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Basic question on p-adic numbers
Clash Royale CLAN TAG#URR8PPP
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I have a very basic question about $p$-adic numbers that arose while reading about $p$-divisible group-schemes. On the very first page of this paper for instance, as well as in other sources, it is suggested that
$$lim_rightarrowmathbbZ/p^nmathbbZsimeqmathbbQ_p/mathbbZ_p$$
which surprizes me because as far as I know, this inductive limit actually is the definition of $mathbbZ_p$ (as I see it on Wikipedia).
Could someone please confirm whether the above isomorphism is true (ie. exists), and if yes, explain why we have this identity?
I thank you very much.
abstract-algebra p-adic-number-theory
asked Jul 30 at 11:30
Suzet
2,203427
add a comment |Â
up vote
1
down vote
favorite
I have a very basic question about $p$-adic numbers that arose while reading about $p$-divisible group-schemes. On the very first page of this paper for instance, as well as in other sources, it is suggested that
$$lim_rightarrowmathbbZ/p^nmathbbZsimeqmathbbQ_p/mathbbZ_p$$
which surprizes me because as far as I know, this inductive limit actually is the definition of $mathbbZ_p$ (as I see it on Wikipedia).
Could someone please confirm whether the above isomorphism is true (ie. exists), and if yes, explain why we have this identity?
I thank you very much.
abstract-algebra p-adic-number-theory
asked Jul 30 at 11:30
Suzet
2,203427
1
This is what you're looking for.
– ÃgjøgnumMeg
Jul 30 at 11:47
Fantastic, thank you very much @ÃgjøgnumMeg !
– Suzet
Jul 30 at 11:48
2
By the way, the isomorphism you mention here is obtained by mapping $a modp^n$ to $a/p^n$.
– Gal Porat
Jul 30 at 13:32
And here’s a pleasant exercise for you: to show that $textAut(Bbb Q_p/Bbb Z_p)congBbb Z_p$.
– Lubin
Aug 3 at 1:07
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have a very basic question about $p$-adic numbers that arose while reading about $p$-divisible group-schemes. On the very first page of this paper for instance, as well as in other sources, it is suggested that
$$lim_rightarrowmathbbZ/p^nmathbbZsimeqmathbbQ_p/mathbbZ_p$$
which surprizes me because as far as I know, this inductive limit actually is the definition of $mathbbZ_p$ (as I see it on Wikipedia).
Could someone please confirm whether the above isomorphism is true (ie. exists), and if yes, explain why we have this identity?
I thank you very much.
abstract-algebra p-adic-number-theory
asked Jul 30 at 11:30
Suzet
2,203427
I have a very basic question about $p$-adic numbers that arose while reading about $p$-divisible group-schemes. On the very first page of this paper for instance, as well as in other sources, it is suggested that
$$lim_rightarrowmathbbZ/p^nmathbbZsimeqmathbbQ_p/mathbbZ_p$$
which surprizes me because as far as I know, this inductive limit actually is the definition of $mathbbZ_p$ (as I see it on Wikipedia).
Could someone please confirm whether the above isomorphism is true (ie. exists), and if yes, explain why we have this identity?
I thank you very much.
abstract-algebra p-adic-number-theory
asked Jul 30 at 11:30
Suzet
2,203427
asked Jul 30 at 11:30
Suzet
2,203427
asked Jul 30 at 11:30
Suzet
2,203427
asked Jul 30 at 11:30
Suzet
2,203427
2,203427
1
This is what you're looking for.
– ÃgjøgnumMeg
Jul 30 at 11:47
Fantastic, thank you very much @ÃgjøgnumMeg !
– Suzet
Jul 30 at 11:48
2
By the way, the isomorphism you mention here is obtained by mapping $a modp^n$ to $a/p^n$.
– Gal Porat
Jul 30 at 13:32
And here’s a pleasant exercise for you: to show that $textAut(Bbb Q_p/Bbb Z_p)congBbb Z_p$.
– Lubin
Aug 3 at 1:07
add a comment |Â
1
This is what you're looking for.
– ÃgjøgnumMeg
Jul 30 at 11:47
Fantastic, thank you very much @ÃgjøgnumMeg !
– Suzet
Jul 30 at 11:48
2
By the way, the isomorphism you mention here is obtained by mapping $a modp^n$ to $a/p^n$.
– Gal Porat
Jul 30 at 13:32
And here’s a pleasant exercise for you: to show that $textAut(Bbb Q_p/Bbb Z_p)congBbb Z_p$.
– Lubin
Aug 3 at 1:07
1
1
This is what you're looking for.
– ÃgjøgnumMeg
Jul 30 at 11:47
This is what you're looking for.
– ÃgjøgnumMeg
Jul 30 at 11:47
Fantastic, thank you very much @ÃgjøgnumMeg !
– Suzet
Jul 30 at 11:48
Fantastic, thank you very much @ÃgjøgnumMeg !
– Suzet
Jul 30 at 11:48
2
2
By the way, the isomorphism you mention here is obtained by mapping $a modp^n$ to $a/p^n$.
– Gal Porat
Jul 30 at 13:32
By the way, the isomorphism you mention here is obtained by mapping $a modp^n$ to $a/p^n$.
– Gal Porat
Jul 30 at 13:32
And here’s a pleasant exercise for you: to show that $textAut(Bbb Q_p/Bbb Z_p)congBbb Z_p$.
– Lubin
Aug 3 at 1:07
And here’s a pleasant exercise for you: to show that $textAut(Bbb Q_p/Bbb Z_p)congBbb Z_p$.
– Lubin
Aug 3 at 1:07
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
The inductive limit is indeed what you said. First you need to say how do you send $mathbbZ/p^nmathbbZ$ to $mathbbZ/p^n+1mathbbZ$. This is with a map as abelian groups (not as a rings) given by sending $1$ to $p$. So we identify $mathbbZ/p^nmathbbZ$ with the subgroup $pmathbbZ/p^n+1mathbbZ$.
Now the inductive limit becomes the union when you put everything inside the correct place: $mathbbQ_p/mathbbZ_p$. This is done by sending $mathbbZ/p^nmathbbZ$ to the subgroup $frac 1p^nmathbbZ_p/mathbbZ_p$ form by the elements killed for multiplication by $p^n$, sending $1$ to the class of this $p$-adic number $frac 1p^n$. Clearly this subgroup has order $p^n$, so the map is an isomorphism. Finally, any element of $mathbbQ_p/mathbbZ_p$ has order $p^n$ for some $nge 0$, so
$$mathbbQ_p/mathbbZ_p=bigcup_nge 1 frac 1p^nmathbbZ_p/mathbbZ_pconglim_to mathbbZ/p^nmathbbZ$$
answered Jul 30 at 15:40
xarles
83558
1
Thank you so much for this very clear answer, @xarles .
– Suzet
Jul 31 at 0:03
add a comment |Â
up vote
4
down vote
I don't know about the direct (inductive) limit, but the definition of $mathbf Z_p$ is that it is the inverse or projective limit:
$$mathbf Z_p=varprojlim mathbf Z/p^n mathbf Z=Bigl(bar x_n)inprod_nmathbf Z/p^n mathbf Zmid x_n+1equiv x_nbmod p^n Bigr$$
answered Jul 30 at 11:43
Bernard
110k635102
You are right, thank you very much for pointing this out.
– Suzet
Jul 30 at 11:48
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The inductive limit is indeed what you said. First you need to say how do you send $mathbbZ/p^nmathbbZ$ to $mathbbZ/p^n+1mathbbZ$. This is with a map as abelian groups (not as a rings) given by sending $1$ to $p$. So we identify $mathbbZ/p^nmathbbZ$ with the subgroup $pmathbbZ/p^n+1mathbbZ$.
Now the inductive limit becomes the union when you put everything inside the correct place: $mathbbQ_p/mathbbZ_p$. This is done by sending $mathbbZ/p^nmathbbZ$ to the subgroup $frac 1p^nmathbbZ_p/mathbbZ_p$ form by the elements killed for multiplication by $p^n$, sending $1$ to the class of this $p$-adic number $frac 1p^n$. Clearly this subgroup has order $p^n$, so the map is an isomorphism. Finally, any element of $mathbbQ_p/mathbbZ_p$ has order $p^n$ for some $nge 0$, so
$$mathbbQ_p/mathbbZ_p=bigcup_nge 1 frac 1p^nmathbbZ_p/mathbbZ_pconglim_to mathbbZ/p^nmathbbZ$$
answered Jul 30 at 15:40
xarles
83558
1
Thank you so much for this very clear answer, @xarles .
– Suzet
Jul 31 at 0:03
add a comment |Â
up vote
2
down vote
accepted
The inductive limit is indeed what you said. First you need to say how do you send $mathbbZ/p^nmathbbZ$ to $mathbbZ/p^n+1mathbbZ$. This is with a map as abelian groups (not as a rings) given by sending $1$ to $p$. So we identify $mathbbZ/p^nmathbbZ$ with the subgroup $pmathbbZ/p^n+1mathbbZ$.
Now the inductive limit becomes the union when you put everything inside the correct place: $mathbbQ_p/mathbbZ_p$. This is done by sending $mathbbZ/p^nmathbbZ$ to the subgroup $frac 1p^nmathbbZ_p/mathbbZ_p$ form by the elements killed for multiplication by $p^n$, sending $1$ to the class of this $p$-adic number $frac 1p^n$. Clearly this subgroup has order $p^n$, so the map is an isomorphism. Finally, any element of $mathbbQ_p/mathbbZ_p$ has order $p^n$ for some $nge 0$, so
$$mathbbQ_p/mathbbZ_p=bigcup_nge 1 frac 1p^nmathbbZ_p/mathbbZ_pconglim_to mathbbZ/p^nmathbbZ$$
answered Jul 30 at 15:40
xarles
83558
1
Thank you so much for this very clear answer, @xarles .
– Suzet
Jul 31 at 0:03
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The inductive limit is indeed what you said. First you need to say how do you send $mathbbZ/p^nmathbbZ$ to $mathbbZ/p^n+1mathbbZ$. This is with a map as abelian groups (not as a rings) given by sending $1$ to $p$. So we identify $mathbbZ/p^nmathbbZ$ with the subgroup $pmathbbZ/p^n+1mathbbZ$.
Now the inductive limit becomes the union when you put everything inside the correct place: $mathbbQ_p/mathbbZ_p$. This is done by sending $mathbbZ/p^nmathbbZ$ to the subgroup $frac 1p^nmathbbZ_p/mathbbZ_p$ form by the elements killed for multiplication by $p^n$, sending $1$ to the class of this $p$-adic number $frac 1p^n$. Clearly this subgroup has order $p^n$, so the map is an isomorphism. Finally, any element of $mathbbQ_p/mathbbZ_p$ has order $p^n$ for some $nge 0$, so
$$mathbbQ_p/mathbbZ_p=bigcup_nge 1 frac 1p^nmathbbZ_p/mathbbZ_pconglim_to mathbbZ/p^nmathbbZ$$
answered Jul 30 at 15:40
xarles
83558
The inductive limit is indeed what you said. First you need to say how do you send $mathbbZ/p^nmathbbZ$ to $mathbbZ/p^n+1mathbbZ$. This is with a map as abelian groups (not as a rings) given by sending $1$ to $p$. So we identify $mathbbZ/p^nmathbbZ$ with the subgroup $pmathbbZ/p^n+1mathbbZ$.
Now the inductive limit becomes the union when you put everything inside the correct place: $mathbbQ_p/mathbbZ_p$. This is done by sending $mathbbZ/p^nmathbbZ$ to the subgroup $frac 1p^nmathbbZ_p/mathbbZ_p$ form by the elements killed for multiplication by $p^n$, sending $1$ to the class of this $p$-adic number $frac 1p^n$. Clearly this subgroup has order $p^n$, so the map is an isomorphism. Finally, any element of $mathbbQ_p/mathbbZ_p$ has order $p^n$ for some $nge 0$, so
$$mathbbQ_p/mathbbZ_p=bigcup_nge 1 frac 1p^nmathbbZ_p/mathbbZ_pconglim_to mathbbZ/p^nmathbbZ$$
answered Jul 30 at 15:40
xarles
83558
edited Jul 30 at 16:22
answered Jul 30 at 15:40
xarles
83558
answered Jul 30 at 15:40
xarles
83558
answered Jul 30 at 15:40
xarles
83558
83558
1
Thank you so much for this very clear answer, @xarles .
– Suzet
Jul 31 at 0:03
add a comment |Â
1
Thank you so much for this very clear answer, @xarles .
– Suzet
Jul 31 at 0:03
1
1
Thank you so much for this very clear answer, @xarles .
– Suzet
Jul 31 at 0:03
Thank you so much for this very clear answer, @xarles .
– Suzet
Jul 31 at 0:03
add a comment |Â
up vote
4
down vote
I don't know about the direct (inductive) limit, but the definition of $mathbf Z_p$ is that it is the inverse or projective limit:
$$mathbf Z_p=varprojlim mathbf Z/p^n mathbf Z=Bigl(bar x_n)inprod_nmathbf Z/p^n mathbf Zmid x_n+1equiv x_nbmod p^n Bigr$$
answered Jul 30 at 11:43
Bernard
110k635102
You are right, thank you very much for pointing this out.
– Suzet
Jul 30 at 11:48
add a comment |Â
up vote
4
down vote
I don't know about the direct (inductive) limit, but the definition of $mathbf Z_p$ is that it is the inverse or projective limit:
$$mathbf Z_p=varprojlim mathbf Z/p^n mathbf Z=Bigl(bar x_n)inprod_nmathbf Z/p^n mathbf Zmid x_n+1equiv x_nbmod p^n Bigr$$
answered Jul 30 at 11:43
Bernard
110k635102
You are right, thank you very much for pointing this out.
– Suzet
Jul 30 at 11:48
add a comment |Â
up vote
4
down vote
up vote
4
down vote
I don't know about the direct (inductive) limit, but the definition of $mathbf Z_p$ is that it is the inverse or projective limit:
$$mathbf Z_p=varprojlim mathbf Z/p^n mathbf Z=Bigl(bar x_n)inprod_nmathbf Z/p^n mathbf Zmid x_n+1equiv x_nbmod p^n Bigr$$
answered Jul 30 at 11:43
Bernard
110k635102
I don't know about the direct (inductive) limit, but the definition of $mathbf Z_p$ is that it is the inverse or projective limit:
$$mathbf Z_p=varprojlim mathbf Z/p^n mathbf Z=Bigl(bar x_n)inprod_nmathbf Z/p^n mathbf Zmid x_n+1equiv x_nbmod p^n Bigr$$
answered Jul 30 at 11:43
Bernard
110k635102
edited Jul 30 at 11:52
answered Jul 30 at 11:43
Bernard
110k635102
answered Jul 30 at 11:43
Bernard
110k635102
answered Jul 30 at 11:43
Bernard
110k635102
110k635102
You are right, thank you very much for pointing this out.
– Suzet
Jul 30 at 11:48
add a comment |Â
You are right, thank you very much for pointing this out.
– Suzet
Jul 30 at 11:48
You are right, thank you very much for pointing this out.
– Suzet
Jul 30 at 11:48
You are right, thank you very much for pointing this out.
– Suzet
Jul 30 at 11:48
add a comment |Â
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1
This is what you're looking for.
– ÃgjøgnumMeg
Jul 30 at 11:47
Fantastic, thank you very much @ÃgjøgnumMeg !
– Suzet
Jul 30 at 11:48
2
By the way, the isomorphism you mention here is obtained by mapping $a modp^n$ to $a/p^n$.
– Gal Porat
Jul 30 at 13:32
And here’s a pleasant exercise for you: to show that $textAut(Bbb Q_p/Bbb Z_p)congBbb Z_p$.
– Lubin
Aug 3 at 1:07