Localization and p-adic completion of Integers coincide?
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
asked Nov 24 at 2:29
Elvis Torres Pérez
382
add a comment |
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
asked Nov 24 at 2:29
Elvis Torres Pérez
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
add a comment |
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
asked Nov 24 at 2:29
Elvis Torres Pérez
382
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
asked Nov 24 at 2:29
Elvis Torres Pérez
382
asked Nov 24 at 2:29
Elvis Torres Pérez
382
asked Nov 24 at 2:29
Elvis Torres Pérez
382
asked Nov 24 at 2:29
Elvis Torres Pérez
382
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
add a comment |
$mathbb{Z}_{(p)}$ 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
add a comment |
2 Answers
2
active
oldest
votes
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.
answered Nov 24 at 2:32
Eric Wofsey
179k12204331
add a comment |
$mathbb{Z}_{(p)}$ 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 $mathbb{Z}_{p}$.
answered Nov 24 at 3:12
Seewoo Lee
6,165826
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered Nov 24 at 2:32
Eric Wofsey
179k12204331
add a comment |
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.
answered Nov 24 at 2:32
Eric Wofsey
179k12204331
add a comment |
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.
answered Nov 24 at 2:32
Eric Wofsey
179k12204331
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.
answered Nov 24 at 2:32
Eric Wofsey
179k12204331
answered Nov 24 at 2:32
Eric Wofsey
179k12204331
answered Nov 24 at 2:32
Eric Wofsey
179k12204331
answered Nov 24 at 2:32
Eric Wofsey
179k12204331
179k12204331
add a comment |
add a comment |
$mathbb{Z}_{(p)}$ 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 $mathbb{Z}_{p}$.
answered Nov 24 at 3:12
Seewoo Lee
6,165826
add a comment |
$mathbb{Z}_{(p)}$ 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 $mathbb{Z}_{p}$.
answered Nov 24 at 3:12
Seewoo Lee
6,165826
add a comment |
$mathbb{Z}_{(p)}$ 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 $mathbb{Z}_{p}$.
answered Nov 24 at 3:12
Seewoo Lee
6,165826
$mathbb{Z}_{(p)}$ 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 $mathbb{Z}_{p}$.
answered Nov 24 at 3:12
Seewoo Lee
6,165826
answered Nov 24 at 3:12
Seewoo Lee
6,165826
answered Nov 24 at 3:12
Seewoo Lee
6,165826
answered Nov 24 at 3:12
Seewoo Lee
6,165826
6,165826
add a comment |
add a comment |
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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