Show that if $ a in mathbb{Z}_p $, then $ frac{a(a-1) cdots (a-n+1)}{n!} in mathbb{Z}_p$ , p-adic numbers...
p-adic numbers:
Consider the ring of p-adic integers $ mathbb{Z}_p $.
Show that if $ a in mathbb{Z}_p $, then $ frac{a(a-1) cdots (a-n+1)}{n!} in mathbb{Z}_p$ also, where $n$ is natural number.
Answer:
To show that $frac{a(a-1) cdots (a-n+1)}{n!} in mathbb{Z}_p$, we have to show that $ left|frac{a(a-1) cdots (a-n+1)}{n!} right|_p leq 1$.
I have seen some hintz in the book of "p-adic numbers, p-adic analysis and zet-function" of the autor "Neal Koblitz". The hintz is given below:
I did not understand how they have chosen $N$ and what does $ord_p(a-a_0)>N$ mean?
please help to understand this.
or
Any other method to show that $ a in mathbb{Z}_p Rightarrow frac{a(a-1) cdots (a-n+1)}{n!} in mathbb{Z}_p$ .
p-adic-number-theory
closed as off-topic by amWhy, Saad, user26857, Gibbs, José Carlos Santos Nov 22 at 11:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Saad, user26857, Gibbs, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
p-adic numbers:
Consider the ring of p-adic integers $ mathbb{Z}_p $.
Show that if $ a in mathbb{Z}_p $, then $ frac{a(a-1) cdots (a-n+1)}{n!} in mathbb{Z}_p$ also, where $n$ is natural number.
Answer:
To show that $frac{a(a-1) cdots (a-n+1)}{n!} in mathbb{Z}_p$, we have to show that $ left|frac{a(a-1) cdots (a-n+1)}{n!} right|_p leq 1$.
I have seen some hintz in the book of "p-adic numbers, p-adic analysis and zet-function" of the autor "Neal Koblitz". The hintz is given below:
I did not understand how they have chosen $N$ and what does $ord_p(a-a_0)>N$ mean?
please help to understand this.
or
Any other method to show that $ a in mathbb{Z}_p Rightarrow frac{a(a-1) cdots (a-n+1)}{n!} in mathbb{Z}_p$ .
p-adic-number-theory
closed as off-topic by amWhy, Saad, user26857, Gibbs, José Carlos Santos Nov 22 at 11:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Saad, user26857, Gibbs, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
Please do not vandalise your posts.
– Asaf Karagila♦
Nov 22 at 17:59
add a comment |
p-adic numbers:
Consider the ring of p-adic integers $ mathbb{Z}_p $.
Show that if $ a in mathbb{Z}_p $, then $ frac{a(a-1) cdots (a-n+1)}{n!} in mathbb{Z}_p$ also, where $n$ is natural number.
Answer:
To show that $frac{a(a-1) cdots (a-n+1)}{n!} in mathbb{Z}_p$, we have to show that $ left|frac{a(a-1) cdots (a-n+1)}{n!} right|_p leq 1$.
I have seen some hintz in the book of "p-adic numbers, p-adic analysis and zet-function" of the autor "Neal Koblitz". The hintz is given below:
I did not understand how they have chosen $N$ and what does $ord_p(a-a_0)>N$ mean?
please help to understand this.
or
Any other method to show that $ a in mathbb{Z}_p Rightarrow frac{a(a-1) cdots (a-n+1)}{n!} in mathbb{Z}_p$ .
p-adic-number-theory
p-adic numbers:
Consider the ring of p-adic integers $ mathbb{Z}_p $.
Show that if $ a in mathbb{Z}_p $, then $ frac{a(a-1) cdots (a-n+1)}{n!} in mathbb{Z}_p$ also, where $n$ is natural number.
Answer:
To show that $frac{a(a-1) cdots (a-n+1)}{n!} in mathbb{Z}_p$, we have to show that $ left|frac{a(a-1) cdots (a-n+1)}{n!} right|_p leq 1$.
I have seen some hintz in the book of "p-adic numbers, p-adic analysis and zet-function" of the autor "Neal Koblitz". The hintz is given below:
I did not understand how they have chosen $N$ and what does $ord_p(a-a_0)>N$ mean?
please help to understand this.
or
Any other method to show that $ a in mathbb{Z}_p Rightarrow frac{a(a-1) cdots (a-n+1)}{n!} in mathbb{Z}_p$ .
p-adic-number-theory
p-adic-number-theory
edited Nov 22 at 17:59
Asaf Karagila♦
301k32424755
301k32424755
asked Oct 29 at 13:51
M. A. SARKAR
2,1721619
2,1721619
closed as off-topic by amWhy, Saad, user26857, Gibbs, José Carlos Santos Nov 22 at 11:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Saad, user26857, Gibbs, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Saad, user26857, Gibbs, José Carlos Santos Nov 22 at 11:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Saad, user26857, Gibbs, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
Please do not vandalise your posts.
– Asaf Karagila♦
Nov 22 at 17:59
add a comment |
Please do not vandalise your posts.
– Asaf Karagila♦
Nov 22 at 17:59
Please do not vandalise your posts.
– Asaf Karagila♦
Nov 22 at 17:59
Please do not vandalise your posts.
– Asaf Karagila♦
Nov 22 at 17:59
add a comment |
1 Answer
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Because of the denominator $n!$, the point is that $a(a-1)...(a-n+1)/n!$ lives a priori in $mathbf Q_p$, not necessarily in $mathbf Z_p$. Whatever your construction of $mathbf Z_p$ may be, you must think that an element $ain mathbf Z_p$, by definition, is"$p$-adically approximated" at any order $N$ by an $a_0 in mathbf Z$. This locution means that for any integer $N$, there exists $a_0 in mathbf Z$ s.t. $ord_p (a-a_0)>N$, which means in turn (by definition), that in $mathbf Z_p$, you can write $a-a_o=up^M$, where $uin mathbf Z^*_p$ and $M>N$. Consequently, given $n$, the fraction $a(a-1)...(a-n+1)/n!in mathbf Q_p$ can be approximated at any order by $a_0(a_0-1)...(a_0-n+1)/n!$ (since $n$ is fixed, just take $N$ large enough). The latter fraction lives in $mathbf Z$ because it expresses the number of combinations of $a_0$ objects among $n$, so the given fraction lives in $mathbf Z_p$ .
EDIT: In case the final argument is not clear enough (see the OP query), I develop it in detail. For $N$ large enough, the two fractions (involving resp. $a$ and $a_0$) are $p$-adically close. But since the $p$-adic valuation is discrete, this means that the two fractions have the same $p$-adic absolute value, and we are done.
1
Oh ! I thought that the question was in the...question alone. But the extra-assertion comes for free as soon as we have shown that $frac {a(a-1)...(a-n+1)}{n!}in mathbf Z_p$
– nguyen quang do
Oct 29 at 18:13
No, there is no inaccuracy at all, I explained the details in my edit. The fraction involving $a_0$ lives in $mathbf Z$, so the one involving $a$, which is $p$-adically close, lives in $mathbf Z_p$.
– nguyen quang do
Oct 30 at 12:58
1
Oh! In the definition of the number of combinations of $a_0$ objects among $n$ ? OK, this was an obvious misprint (I edit it), I said right after: "the latter fraction", i.e. with $a_0$. I thought your objection was about the $p$-adic argument.
– nguyen quang do
Oct 30 at 15:08
1
No problem. After all, reputation points are just a kind of game. The important thing is the correctness or not of the answer.
– nguyen quang do
Nov 5 at 17:40
1
As I said, the classical book "Class Field Theory" by Artin & Tate.
– nguyen quang do
Nov 24 at 17:05
|
show 8 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Because of the denominator $n!$, the point is that $a(a-1)...(a-n+1)/n!$ lives a priori in $mathbf Q_p$, not necessarily in $mathbf Z_p$. Whatever your construction of $mathbf Z_p$ may be, you must think that an element $ain mathbf Z_p$, by definition, is"$p$-adically approximated" at any order $N$ by an $a_0 in mathbf Z$. This locution means that for any integer $N$, there exists $a_0 in mathbf Z$ s.t. $ord_p (a-a_0)>N$, which means in turn (by definition), that in $mathbf Z_p$, you can write $a-a_o=up^M$, where $uin mathbf Z^*_p$ and $M>N$. Consequently, given $n$, the fraction $a(a-1)...(a-n+1)/n!in mathbf Q_p$ can be approximated at any order by $a_0(a_0-1)...(a_0-n+1)/n!$ (since $n$ is fixed, just take $N$ large enough). The latter fraction lives in $mathbf Z$ because it expresses the number of combinations of $a_0$ objects among $n$, so the given fraction lives in $mathbf Z_p$ .
EDIT: In case the final argument is not clear enough (see the OP query), I develop it in detail. For $N$ large enough, the two fractions (involving resp. $a$ and $a_0$) are $p$-adically close. But since the $p$-adic valuation is discrete, this means that the two fractions have the same $p$-adic absolute value, and we are done.
1
Oh ! I thought that the question was in the...question alone. But the extra-assertion comes for free as soon as we have shown that $frac {a(a-1)...(a-n+1)}{n!}in mathbf Z_p$
– nguyen quang do
Oct 29 at 18:13
No, there is no inaccuracy at all, I explained the details in my edit. The fraction involving $a_0$ lives in $mathbf Z$, so the one involving $a$, which is $p$-adically close, lives in $mathbf Z_p$.
– nguyen quang do
Oct 30 at 12:58
1
Oh! In the definition of the number of combinations of $a_0$ objects among $n$ ? OK, this was an obvious misprint (I edit it), I said right after: "the latter fraction", i.e. with $a_0$. I thought your objection was about the $p$-adic argument.
– nguyen quang do
Oct 30 at 15:08
1
No problem. After all, reputation points are just a kind of game. The important thing is the correctness or not of the answer.
– nguyen quang do
Nov 5 at 17:40
1
As I said, the classical book "Class Field Theory" by Artin & Tate.
– nguyen quang do
Nov 24 at 17:05
|
show 8 more comments
Because of the denominator $n!$, the point is that $a(a-1)...(a-n+1)/n!$ lives a priori in $mathbf Q_p$, not necessarily in $mathbf Z_p$. Whatever your construction of $mathbf Z_p$ may be, you must think that an element $ain mathbf Z_p$, by definition, is"$p$-adically approximated" at any order $N$ by an $a_0 in mathbf Z$. This locution means that for any integer $N$, there exists $a_0 in mathbf Z$ s.t. $ord_p (a-a_0)>N$, which means in turn (by definition), that in $mathbf Z_p$, you can write $a-a_o=up^M$, where $uin mathbf Z^*_p$ and $M>N$. Consequently, given $n$, the fraction $a(a-1)...(a-n+1)/n!in mathbf Q_p$ can be approximated at any order by $a_0(a_0-1)...(a_0-n+1)/n!$ (since $n$ is fixed, just take $N$ large enough). The latter fraction lives in $mathbf Z$ because it expresses the number of combinations of $a_0$ objects among $n$, so the given fraction lives in $mathbf Z_p$ .
EDIT: In case the final argument is not clear enough (see the OP query), I develop it in detail. For $N$ large enough, the two fractions (involving resp. $a$ and $a_0$) are $p$-adically close. But since the $p$-adic valuation is discrete, this means that the two fractions have the same $p$-adic absolute value, and we are done.
1
Oh ! I thought that the question was in the...question alone. But the extra-assertion comes for free as soon as we have shown that $frac {a(a-1)...(a-n+1)}{n!}in mathbf Z_p$
– nguyen quang do
Oct 29 at 18:13
No, there is no inaccuracy at all, I explained the details in my edit. The fraction involving $a_0$ lives in $mathbf Z$, so the one involving $a$, which is $p$-adically close, lives in $mathbf Z_p$.
– nguyen quang do
Oct 30 at 12:58
1
Oh! In the definition of the number of combinations of $a_0$ objects among $n$ ? OK, this was an obvious misprint (I edit it), I said right after: "the latter fraction", i.e. with $a_0$. I thought your objection was about the $p$-adic argument.
– nguyen quang do
Oct 30 at 15:08
1
No problem. After all, reputation points are just a kind of game. The important thing is the correctness or not of the answer.
– nguyen quang do
Nov 5 at 17:40
1
As I said, the classical book "Class Field Theory" by Artin & Tate.
– nguyen quang do
Nov 24 at 17:05
|
show 8 more comments
Because of the denominator $n!$, the point is that $a(a-1)...(a-n+1)/n!$ lives a priori in $mathbf Q_p$, not necessarily in $mathbf Z_p$. Whatever your construction of $mathbf Z_p$ may be, you must think that an element $ain mathbf Z_p$, by definition, is"$p$-adically approximated" at any order $N$ by an $a_0 in mathbf Z$. This locution means that for any integer $N$, there exists $a_0 in mathbf Z$ s.t. $ord_p (a-a_0)>N$, which means in turn (by definition), that in $mathbf Z_p$, you can write $a-a_o=up^M$, where $uin mathbf Z^*_p$ and $M>N$. Consequently, given $n$, the fraction $a(a-1)...(a-n+1)/n!in mathbf Q_p$ can be approximated at any order by $a_0(a_0-1)...(a_0-n+1)/n!$ (since $n$ is fixed, just take $N$ large enough). The latter fraction lives in $mathbf Z$ because it expresses the number of combinations of $a_0$ objects among $n$, so the given fraction lives in $mathbf Z_p$ .
EDIT: In case the final argument is not clear enough (see the OP query), I develop it in detail. For $N$ large enough, the two fractions (involving resp. $a$ and $a_0$) are $p$-adically close. But since the $p$-adic valuation is discrete, this means that the two fractions have the same $p$-adic absolute value, and we are done.
Because of the denominator $n!$, the point is that $a(a-1)...(a-n+1)/n!$ lives a priori in $mathbf Q_p$, not necessarily in $mathbf Z_p$. Whatever your construction of $mathbf Z_p$ may be, you must think that an element $ain mathbf Z_p$, by definition, is"$p$-adically approximated" at any order $N$ by an $a_0 in mathbf Z$. This locution means that for any integer $N$, there exists $a_0 in mathbf Z$ s.t. $ord_p (a-a_0)>N$, which means in turn (by definition), that in $mathbf Z_p$, you can write $a-a_o=up^M$, where $uin mathbf Z^*_p$ and $M>N$. Consequently, given $n$, the fraction $a(a-1)...(a-n+1)/n!in mathbf Q_p$ can be approximated at any order by $a_0(a_0-1)...(a_0-n+1)/n!$ (since $n$ is fixed, just take $N$ large enough). The latter fraction lives in $mathbf Z$ because it expresses the number of combinations of $a_0$ objects among $n$, so the given fraction lives in $mathbf Z_p$ .
EDIT: In case the final argument is not clear enough (see the OP query), I develop it in detail. For $N$ large enough, the two fractions (involving resp. $a$ and $a_0$) are $p$-adically close. But since the $p$-adic valuation is discrete, this means that the two fractions have the same $p$-adic absolute value, and we are done.
edited Oct 30 at 15:12
answered Oct 29 at 14:39
nguyen quang do
8,3611722
8,3611722
1
Oh ! I thought that the question was in the...question alone. But the extra-assertion comes for free as soon as we have shown that $frac {a(a-1)...(a-n+1)}{n!}in mathbf Z_p$
– nguyen quang do
Oct 29 at 18:13
No, there is no inaccuracy at all, I explained the details in my edit. The fraction involving $a_0$ lives in $mathbf Z$, so the one involving $a$, which is $p$-adically close, lives in $mathbf Z_p$.
– nguyen quang do
Oct 30 at 12:58
1
Oh! In the definition of the number of combinations of $a_0$ objects among $n$ ? OK, this was an obvious misprint (I edit it), I said right after: "the latter fraction", i.e. with $a_0$. I thought your objection was about the $p$-adic argument.
– nguyen quang do
Oct 30 at 15:08
1
No problem. After all, reputation points are just a kind of game. The important thing is the correctness or not of the answer.
– nguyen quang do
Nov 5 at 17:40
1
As I said, the classical book "Class Field Theory" by Artin & Tate.
– nguyen quang do
Nov 24 at 17:05
|
show 8 more comments
1
Oh ! I thought that the question was in the...question alone. But the extra-assertion comes for free as soon as we have shown that $frac {a(a-1)...(a-n+1)}{n!}in mathbf Z_p$
– nguyen quang do
Oct 29 at 18:13
No, there is no inaccuracy at all, I explained the details in my edit. The fraction involving $a_0$ lives in $mathbf Z$, so the one involving $a$, which is $p$-adically close, lives in $mathbf Z_p$.
– nguyen quang do
Oct 30 at 12:58
1
Oh! In the definition of the number of combinations of $a_0$ objects among $n$ ? OK, this was an obvious misprint (I edit it), I said right after: "the latter fraction", i.e. with $a_0$. I thought your objection was about the $p$-adic argument.
– nguyen quang do
Oct 30 at 15:08
1
No problem. After all, reputation points are just a kind of game. The important thing is the correctness or not of the answer.
– nguyen quang do
Nov 5 at 17:40
1
As I said, the classical book "Class Field Theory" by Artin & Tate.
– nguyen quang do
Nov 24 at 17:05
1
1
Oh ! I thought that the question was in the...question alone. But the extra-assertion comes for free as soon as we have shown that $frac {a(a-1)...(a-n+1)}{n!}in mathbf Z_p$
– nguyen quang do
Oct 29 at 18:13
Oh ! I thought that the question was in the...question alone. But the extra-assertion comes for free as soon as we have shown that $frac {a(a-1)...(a-n+1)}{n!}in mathbf Z_p$
– nguyen quang do
Oct 29 at 18:13
No, there is no inaccuracy at all, I explained the details in my edit. The fraction involving $a_0$ lives in $mathbf Z$, so the one involving $a$, which is $p$-adically close, lives in $mathbf Z_p$.
– nguyen quang do
Oct 30 at 12:58
No, there is no inaccuracy at all, I explained the details in my edit. The fraction involving $a_0$ lives in $mathbf Z$, so the one involving $a$, which is $p$-adically close, lives in $mathbf Z_p$.
– nguyen quang do
Oct 30 at 12:58
1
1
Oh! In the definition of the number of combinations of $a_0$ objects among $n$ ? OK, this was an obvious misprint (I edit it), I said right after: "the latter fraction", i.e. with $a_0$. I thought your objection was about the $p$-adic argument.
– nguyen quang do
Oct 30 at 15:08
Oh! In the definition of the number of combinations of $a_0$ objects among $n$ ? OK, this was an obvious misprint (I edit it), I said right after: "the latter fraction", i.e. with $a_0$. I thought your objection was about the $p$-adic argument.
– nguyen quang do
Oct 30 at 15:08
1
1
No problem. After all, reputation points are just a kind of game. The important thing is the correctness or not of the answer.
– nguyen quang do
Nov 5 at 17:40
No problem. After all, reputation points are just a kind of game. The important thing is the correctness or not of the answer.
– nguyen quang do
Nov 5 at 17:40
1
1
As I said, the classical book "Class Field Theory" by Artin & Tate.
– nguyen quang do
Nov 24 at 17:05
As I said, the classical book "Class Field Theory" by Artin & Tate.
– nguyen quang do
Nov 24 at 17:05
|
show 8 more comments
Please do not vandalise your posts.
– Asaf Karagila♦
Nov 22 at 17:59