Chracterisation of Projective Line $mathbbP^1_k$ The 2019 Stack Overflow Developer Survey Results Are InDirect image of principal divisor on one-dimensional scheme is 0Is morphism from projective space projective morphism？The importance of the structural morphism of a projective variety.Coproduct of projective schemesDegree of $f:mathbbP^1_krightarrow mathbbP^1_k$Properties of the morphism $mathbb A^1 rightarrow mathbb A^2$ given by $tmapsto (t^2,t^3)$What is the intuition behind the unit/counit for the Spec/GlobalSections adjunction?Extending a morphism to a projective schemeFamilies of genus zero curves vs Family of deformations of $mathbbP_k^1$.Set $X(mathbbQ_p)$ of $mathbbQ_p$-valued Points not Empty
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Chracterisation of Projective Line $mathbbP^1_k$
The 2019 Stack Overflow Developer Survey Results Are InDirect image of principal divisor on one-dimensional scheme is 0Is morphism from projective space projective morphism？The importance of the structural morphism of a projective variety.Coproduct of projective schemesDegree of $f:mathbbP^1_krightarrow mathbbP^1_k$Properties of the morphism $mathbb A^1 rightarrow mathbb A^2$ given by $tmapsto (t^2,t^3)$What is the intuition behind the unit/counit for the Spec/GlobalSections adjunction?Extending a morphism to a projective schemeFamilies of genus zero curves vs Family of deformations of $mathbbP_k^1$.Set $X(mathbbQ_p)$ of $mathbbQ_p$-valued Points not Empty
$begingroup$
I have a question about an argument used in Prop 50.10.4 for chracterisation of projective line $mathbbP^1_k$ from https://stacks.math.columbia.edu/tag/0C6L
Let $k$ be a field and $X$ be proper.
Consider the implication (6) -> (1)
According to the source this is a consequence from 50.10.2: https://stacks.math.columbia.edu/tag/0C6T
The problem is that 50.10.2 provides only a closed immersion $X to mathbbP^1_k$.
Why is it an isomorphism of schemes?
My considerations: Since both are irreducible curves of dimension $1$ we conclude that $X$ is homeomorphic to $mathbbP^1_k$ (as topological space) and by 50.10.2 a closed immersion as scheme morphism.
But does this already imply that it is an iso of schemes?
algebraic-geometry
asked Mar 24 at 4:01
KarlPeterKarlPeter
6691416
$endgroup$
add a comment |
$begingroup$
I have a question about an argument used in Prop 50.10.4 for chracterisation of projective line $mathbbP^1_k$ from https://stacks.math.columbia.edu/tag/0C6L
Let $k$ be a field and $X$ be proper.
Consider the implication (6) -> (1)
According to the source this is a consequence from 50.10.2: https://stacks.math.columbia.edu/tag/0C6T
The problem is that 50.10.2 provides only a closed immersion $X to mathbbP^1_k$.
Why is it an isomorphism of schemes?
My considerations: Since both are irreducible curves of dimension $1$ we conclude that $X$ is homeomorphic to $mathbbP^1_k$ (as topological space) and by 50.10.2 a closed immersion as scheme morphism.
But does this already imply that it is an iso of schemes?
algebraic-geometry
asked Mar 24 at 4:01
KarlPeterKarlPeter
6691416
$endgroup$
$begingroup$
It’s dominant, injective, and has closed image. Remember that curves are varieties in the Stacks Project so they’re reduced and irreducible.
$endgroup$
– Samir Canning
Mar 24 at 6:19
$begingroup$
@Samir Canning: So essentially (after having our closed immersion by 50.10.2) this problem indeed reduces to following statement: $R to R/I$ is a ring map and $I=(0)$ iff $dim R = dim R/I$ (as Krull dimension),right? Since irreducible here imply that $I$ MUST be a prime
$endgroup$
– KarlPeter
Mar 24 at 13:15
$begingroup$
You also need reduced to get that $(0)$ is prime. More geometrically, just note that the map is dominant and injective so it’s birational. Birational maps between curves are isomorphisms (this comes earlier in the stacks project).
$endgroup$
– Samir Canning
Mar 24 at 16:35
$begingroup$
@SamirCanning: In your last statement you refer to 50.2.6 from stacks.math.columbia.edu/tag/0BXX ?
$endgroup$
– KarlPeter
Mar 25 at 14:58
$begingroup$
Yes that is what I had in mind.
$endgroup$
– Samir Canning
Mar 25 at 17:13
add a comment |
$begingroup$
I have a question about an argument used in Prop 50.10.4 for chracterisation of projective line $mathbbP^1_k$ from https://stacks.math.columbia.edu/tag/0C6L
Let $k$ be a field and $X$ be proper.
Consider the implication (6) -> (1)
According to the source this is a consequence from 50.10.2: https://stacks.math.columbia.edu/tag/0C6T
The problem is that 50.10.2 provides only a closed immersion $X to mathbbP^1_k$.
Why is it an isomorphism of schemes?
My considerations: Since both are irreducible curves of dimension $1$ we conclude that $X$ is homeomorphic to $mathbbP^1_k$ (as topological space) and by 50.10.2 a closed immersion as scheme morphism.
But does this already imply that it is an iso of schemes?
algebraic-geometry
asked Mar 24 at 4:01
KarlPeterKarlPeter
6691416
$endgroup$
I have a question about an argument used in Prop 50.10.4 for chracterisation of projective line $mathbbP^1_k$ from https://stacks.math.columbia.edu/tag/0C6L
Let $k$ be a field and $X$ be proper.
Consider the implication (6) -> (1)
According to the source this is a consequence from 50.10.2: https://stacks.math.columbia.edu/tag/0C6T
The problem is that 50.10.2 provides only a closed immersion $X to mathbbP^1_k$.
Why is it an isomorphism of schemes?
My considerations: Since both are irreducible curves of dimension $1$ we conclude that $X$ is homeomorphic to $mathbbP^1_k$ (as topological space) and by 50.10.2 a closed immersion as scheme morphism.
But does this already imply that it is an iso of schemes?
algebraic-geometry
algebraic-geometry
asked Mar 24 at 4:01
KarlPeterKarlPeter
6691416
asked Mar 24 at 4:01
KarlPeterKarlPeter
6691416
asked Mar 24 at 4:01
KarlPeterKarlPeter
6691416
asked Mar 24 at 4:01
KarlPeterKarlPeter
6691416
asked Mar 24 at 4:01
KarlPeterKarlPeter
6691416
6691416
$begingroup$
It’s dominant, injective, and has closed image. Remember that curves are varieties in the Stacks Project so they’re reduced and irreducible.
$endgroup$
– Samir Canning
Mar 24 at 6:19
$begingroup$
@Samir Canning: So essentially (after having our closed immersion by 50.10.2) this problem indeed reduces to following statement: $R to R/I$ is a ring map and $I=(0)$ iff $dim R = dim R/I$ (as Krull dimension),right? Since irreducible here imply that $I$ MUST be a prime
$endgroup$
– KarlPeter
Mar 24 at 13:15
$begingroup$
You also need reduced to get that $(0)$ is prime. More geometrically, just note that the map is dominant and injective so it’s birational. Birational maps between curves are isomorphisms (this comes earlier in the stacks project).
$endgroup$
– Samir Canning
Mar 24 at 16:35
$begingroup$
@SamirCanning: In your last statement you refer to 50.2.6 from stacks.math.columbia.edu/tag/0BXX ?
$endgroup$
– KarlPeter
Mar 25 at 14:58
$begingroup$
Yes that is what I had in mind.
$endgroup$
– Samir Canning
Mar 25 at 17:13
add a comment |
$begingroup$
It’s dominant, injective, and has closed image. Remember that curves are varieties in the Stacks Project so they’re reduced and irreducible.
$endgroup$
– Samir Canning
Mar 24 at 6:19
$begingroup$
@Samir Canning: So essentially (after having our closed immersion by 50.10.2) this problem indeed reduces to following statement: $R to R/I$ is a ring map and $I=(0)$ iff $dim R = dim R/I$ (as Krull dimension),right? Since irreducible here imply that $I$ MUST be a prime
$endgroup$
– KarlPeter
Mar 24 at 13:15
$begingroup$
You also need reduced to get that $(0)$ is prime. More geometrically, just note that the map is dominant and injective so it’s birational. Birational maps between curves are isomorphisms (this comes earlier in the stacks project).
$endgroup$
– Samir Canning
Mar 24 at 16:35
$begingroup$
@SamirCanning: In your last statement you refer to 50.2.6 from stacks.math.columbia.edu/tag/0BXX ?
$endgroup$
– KarlPeter
Mar 25 at 14:58
$begingroup$
Yes that is what I had in mind.
$endgroup$
– Samir Canning
Mar 25 at 17:13
$begingroup$
It’s dominant, injective, and has closed image. Remember that curves are varieties in the Stacks Project so they’re reduced and irreducible.
$endgroup$
– Samir Canning
Mar 24 at 6:19
$begingroup$
It’s dominant, injective, and has closed image. Remember that curves are varieties in the Stacks Project so they’re reduced and irreducible.
$endgroup$
– Samir Canning
Mar 24 at 6:19
$begingroup$
@Samir Canning: So essentially (after having our closed immersion by 50.10.2) this problem indeed reduces to following statement: $R to R/I$ is a ring map and $I=(0)$ iff $dim R = dim R/I$ (as Krull dimension),right? Since irreducible here imply that $I$ MUST be a prime
$endgroup$
– KarlPeter
Mar 24 at 13:15
$begingroup$
@Samir Canning: So essentially (after having our closed immersion by 50.10.2) this problem indeed reduces to following statement: $R to R/I$ is a ring map and $I=(0)$ iff $dim R = dim R/I$ (as Krull dimension),right? Since irreducible here imply that $I$ MUST be a prime
$endgroup$
– KarlPeter
Mar 24 at 13:15
$begingroup$
You also need reduced to get that $(0)$ is prime. More geometrically, just note that the map is dominant and injective so it’s birational. Birational maps between curves are isomorphisms (this comes earlier in the stacks project).
$endgroup$
– Samir Canning
Mar 24 at 16:35
$begingroup$
You also need reduced to get that $(0)$ is prime. More geometrically, just note that the map is dominant and injective so it’s birational. Birational maps between curves are isomorphisms (this comes earlier in the stacks project).
$endgroup$
– Samir Canning
Mar 24 at 16:35
$begingroup$
@SamirCanning: In your last statement you refer to 50.2.6 from stacks.math.columbia.edu/tag/0BXX ?
$endgroup$
– KarlPeter
Mar 25 at 14:58
$begingroup$
@SamirCanning: In your last statement you refer to 50.2.6 from stacks.math.columbia.edu/tag/0BXX ?
$endgroup$
– KarlPeter
Mar 25 at 14:58
$begingroup$
Yes that is what I had in mind.
$endgroup$
– Samir Canning
Mar 25 at 17:13
$begingroup$
Yes that is what I had in mind.
$endgroup$
– Samir Canning
Mar 25 at 17:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A closed immersion of schemes $f:Xto mathbbP^1_k$ means that on an affine open $U=Spec(R)subset mathbbP^1_k$, we have $f^-1(U)$ isomorphic to $Spec(R/I)$ for an ideal $Isubset R$. Since $X$ is one dimensional, and since closed subsets of $mathbbP_k^1$ are either finite or the entire space, this should give what you want.
answered Mar 24 at 4:17
TomGrubbTomGrubb
11.2k11639
$endgroup$
$begingroup$
I don't see how this information provides the desired result. Indeed locally on the ring level we have quotient maps $R to R/I$ and obviously that suffice to verify that the corresponding ideal sheaf $mathcalI$ is trivial/zero sheaf. Do you mean it in the sense that $1= dim X = dim U$ for every open non trivial set and then apply Krull dimension argumen for affine open sets? Or did I misunderstood you?
$endgroup$
– KarlPeter
Mar 24 at 4:29
$begingroup$
The problem that I see in your argument is that it might be possible that a scheme $Z$ can have different structure sheaf (for example $O_X$ but also $O_X/I$ with $I= rad((0))$ but have the same underlying topological space...
$endgroup$
– KarlPeter
Mar 24 at 5:02
add a comment |
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1 Answer
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1 Answer
1
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votes
$begingroup$
A closed immersion of schemes $f:Xto mathbbP^1_k$ means that on an affine open $U=Spec(R)subset mathbbP^1_k$, we have $f^-1(U)$ isomorphic to $Spec(R/I)$ for an ideal $Isubset R$. Since $X$ is one dimensional, and since closed subsets of $mathbbP_k^1$ are either finite or the entire space, this should give what you want.
answered Mar 24 at 4:17
TomGrubbTomGrubb
11.2k11639
$endgroup$
$begingroup$
I don't see how this information provides the desired result. Indeed locally on the ring level we have quotient maps $R to R/I$ and obviously that suffice to verify that the corresponding ideal sheaf $mathcalI$ is trivial/zero sheaf. Do you mean it in the sense that $1= dim X = dim U$ for every open non trivial set and then apply Krull dimension argumen for affine open sets? Or did I misunderstood you?
$endgroup$
– KarlPeter
Mar 24 at 4:29
$begingroup$
The problem that I see in your argument is that it might be possible that a scheme $Z$ can have different structure sheaf (for example $O_X$ but also $O_X/I$ with $I= rad((0))$ but have the same underlying topological space...
$endgroup$
– KarlPeter
Mar 24 at 5:02
add a comment |
$begingroup$
A closed immersion of schemes $f:Xto mathbbP^1_k$ means that on an affine open $U=Spec(R)subset mathbbP^1_k$, we have $f^-1(U)$ isomorphic to $Spec(R/I)$ for an ideal $Isubset R$. Since $X$ is one dimensional, and since closed subsets of $mathbbP_k^1$ are either finite or the entire space, this should give what you want.
answered Mar 24 at 4:17
TomGrubbTomGrubb
11.2k11639
$endgroup$
$begingroup$
I don't see how this information provides the desired result. Indeed locally on the ring level we have quotient maps $R to R/I$ and obviously that suffice to verify that the corresponding ideal sheaf $mathcalI$ is trivial/zero sheaf. Do you mean it in the sense that $1= dim X = dim U$ for every open non trivial set and then apply Krull dimension argumen for affine open sets? Or did I misunderstood you?
$endgroup$
– KarlPeter
Mar 24 at 4:29
$begingroup$
The problem that I see in your argument is that it might be possible that a scheme $Z$ can have different structure sheaf (for example $O_X$ but also $O_X/I$ with $I= rad((0))$ but have the same underlying topological space...
$endgroup$
– KarlPeter
Mar 24 at 5:02
add a comment |
$begingroup$
A closed immersion of schemes $f:Xto mathbbP^1_k$ means that on an affine open $U=Spec(R)subset mathbbP^1_k$, we have $f^-1(U)$ isomorphic to $Spec(R/I)$ for an ideal $Isubset R$. Since $X$ is one dimensional, and since closed subsets of $mathbbP_k^1$ are either finite or the entire space, this should give what you want.
answered Mar 24 at 4:17
TomGrubbTomGrubb
11.2k11639
$endgroup$
A closed immersion of schemes $f:Xto mathbbP^1_k$ means that on an affine open $U=Spec(R)subset mathbbP^1_k$, we have $f^-1(U)$ isomorphic to $Spec(R/I)$ for an ideal $Isubset R$. Since $X$ is one dimensional, and since closed subsets of $mathbbP_k^1$ are either finite or the entire space, this should give what you want.
answered Mar 24 at 4:17
TomGrubbTomGrubb
11.2k11639
answered Mar 24 at 4:17
TomGrubbTomGrubb
11.2k11639
answered Mar 24 at 4:17
TomGrubbTomGrubb
11.2k11639
answered Mar 24 at 4:17
TomGrubbTomGrubb
11.2k11639
11.2k11639
$begingroup$
I don't see how this information provides the desired result. Indeed locally on the ring level we have quotient maps $R to R/I$ and obviously that suffice to verify that the corresponding ideal sheaf $mathcalI$ is trivial/zero sheaf. Do you mean it in the sense that $1= dim X = dim U$ for every open non trivial set and then apply Krull dimension argumen for affine open sets? Or did I misunderstood you?
$endgroup$
– KarlPeter
Mar 24 at 4:29
$begingroup$
The problem that I see in your argument is that it might be possible that a scheme $Z$ can have different structure sheaf (for example $O_X$ but also $O_X/I$ with $I= rad((0))$ but have the same underlying topological space...
$endgroup$
– KarlPeter
Mar 24 at 5:02
add a comment |
$begingroup$
I don't see how this information provides the desired result. Indeed locally on the ring level we have quotient maps $R to R/I$ and obviously that suffice to verify that the corresponding ideal sheaf $mathcalI$ is trivial/zero sheaf. Do you mean it in the sense that $1= dim X = dim U$ for every open non trivial set and then apply Krull dimension argumen for affine open sets? Or did I misunderstood you?
$endgroup$
– KarlPeter
Mar 24 at 4:29
$begingroup$
The problem that I see in your argument is that it might be possible that a scheme $Z$ can have different structure sheaf (for example $O_X$ but also $O_X/I$ with $I= rad((0))$ but have the same underlying topological space...
$endgroup$
– KarlPeter
Mar 24 at 5:02
$begingroup$
I don't see how this information provides the desired result. Indeed locally on the ring level we have quotient maps $R to R/I$ and obviously that suffice to verify that the corresponding ideal sheaf $mathcalI$ is trivial/zero sheaf. Do you mean it in the sense that $1= dim X = dim U$ for every open non trivial set and then apply Krull dimension argumen for affine open sets? Or did I misunderstood you?
$endgroup$
– KarlPeter
Mar 24 at 4:29
$begingroup$
I don't see how this information provides the desired result. Indeed locally on the ring level we have quotient maps $R to R/I$ and obviously that suffice to verify that the corresponding ideal sheaf $mathcalI$ is trivial/zero sheaf. Do you mean it in the sense that $1= dim X = dim U$ for every open non trivial set and then apply Krull dimension argumen for affine open sets? Or did I misunderstood you?
$endgroup$
– KarlPeter
Mar 24 at 4:29
$begingroup$
The problem that I see in your argument is that it might be possible that a scheme $Z$ can have different structure sheaf (for example $O_X$ but also $O_X/I$ with $I= rad((0))$ but have the same underlying topological space...
$endgroup$
– KarlPeter
Mar 24 at 5:02
$begingroup$
The problem that I see in your argument is that it might be possible that a scheme $Z$ can have different structure sheaf (for example $O_X$ but also $O_X/I$ with $I= rad((0))$ but have the same underlying topological space...
$endgroup$
– KarlPeter
Mar 24 at 5:02
add a comment |
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$begingroup$
It’s dominant, injective, and has closed image. Remember that curves are varieties in the Stacks Project so they’re reduced and irreducible.
$endgroup$
– Samir Canning
Mar 24 at 6:19
$begingroup$
@Samir Canning: So essentially (after having our closed immersion by 50.10.2) this problem indeed reduces to following statement: $R to R/I$ is a ring map and $I=(0)$ iff $dim R = dim R/I$ (as Krull dimension),right? Since irreducible here imply that $I$ MUST be a prime
$endgroup$
– KarlPeter
Mar 24 at 13:15
$begingroup$
You also need reduced to get that $(0)$ is prime. More geometrically, just note that the map is dominant and injective so it’s birational. Birational maps between curves are isomorphisms (this comes earlier in the stacks project).
$endgroup$
– Samir Canning
Mar 24 at 16:35
$begingroup$
@SamirCanning: In your last statement you refer to 50.2.6 from stacks.math.columbia.edu/tag/0BXX ?
$endgroup$
– KarlPeter
Mar 25 at 14:58
$begingroup$
Yes that is what I had in mind.
$endgroup$
– Samir Canning
Mar 25 at 17:13