Localization and p-adic completion of Integers coincide?












3














I want to know if $mathbb{Z}_{(p)}$ (localization by a prime ideal) and $mathbb{Z}_p$ (the completion of p-adic integers) are isomorphic. It seems true, but i don't know how to prove it. Does it holds for every PID?



Thanks.






number-theory commutative-algebra







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  • $mathbb{Z}_{(p)}$ is dense in $mathbb{Z}_p$ and in both $(p^n)$ are the only ideals
    – reuns
    Nov 24 at 2:40
















3














I want to know if $mathbb{Z}_{(p)}$ (localization by a prime ideal) and $mathbb{Z}_p$ (the completion of p-adic integers) are isomorphic. It seems true, but i don't know how to prove it. Does it holds for every PID?



Thanks.






number-theory commutative-algebra







share|cite|improve this question






















  • $mathbb{Z}_{(p)}$ is dense in $mathbb{Z}_p$ and in both $(p^n)$ are the only ideals
    – reuns
    Nov 24 at 2:40














3












3








3







I want to know if $mathbb{Z}_{(p)}$ (localization by a prime ideal) and $mathbb{Z}_p$ (the completion of p-adic integers) are isomorphic. It seems true, but i don't know how to prove it. Does it holds for every PID?



Thanks.






number-theory commutative-algebra







share|cite|improve this question













I want to know if $mathbb{Z}_{(p)}$ (localization by a prime ideal) and $mathbb{Z}_p$ (the completion of p-adic integers) are isomorphic. It seems true, but i don't know how to prove it. Does it holds for every PID?



Thanks.






number-theory commutative-algebra




number-theory commutative-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 24 at 2:29









Elvis Torres Pérez

382




382












  • $mathbb{Z}_{(p)}$ is dense in $mathbb{Z}_p$ and in both $(p^n)$ are the only ideals
    – reuns
    Nov 24 at 2:40


















  • $mathbb{Z}_{(p)}$ is dense in $mathbb{Z}_p$ and in both $(p^n)$ are the only ideals
    – reuns
    Nov 24 at 2:40
















$mathbb{Z}_{(p)}$ is dense in $mathbb{Z}_p$ and in both $(p^n)$ are the only ideals
– reuns
Nov 24 at 2:40




$mathbb{Z}_{(p)}$ is dense in $mathbb{Z}_p$ and in both $(p^n)$ are the only ideals
– reuns
Nov 24 at 2:40










2 Answers
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No, $mathbb{Z}_p$ is much larger than $mathbb{Z}_{(p)}$. Indeed, $mathbb{Z}_p$ is uncountable, since it has an element $sum a_np^n$ for any sequence of coefficients $a_nin{0,1,dots,p-1}$. On the other hand, $mathbb{Z}_{(p)}$ is a subring of $mathbb{Q}$ (the rationals with denominator not divisible by $p$), so it is countable.






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    1














    $mathbb{Z}_{(p)}$ is a proper subring of $mathbb{Z}_{p}$, and the latter one is complete discrete valuation ring, but $mathbb{Z}_{(p)}$ is not complete (but still DVR with respect to the same valuation). However, if you take completion of $mathbb{Z}_{(p)}$ with respect to the $p$-adic norm, you get $mathbb{Z}_{p}$.






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      2 Answers
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      2 Answers
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      3














      No, $mathbb{Z}_p$ is much larger than $mathbb{Z}_{(p)}$. Indeed, $mathbb{Z}_p$ is uncountable, since it has an element $sum a_np^n$ for any sequence of coefficients $a_nin{0,1,dots,p-1}$. On the other hand, $mathbb{Z}_{(p)}$ is a subring of $mathbb{Q}$ (the rationals with denominator not divisible by $p$), so it is countable.






      share|cite|improve this answer


























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        No, $mathbb{Z}_p$ is much larger than $mathbb{Z}_{(p)}$. Indeed, $mathbb{Z}_p$ is uncountable, since it has an element $sum a_np^n$ for any sequence of coefficients $a_nin{0,1,dots,p-1}$. On the other hand, $mathbb{Z}_{(p)}$ is a subring of $mathbb{Q}$ (the rationals with denominator not divisible by $p$), so it is countable.






        share|cite|improve this answer
























          3












          3








          3






          No, $mathbb{Z}_p$ is much larger than $mathbb{Z}_{(p)}$. Indeed, $mathbb{Z}_p$ is uncountable, since it has an element $sum a_np^n$ for any sequence of coefficients $a_nin{0,1,dots,p-1}$. On the other hand, $mathbb{Z}_{(p)}$ is a subring of $mathbb{Q}$ (the rationals with denominator not divisible by $p$), so it is countable.






          share|cite|improve this answer












          No, $mathbb{Z}_p$ is much larger than $mathbb{Z}_{(p)}$. Indeed, $mathbb{Z}_p$ is uncountable, since it has an element $sum a_np^n$ for any sequence of coefficients $a_nin{0,1,dots,p-1}$. On the other hand, $mathbb{Z}_{(p)}$ is a subring of $mathbb{Q}$ (the rationals with denominator not divisible by $p$), so it is countable.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 24 at 2:32









          Eric Wofsey

          178k12202330




          178k12202330























              1














              $mathbb{Z}_{(p)}$ is a proper subring of $mathbb{Z}_{p}$, and the latter one is complete discrete valuation ring, but $mathbb{Z}_{(p)}$ is not complete (but still DVR with respect to the same valuation). However, if you take completion of $mathbb{Z}_{(p)}$ with respect to the $p$-adic norm, you get $mathbb{Z}_{p}$.






              share|cite|improve this answer


























                1














                $mathbb{Z}_{(p)}$ is a proper subring of $mathbb{Z}_{p}$, and the latter one is complete discrete valuation ring, but $mathbb{Z}_{(p)}$ is not complete (but still DVR with respect to the same valuation). However, if you take completion of $mathbb{Z}_{(p)}$ with respect to the $p$-adic norm, you get $mathbb{Z}_{p}$.






                share|cite|improve this answer
























                  1












                  1








                  1






                  $mathbb{Z}_{(p)}$ is a proper subring of $mathbb{Z}_{p}$, and the latter one is complete discrete valuation ring, but $mathbb{Z}_{(p)}$ is not complete (but still DVR with respect to the same valuation). However, if you take completion of $mathbb{Z}_{(p)}$ with respect to the $p$-adic norm, you get $mathbb{Z}_{p}$.






                  share|cite|improve this answer












                  $mathbb{Z}_{(p)}$ is a proper subring of $mathbb{Z}_{p}$, and the latter one is complete discrete valuation ring, but $mathbb{Z}_{(p)}$ is not complete (but still DVR with respect to the same valuation). However, if you take completion of $mathbb{Z}_{(p)}$ with respect to the $p$-adic norm, you get $mathbb{Z}_{p}$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 24 at 3:12









                  Seewoo Lee

                  6,110826




                  6,110826






























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