Do abelian varieties have Neron models over arbitrary valuation rings?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
5
down vote

favorite
1












Let $mathcalO_K$ be a valuation ring with fraction field $K$. Let $A$ be an abelian variety over $K$. Does $A$ have a Neron model?



If $mathcalO_K$ is a discrete valuation ring, then this is proven in the book of Bosch-Lutkebohmert-Raynaud on Neron models.




ag.algebraic-geometry complex-geometry abelian-varieties neron-models




share|cite|improve this question















  • 1




    An easy way to convince yourself that it may not exist is the following example. Consider an elliptic curve $E$ over a field $K:=operatornameFrac(R)$ (where $R$ is a dvr) with a bad multiplicative reduction. And assume that a special fiber of the Neron model $mathcal E$ has $n$ connected components. Now note that for any integer m we have $E[m](K)=mathcal E[m](mathcal O_K)$ and $mathcal E[m]$ is etale whenever $m$ is coprime with char. $k$. Thus $|mathcal E[m](mathcal O_K)| leq |mathcal E_0[m](k)|leq nm$ where $E_0$ is the special fibre of $mathcal E$.
    – gdb
    5 hours ago






  • 1




    This implies that a semi-stable elliptic curve over $K$ with at most $n$ connected components in the special fibre of the Neron model cannot have more than $nm$ $m$-torsion points (for $m$ coprime with char of $k$). So, consider any valuation ring with alg. closed fraction field (for example, $mathcal O_mathbf C_p$), any semi-stable elliptic curve $E$ over $K$ has full $m^2$ rational $m$-torsion points for any $m$ coprime with $K$. Hence, the arg.above shows that if the Neron model of $E$ exists its special fibre must have infinitely many connected components.
    – gdb
    4 hours ago







  • 1




    In particular, it can't be a finite type scheme (it is a part of the definition of Neron model to be a finite type $R$-scheme)
    – gdb
    4 hours ago










  • @gdb Thank you. This is very convincing.
    – Kriss
    4 hours ago














up vote
5
down vote

favorite
1












Let $mathcalO_K$ be a valuation ring with fraction field $K$. Let $A$ be an abelian variety over $K$. Does $A$ have a Neron model?



If $mathcalO_K$ is a discrete valuation ring, then this is proven in the book of Bosch-Lutkebohmert-Raynaud on Neron models.




ag.algebraic-geometry complex-geometry abelian-varieties neron-models




share|cite|improve this question















  • 1




    An easy way to convince yourself that it may not exist is the following example. Consider an elliptic curve $E$ over a field $K:=operatornameFrac(R)$ (where $R$ is a dvr) with a bad multiplicative reduction. And assume that a special fiber of the Neron model $mathcal E$ has $n$ connected components. Now note that for any integer m we have $E[m](K)=mathcal E[m](mathcal O_K)$ and $mathcal E[m]$ is etale whenever $m$ is coprime with char. $k$. Thus $|mathcal E[m](mathcal O_K)| leq |mathcal E_0[m](k)|leq nm$ where $E_0$ is the special fibre of $mathcal E$.
    – gdb
    5 hours ago






  • 1




    This implies that a semi-stable elliptic curve over $K$ with at most $n$ connected components in the special fibre of the Neron model cannot have more than $nm$ $m$-torsion points (for $m$ coprime with char of $k$). So, consider any valuation ring with alg. closed fraction field (for example, $mathcal O_mathbf C_p$), any semi-stable elliptic curve $E$ over $K$ has full $m^2$ rational $m$-torsion points for any $m$ coprime with $K$. Hence, the arg.above shows that if the Neron model of $E$ exists its special fibre must have infinitely many connected components.
    – gdb
    4 hours ago







  • 1




    In particular, it can't be a finite type scheme (it is a part of the definition of Neron model to be a finite type $R$-scheme)
    – gdb
    4 hours ago










  • @gdb Thank you. This is very convincing.
    – Kriss
    4 hours ago












up vote
5
down vote

favorite
1









up vote
5
down vote

favorite
1






1





Let $mathcalO_K$ be a valuation ring with fraction field $K$. Let $A$ be an abelian variety over $K$. Does $A$ have a Neron model?



If $mathcalO_K$ is a discrete valuation ring, then this is proven in the book of Bosch-Lutkebohmert-Raynaud on Neron models.




ag.algebraic-geometry complex-geometry abelian-varieties neron-models




share|cite|improve this question











Let $mathcalO_K$ be a valuation ring with fraction field $K$. Let $A$ be an abelian variety over $K$. Does $A$ have a Neron model?



If $mathcalO_K$ is a discrete valuation ring, then this is proven in the book of Bosch-Lutkebohmert-Raynaud on Neron models.





ag.algebraic-geometry complex-geometry abelian-varieties neron-models





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked 9 hours ago









Kriss

361




361







  • 1




    An easy way to convince yourself that it may not exist is the following example. Consider an elliptic curve $E$ over a field $K:=operatornameFrac(R)$ (where $R$ is a dvr) with a bad multiplicative reduction. And assume that a special fiber of the Neron model $mathcal E$ has $n$ connected components. Now note that for any integer m we have $E[m](K)=mathcal E[m](mathcal O_K)$ and $mathcal E[m]$ is etale whenever $m$ is coprime with char. $k$. Thus $|mathcal E[m](mathcal O_K)| leq |mathcal E_0[m](k)|leq nm$ where $E_0$ is the special fibre of $mathcal E$.
    – gdb
    5 hours ago






  • 1




    This implies that a semi-stable elliptic curve over $K$ with at most $n$ connected components in the special fibre of the Neron model cannot have more than $nm$ $m$-torsion points (for $m$ coprime with char of $k$). So, consider any valuation ring with alg. closed fraction field (for example, $mathcal O_mathbf C_p$), any semi-stable elliptic curve $E$ over $K$ has full $m^2$ rational $m$-torsion points for any $m$ coprime with $K$. Hence, the arg.above shows that if the Neron model of $E$ exists its special fibre must have infinitely many connected components.
    – gdb
    4 hours ago







  • 1




    In particular, it can't be a finite type scheme (it is a part of the definition of Neron model to be a finite type $R$-scheme)
    – gdb
    4 hours ago










  • @gdb Thank you. This is very convincing.
    – Kriss
    4 hours ago












  • 1




    An easy way to convince yourself that it may not exist is the following example. Consider an elliptic curve $E$ over a field $K:=operatornameFrac(R)$ (where $R$ is a dvr) with a bad multiplicative reduction. And assume that a special fiber of the Neron model $mathcal E$ has $n$ connected components. Now note that for any integer m we have $E[m](K)=mathcal E[m](mathcal O_K)$ and $mathcal E[m]$ is etale whenever $m$ is coprime with char. $k$. Thus $|mathcal E[m](mathcal O_K)| leq |mathcal E_0[m](k)|leq nm$ where $E_0$ is the special fibre of $mathcal E$.
    – gdb
    5 hours ago






  • 1




    This implies that a semi-stable elliptic curve over $K$ with at most $n$ connected components in the special fibre of the Neron model cannot have more than $nm$ $m$-torsion points (for $m$ coprime with char of $k$). So, consider any valuation ring with alg. closed fraction field (for example, $mathcal O_mathbf C_p$), any semi-stable elliptic curve $E$ over $K$ has full $m^2$ rational $m$-torsion points for any $m$ coprime with $K$. Hence, the arg.above shows that if the Neron model of $E$ exists its special fibre must have infinitely many connected components.
    – gdb
    4 hours ago







  • 1




    In particular, it can't be a finite type scheme (it is a part of the definition of Neron model to be a finite type $R$-scheme)
    – gdb
    4 hours ago










  • @gdb Thank you. This is very convincing.
    – Kriss
    4 hours ago







1




1




An easy way to convince yourself that it may not exist is the following example. Consider an elliptic curve $E$ over a field $K:=operatornameFrac(R)$ (where $R$ is a dvr) with a bad multiplicative reduction. And assume that a special fiber of the Neron model $mathcal E$ has $n$ connected components. Now note that for any integer m we have $E[m](K)=mathcal E[m](mathcal O_K)$ and $mathcal E[m]$ is etale whenever $m$ is coprime with char. $k$. Thus $|mathcal E[m](mathcal O_K)| leq |mathcal E_0[m](k)|leq nm$ where $E_0$ is the special fibre of $mathcal E$.
– gdb
5 hours ago




An easy way to convince yourself that it may not exist is the following example. Consider an elliptic curve $E$ over a field $K:=operatornameFrac(R)$ (where $R$ is a dvr) with a bad multiplicative reduction. And assume that a special fiber of the Neron model $mathcal E$ has $n$ connected components. Now note that for any integer m we have $E[m](K)=mathcal E[m](mathcal O_K)$ and $mathcal E[m]$ is etale whenever $m$ is coprime with char. $k$. Thus $|mathcal E[m](mathcal O_K)| leq |mathcal E_0[m](k)|leq nm$ where $E_0$ is the special fibre of $mathcal E$.
– gdb
5 hours ago




1




1




This implies that a semi-stable elliptic curve over $K$ with at most $n$ connected components in the special fibre of the Neron model cannot have more than $nm$ $m$-torsion points (for $m$ coprime with char of $k$). So, consider any valuation ring with alg. closed fraction field (for example, $mathcal O_mathbf C_p$), any semi-stable elliptic curve $E$ over $K$ has full $m^2$ rational $m$-torsion points for any $m$ coprime with $K$. Hence, the arg.above shows that if the Neron model of $E$ exists its special fibre must have infinitely many connected components.
– gdb
4 hours ago





This implies that a semi-stable elliptic curve over $K$ with at most $n$ connected components in the special fibre of the Neron model cannot have more than $nm$ $m$-torsion points (for $m$ coprime with char of $k$). So, consider any valuation ring with alg. closed fraction field (for example, $mathcal O_mathbf C_p$), any semi-stable elliptic curve $E$ over $K$ has full $m^2$ rational $m$-torsion points for any $m$ coprime with $K$. Hence, the arg.above shows that if the Neron model of $E$ exists its special fibre must have infinitely many connected components.
– gdb
4 hours ago





1




1




In particular, it can't be a finite type scheme (it is a part of the definition of Neron model to be a finite type $R$-scheme)
– gdb
4 hours ago




In particular, it can't be a finite type scheme (it is a part of the definition of Neron model to be a finite type $R$-scheme)
– gdb
4 hours ago












@gdb Thank you. This is very convincing.
– Kriss
4 hours ago




@gdb Thank you. This is very convincing.
– Kriss
4 hours ago










1 Answer
1






active

oldest

votes

















up vote
3
down vote













No, they needn't.



See: David Holmes: Neron models of jacobians over base schemes of dimension greater than 1., to appear in Journal fur die reine und angewandte Mathematik.



and



Giulio Orecchia: A criterion for existence of Néron models of jacobians .






share|cite|improve this answer



















  • 1




    I'm not sure where to find the answer in these two papers. Do these papers not assume the base scheme is locally noetherian (and thereby exclude certain valuation rings)?
    – Kriss
    4 hours ago










Your Answer




StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "504"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);








 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f307603%2fdo-abelian-varieties-have-neron-models-over-arbitrary-valuation-rings%23new-answer', 'question_page');

);

Post as a guest

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote













No, they needn't.



See: David Holmes: Neron models of jacobians over base schemes of dimension greater than 1., to appear in Journal fur die reine und angewandte Mathematik.



and



Giulio Orecchia: A criterion for existence of Néron models of jacobians .






share|cite|improve this answer



















  • 1




    I'm not sure where to find the answer in these two papers. Do these papers not assume the base scheme is locally noetherian (and thereby exclude certain valuation rings)?
    – Kriss
    4 hours ago














up vote
3
down vote













No, they needn't.



See: David Holmes: Neron models of jacobians over base schemes of dimension greater than 1., to appear in Journal fur die reine und angewandte Mathematik.



and



Giulio Orecchia: A criterion for existence of Néron models of jacobians .






share|cite|improve this answer



















  • 1




    I'm not sure where to find the answer in these two papers. Do these papers not assume the base scheme is locally noetherian (and thereby exclude certain valuation rings)?
    – Kriss
    4 hours ago












up vote
3
down vote










up vote
3
down vote









No, they needn't.



See: David Holmes: Neron models of jacobians over base schemes of dimension greater than 1., to appear in Journal fur die reine und angewandte Mathematik.



and



Giulio Orecchia: A criterion for existence of Néron models of jacobians .






share|cite|improve this answer















No, they needn't.



See: David Holmes: Neron models of jacobians over base schemes of dimension greater than 1., to appear in Journal fur die reine und angewandte Mathematik.



and



Giulio Orecchia: A criterion for existence of Néron models of jacobians .







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited 5 hours ago









Glorfindel

1,10031020




1,10031020











answered 7 hours ago









anon

311




311







  • 1




    I'm not sure where to find the answer in these two papers. Do these papers not assume the base scheme is locally noetherian (and thereby exclude certain valuation rings)?
    – Kriss
    4 hours ago












  • 1




    I'm not sure where to find the answer in these two papers. Do these papers not assume the base scheme is locally noetherian (and thereby exclude certain valuation rings)?
    – Kriss
    4 hours ago







1




1




I'm not sure where to find the answer in these two papers. Do these papers not assume the base scheme is locally noetherian (and thereby exclude certain valuation rings)?
– Kriss
4 hours ago




I'm not sure where to find the answer in these two papers. Do these papers not assume the base scheme is locally noetherian (and thereby exclude certain valuation rings)?
– Kriss
4 hours ago












 

draft saved


draft discarded


























 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f307603%2fdo-abelian-varieties-have-neron-models-over-arbitrary-valuation-rings%23new-answer', 'question_page');

);

Post as a guest
































































Comments

Post a Comment

Popular posts from this blog

Color the edges and diagonals of a regular polygon

Image
Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne
Read more

Relationship between determinant of matrix and determinant of adjoint?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/… – Hector Blandin Nov 17 &
Read more

What is the equation of a 3D cone with generalised tilt?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an
Read more