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Completion of the local ring at a point on arithmetic surfaces.

2

$begingroup$

Let $K$ be a number field and consider a arithmetic surface $Xto B=operatornameSpec O_K$, i.e. $X$ is integral, regular, flat, proper over $O_K$ and it has dimension $2$.

Now pick a closed point $xin X$ such that $xmapsto bin B$ and consider $widehatmathcal O_X,x$. In other words the completion of the local ring $mathcal O_X,x$ with respect to its maximal ideal.

Can we express $widehatmathcal O_X,x= A[[t]]$? What is $A$ in
this case? Do we have $A=O_L$ where $L$ is a complete discrete
valuation field and a finite extension of $K_b$ (here $K_b$ is the completion of $K$ at $b$)?

For sure we have an embedding $mathcal O_B,b[t]hookrightarrowmathcal
O_X,x$

Thanks in advance

algebraic-geometry power-series surfaces local-rings

share|cite|improve this question

edited Feb 18 at 15:08

notsure

asked Feb 17 at 15:52

notsurenotsure

547

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I asked a similar question in Mathoverflow some time ago but I didn't get an answer: Completion of a local ring of an arithmetic surface
  $endgroup$
  – yamete kudasai
  Feb 18 at 15:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Take a look at Liu's book exercise 4, 3.35 (d) at page 149. But it assumes the existence of a section.
  $endgroup$
  – notsure
  Feb 18 at 15:03
  

  
  
  
  

add a comment | 

2

$begingroup$

Let $K$ be a number field and consider a arithmetic surface $Xto B=operatornameSpec O_K$, i.e. $X$ is integral, regular, flat, proper over $O_K$ and it has dimension $2$.

Now pick a closed point $xin X$ such that $xmapsto bin B$ and consider $widehatmathcal O_X,x$. In other words the completion of the local ring $mathcal O_X,x$ with respect to its maximal ideal.

Can we express $widehatmathcal O_X,x= A[[t]]$? What is $A$ in
this case? Do we have $A=O_L$ where $L$ is a complete discrete
valuation field and a finite extension of $K_b$ (here $K_b$ is the completion of $K$ at $b$)?

For sure we have an embedding $mathcal O_B,b[t]hookrightarrowmathcal
O_X,x$

Thanks in advance

algebraic-geometry power-series surfaces local-rings

share|cite|improve this question

edited Feb 18 at 15:08

notsure

asked Feb 17 at 15:52

notsurenotsure

547

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I asked a similar question in Mathoverflow some time ago but I didn't get an answer: Completion of a local ring of an arithmetic surface
  $endgroup$
  – yamete kudasai
  Feb 18 at 15:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Take a look at Liu's book exercise 4, 3.35 (d) at page 149. But it assumes the existence of a section.
  $endgroup$
  – notsure
  Feb 18 at 15:03
  

  
  
  
  

add a comment | 

$begingroup$

Let $K$ be a number field and consider a arithmetic surface $Xto B=operatornameSpec O_K$, i.e. $X$ is integral, regular, flat, proper over $O_K$ and it has dimension $2$.

Now pick a closed point $xin X$ such that $xmapsto bin B$ and consider $widehatmathcal O_X,x$. In other words the completion of the local ring $mathcal O_X,x$ with respect to its maximal ideal.

Can we express $widehatmathcal O_X,x= A[[t]]$? What is $A$ in
this case? Do we have $A=O_L$ where $L$ is a complete discrete
valuation field and a finite extension of $K_b$ (here $K_b$ is the completion of $K$ at $b$)?

For sure we have an embedding $mathcal O_B,b[t]hookrightarrowmathcal
O_X,x$

Thanks in advance

algebraic-geometry power-series surfaces local-rings

share|cite|improve this question

edited Feb 18 at 15:08

notsure

asked Feb 17 at 15:52

notsurenotsure

547

$endgroup$

share|cite|improve this question

edited Feb 18 at 15:08

notsure

asked Feb 17 at 15:52

notsurenotsure

547

share|cite|improve this question

edited Feb 18 at 15:08

notsure

asked Feb 17 at 15:52

notsurenotsure

547

asked Feb 17 at 15:52

notsurenotsure

547

asked Feb 17 at 15:52

notsurenotsure

547