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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 $\Bbb C[[x]]/(x-p)$?
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Very generally, if $R$ is a commutative ring and $f\in R[[x]]$, 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 $\mathbb{Z}[[x]]/(x-p)$ and formally inverted $p$, then you would get the $p$-adic rationals. The difference between that and $\mathbb{Q}[[x]]/(x-p)$ is that in $\mathbb{Q}[[x]]$ 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 $\mathbb{Z}$ divided by any fixed power of $p$. This is in fact exactly what happens with the inverse of $x-p$.

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  • $\begingroup$ 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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