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...












-4














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:



enter image description here



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$ .










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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
















-4














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:



enter image description here



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$ .










share|cite|improve this 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














-4












-4








-4


0





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:



enter image description here



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$ .










share|cite|improve this question















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:



enter image description here



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






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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


















  • 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










1 Answer
1






active

oldest

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2














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.






share|cite|improve this answer



















  • 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 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














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.






share|cite|improve this answer



















  • 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
















2














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.






share|cite|improve this answer



















  • 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














2












2








2






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.






share|cite|improve this answer














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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








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














  • 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



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