Consider the p-adic field $ \ \mathbb{Q}_p \ $ . Define $ \operatorname{ord}_p(x) \ $ to be the p-adic valuation of $ \ x \ $

Question

Consider the p-adic field $ \ \mathbb{Q}_p \ $ . Define $ \operatorname{ord}_p(x) \ $ to be the $p$-adic valuation of $ \ x \ $ by $ \operatorname{ord}_p(x)=\max \{r : \ p^r \ \text{ divides } \ x \} \ $.

The inequality $$ 1 \leq p^{\operatorname{ord}_p(n)} \leq n $$ holds good for every $ \ n \ $.

My question is -

Does the inequality $ \ 1 \leq p^{\operatorname{ord}_p(\large n^3)} \leq n \ $ holds?

That is , we are replacing $ \ n^3 \ $ by $ \ n \ $.

Answer:

Let $\ \ n=p^r \cdot \frac{a}{b} , \ b \neq 0 , \ p \ \text{ does not divide}\ a,b \ $

Then

$ \operatorname{ord}_p(n)=r \ $

So,

$ n^3=(p^r \cdot \frac{a}{b}) (p^r \cdot \frac{a}{b} ) (p^r \cdot \frac{a}{b}) =p^{3r} \left( \frac{a}{b} \right)^3 \ $

Thus,

$ ord_p(n^3)=3r \ $

So we have,

$ 1 \leq p^{\operatorname{ord}_p(n)} \leq p^{\operatorname{ord}_p(n^3)} \leq n \ $

Am I right?

Please kindly check my work and correct it if necessary . Thanks.

Answer 1

Presumably, for the question at hand, $n$ is required to be a positive integer.

But even with that restriction, the inequality $p^{\text{ord}_p(n^3)}\le n$ is not always true.

For example, if $p$ is prime, and $n=p$, we get $$p^{\text{ord}_p(n^3)}=p^{\text{ord}_p(p^3)}=p^3 > p = n$$