Polynomial analogue of $p$-adic numbers

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Consider a monic irreducible polynomial $P in Bbb Z[X]$ (equivalently, by Gauss lemma, it is irreducible in $Bbb Q[X]$).
What do the rings
$$
R = varprojlim_n Bbb Z[X] / (P^n) qquadtextandqquad
S = varprojlim_n Bbb Q[X] / (P^n)
$$
look like? How are they related to the $p$-adic numbers (for some prime $p$)? How do they relate to $Bbb Z[[X]]$ ? I first saw this construction here (hem!).



If $P=X$, then we have $R = Bbb Z[[X]], S = Bbb Q[[X]]$ (in particular, it is not true that $S = R otimes_Bbb Z Bbb Q$), and $R/(X-p) cong Bbb Z_p$ for every prime $p$. But what do we get if for instance $P = X^2 + 1$ ? Do we get someting like $R = Bbb Z[i][[X]]$ ?



Thank you!




abstract-algebra polynomials commutative-algebra




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  • Seems to look close to this (but not exactly); in the chapter 4 of this book it is even exposed as a starting construction (bypassing the ordinary $p$-adic one).
    – metamorphy
    Aug 21 at 10:29











  • It is analogue of $p-$adic expansion. This is more or less formal taylor series around the solution of $P$.(Pick whatever your solution and this is a ring allowing you to do taylor series at that solution.) $Z[|x|]$ is a special case of this.
    – user45765
    Aug 21 at 11:43










  • If we consider $Bbb F_q[X]$ instead of $Bbb Z[X]$, then I think that we get some local field of characteristic $p>0$.
    – Alphonse
    Aug 21 at 12:35














up vote
1
down vote

favorite












Consider a monic irreducible polynomial $P in Bbb Z[X]$ (equivalently, by Gauss lemma, it is irreducible in $Bbb Q[X]$).
What do the rings
$$
R = varprojlim_n Bbb Z[X] / (P^n) qquadtextandqquad
S = varprojlim_n Bbb Q[X] / (P^n)
$$
look like? How are they related to the $p$-adic numbers (for some prime $p$)? How do they relate to $Bbb Z[[X]]$ ? I first saw this construction here (hem!).



If $P=X$, then we have $R = Bbb Z[[X]], S = Bbb Q[[X]]$ (in particular, it is not true that $S = R otimes_Bbb Z Bbb Q$), and $R/(X-p) cong Bbb Z_p$ for every prime $p$. But what do we get if for instance $P = X^2 + 1$ ? Do we get someting like $R = Bbb Z[i][[X]]$ ?



Thank you!




abstract-algebra polynomials commutative-algebra




share|cite|improve this question




















  • Seems to look close to this (but not exactly); in the chapter 4 of this book it is even exposed as a starting construction (bypassing the ordinary $p$-adic one).
    – metamorphy
    Aug 21 at 10:29











  • It is analogue of $p-$adic expansion. This is more or less formal taylor series around the solution of $P$.(Pick whatever your solution and this is a ring allowing you to do taylor series at that solution.) $Z[|x|]$ is a special case of this.
    – user45765
    Aug 21 at 11:43










  • If we consider $Bbb F_q[X]$ instead of $Bbb Z[X]$, then I think that we get some local field of characteristic $p>0$.
    – Alphonse
    Aug 21 at 12:35












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Consider a monic irreducible polynomial $P in Bbb Z[X]$ (equivalently, by Gauss lemma, it is irreducible in $Bbb Q[X]$).
What do the rings
$$
R = varprojlim_n Bbb Z[X] / (P^n) qquadtextandqquad
S = varprojlim_n Bbb Q[X] / (P^n)
$$
look like? How are they related to the $p$-adic numbers (for some prime $p$)? How do they relate to $Bbb Z[[X]]$ ? I first saw this construction here (hem!).



If $P=X$, then we have $R = Bbb Z[[X]], S = Bbb Q[[X]]$ (in particular, it is not true that $S = R otimes_Bbb Z Bbb Q$), and $R/(X-p) cong Bbb Z_p$ for every prime $p$. But what do we get if for instance $P = X^2 + 1$ ? Do we get someting like $R = Bbb Z[i][[X]]$ ?



Thank you!




abstract-algebra polynomials commutative-algebra




share|cite|improve this question












Consider a monic irreducible polynomial $P in Bbb Z[X]$ (equivalently, by Gauss lemma, it is irreducible in $Bbb Q[X]$).
What do the rings
$$
R = varprojlim_n Bbb Z[X] / (P^n) qquadtextandqquad
S = varprojlim_n Bbb Q[X] / (P^n)
$$
look like? How are they related to the $p$-adic numbers (for some prime $p$)? How do they relate to $Bbb Z[[X]]$ ? I first saw this construction here (hem!).



If $P=X$, then we have $R = Bbb Z[[X]], S = Bbb Q[[X]]$ (in particular, it is not true that $S = R otimes_Bbb Z Bbb Q$), and $R/(X-p) cong Bbb Z_p$ for every prime $p$. But what do we get if for instance $P = X^2 + 1$ ? Do we get someting like $R = Bbb Z[i][[X]]$ ?



Thank you!





abstract-algebra polynomials commutative-algebra





share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 21 at 10:04









Alphonse

1,810622




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  • Seems to look close to this (but not exactly); in the chapter 4 of this book it is even exposed as a starting construction (bypassing the ordinary $p$-adic one).
    – metamorphy
    Aug 21 at 10:29











  • It is analogue of $p-$adic expansion. This is more or less formal taylor series around the solution of $P$.(Pick whatever your solution and this is a ring allowing you to do taylor series at that solution.) $Z[|x|]$ is a special case of this.
    – user45765
    Aug 21 at 11:43










  • If we consider $Bbb F_q[X]$ instead of $Bbb Z[X]$, then I think that we get some local field of characteristic $p>0$.
    – Alphonse
    Aug 21 at 12:35
















  • Seems to look close to this (but not exactly); in the chapter 4 of this book it is even exposed as a starting construction (bypassing the ordinary $p$-adic one).
    – metamorphy
    Aug 21 at 10:29











  • It is analogue of $p-$adic expansion. This is more or less formal taylor series around the solution of $P$.(Pick whatever your solution and this is a ring allowing you to do taylor series at that solution.) $Z[|x|]$ is a special case of this.
    – user45765
    Aug 21 at 11:43










  • If we consider $Bbb F_q[X]$ instead of $Bbb Z[X]$, then I think that we get some local field of characteristic $p>0$.
    – Alphonse
    Aug 21 at 12:35















Seems to look close to this (but not exactly); in the chapter 4 of this book it is even exposed as a starting construction (bypassing the ordinary $p$-adic one).
– metamorphy
Aug 21 at 10:29





Seems to look close to this (but not exactly); in the chapter 4 of this book it is even exposed as a starting construction (bypassing the ordinary $p$-adic one).
– metamorphy
Aug 21 at 10:29













It is analogue of $p-$adic expansion. This is more or less formal taylor series around the solution of $P$.(Pick whatever your solution and this is a ring allowing you to do taylor series at that solution.) $Z[|x|]$ is a special case of this.
– user45765
Aug 21 at 11:43




It is analogue of $p-$adic expansion. This is more or less formal taylor series around the solution of $P$.(Pick whatever your solution and this is a ring allowing you to do taylor series at that solution.) $Z[|x|]$ is a special case of this.
– user45765
Aug 21 at 11:43












If we consider $Bbb F_q[X]$ instead of $Bbb Z[X]$, then I think that we get some local field of characteristic $p>0$.
– Alphonse
Aug 21 at 12:35




If we consider $Bbb F_q[X]$ instead of $Bbb Z[X]$, then I think that we get some local field of characteristic $p>0$.
– Alphonse
Aug 21 at 12:35















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