The underlying space of a scheme remembers its affineness? Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Can any topological space be the result of a scheme?Does reduced+Noetherian space imply Noetherian schemeWhich local ringed spaces are schemes?Surjectivity of tangent spaces induced by smooth morphismSections of morphisms up to fppf coveringfunctoriality of hilbert schemeExplicit description of the scheme obtained by relative gluing data over a base schemeWhen does the image of a morphism of schemes support scheme structure?Schemes monomorphing into affine scheme of dimension 1Epimorphisms from an affine scheme?
The underlying space of a scheme remembers its affineness?
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Can any topological space be the result of a scheme?Does reduced+Noetherian space imply Noetherian schemeWhich local ringed spaces are schemes?Surjectivity of tangent spaces induced by smooth morphismSections of morphisms up to fppf coveringfunctoriality of hilbert schemeExplicit description of the scheme obtained by relative gluing data over a base schemeWhen does the image of a morphism of schemes support scheme structure?Schemes monomorphing into affine scheme of dimension 1Epimorphisms from an affine scheme?
$begingroup$
Let $f:Xrightarrow Y$ be a morphism of schemes. We know that if $Y$ is affine and $f$ induces homeomorphism on the underlying spaces then $X$ is affine. Is it true that if $X$ is affine and $f$ induces a homeomorphism on the underlying spaces then $Y$ is affine?
More generally, is it true that a scheme whose underlying space is homeomorphic (possibly via a homeomorphism that is not induced by a morphism of schemes) to the underlying space of an affine scheme is affine? EDIT: actually, the answer to the last question is a very strong "NO" as the underlying space of any scheme is sober and Noetherian sober spaces are spectral (i.e. homeomorphic to the underlying space of an affine scheme).
ag.algebraic-geometry schemes
$endgroup$
This question has an open bounty worth +200
reputation from Community♦ ending ending at 2019-04-24 08:50:15Z">in 2 days.
This question has not received enough attention.
Desired is an example of a $mathbbC$-morphism of schemes inducing a homeomorphism on the underlying topological spaces, with the source being a connected affine scheme smooth over $mathbbC$, and the target being an integral separated scheme of finite type over $mathbbC$ that is not affine. It would be better if the target is an integral scheme projective over $mathbbC$ that is not affine. I believe that the target can not be normal (mathoverflow.net/q/264204/137767).
|
show 3 more comments
$begingroup$
Let $f:Xrightarrow Y$ be a morphism of schemes. We know that if $Y$ is affine and $f$ induces homeomorphism on the underlying spaces then $X$ is affine. Is it true that if $X$ is affine and $f$ induces a homeomorphism on the underlying spaces then $Y$ is affine?
More generally, is it true that a scheme whose underlying space is homeomorphic (possibly via a homeomorphism that is not induced by a morphism of schemes) to the underlying space of an affine scheme is affine? EDIT: actually, the answer to the last question is a very strong "NO" as the underlying space of any scheme is sober and Noetherian sober spaces are spectral (i.e. homeomorphic to the underlying space of an affine scheme).
ag.algebraic-geometry schemes
$endgroup$
This question has an open bounty worth +200
reputation from Community♦ ending ending at 2019-04-24 08:50:15Z">in 2 days.
This question has not received enough attention.
Desired is an example of a $mathbbC$-morphism of schemes inducing a homeomorphism on the underlying topological spaces, with the source being a connected affine scheme smooth over $mathbbC$, and the target being an integral separated scheme of finite type over $mathbbC$ that is not affine. It would be better if the target is an integral scheme projective over $mathbbC$ that is not affine. I believe that the target can not be normal (mathoverflow.net/q/264204/137767).
9
$begingroup$
The answer to the second question is no: over an infinite field, $mathbbA^1$ and $mathbbP^1$ are homeomorphic.
$endgroup$
– Julian Rosen
Apr 13 at 2:40
1
$begingroup$
The following argument may lead to a proof when $Y$ is quasi-projective. If $Y$ is quasi-projective, then it is completely described by "patching" of affine open sets: an affine cover $U_i$ such that $U_icap U_j$ is affine. It then follows from the quoted result that $f^-1(U_i)$ and $f^-1(U_icap U_j)$ are affine. So $X$ is defined by the same patching.
$endgroup$
– Kapil
Apr 13 at 3:04
$begingroup$
@Kapil maybe I am missing something but do you prove that $X$ is affine (which is an assumption)?
$endgroup$
– user137767
Apr 13 at 3:07
1
$begingroup$
@TabesBridges you can not have a bijective morphism between integral separated schemes of finite type over $mathbbC$ with target normal and source connected that is not an isomorphism (as shown in the link in the bounty description), so results about normal varieties should be useless. I do not understand what you mean by "curves are hopeless". The answer to the second question is no whether you are talking about 1-dimensional schemes or not (as is shown in the second link in the answer).
$endgroup$
– schematic_boi
15 hours ago
1
$begingroup$
Nobody has yet given an answer to the first question with the target separated (if you can prove that there can not be such example in dimension 1, please write it up as an answer, I would gladly upvote that).
$endgroup$
– schematic_boi
15 hours ago
|
show 3 more comments
$begingroup$
Let $f:Xrightarrow Y$ be a morphism of schemes. We know that if $Y$ is affine and $f$ induces homeomorphism on the underlying spaces then $X$ is affine. Is it true that if $X$ is affine and $f$ induces a homeomorphism on the underlying spaces then $Y$ is affine?
More generally, is it true that a scheme whose underlying space is homeomorphic (possibly via a homeomorphism that is not induced by a morphism of schemes) to the underlying space of an affine scheme is affine? EDIT: actually, the answer to the last question is a very strong "NO" as the underlying space of any scheme is sober and Noetherian sober spaces are spectral (i.e. homeomorphic to the underlying space of an affine scheme).
ag.algebraic-geometry schemes
$endgroup$
Let $f:Xrightarrow Y$ be a morphism of schemes. We know that if $Y$ is affine and $f$ induces homeomorphism on the underlying spaces then $X$ is affine. Is it true that if $X$ is affine and $f$ induces a homeomorphism on the underlying spaces then $Y$ is affine?
More generally, is it true that a scheme whose underlying space is homeomorphic (possibly via a homeomorphism that is not induced by a morphism of schemes) to the underlying space of an affine scheme is affine? EDIT: actually, the answer to the last question is a very strong "NO" as the underlying space of any scheme is sober and Noetherian sober spaces are spectral (i.e. homeomorphic to the underlying space of an affine scheme).
ag.algebraic-geometry schemes
ag.algebraic-geometry schemes
edited Apr 13 at 21:35
asked Apr 13 at 2:11
user137767
This question has an open bounty worth +200
reputation from Community♦ ending ending at 2019-04-24 08:50:15Z">in 2 days.
This question has not received enough attention.
Desired is an example of a $mathbbC$-morphism of schemes inducing a homeomorphism on the underlying topological spaces, with the source being a connected affine scheme smooth over $mathbbC$, and the target being an integral separated scheme of finite type over $mathbbC$ that is not affine. It would be better if the target is an integral scheme projective over $mathbbC$ that is not affine. I believe that the target can not be normal (mathoverflow.net/q/264204/137767).
This question has an open bounty worth +200
reputation from Community♦ ending ending at 2019-04-24 08:50:15Z">in 2 days.
This question has not received enough attention.
Desired is an example of a $mathbbC$-morphism of schemes inducing a homeomorphism on the underlying topological spaces, with the source being a connected affine scheme smooth over $mathbbC$, and the target being an integral separated scheme of finite type over $mathbbC$ that is not affine. It would be better if the target is an integral scheme projective over $mathbbC$ that is not affine. I believe that the target can not be normal (mathoverflow.net/q/264204/137767).
9
$begingroup$
The answer to the second question is no: over an infinite field, $mathbbA^1$ and $mathbbP^1$ are homeomorphic.
$endgroup$
– Julian Rosen
Apr 13 at 2:40
1
$begingroup$
The following argument may lead to a proof when $Y$ is quasi-projective. If $Y$ is quasi-projective, then it is completely described by "patching" of affine open sets: an affine cover $U_i$ such that $U_icap U_j$ is affine. It then follows from the quoted result that $f^-1(U_i)$ and $f^-1(U_icap U_j)$ are affine. So $X$ is defined by the same patching.
$endgroup$
– Kapil
Apr 13 at 3:04
$begingroup$
@Kapil maybe I am missing something but do you prove that $X$ is affine (which is an assumption)?
$endgroup$
– user137767
Apr 13 at 3:07
1
$begingroup$
@TabesBridges you can not have a bijective morphism between integral separated schemes of finite type over $mathbbC$ with target normal and source connected that is not an isomorphism (as shown in the link in the bounty description), so results about normal varieties should be useless. I do not understand what you mean by "curves are hopeless". The answer to the second question is no whether you are talking about 1-dimensional schemes or not (as is shown in the second link in the answer).
$endgroup$
– schematic_boi
15 hours ago
1
$begingroup$
Nobody has yet given an answer to the first question with the target separated (if you can prove that there can not be such example in dimension 1, please write it up as an answer, I would gladly upvote that).
$endgroup$
– schematic_boi
15 hours ago
|
show 3 more comments
9
$begingroup$
The answer to the second question is no: over an infinite field, $mathbbA^1$ and $mathbbP^1$ are homeomorphic.
$endgroup$
– Julian Rosen
Apr 13 at 2:40
1
$begingroup$
The following argument may lead to a proof when $Y$ is quasi-projective. If $Y$ is quasi-projective, then it is completely described by "patching" of affine open sets: an affine cover $U_i$ such that $U_icap U_j$ is affine. It then follows from the quoted result that $f^-1(U_i)$ and $f^-1(U_icap U_j)$ are affine. So $X$ is defined by the same patching.
$endgroup$
– Kapil
Apr 13 at 3:04
$begingroup$
@Kapil maybe I am missing something but do you prove that $X$ is affine (which is an assumption)?
$endgroup$
– user137767
Apr 13 at 3:07
1
$begingroup$
@TabesBridges you can not have a bijective morphism between integral separated schemes of finite type over $mathbbC$ with target normal and source connected that is not an isomorphism (as shown in the link in the bounty description), so results about normal varieties should be useless. I do not understand what you mean by "curves are hopeless". The answer to the second question is no whether you are talking about 1-dimensional schemes or not (as is shown in the second link in the answer).
$endgroup$
– schematic_boi
15 hours ago
1
$begingroup$
Nobody has yet given an answer to the first question with the target separated (if you can prove that there can not be such example in dimension 1, please write it up as an answer, I would gladly upvote that).
$endgroup$
– schematic_boi
15 hours ago
9
9
$begingroup$
The answer to the second question is no: over an infinite field, $mathbbA^1$ and $mathbbP^1$ are homeomorphic.
$endgroup$
– Julian Rosen
Apr 13 at 2:40
$begingroup$
The answer to the second question is no: over an infinite field, $mathbbA^1$ and $mathbbP^1$ are homeomorphic.
$endgroup$
– Julian Rosen
Apr 13 at 2:40
1
1
$begingroup$
The following argument may lead to a proof when $Y$ is quasi-projective. If $Y$ is quasi-projective, then it is completely described by "patching" of affine open sets: an affine cover $U_i$ such that $U_icap U_j$ is affine. It then follows from the quoted result that $f^-1(U_i)$ and $f^-1(U_icap U_j)$ are affine. So $X$ is defined by the same patching.
$endgroup$
– Kapil
Apr 13 at 3:04
$begingroup$
The following argument may lead to a proof when $Y$ is quasi-projective. If $Y$ is quasi-projective, then it is completely described by "patching" of affine open sets: an affine cover $U_i$ such that $U_icap U_j$ is affine. It then follows from the quoted result that $f^-1(U_i)$ and $f^-1(U_icap U_j)$ are affine. So $X$ is defined by the same patching.
$endgroup$
– Kapil
Apr 13 at 3:04
$begingroup$
@Kapil maybe I am missing something but do you prove that $X$ is affine (which is an assumption)?
$endgroup$
– user137767
Apr 13 at 3:07
$begingroup$
@Kapil maybe I am missing something but do you prove that $X$ is affine (which is an assumption)?
$endgroup$
– user137767
Apr 13 at 3:07
1
1
$begingroup$
@TabesBridges you can not have a bijective morphism between integral separated schemes of finite type over $mathbbC$ with target normal and source connected that is not an isomorphism (as shown in the link in the bounty description), so results about normal varieties should be useless. I do not understand what you mean by "curves are hopeless". The answer to the second question is no whether you are talking about 1-dimensional schemes or not (as is shown in the second link in the answer).
$endgroup$
– schematic_boi
15 hours ago
$begingroup$
@TabesBridges you can not have a bijective morphism between integral separated schemes of finite type over $mathbbC$ with target normal and source connected that is not an isomorphism (as shown in the link in the bounty description), so results about normal varieties should be useless. I do not understand what you mean by "curves are hopeless". The answer to the second question is no whether you are talking about 1-dimensional schemes or not (as is shown in the second link in the answer).
$endgroup$
– schematic_boi
15 hours ago
1
1
$begingroup$
Nobody has yet given an answer to the first question with the target separated (if you can prove that there can not be such example in dimension 1, please write it up as an answer, I would gladly upvote that).
$endgroup$
– schematic_boi
15 hours ago
$begingroup$
Nobody has yet given an answer to the first question with the target separated (if you can prove that there can not be such example in dimension 1, please write it up as an answer, I would gladly upvote that).
$endgroup$
– schematic_boi
15 hours ago
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Here is a counterexample. Fix a field $k$, and let $Y$ be built from two copies of the affine nodal curve $y^2=x^3+x^2$, glued together on the complement of the singular point. In other words $Y$ is a nodal curve with doubled singular point. Then $Y$ is not affine because it is not separated. However, there is a homeomorphism $mathbbA^1to Y$, which is built from the usual parameterization of the nodal curve by $mathbbA^1$ (which passes through the singular point twice) by choosing one of version of the singular point the first time through and the other version the second time through.
answered Apr 13 at 3:19
Julian RosenJulian Rosen
6,56423048
$endgroup$
$begingroup$
do you think there is a counter-example with $Y$ separated?
$endgroup$
– user137767
Apr 13 at 3:24
$begingroup$
I don't know of a separated counterexample, but I wouldn't be surprised if there is one.
$endgroup$
– Julian Rosen
Apr 13 at 13:55
add a comment |
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1 Answer
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1 Answer
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oldest
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active
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$begingroup$
Here is a counterexample. Fix a field $k$, and let $Y$ be built from two copies of the affine nodal curve $y^2=x^3+x^2$, glued together on the complement of the singular point. In other words $Y$ is a nodal curve with doubled singular point. Then $Y$ is not affine because it is not separated. However, there is a homeomorphism $mathbbA^1to Y$, which is built from the usual parameterization of the nodal curve by $mathbbA^1$ (which passes through the singular point twice) by choosing one of version of the singular point the first time through and the other version the second time through.
answered Apr 13 at 3:19
Julian RosenJulian Rosen
6,56423048
$endgroup$
$begingroup$
do you think there is a counter-example with $Y$ separated?
$endgroup$
– user137767
Apr 13 at 3:24
$begingroup$
I don't know of a separated counterexample, but I wouldn't be surprised if there is one.
$endgroup$
– Julian Rosen
Apr 13 at 13:55
add a comment |
$begingroup$
Here is a counterexample. Fix a field $k$, and let $Y$ be built from two copies of the affine nodal curve $y^2=x^3+x^2$, glued together on the complement of the singular point. In other words $Y$ is a nodal curve with doubled singular point. Then $Y$ is not affine because it is not separated. However, there is a homeomorphism $mathbbA^1to Y$, which is built from the usual parameterization of the nodal curve by $mathbbA^1$ (which passes through the singular point twice) by choosing one of version of the singular point the first time through and the other version the second time through.
answered Apr 13 at 3:19
Julian RosenJulian Rosen
6,56423048
$endgroup$
$begingroup$
do you think there is a counter-example with $Y$ separated?
$endgroup$
– user137767
Apr 13 at 3:24
$begingroup$
I don't know of a separated counterexample, but I wouldn't be surprised if there is one.
$endgroup$
– Julian Rosen
Apr 13 at 13:55
add a comment |
$begingroup$
Here is a counterexample. Fix a field $k$, and let $Y$ be built from two copies of the affine nodal curve $y^2=x^3+x^2$, glued together on the complement of the singular point. In other words $Y$ is a nodal curve with doubled singular point. Then $Y$ is not affine because it is not separated. However, there is a homeomorphism $mathbbA^1to Y$, which is built from the usual parameterization of the nodal curve by $mathbbA^1$ (which passes through the singular point twice) by choosing one of version of the singular point the first time through and the other version the second time through.
answered Apr 13 at 3:19
Julian RosenJulian Rosen
6,56423048
$endgroup$
Here is a counterexample. Fix a field $k$, and let $Y$ be built from two copies of the affine nodal curve $y^2=x^3+x^2$, glued together on the complement of the singular point. In other words $Y$ is a nodal curve with doubled singular point. Then $Y$ is not affine because it is not separated. However, there is a homeomorphism $mathbbA^1to Y$, which is built from the usual parameterization of the nodal curve by $mathbbA^1$ (which passes through the singular point twice) by choosing one of version of the singular point the first time through and the other version the second time through.
answered Apr 13 at 3:19
Julian RosenJulian Rosen
6,56423048
answered Apr 13 at 3:19
Julian RosenJulian Rosen
6,56423048
answered Apr 13 at 3:19
Julian RosenJulian Rosen
6,56423048
answered Apr 13 at 3:19
Julian RosenJulian Rosen
6,56423048
6,56423048
$begingroup$
do you think there is a counter-example with $Y$ separated?
$endgroup$
– user137767
Apr 13 at 3:24
$begingroup$
I don't know of a separated counterexample, but I wouldn't be surprised if there is one.
$endgroup$
– Julian Rosen
Apr 13 at 13:55
add a comment |
$begingroup$
do you think there is a counter-example with $Y$ separated?
$endgroup$
– user137767
Apr 13 at 3:24
$begingroup$
I don't know of a separated counterexample, but I wouldn't be surprised if there is one.
$endgroup$
– Julian Rosen
Apr 13 at 13:55
$begingroup$
do you think there is a counter-example with $Y$ separated?
$endgroup$
– user137767
Apr 13 at 3:24
$begingroup$
do you think there is a counter-example with $Y$ separated?
$endgroup$
– user137767
Apr 13 at 3:24
$begingroup$
I don't know of a separated counterexample, but I wouldn't be surprised if there is one.
$endgroup$
– Julian Rosen
Apr 13 at 13:55
$begingroup$
I don't know of a separated counterexample, but I wouldn't be surprised if there is one.
$endgroup$
– Julian Rosen
Apr 13 at 13:55
add a comment |
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9
$begingroup$
The answer to the second question is no: over an infinite field, $mathbbA^1$ and $mathbbP^1$ are homeomorphic.
$endgroup$
– Julian Rosen
Apr 13 at 2:40
1
$begingroup$
The following argument may lead to a proof when $Y$ is quasi-projective. If $Y$ is quasi-projective, then it is completely described by "patching" of affine open sets: an affine cover $U_i$ such that $U_icap U_j$ is affine. It then follows from the quoted result that $f^-1(U_i)$ and $f^-1(U_icap U_j)$ are affine. So $X$ is defined by the same patching.
$endgroup$
– Kapil
Apr 13 at 3:04
$begingroup$
@Kapil maybe I am missing something but do you prove that $X$ is affine (which is an assumption)?
$endgroup$
– user137767
Apr 13 at 3:07
1
$begingroup$
@TabesBridges you can not have a bijective morphism between integral separated schemes of finite type over $mathbbC$ with target normal and source connected that is not an isomorphism (as shown in the link in the bounty description), so results about normal varieties should be useless. I do not understand what you mean by "curves are hopeless". The answer to the second question is no whether you are talking about 1-dimensional schemes or not (as is shown in the second link in the answer).
$endgroup$
– schematic_boi
15 hours ago
1
$begingroup$
Nobody has yet given an answer to the first question with the target separated (if you can prove that there can not be such example in dimension 1, please write it up as an answer, I would gladly upvote that).
$endgroup$
– schematic_boi
15 hours ago