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Let $p$ be a rational prime, and let $x$ be an element of $\mathbb{Q}_p$ with $|1-x|_p < 1$. I want to show that $|1-x^{p^n}|_p \leq p^{-p^n}$ for all positive integers $n$, but I'm having a hard time. My idea was to write $$ (1-x)^{p^n} = \sum_{k=0}^{p^n} (-1)^k \binom{p^n}{k}x^k = 1-x^{p^n} + \sum_{k=1}^{p^n - 1} (-1)^k \binom{p^n}{k}x^k $$ so that $$ |1-x^{p^n}|_p \leq \max\left\{|(1-x)^{p^n}|_p, \left|\sum_{k=1}^{p^n - 1} (-1)^k \binom{p^n}{k}x^k \right|_p \right\} = \max\left\{ p^{-p^n}, \left|\sum_{k=1}^{p^n - 1} (-1)^k \binom{p^n}{k}x^k \right|_p \right\}. $$ But I'm stuck there. Can someone please help me out?

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$p^{-p^n}$ decreases too fast. I don't think this is true even for $n = 1$. For example, take $x = 4$, $p = 3$, $n = 1$. Then $|-63|_3 = 1/9$, but $3^{-3} = 1/27$. – Soarer Aug 5 '11 at 2:01
On the other hand, you can prove that $|1-x| \leq 1$ implies $|1 - x^{p^n}| \leq p^{-n}$. To show this, use induction and show that $|1-x| \leq p^{-k}$ implies $|1-x^p| \leq p^{-(k+1)}$. This can be easily proven by letting $y = 1-x$, express $1-x^p$ in terms of $y$, and ultrametric inequality – Soarer Aug 5 '11 at 2:04

Lemma: let $p$ be a prime and let $n\in 1 +p\mathbb{Z}_p$ if $p$ is odd and

$n\in 1+4\mathbb{Z}_p$ if $p=2$. Then

$$ v_p(x^s - 1) = v_p(x - 1) + v_p(s)$$ for any positive integer $s$, where $v_p$ is the normalised $p$-adic


The easiest way to prove this is to use the isomorphism $\exp: q\mathbb{Z}_p\rightarrow 1 + q\mathbb{Z}_p$, where $q=p$ if $p$ is odd and $q=4$ if $p=2$. The points is that this isomorphism preserves the filtrations given by the valuation on both sides, i.e. $p^s \mathbb{Z}_p$ goes to $1+p^s \mathbb{Z}_p$. The claim is now clear, since raising to the $s$-th power on the right hand side corresponds to multiplying by $s$ on the left hand side.

Your statement is a special case of the lemma. The only modification you need to make is when $p=2$ and $x\equiv 3\pmod 4$, in which case you should work with $x^2$ instead: $x^{2^n}=(x^2)^{2^{n-1}}$.

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Yes, this gives $\nu_p(1-x^{p^n}) \geq n+1$, but the question seems to make a stronger claim than that, doesnt' it (too strong, in fact). – Geoff Robinson Aug 5 '11 at 7:33
@Geoff: yes, I guess I should say "the correct version of your statement..." instead of "your statement". – Alex B. Aug 5 '11 at 9:07

Let me speak in terms of the $p$-adic valuation instead of the $p$-adic norm: for $x \in \mathbb{Q}_p$, $|x|_p = p^{-v_p(x)}$. Note that for $x \in \mathbb{Q}_p$, the running hypothesis $v_p(x-1) \geq 1$ implies $x \in \mathbb{Z}_p$. In fact, unlike Alex B.'s nice "analytic" solution, everything I'm doing here is based on congruential considerations. (It actually suffices to take $x \in \mathbb{Z}$: since $\mathbb{Z}$ is dense in $\mathbb{Z}_p$, all the inequalities and equalities below hold for all $x \in \mathbb{Z}_p$ as soon as they hold for all $x \in \mathbb{Z}$.)

As others have said in the comments, the inequality you want is too strong (there may just be a typo / slight confusion about norms versus valuations). Here is a repaired statement:

For all primes $p$ and all $n \in \mathbb{Z}^+$, $v_p(x -1) \geq 1 \implies v_p(x^{p^n}-1) \geq v_p(x-1) + n$.

In fact we can establish the inequality by computing $v_p(x^{p^n}-1)- v_p(x^p-1)$ exactly in all cases and verifying that it is non-negative. First:

For all $p > 2$ and all $n \in \mathbb{Z}^+$, $v_p(x-1) \geq 1 \implies v_p(x^{p^n}-1) - v_p(x-1) = n$.

For a proof, see Lemma 3 of these notes.

Now let's consider the case $p = 2$. Let $x$ be such that $v_2(x-1) \geq 1$. Then

$v_2(x^2-1) = v_2(x-1) + v_2(x+1)$.

Now, $v_2(x-1) \geq 2 \iff v_2(x+1) = 1$. So in the case $v_2(x-1) \geq 2$ we simply get $v_2(x^2-1) = v_2(x-1) + 1$ as above, and then induction immediately gives:

For all $n \in \mathbb{Z}^+$, $v_2(x-1) \geq 2 \implies v_2(x^{2^n}-1) - v_2(x-1) = n$.

Consider now the case $v_2(x-1) = 1$. Then all we know about $v_2(x+1)$ is that it is at least $2$. We obviously cannot express it directly in terms of $v_2(x-1) = 1$, so we will have to include it in our formula:

$v_2(x^2-1) - v_2(x-1) = v_2(x+1)$.

Now consider $x^4 - 1 = (x^2)^2 - 1$. Then $v_2(x^2-1) \geq 2$, and thus

$v_2(x^4 - 1) - v_2(x-1) = v_2((x^2)^2-1) - v_2(x-1) = (v_2(x^2-1)+1) - v_2(x-1)$

$= v_2(x+1) + v_2(x-1) + 1 - v_2(x-1) = v_2(x+1) + 1$.

From here a simple inductive argument establishes:

For all $n \in \mathbb{Z}^+$, $v_2(x-1) \geq 1 \implies v_2( x^{2^n} - 1) - v_2(x-1) = n-1 + v_2(x+1) \geq n$.

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If $|1 - x|_p < p^{-k}$ for some $k \geq 0$, then one can write $x = 1 + p^{k+1}y$ for some $y \in {\mathbb Z}_p$. Then raising both sides to the $p$th power, using the binomial theorem you have $x^p = 1 + p p^{k+1}y + ... = 1 + p^{k+2}z$ for some $z \in {\mathbb Z}_p$. So $|1 - x^p|_p < p^{-k-1}$

So applying the above inductively $n$ times, you get that if $|1 - x|_p < 1$, then $|1 - x^{p^n}| < p^{-n}$, which is the corrected version of the question.

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