup$
– Watson
Dec 8 '18 at 10:00