Prove if $ord_p(d) < ord_p(n)$ then d divides n

Question

I have to prove that $d$ divides $n$ if and only if $ord_p(d)\leq ord_p(n)$

I have already proved that $ord_p(d)\leq ord_p(n)$ if $d$ divides $n$ but I am struggling to prove the converse. Can anyone give any help?

Answer 1

Let $d=\prod p^{a_p}$ and $n=\prod p^{b_p},$ where in each case the product is over all primes $p$ and the exponents are all $0$ beyond some point (and may be $0$ for lower primes also).

Then since $ord_p(d) \le ord_p(n)$ we have $a_p \le b_p$ at each prime $p.$ From this it's easy to see that $d|n$, in fact $n=d\cdot k$ where $k=\prod p^{b_p-a_p}.$

Answer 2

Write: $$n=p_{1}^{r_{1}}\cdots p_{k}^{r_{k}}$$ where the $p_{i}$ are distinct primes and the $r_{i}$ are nonnegative integers.

Then $\operatorname{ord}_{p}\left(d\right)\leq\operatorname{ord}_{p}\left(n\right)$ for each prime $p$ implies that we can write: $$d=p_{1}^{s_{1}}\cdots p_{k}^{s_{k}}$$ where the $s_i$ are integers that satisfy $0\leq s_{i}\leq r_{i}$ for $i=1,\dots,k$.

Then $n=md$ for $m=p_{1}^{r_{1}-s_{1}}\cdots p_{k}^{r_{k}-s_{k}}$.