Mixing
4-adic numbers and zero divisors
Clash Royale CLAN TAG#URR8PPP
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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 3 hours ago
badjohn
3,4651618
add a comment |Â
up vote
2
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 3 hours ago
badjohn
3,4651618
1
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
3 hours ago
@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
3 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
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 3 hours ago
badjohn
3,4651618
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 3 hours ago
badjohn
3,4651618
asked 3 hours ago
badjohn
3,4651618
asked 3 hours ago
badjohn
3,4651618
asked 3 hours ago
badjohn
3,4651618
asked 3 hours ago
badjohn
3,4651618
3,4651618
1
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
3 hours ago
@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
3 hours ago
add a comment |Â
1
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
3 hours ago
@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
3 hours ago
1
1
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
3 hours ago
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
3 hours ago
@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
3 hours ago
@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
3 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
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 3 hours ago
Lord Shark the Unknown
90.1k955117
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
3 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
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 3 hours ago
Lord Shark the Unknown
90.1k955117
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
3 hours ago
add a comment |Â
up vote
4
down vote
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 3 hours ago
Lord Shark the Unknown
90.1k955117
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
3 hours ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
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 3 hours ago
Lord Shark the Unknown
90.1k955117
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 3 hours ago
Lord Shark the Unknown
90.1k955117
answered 3 hours ago
Lord Shark the Unknown
90.1k955117
answered 3 hours ago
Lord Shark the Unknown
90.1k955117
answered 3 hours ago
Lord Shark the Unknown
90.1k955117
90.1k955117
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
3 hours ago
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
3 hours ago
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
3 hours ago
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
3 hours ago
add a comment |Â
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1
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
3 hours ago
@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
3 hours ago