Krdytkyu

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$ .

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.