4-adic numbers and zero divisors
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The $p$-adic numbers form an integral domain provided that $p$ is prime.
Let's look at the $n$-adic numbers when $n$ is not prime.
Case $n = 10$
There are zero divisors. See this previous question.
Case $n = pq$ where $p$ and $q$ are coprime (not necessarily prime but not $1$).
There are also zero divisors. A similar construction works.
Case $n = p^k$ where $p$ is prime and $k > 1$
I have not figured this one out yet, not even the simplest case of $n = 4$. The construction in the previous question above does not work and I have not found an alternative yet. Looking at approximations in $mathbbZ_4$, $mathbbZ_16$,
$mathbbZ_64$, etc just leads me to zeros divisors ending in zeroes suggesting, but not proving, that there are none.
Note that I am using $mathbbZ_n$ for the integers modulo $n$ and not the $n$-adic numbers. I think that I have seen it used for both. What is usual if you want discuss both at the same time?
Another previous question asks why $4$-adic numbers are not possible. The answer seems to be that they are possible but a norm cannot be defined. So, it leaves the existence of zero divisors open.
Are there zero divisors in the $4$-adic numbers?
Are there idempotents in the $4$-adic numbers?
I have not looked at $9$-adic or other prime powers yet.
Please don't answer directly but some hints would be appreciated.
number-theory p-adic-number-theory
asked Sep 22 at 7:11
badjohn
3,5171618
add a comment |Â
up vote
4
down vote
favorite
The $p$-adic numbers form an integral domain provided that $p$ is prime.
Let's look at the $n$-adic numbers when $n$ is not prime.
Case $n = 10$
There are zero divisors. See this previous question.
Case $n = pq$ where $p$ and $q$ are coprime (not necessarily prime but not $1$).
There are also zero divisors. A similar construction works.
Case $n = p^k$ where $p$ is prime and $k > 1$
I have not figured this one out yet, not even the simplest case of $n = 4$. The construction in the previous question above does not work and I have not found an alternative yet. Looking at approximations in $mathbbZ_4$, $mathbbZ_16$,
$mathbbZ_64$, etc just leads me to zeros divisors ending in zeroes suggesting, but not proving, that there are none.
Note that I am using $mathbbZ_n$ for the integers modulo $n$ and not the $n$-adic numbers. I think that I have seen it used for both. What is usual if you want discuss both at the same time?
Another previous question asks why $4$-adic numbers are not possible. The answer seems to be that they are possible but a norm cannot be defined. So, it leaves the existence of zero divisors open.
Are there zero divisors in the $4$-adic numbers?
Are there idempotents in the $4$-adic numbers?
I have not looked at $9$-adic or other prime powers yet.
Please don't answer directly but some hints would be appreciated.
number-theory p-adic-number-theory
asked Sep 22 at 7:11
badjohn
3,5171618
2
I would expect that we have something like $mathbbZ_p^k = mathbbZ_p$ and $mathbbZ_ab = mathbbZ_a times mathbbZ_b$, but I didn't check it.
– Paul K
Sep 22 at 7:48
@JyrkiLahtonen That suggests another question. What are the possibilities? To avoid additional questions and answers hidden in comments, I will post another question unless you can point me to an existing answer.
– badjohn
Sep 22 at 7:52
as a hint, if you imagine converting from base 4 to base 2, not much changes.
– Steven
Sep 23 at 10:16
1
@Steven Thanks. I figured that out now. I briefly considered the 100-adic numbers and it was obviously that they were essentially the same as the 10-adic.
– badjohn
Sep 23 at 10:56
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
The $p$-adic numbers form an integral domain provided that $p$ is prime.
Let's look at the $n$-adic numbers when $n$ is not prime.
Case $n = 10$
There are zero divisors. See this previous question.
Case $n = pq$ where $p$ and $q$ are coprime (not necessarily prime but not $1$).
There are also zero divisors. A similar construction works.
Case $n = p^k$ where $p$ is prime and $k > 1$
I have not figured this one out yet, not even the simplest case of $n = 4$. The construction in the previous question above does not work and I have not found an alternative yet. Looking at approximations in $mathbbZ_4$, $mathbbZ_16$,
$mathbbZ_64$, etc just leads me to zeros divisors ending in zeroes suggesting, but not proving, that there are none.
Note that I am using $mathbbZ_n$ for the integers modulo $n$ and not the $n$-adic numbers. I think that I have seen it used for both. What is usual if you want discuss both at the same time?
Another previous question asks why $4$-adic numbers are not possible. The answer seems to be that they are possible but a norm cannot be defined. So, it leaves the existence of zero divisors open.
Are there zero divisors in the $4$-adic numbers?
Are there idempotents in the $4$-adic numbers?
I have not looked at $9$-adic or other prime powers yet.
Please don't answer directly but some hints would be appreciated.
number-theory p-adic-number-theory
asked Sep 22 at 7:11
badjohn
3,5171618
The $p$-adic numbers form an integral domain provided that $p$ is prime.
Let's look at the $n$-adic numbers when $n$ is not prime.
Case $n = 10$
There are zero divisors. See this previous question.
Case $n = pq$ where $p$ and $q$ are coprime (not necessarily prime but not $1$).
There are also zero divisors. A similar construction works.
Case $n = p^k$ where $p$ is prime and $k > 1$
I have not figured this one out yet, not even the simplest case of $n = 4$. The construction in the previous question above does not work and I have not found an alternative yet. Looking at approximations in $mathbbZ_4$, $mathbbZ_16$,
$mathbbZ_64$, etc just leads me to zeros divisors ending in zeroes suggesting, but not proving, that there are none.
Note that I am using $mathbbZ_n$ for the integers modulo $n$ and not the $n$-adic numbers. I think that I have seen it used for both. What is usual if you want discuss both at the same time?
Another previous question asks why $4$-adic numbers are not possible. The answer seems to be that they are possible but a norm cannot be defined. So, it leaves the existence of zero divisors open.
Are there zero divisors in the $4$-adic numbers?
Are there idempotents in the $4$-adic numbers?
I have not looked at $9$-adic or other prime powers yet.
Please don't answer directly but some hints would be appreciated.
number-theory p-adic-number-theory
number-theory p-adic-number-theory
asked Sep 22 at 7:11
badjohn
3,5171618
asked Sep 22 at 7:11
badjohn
3,5171618
asked Sep 22 at 7:11
badjohn
3,5171618
asked Sep 22 at 7:11
badjohn
3,5171618
asked Sep 22 at 7:11
badjohn
3,5171618
3,5171618
2
I would expect that we have something like $mathbbZ_p^k = mathbbZ_p$ and $mathbbZ_ab = mathbbZ_a times mathbbZ_b$, but I didn't check it.
– Paul K
Sep 22 at 7:48
@JyrkiLahtonen That suggests another question. What are the possibilities? To avoid additional questions and answers hidden in comments, I will post another question unless you can point me to an existing answer.
– badjohn
Sep 22 at 7:52
as a hint, if you imagine converting from base 4 to base 2, not much changes.
– Steven
Sep 23 at 10:16
1
@Steven Thanks. I figured that out now. I briefly considered the 100-adic numbers and it was obviously that they were essentially the same as the 10-adic.
– badjohn
Sep 23 at 10:56
add a comment |Â
2
I would expect that we have something like $mathbbZ_p^k = mathbbZ_p$ and $mathbbZ_ab = mathbbZ_a times mathbbZ_b$, but I didn't check it.
– Paul K
Sep 22 at 7:48
@JyrkiLahtonen That suggests another question. What are the possibilities? To avoid additional questions and answers hidden in comments, I will post another question unless you can point me to an existing answer.
– badjohn
Sep 22 at 7:52
as a hint, if you imagine converting from base 4 to base 2, not much changes.
– Steven
Sep 23 at 10:16
1
@Steven Thanks. I figured that out now. I briefly considered the 100-adic numbers and it was obviously that they were essentially the same as the 10-adic.
– badjohn
Sep 23 at 10:56
2
2
I would expect that we have something like $mathbbZ_p^k = mathbbZ_p$ and $mathbbZ_ab = mathbbZ_a times mathbbZ_b$, but I didn't check it.
– Paul K
Sep 22 at 7:48
I would expect that we have something like $mathbbZ_p^k = mathbbZ_p$ and $mathbbZ_ab = mathbbZ_a times mathbbZ_b$, but I didn't check it.
– Paul K
Sep 22 at 7:48
@JyrkiLahtonen That suggests another question. What are the possibilities? To avoid additional questions and answers hidden in comments, I will post another question unless you can point me to an existing answer.
– badjohn
Sep 22 at 7:52
@JyrkiLahtonen That suggests another question. What are the possibilities? To avoid additional questions and answers hidden in comments, I will post another question unless you can point me to an existing answer.
– badjohn
Sep 22 at 7:52
as a hint, if you imagine converting from base 4 to base 2, not much changes.
– Steven
Sep 23 at 10:16
as a hint, if you imagine converting from base 4 to base 2, not much changes.
– Steven
Sep 23 at 10:16
1
1
@Steven Thanks. I figured that out now. I briefly considered the 100-adic numbers and it was obviously that they were essentially the same as the 10-adic.
– badjohn
Sep 23 at 10:56
@Steven Thanks. I figured that out now. I briefly considered the 100-adic numbers and it was obviously that they were essentially the same as the 10-adic.
– badjohn
Sep 23 at 10:56
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
6
down vote
accepted
If one defines the $4$-adic numbers as the inverse limit
$$Bbb Z_4conglim_longleftarrow(Bbb Z/4^nBbb Z)$$
then $Bbb Z_4congBbb Z_2$, the $2$-adic numbers.
In general $Bbb Z_p^kcongBbb Z_p$.
answered Sep 22 at 7:21
Lord Shark the Unknown
91.2k955117
Thanks - that's an answer that I had not suspected yet. It explains why I could not find any zero divisors. Also, reviewing my attempts at finding them, I see that maybe I should have suspected this.
– badjohn
Sep 22 at 7:54
1
It’s fairly simply seen: the families $p^nBbb Z$ and $p^2nBbb Z$ are coterminal: each member of the one family contains a member of the other; and vice versa.
– Lubin
Sep 22 at 18:38
add a comment |Â
up vote
1
down vote
In "p-adic Analysis Compared with Real" from Svetlana Katok it is proven that for distinct primes $p_1,ldots,p_k$ we have $mathbbQ_g = mathbbQ_p_1 oplus ldots oplus mathbbQ_p_k$. She doesn't mention the case of non-distinct primes. Moreover she mentions that Hensel proved the above fact.
answered yesterday
Paul K
2,655315
Thanks. We also have, from Lord Shark, $Bbb Z_p^kcongBbb Z_p$ so between the two results, we have the whole story.
– badjohn
yesterday
I would pay a bit attention there. The $g$-adic numbers in Katok are defined via a pseudo-norm which is defined analogously to the $p$-adic norm. I am not sure if one can obtain the $g$-adic integers like this if you define them via the inverse limit.
– Paul K
yesterday
Thanks. I have some more study to do.
– badjohn
yesterday
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
If one defines the $4$-adic numbers as the inverse limit
$$Bbb Z_4conglim_longleftarrow(Bbb Z/4^nBbb Z)$$
then $Bbb Z_4congBbb Z_2$, the $2$-adic numbers.
In general $Bbb Z_p^kcongBbb Z_p$.
answered Sep 22 at 7:21
Lord Shark the Unknown
91.2k955117
Thanks - that's an answer that I had not suspected yet. It explains why I could not find any zero divisors. Also, reviewing my attempts at finding them, I see that maybe I should have suspected this.
– badjohn
Sep 22 at 7:54
1
It’s fairly simply seen: the families $p^nBbb Z$ and $p^2nBbb Z$ are coterminal: each member of the one family contains a member of the other; and vice versa.
– Lubin
Sep 22 at 18:38
add a comment |Â
up vote
6
down vote
accepted
If one defines the $4$-adic numbers as the inverse limit
$$Bbb Z_4conglim_longleftarrow(Bbb Z/4^nBbb Z)$$
then $Bbb Z_4congBbb Z_2$, the $2$-adic numbers.
In general $Bbb Z_p^kcongBbb Z_p$.
answered Sep 22 at 7:21
Lord Shark the Unknown
91.2k955117
Thanks - that's an answer that I had not suspected yet. It explains why I could not find any zero divisors. Also, reviewing my attempts at finding them, I see that maybe I should have suspected this.
– badjohn
Sep 22 at 7:54
1
It’s fairly simply seen: the families $p^nBbb Z$ and $p^2nBbb Z$ are coterminal: each member of the one family contains a member of the other; and vice versa.
– Lubin
Sep 22 at 18:38
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
If one defines the $4$-adic numbers as the inverse limit
$$Bbb Z_4conglim_longleftarrow(Bbb Z/4^nBbb Z)$$
then $Bbb Z_4congBbb Z_2$, the $2$-adic numbers.
In general $Bbb Z_p^kcongBbb Z_p$.
answered Sep 22 at 7:21
Lord Shark the Unknown
91.2k955117
If one defines the $4$-adic numbers as the inverse limit
$$Bbb Z_4conglim_longleftarrow(Bbb Z/4^nBbb Z)$$
then $Bbb Z_4congBbb Z_2$, the $2$-adic numbers.
In general $Bbb Z_p^kcongBbb Z_p$.
answered Sep 22 at 7:21
Lord Shark the Unknown
91.2k955117
answered Sep 22 at 7:21
Lord Shark the Unknown
91.2k955117
answered Sep 22 at 7:21
Lord Shark the Unknown
91.2k955117
answered Sep 22 at 7:21
Lord Shark the Unknown
91.2k955117
91.2k955117
Thanks - that's an answer that I had not suspected yet. It explains why I could not find any zero divisors. Also, reviewing my attempts at finding them, I see that maybe I should have suspected this.
– badjohn
Sep 22 at 7:54
1
It’s fairly simply seen: the families $p^nBbb Z$ and $p^2nBbb Z$ are coterminal: each member of the one family contains a member of the other; and vice versa.
– Lubin
Sep 22 at 18:38
add a comment |Â
Thanks - that's an answer that I had not suspected yet. It explains why I could not find any zero divisors. Also, reviewing my attempts at finding them, I see that maybe I should have suspected this.
– badjohn
Sep 22 at 7:54
1
It’s fairly simply seen: the families $p^nBbb Z$ and $p^2nBbb Z$ are coterminal: each member of the one family contains a member of the other; and vice versa.
– Lubin
Sep 22 at 18:38
Thanks - that's an answer that I had not suspected yet. It explains why I could not find any zero divisors. Also, reviewing my attempts at finding them, I see that maybe I should have suspected this.
– badjohn
Sep 22 at 7:54
Thanks - that's an answer that I had not suspected yet. It explains why I could not find any zero divisors. Also, reviewing my attempts at finding them, I see that maybe I should have suspected this.
– badjohn
Sep 22 at 7:54
1
1
It’s fairly simply seen: the families $p^nBbb Z$ and $p^2nBbb Z$ are coterminal: each member of the one family contains a member of the other; and vice versa.
– Lubin
Sep 22 at 18:38
It’s fairly simply seen: the families $p^nBbb Z$ and $p^2nBbb Z$ are coterminal: each member of the one family contains a member of the other; and vice versa.
– Lubin
Sep 22 at 18:38
add a comment |Â
up vote
1
down vote
In "p-adic Analysis Compared with Real" from Svetlana Katok it is proven that for distinct primes $p_1,ldots,p_k$ we have $mathbbQ_g = mathbbQ_p_1 oplus ldots oplus mathbbQ_p_k$. She doesn't mention the case of non-distinct primes. Moreover she mentions that Hensel proved the above fact.
answered yesterday
Paul K
2,655315
Thanks. We also have, from Lord Shark, $Bbb Z_p^kcongBbb Z_p$ so between the two results, we have the whole story.
– badjohn
yesterday
I would pay a bit attention there. The $g$-adic numbers in Katok are defined via a pseudo-norm which is defined analogously to the $p$-adic norm. I am not sure if one can obtain the $g$-adic integers like this if you define them via the inverse limit.
– Paul K
yesterday
Thanks. I have some more study to do.
– badjohn
yesterday
add a comment |Â
up vote
1
down vote
In "p-adic Analysis Compared with Real" from Svetlana Katok it is proven that for distinct primes $p_1,ldots,p_k$ we have $mathbbQ_g = mathbbQ_p_1 oplus ldots oplus mathbbQ_p_k$. She doesn't mention the case of non-distinct primes. Moreover she mentions that Hensel proved the above fact.
answered yesterday
Paul K
2,655315
Thanks. We also have, from Lord Shark, $Bbb Z_p^kcongBbb Z_p$ so between the two results, we have the whole story.
– badjohn
yesterday
I would pay a bit attention there. The $g$-adic numbers in Katok are defined via a pseudo-norm which is defined analogously to the $p$-adic norm. I am not sure if one can obtain the $g$-adic integers like this if you define them via the inverse limit.
– Paul K
yesterday
Thanks. I have some more study to do.
– badjohn
yesterday
add a comment |Â
up vote
1
down vote
up vote
1
down vote
In "p-adic Analysis Compared with Real" from Svetlana Katok it is proven that for distinct primes $p_1,ldots,p_k$ we have $mathbbQ_g = mathbbQ_p_1 oplus ldots oplus mathbbQ_p_k$. She doesn't mention the case of non-distinct primes. Moreover she mentions that Hensel proved the above fact.
answered yesterday
Paul K
2,655315
In "p-adic Analysis Compared with Real" from Svetlana Katok it is proven that for distinct primes $p_1,ldots,p_k$ we have $mathbbQ_g = mathbbQ_p_1 oplus ldots oplus mathbbQ_p_k$. She doesn't mention the case of non-distinct primes. Moreover she mentions that Hensel proved the above fact.
answered yesterday
Paul K
2,655315
answered yesterday
Paul K
2,655315
answered yesterday
Paul K
2,655315
answered yesterday
Paul K
2,655315
2,655315
Thanks. We also have, from Lord Shark, $Bbb Z_p^kcongBbb Z_p$ so between the two results, we have the whole story.
– badjohn
yesterday
I would pay a bit attention there. The $g$-adic numbers in Katok are defined via a pseudo-norm which is defined analogously to the $p$-adic norm. I am not sure if one can obtain the $g$-adic integers like this if you define them via the inverse limit.
– Paul K
yesterday
Thanks. I have some more study to do.
– badjohn
yesterday
add a comment |Â
Thanks. We also have, from Lord Shark, $Bbb Z_p^kcongBbb Z_p$ so between the two results, we have the whole story.
– badjohn
yesterday
I would pay a bit attention there. The $g$-adic numbers in Katok are defined via a pseudo-norm which is defined analogously to the $p$-adic norm. I am not sure if one can obtain the $g$-adic integers like this if you define them via the inverse limit.
– Paul K
yesterday
Thanks. I have some more study to do.
– badjohn
yesterday
Thanks. We also have, from Lord Shark, $Bbb Z_p^kcongBbb Z_p$ so between the two results, we have the whole story.
– badjohn
yesterday
Thanks. We also have, from Lord Shark, $Bbb Z_p^kcongBbb Z_p$ so between the two results, we have the whole story.
– badjohn
yesterday
I would pay a bit attention there. The $g$-adic numbers in Katok are defined via a pseudo-norm which is defined analogously to the $p$-adic norm. I am not sure if one can obtain the $g$-adic integers like this if you define them via the inverse limit.
– Paul K
yesterday
I would pay a bit attention there. The $g$-adic numbers in Katok are defined via a pseudo-norm which is defined analogously to the $p$-adic norm. I am not sure if one can obtain the $g$-adic integers like this if you define them via the inverse limit.
– Paul K
yesterday
Thanks. I have some more study to do.
– badjohn
yesterday
Thanks. I have some more study to do.
– badjohn
yesterday
add a comment |Â
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2
I would expect that we have something like $mathbbZ_p^k = mathbbZ_p$ and $mathbbZ_ab = mathbbZ_a times mathbbZ_b$, but I didn't check it.
– Paul K
Sep 22 at 7:48
@JyrkiLahtonen That suggests another question. What are the possibilities? To avoid additional questions and answers hidden in comments, I will post another question unless you can point me to an existing answer.
– badjohn
Sep 22 at 7:52
as a hint, if you imagine converting from base 4 to base 2, not much changes.
– Steven
Sep 23 at 10:16
1
@Steven Thanks. I figured that out now. I briefly considered the 100-adic numbers and it was obviously that they were essentially the same as the 10-adic.
– badjohn
Sep 23 at 10:56