How can we have he similar decomposition of $ mathbb{Q}_p$ and $mathbb{Q}_p(zeta_p)$?
$begingroup$
We have,
$mathbb{Q}_p=$p-adic field, $mathbb{Z}_p=$ring of p-adic integers, $mathbb{Z}_p^{times}=$multiplicative group of units in $mathbb{Z}_p$.
We have the following decompositions:
$mathbb{Z}_p^{times}=mu_{p-1} times (1+pmathbb{Z}_p)$,
where $mu_{p-1}$ is the group of roots of unity of order $ p-1$,
$mathbb{Q}_p^{times}=p^{mathbb{Z}} times mu_{p-1} times (1+pmathbb{Z}_p) simeq mathbb{Z} times mathbb{Z} /(p-1) mathbb{Z} times mathbb{Z}_p$.
The question is-
How can we have he similar decomposition of $ mathbb{Q}_p[zeta_p]^{times}$ and $mathbb{Q}_p(zeta_p)$ ?
commutative-algebra p-adic-number-theory
asked Jan 23 at 16:37
M. A. SARKARM. A. SARKAR
2,3071719
$endgroup$
add a comment |
$begingroup$
We have,
$mathbb{Q}_p=$p-adic field, $mathbb{Z}_p=$ring of p-adic integers, $mathbb{Z}_p^{times}=$multiplicative group of units in $mathbb{Z}_p$.
We have the following decompositions:
$mathbb{Z}_p^{times}=mu_{p-1} times (1+pmathbb{Z}_p)$,
where $mu_{p-1}$ is the group of roots of unity of order $ p-1$,
$mathbb{Q}_p^{times}=p^{mathbb{Z}} times mu_{p-1} times (1+pmathbb{Z}_p) simeq mathbb{Z} times mathbb{Z} /(p-1) mathbb{Z} times mathbb{Z}_p$.
The question is-
How can we have he similar decomposition of $ mathbb{Q}_p[zeta_p]^{times}$ and $mathbb{Q}_p(zeta_p)$ ?
commutative-algebra p-adic-number-theory
asked Jan 23 at 16:37
M. A. SARKARM. A. SARKAR
2,3071719
$endgroup$
2
$begingroup$
It is obvious how to go from $mathbb{Z}_p[zeta_p]^times$ to $mathbb{Q}_p(zeta_p)^times$. I don't see what you mean with a decomposition of $mathbb{Q}_p$. And my previous comment was incorrect. $zeta_p-1$ is an uniformizer. Is there some $pi = a (zeta_p-1), a in mathbb{Z}_p[pi]^times$ such that $(1+pi)^{mathbb{Z}}$ is dense in $1+pi mathbb{Z}_p[pi]$ ? If it doesn't depend on $a$ then no since $a = 1$ gives $(1+pi)^{mathbb{Z}} = zeta_p^{mathbb{Z}}$.
$endgroup$
– reuns
Jan 23 at 17:36
$begingroup$
@reuns, what is decomposition of $ mathbb{Z}_p[zeta_p]^{times}$ and $ mathbb{Q}_p[zeta_p]^{times}$
$endgroup$
– M. A. SARKAR
Jan 23 at 17:52
2
$begingroup$
I think the best way to start is then with the $p$-adic $log,exp$, showing for some $k$ that $log,exp$ is a pair of isomomorphism $1+pi^k mathbb{Z}_p[pi] leftrightarrow pi^k mathbb{Z}_p[pi]$. Then use $mathbb{Z}_p[pi]^times/(1+pi^k mathbb{Z}_p[pi] )$ is finite. This is also a proof $(1+pi)^mathbb{Z}$ is never dense in $1+pimathbb{Z}_p[pi] $
$endgroup$
– reuns
Jan 23 at 18:19
add a comment |
$begingroup$
We have,
$mathbb{Q}_p=$p-adic field, $mathbb{Z}_p=$ring of p-adic integers, $mathbb{Z}_p^{times}=$multiplicative group of units in $mathbb{Z}_p$.
We have the following decompositions:
$mathbb{Z}_p^{times}=mu_{p-1} times (1+pmathbb{Z}_p)$,
where $mu_{p-1}$ is the group of roots of unity of order $ p-1$,
$mathbb{Q}_p^{times}=p^{mathbb{Z}} times mu_{p-1} times (1+pmathbb{Z}_p) simeq mathbb{Z} times mathbb{Z} /(p-1) mathbb{Z} times mathbb{Z}_p$.
The question is-
How can we have he similar decomposition of $ mathbb{Q}_p[zeta_p]^{times}$ and $mathbb{Q}_p(zeta_p)$ ?
commutative-algebra p-adic-number-theory
asked Jan 23 at 16:37
M. A. SARKARM. A. SARKAR
2,3071719
$endgroup$
We have,
$mathbb{Q}_p=$p-adic field, $mathbb{Z}_p=$ring of p-adic integers, $mathbb{Z}_p^{times}=$multiplicative group of units in $mathbb{Z}_p$.
We have the following decompositions:
$mathbb{Z}_p^{times}=mu_{p-1} times (1+pmathbb{Z}_p)$,
where $mu_{p-1}$ is the group of roots of unity of order $ p-1$,
$mathbb{Q}_p^{times}=p^{mathbb{Z}} times mu_{p-1} times (1+pmathbb{Z}_p) simeq mathbb{Z} times mathbb{Z} /(p-1) mathbb{Z} times mathbb{Z}_p$.
The question is-
How can we have he similar decomposition of $ mathbb{Q}_p[zeta_p]^{times}$ and $mathbb{Q}_p(zeta_p)$ ?
commutative-algebra p-adic-number-theory
commutative-algebra p-adic-number-theory
asked Jan 23 at 16:37
M. A. SARKARM. A. SARKAR
2,3071719
asked Jan 23 at 16:37
M. A. SARKARM. A. SARKAR
2,3071719
edited Jan 23 at 18:13
M. A. SARKAR
asked Jan 23 at 16:37
M. A. SARKARM. A. SARKAR
2,3071719
asked Jan 23 at 16:37
M. A. SARKARM. A. SARKAR
2,3071719
asked Jan 23 at 16:37
M. A. SARKARM. A. SARKAR
2,3071719
2,3071719
2
$begingroup$
It is obvious how to go from $mathbb{Z}_p[zeta_p]^times$ to $mathbb{Q}_p(zeta_p)^times$. I don't see what you mean with a decomposition of $mathbb{Q}_p$. And my previous comment was incorrect. $zeta_p-1$ is an uniformizer. Is there some $pi = a (zeta_p-1), a in mathbb{Z}_p[pi]^times$ such that $(1+pi)^{mathbb{Z}}$ is dense in $1+pi mathbb{Z}_p[pi]$ ? If it doesn't depend on $a$ then no since $a = 1$ gives $(1+pi)^{mathbb{Z}} = zeta_p^{mathbb{Z}}$.
$endgroup$
– reuns
Jan 23 at 17:36
$begingroup$
@reuns, what is decomposition of $ mathbb{Z}_p[zeta_p]^{times}$ and $ mathbb{Q}_p[zeta_p]^{times}$
$endgroup$
– M. A. SARKAR
Jan 23 at 17:52
2
$begingroup$
I think the best way to start is then with the $p$-adic $log,exp$, showing for some $k$ that $log,exp$ is a pair of isomomorphism $1+pi^k mathbb{Z}_p[pi] leftrightarrow pi^k mathbb{Z}_p[pi]$. Then use $mathbb{Z}_p[pi]^times/(1+pi^k mathbb{Z}_p[pi] )$ is finite. This is also a proof $(1+pi)^mathbb{Z}$ is never dense in $1+pimathbb{Z}_p[pi] $
$endgroup$
– reuns
Jan 23 at 18:19
add a comment |
2
$begingroup$
It is obvious how to go from $mathbb{Z}_p[zeta_p]^times$ to $mathbb{Q}_p(zeta_p)^times$. I don't see what you mean with a decomposition of $mathbb{Q}_p$. And my previous comment was incorrect. $zeta_p-1$ is an uniformizer. Is there some $pi = a (zeta_p-1), a in mathbb{Z}_p[pi]^times$ such that $(1+pi)^{mathbb{Z}}$ is dense in $1+pi mathbb{Z}_p[pi]$ ? If it doesn't depend on $a$ then no since $a = 1$ gives $(1+pi)^{mathbb{Z}} = zeta_p^{mathbb{Z}}$.
$endgroup$
– reuns
Jan 23 at 17:36
$begingroup$
@reuns, what is decomposition of $ mathbb{Z}_p[zeta_p]^{times}$ and $ mathbb{Q}_p[zeta_p]^{times}$
$endgroup$
– M. A. SARKAR
Jan 23 at 17:52
2
$begingroup$
I think the best way to start is then with the $p$-adic $log,exp$, showing for some $k$ that $log,exp$ is a pair of isomomorphism $1+pi^k mathbb{Z}_p[pi] leftrightarrow pi^k mathbb{Z}_p[pi]$. Then use $mathbb{Z}_p[pi]^times/(1+pi^k mathbb{Z}_p[pi] )$ is finite. This is also a proof $(1+pi)^mathbb{Z}$ is never dense in $1+pimathbb{Z}_p[pi] $
$endgroup$
– reuns
Jan 23 at 18:19
2
2
$begingroup$
It is obvious how to go from $mathbb{Z}_p[zeta_p]^times$ to $mathbb{Q}_p(zeta_p)^times$. I don't see what you mean with a decomposition of $mathbb{Q}_p$. And my previous comment was incorrect. $zeta_p-1$ is an uniformizer. Is there some $pi = a (zeta_p-1), a in mathbb{Z}_p[pi]^times$ such that $(1+pi)^{mathbb{Z}}$ is dense in $1+pi mathbb{Z}_p[pi]$ ? If it doesn't depend on $a$ then no since $a = 1$ gives $(1+pi)^{mathbb{Z}} = zeta_p^{mathbb{Z}}$.
$endgroup$
– reuns
Jan 23 at 17:36
$begingroup$
It is obvious how to go from $mathbb{Z}_p[zeta_p]^times$ to $mathbb{Q}_p(zeta_p)^times$. I don't see what you mean with a decomposition of $mathbb{Q}_p$. And my previous comment was incorrect. $zeta_p-1$ is an uniformizer. Is there some $pi = a (zeta_p-1), a in mathbb{Z}_p[pi]^times$ such that $(1+pi)^{mathbb{Z}}$ is dense in $1+pi mathbb{Z}_p[pi]$ ? If it doesn't depend on $a$ then no since $a = 1$ gives $(1+pi)^{mathbb{Z}} = zeta_p^{mathbb{Z}}$.
$endgroup$
– reuns
Jan 23 at 17:36
$begingroup$
@reuns, what is decomposition of $ mathbb{Z}_p[zeta_p]^{times}$ and $ mathbb{Q}_p[zeta_p]^{times}$
$endgroup$
– M. A. SARKAR
Jan 23 at 17:52
$begingroup$
@reuns, what is decomposition of $ mathbb{Z}_p[zeta_p]^{times}$ and $ mathbb{Q}_p[zeta_p]^{times}$
$endgroup$
– M. A. SARKAR
Jan 23 at 17:52
2
2
$begingroup$
I think the best way to start is then with the $p$-adic $log,exp$, showing for some $k$ that $log,exp$ is a pair of isomomorphism $1+pi^k mathbb{Z}_p[pi] leftrightarrow pi^k mathbb{Z}_p[pi]$. Then use $mathbb{Z}_p[pi]^times/(1+pi^k mathbb{Z}_p[pi] )$ is finite. This is also a proof $(1+pi)^mathbb{Z}$ is never dense in $1+pimathbb{Z}_p[pi] $
$endgroup$
– reuns
Jan 23 at 18:19
$begingroup$
I think the best way to start is then with the $p$-adic $log,exp$, showing for some $k$ that $log,exp$ is a pair of isomomorphism $1+pi^k mathbb{Z}_p[pi] leftrightarrow pi^k mathbb{Z}_p[pi]$. Then use $mathbb{Z}_p[pi]^times/(1+pi^k mathbb{Z}_p[pi] )$ is finite. This is also a proof $(1+pi)^mathbb{Z}$ is never dense in $1+pimathbb{Z}_p[pi] $
$endgroup$
– reuns
Jan 23 at 18:19
add a comment |
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$begingroup$
It is obvious how to go from $mathbb{Z}_p[zeta_p]^times$ to $mathbb{Q}_p(zeta_p)^times$. I don't see what you mean with a decomposition of $mathbb{Q}_p$. And my previous comment was incorrect. $zeta_p-1$ is an uniformizer. Is there some $pi = a (zeta_p-1), a in mathbb{Z}_p[pi]^times$ such that $(1+pi)^{mathbb{Z}}$ is dense in $1+pi mathbb{Z}_p[pi]$ ? If it doesn't depend on $a$ then no since $a = 1$ gives $(1+pi)^{mathbb{Z}} = zeta_p^{mathbb{Z}}$.
$endgroup$
– reuns
Jan 23 at 17:36
$begingroup$
@reuns, what is decomposition of $ mathbb{Z}_p[zeta_p]^{times}$ and $ mathbb{Q}_p[zeta_p]^{times}$
$endgroup$
– M. A. SARKAR
Jan 23 at 17:52
2
$begingroup$
I think the best way to start is then with the $p$-adic $log,exp$, showing for some $k$ that $log,exp$ is a pair of isomomorphism $1+pi^k mathbb{Z}_p[pi] leftrightarrow pi^k mathbb{Z}_p[pi]$. Then use $mathbb{Z}_p[pi]^times/(1+pi^k mathbb{Z}_p[pi] )$ is finite. This is also a proof $(1+pi)^mathbb{Z}$ is never dense in $1+pimathbb{Z}_p[pi] $
$endgroup$
– reuns
Jan 23 at 18:19