Do abelian varieties have Neron models over arbitrary valuation rings?

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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




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    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
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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










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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




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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












 

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