The underlying space of a scheme remembers its affineness? Announcing the arrival of Valued...
The underlying space of a scheme remembers its affineness?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)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
asked 2 days ago
Stepan BanachStepan Banach
16011
$endgroup$
add a comment |
$begingroup$
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
asked 2 days ago
Stepan BanachStepan Banach
16011
$endgroup$
9
$begingroup$
The answer to the second question is no: over an infinite field, $mathbb{A}^1$ and $mathbb{P}^1$ are homeomorphic.
$endgroup$
– Julian Rosen
2 days ago
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
yesterday
$begingroup$
@Kapil maybe I am missing something but do you prove that $X$ is affine (which is an assumption)?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
No. At this point one can use that $X$ is affine to describe $U_i$ in terms of the co-ordinate ring of $X$. Anyway, it is not a completely worked out idea.
$endgroup$
– Kapil
yesterday
add a comment |
$begingroup$
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
asked 2 days ago
Stepan BanachStepan Banach
16011
$endgroup$
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
asked 2 days ago
Stepan BanachStepan Banach
16011
asked 2 days ago
Stepan BanachStepan Banach
16011
edited yesterday
Stepan Banach
asked 2 days ago
Stepan BanachStepan Banach
16011
asked 2 days ago
Stepan BanachStepan Banach
16011
asked 2 days ago
Stepan BanachStepan Banach
16011
16011
9
$begingroup$
The answer to the second question is no: over an infinite field, $mathbb{A}^1$ and $mathbb{P}^1$ are homeomorphic.
$endgroup$
– Julian Rosen
2 days ago
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
yesterday
$begingroup$
@Kapil maybe I am missing something but do you prove that $X$ is affine (which is an assumption)?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
No. At this point one can use that $X$ is affine to describe $U_i$ in terms of the co-ordinate ring of $X$. Anyway, it is not a completely worked out idea.
$endgroup$
– Kapil
yesterday
add a comment |
9
$begingroup$
The answer to the second question is no: over an infinite field, $mathbb{A}^1$ and $mathbb{P}^1$ are homeomorphic.
$endgroup$
– Julian Rosen
2 days ago
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
yesterday
$begingroup$
@Kapil maybe I am missing something but do you prove that $X$ is affine (which is an assumption)?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
No. At this point one can use that $X$ is affine to describe $U_i$ in terms of the co-ordinate ring of $X$. Anyway, it is not a completely worked out idea.
$endgroup$
– Kapil
yesterday
9
9
$begingroup$
The answer to the second question is no: over an infinite field, $mathbb{A}^1$ and $mathbb{P}^1$ are homeomorphic.
$endgroup$
– Julian Rosen
2 days ago
$begingroup$
The answer to the second question is no: over an infinite field, $mathbb{A}^1$ and $mathbb{P}^1$ are homeomorphic.
$endgroup$
– Julian Rosen
2 days ago
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
yesterday
$begingroup$
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
yesterday
$begingroup$
@Kapil maybe I am missing something but do you prove that $X$ is affine (which is an assumption)?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
@Kapil maybe I am missing something but do you prove that $X$ is affine (which is an assumption)?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
No. At this point one can use that $X$ is affine to describe $U_i$ in terms of the co-ordinate ring of $X$. Anyway, it is not a completely worked out idea.
$endgroup$
– Kapil
yesterday
$begingroup$
No. At this point one can use that $X$ is affine to describe $U_i$ in terms of the co-ordinate ring of $X$. Anyway, it is not a completely worked out idea.
$endgroup$
– Kapil
yesterday
add a comment |
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 $mathbb{A}^1to Y$, which is built from the usual parameterization of the nodal curve by $mathbb{A}^1$ (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 yesterday
Julian RosenJulian Rosen
6,54423048
$endgroup$
$begingroup$
do you think there is a counter-example with $Y$ separated?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
I don't know of a separated counterexample, but I wouldn't be surprised if there is one.
$endgroup$
– Julian Rosen
yesterday
add a comment |
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1 Answer
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1 Answer
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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 $mathbb{A}^1to Y$, which is built from the usual parameterization of the nodal curve by $mathbb{A}^1$ (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 yesterday
Julian RosenJulian Rosen
6,54423048
$endgroup$
$begingroup$
do you think there is a counter-example with $Y$ separated?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
I don't know of a separated counterexample, but I wouldn't be surprised if there is one.
$endgroup$
– Julian Rosen
yesterday
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 $mathbb{A}^1to Y$, which is built from the usual parameterization of the nodal curve by $mathbb{A}^1$ (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 yesterday
Julian RosenJulian Rosen
6,54423048
$endgroup$
$begingroup$
do you think there is a counter-example with $Y$ separated?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
I don't know of a separated counterexample, but I wouldn't be surprised if there is one.
$endgroup$
– Julian Rosen
yesterday
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 $mathbb{A}^1to Y$, which is built from the usual parameterization of the nodal curve by $mathbb{A}^1$ (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 yesterday
Julian RosenJulian Rosen
6,54423048
$endgroup$
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 $mathbb{A}^1to Y$, which is built from the usual parameterization of the nodal curve by $mathbb{A}^1$ (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 yesterday
Julian RosenJulian Rosen
6,54423048
answered yesterday
Julian RosenJulian Rosen
6,54423048
answered yesterday
Julian RosenJulian Rosen
6,54423048
answered yesterday
Julian RosenJulian Rosen
6,54423048
6,54423048
$begingroup$
do you think there is a counter-example with $Y$ separated?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
I don't know of a separated counterexample, but I wouldn't be surprised if there is one.
$endgroup$
– Julian Rosen
yesterday
add a comment |
$begingroup$
do you think there is a counter-example with $Y$ separated?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
I don't know of a separated counterexample, but I wouldn't be surprised if there is one.
$endgroup$
– Julian Rosen
yesterday
$begingroup$
do you think there is a counter-example with $Y$ separated?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
do you think there is a counter-example with $Y$ separated?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
I don't know of a separated counterexample, but I wouldn't be surprised if there is one.
$endgroup$
– Julian Rosen
yesterday
$begingroup$
I don't know of a separated counterexample, but I wouldn't be surprised if there is one.
$endgroup$
– Julian Rosen
yesterday
add a comment |
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9
$begingroup$
The answer to the second question is no: over an infinite field, $mathbb{A}^1$ and $mathbb{P}^1$ are homeomorphic.
$endgroup$
– Julian Rosen
2 days ago
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
yesterday
$begingroup$
@Kapil maybe I am missing something but do you prove that $X$ is affine (which is an assumption)?
$endgroup$
– Stepan Banach
yesterday
$begingroup$
No. At this point one can use that $X$ is affine to describe $U_i$ in terms of the co-ordinate ring of $X$. Anyway, it is not a completely worked out idea.
$endgroup$
– Kapil
yesterday