Formal power series rings and p-adic solenoids The Next CEO of Stack OverflowQuestion 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
The Next CEO of Stack OverflowQuestion 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 $1+x+x^2+x^3+…$
$begingroup$
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
asked Mar 27 at 20:22
Mike BattagliaMike Battaglia
1,5521128
$endgroup$
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
asked Mar 27 at 20:22
Mike BattagliaMike Battaglia
1,5521128
$endgroup$
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
asked Mar 27 at 20:22
Mike BattagliaMike Battaglia
1,5521128
$endgroup$
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
p-adic-number-theory formal-power-series
asked Mar 27 at 20:22
Mike BattagliaMike Battaglia
1,5521128
asked Mar 27 at 20:22
Mike BattagliaMike Battaglia
1,5521128
asked Mar 27 at 20:22
Mike BattagliaMike Battaglia
1,5521128
asked Mar 27 at 20:22
Mike BattagliaMike Battaglia
1,5521128
asked Mar 27 at 20:22
Mike BattagliaMike Battaglia
1,5521128
1,5521128
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $x-p$.
answered Mar 27 at 20:33
Eric WofseyEric Wofsey
191k14216349
$endgroup$
$begingroup$
Thanks - is the solenoid related to the formal power series in any way then?
$endgroup$
– Mike Battaglia
2 days ago
$begingroup$
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
2 days ago
add a comment |
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1 Answer
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1 Answer
1
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oldest
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active
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$begingroup$
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 $x-p$.
answered Mar 27 at 20:33
Eric WofseyEric Wofsey
191k14216349
$endgroup$
$begingroup$
Thanks - is the solenoid related to the formal power series in any way then?
$endgroup$
– Mike Battaglia
2 days ago
$begingroup$
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
2 days ago
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 $x-p$.
answered Mar 27 at 20:33
Eric WofseyEric Wofsey
191k14216349
$endgroup$
$begingroup$
Thanks - is the solenoid related to the formal power series in any way then?
$endgroup$
– Mike Battaglia
2 days ago
$begingroup$
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
2 days ago
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 $x-p$.
answered Mar 27 at 20:33
Eric WofseyEric Wofsey
191k14216349
$endgroup$
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 $x-p$.
answered Mar 27 at 20:33
Eric WofseyEric Wofsey
191k14216349
edited Mar 28 at 0:07
answered Mar 27 at 20:33
Eric WofseyEric Wofsey
191k14216349
answered Mar 27 at 20:33
Eric WofseyEric Wofsey
191k14216349
answered Mar 27 at 20:33
Eric WofseyEric Wofsey
191k14216349
191k14216349
$begingroup$
Thanks - is the solenoid related to the formal power series in any way then?
$endgroup$
– Mike Battaglia
2 days ago
$begingroup$
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
2 days ago
add a comment |
$begingroup$
Thanks - is the solenoid related to the formal power series in any way then?
$endgroup$
– Mike Battaglia
2 days ago
$begingroup$
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
2 days ago
$begingroup$
Thanks - is the solenoid related to the formal power series in any way then?
$endgroup$
– Mike Battaglia
2 days ago
$begingroup$
Thanks - is the solenoid related to the formal power series in any way then?
$endgroup$
– Mike Battaglia
2 days ago
$begingroup$
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
2 days ago
$begingroup$
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
2 days ago
add a comment |
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