Formal power series rings and p-adic solenoids Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Question about $p$-adic numbers and $p$-adic integersUnique presentation for the field of fractions for formal power seriesAre the $p^n$-adic numbers isomorphic to the $p$-adic numbers?$p$-adic logarithm is a homomorphism, formal power series proofIsomorphisms between quotients of formal power series ringsformal power series ring over field is m-adic completePower Series Arithmetic through Formal Power SeriesRings of formal power series is finitely generated as module.Exponentiate generating functions as formal power seriesRelationship between $p$-adic numbers and analytic continuation of $1+x+x^2+x^3+…$

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Formal power series rings and p-adic solenoids



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Question about $p$-adic numbers and $p$-adic integersUnique presentation for the field of fractions for formal power seriesAre the $p^n$-adic numbers isomorphic to the $p$-adic numbers?$p$-adic logarithm is a homomorphism, formal power series proofIsomorphisms between quotients of formal power series ringsformal power series ring over field is m-adic completePower Series Arithmetic through Formal Power SeriesRings of formal power series is finitely generated as module.Exponentiate generating functions as formal power seriesRelationship between $p$-adic numbers and analytic continuation of $1+x+x^2+x^3+…$










2












$begingroup$


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 $Bbb C[[x]]/(x-p)$?









share|cite|improve this question









$endgroup$
















    2












    $begingroup$


    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 $Bbb C[[x]]/(x-p)$?









    share|cite|improve this question









    $endgroup$














      2












      2








      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 $Bbb C[[x]]/(x-p)$?









      share|cite|improve this question









      $endgroup$




      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













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Mar 27 at 20:22









      Mike BattagliaMike Battaglia

      1,6441230




      1,6441230




















          1 Answer
          1






          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











          $endgroup$












          • $begingroup$
            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











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          1 Answer
          1






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          active

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          active

          oldest

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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











          $endgroup$












          • $begingroup$
            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















          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











          $endgroup$












          • $begingroup$
            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













          3












          3








          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











          $endgroup$



          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














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 28 at 0:07

























          answered Mar 27 at 20:33









          Eric WofseyEric Wofsey

          193k14221352




          193k14221352











          • $begingroup$
            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
















          • $begingroup$
            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















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




          $begingroup$
          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




          $begingroup$
          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

















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