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Formal power series rings and p-adic solenoids

2

0

$begingroup$

For any prime $p$, the ring of $p$-adic integers can be generated as the quotient of the formal power series ring $Bbb Z[[x]]/(x-p)$. My questions:

• If we instead use $Bbb Q[[x]]/(x-p)$, do we get the field of $p$-adic numbers?


• If we instead use $Bbb R[[x]]/(x-p)$, do we get the $p$-adic solenoid?


• Is there any sensible interpretation of $Bbb C[[x]]/(x-p)$?

p-adic-number-theory formal-power-series

share|cite|improve this question

asked Mar 27 at 20:22

Mike BattagliaMike Battaglia

1,6441230

$endgroup$

add a comment | 

2

0

$begingroup$

For any prime $p$, the ring of $p$-adic integers can be generated as the quotient of the formal power series ring $Bbb Z[[x]]/(x-p)$. My questions:

• If we instead use $Bbb Q[[x]]/(x-p)$, do we get the field of $p$-adic numbers?


• If we instead use $Bbb R[[x]]/(x-p)$, do we get the $p$-adic solenoid?


• Is there any sensible interpretation of $Bbb C[[x]]/(x-p)$?

p-adic-number-theory formal-power-series

share|cite|improve this question

asked Mar 27 at 20:22

Mike BattagliaMike Battaglia

1,6441230

$endgroup$

add a comment | 

$begingroup$

For any prime $p$, the ring of $p$-adic integers can be generated as the quotient of the formal power series ring $Bbb Z[[x]]/(x-p)$. My questions:

• If we instead use $Bbb Q[[x]]/(x-p)$, do we get the field of $p$-adic numbers?


• If we instead use $Bbb R[[x]]/(x-p)$, do we get the $p$-adic solenoid?


• Is there any sensible interpretation of $Bbb C[[x]]/(x-p)$?

p-adic-number-theory formal-power-series

share|cite|improve this question

asked Mar 27 at 20:22

Mike BattagliaMike Battaglia

1,6441230

$endgroup$

share|cite|improve this question

asked Mar 27 at 20:22

Mike BattagliaMike Battaglia

1,6441230

share|cite|improve this question

asked Mar 27 at 20:22

Mike BattagliaMike Battaglia

1,6441230

asked Mar 27 at 20:22

Mike BattagliaMike Battaglia

1,6441230

asked Mar 27 at 20:22

Mike BattagliaMike Battaglia

1,6441230

1 Answer
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active

oldest

votes

3

$begingroup$

Very generally, if $R$ is a commutative ring and $fin R[[x]]$, then $f$ is a unit iff the constant term of $f$ is a unit in $R$ (if the constant term is a unit, then you can build the coefficients of an inverse to $f$ one-by-one). So if $p$ is a unit in $R$, then $x-p$ is a unit in $R[[x]]$, and $R[[x]]/(x-p)$ is the zero ring.

Note that if you instead took $mathbbZ[[x]]/(x-p)$ and formally inverted $p$, then you would get the $p$-adic rationals. The difference between that and $mathbbQ[[x]]/(x-p)$ is that in $mathbbQ[[x]]$ you can have a power series which has coefficients with unbounded powers of $p$ in the denominators. Such a power series cannot be written as a power series with coefficients in $mathbbZ$ divided by any fixed power of $p$. This is in fact exactly what happens with the inverse of $x-p$.

share|cite|improve this answer

edited Mar 28 at 0:07

answered Mar 27 at 20:33

Eric WofseyEric Wofsey

193k14221352

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks - is the solenoid related to the formal power series in any way then?
  $endgroup$
  – Mike Battaglia
  Mar 29 at 1:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not that I know of. As far as I know, solenoids do not have any natural ring structure, which makes a connection to formal power series seem unlikely.
  $endgroup$
  – Eric Wofsey
  Mar 29 at 1:53
  

  
  
  
  

add a comment | 

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3

$begingroup$

Very generally, if $R$ is a commutative ring and $fin R[[x]]$, then $f$ is a unit iff the constant term of $f$ is a unit in $R$ (if the constant term is a unit, then you can build the coefficients of an inverse to $f$ one-by-one). So if $p$ is a unit in $R$, then $x-p$ is a unit in $R[[x]]$, and $R[[x]]/(x-p)$ is the zero ring.

Note that if you instead took $mathbbZ[[x]]/(x-p)$ and formally inverted $p$, then you would get the $p$-adic rationals. The difference between that and $mathbbQ[[x]]/(x-p)$ is that in $mathbbQ[[x]]$ you can have a power series which has coefficients with unbounded powers of $p$ in the denominators. Such a power series cannot be written as a power series with coefficients in $mathbbZ$ divided by any fixed power of $p$. This is in fact exactly what happens with the inverse of $x-p$.

share|cite|improve this answer

edited Mar 28 at 0:07

answered Mar 27 at 20:33

Eric WofseyEric Wofsey

193k14221352

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks - is the solenoid related to the formal power series in any way then?
  $endgroup$
  – Mike Battaglia
  Mar 29 at 1:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not that I know of. As far as I know, solenoids do not have any natural ring structure, which makes a connection to formal power series seem unlikely.
  $endgroup$
  – Eric Wofsey
  Mar 29 at 1:53
  

  
  
  
  

add a comment | 

3

$begingroup$

Very generally, if $R$ is a commutative ring and $fin R[[x]]$, then $f$ is a unit iff the constant term of $f$ is a unit in $R$ (if the constant term is a unit, then you can build the coefficients of an inverse to $f$ one-by-one). So if $p$ is a unit in $R$, then $x-p$ is a unit in $R[[x]]$, and $R[[x]]/(x-p)$ is the zero ring.

Note that if you instead took $mathbbZ[[x]]/(x-p)$ and formally inverted $p$, then you would get the $p$-adic rationals. The difference between that and $mathbbQ[[x]]/(x-p)$ is that in $mathbbQ[[x]]$ you can have a power series which has coefficients with unbounded powers of $p$ in the denominators. Such a power series cannot be written as a power series with coefficients in $mathbbZ$ divided by any fixed power of $p$. This is in fact exactly what happens with the inverse of $x-p$.

share|cite|improve this answer

edited Mar 28 at 0:07

answered Mar 27 at 20:33

Eric WofseyEric Wofsey

193k14221352

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks - is the solenoid related to the formal power series in any way then?
  $endgroup$
  – Mike Battaglia
  Mar 29 at 1:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not that I know of. As far as I know, solenoids do not have any natural ring structure, which makes a connection to formal power series seem unlikely.
  $endgroup$
  – Eric Wofsey
  Mar 29 at 1:53
  

  
  
  
  

add a comment | 

$begingroup$

Very generally, if $R$ is a commutative ring and $fin R[[x]]$, then $f$ is a unit iff the constant term of $f$ is a unit in $R$ (if the constant term is a unit, then you can build the coefficients of an inverse to $f$ one-by-one). So if $p$ is a unit in $R$, then $x-p$ is a unit in $R[[x]]$, and $R[[x]]/(x-p)$ is the zero ring.

Note that if you instead took $mathbbZ[[x]]/(x-p)$ and formally inverted $p$, then you would get the $p$-adic rationals. The difference between that and $mathbbQ[[x]]/(x-p)$ is that in $mathbbQ[[x]]$ you can have a power series which has coefficients with unbounded powers of $p$ in the denominators. Such a power series cannot be written as a power series with coefficients in $mathbbZ$ divided by any fixed power of $p$. This is in fact exactly what happens with the inverse of $x-p$.

share|cite|improve this answer

edited Mar 28 at 0:07

answered Mar 27 at 20:33

Eric WofseyEric Wofsey

193k14221352

$endgroup$

share|cite|improve this answer

edited Mar 28 at 0:07

answered Mar 27 at 20:33

Eric WofseyEric Wofsey

193k14221352

answered Mar 27 at 20:33

Eric WofseyEric Wofsey

193k14221352

answered Mar 27 at 20:33

Eric WofseyEric Wofsey

193k14221352