$\Bbb Q$ is not complete w.r.t. $p$-adic norm

Question

I'm trying to show that $\Bbb Q$ is not complete w.r.t. $p$-adic absolute value. The answer is already here. But I have two questions in that answer.

The answer is valid only for prime $p>3$. How can I show it for the case when $p = 2,3$?
During the proof, it says that using strong triangle inequality, it's easy to show that $|x-a|_p<1$. But this is not easy to me. Why is that follow from strong triangle inequality?

$(x-a)=(x-a^{p^n})+(a^{p^n}-a)$. By FLT $|a^{p^n}-a|_p<1$ and for $n$ large enough $|x-a^{p^n}|_p <1$. By ultrametric inequality (or strong triangle inequality) it implies $|x-a|_p<1$. — Aphelli, May 08 '21 at 10:57
@Mindlack How $|a^{p^n}-a|_p<1$ follows by FLT? $a^{\varphi(p^n)}\equiv 1\ (\mod p^{n})$ so $a^{p^{n}-p^{n-1}}\equiv 1\ (\mod p^{n})$. I don't get it. — one potato two potato, May 08 '21 at 12:01
$a^{p^n} \equiv a^{p^{n-1}}$ mod $p$ by FLT. By induction $a^{p^n}-a$ is divisible by $p$, QED. — Aphelli, May 08 '21 at 12:34

score 3 · Answer 1 · answered May 08 '21 at 11:03

There is also another way to show that $\Bbb Q$ is not complete with respect to the $p$-adic metric. For this it suffices to find irrational elements in $\Bbb Q_p$ by using Hensel's lemma. Indeed, for all $p>3$, $\Bbb Q_p$ contains the primitive $(p-1)$-th roots of unity. For $p=2$ we have $\sqrt{-7}\in \Bbb Q_2\setminus \Bbb Q$, and for $p=3$ we have $\sqrt{7}\in \Bbb Q_3\setminus \Bbb Q$.

The post you have cited has an appendix here, for $p=2$ and $p=3$:

Showing that Q is not complete with respect to the 2-adic and 3-adic absolute value

Aphelli · Accepted Answer · 2021-05-08T12:35:28.687

Here’s another explicit idea, different from Dietrich Burde’s very good one (and the other answer that uses the Baire Category Theorem).

Consider the series $\sum_{n \geq 0}{p^{n!}}$. It clearly converges in $\mathbb{Z}_p$. If its limit were a rational number, then one would have, for finite $p$-adic expansions $\sum_{k=0}^m{a_kp^k}$ (this one nonzero) and $\sum_{k=0}^l{b_kp^k}$, the equality $\sum_{k=0}^m{a_kp^k}\sum_{k \geq 0}{p^{k!}}=\sum_{k=0}^l{b_kp^k}$.

Now, it’s easy to write the LHS as $N_0+\sum_{n\geq m+1,0 \leq k \leq m}{a_kp^{n!+k}}$ and thus $\sum_{n \geq m+1,0 \leq k \leq m}{a_kp^{n!+k}}$ is the valid (infinite) $p$-adic expansion of an integer $N$.

If $N$ is positive, that’s impossible because positive integers have a finite $p$-adic expansion. If $N$ is negative, we can write $N=-p^q+r$ with $p^q>r>0$ and $-p^q=\sum_{k \geq q}{(p-1)p^k}$ so the $p$-adic expansion of $-p^q+r$ has all but finitely many $p-1$, which isn’t the case of $\sum_{n \geq m+1,0 \leq k \leq m}{a_kp^{n!+k}}$.

You basically giving an explicit Cauchy sequence in $\Bbb Q$ that does not converge w.r.t $p$-adic norm right? — one potato two potato, May 08 '21 at 12:42

score 2 · Answer 3 · edited May 09 '21 at 09:43

If $\Bbb Q$ is complete then because it is metric space we can use Baire category theorem.

Because there is no isolated points in $\Bbb Q$ (w.r.t. $p$-adic absolute value) ($p^k + q \rightarrow q\ \ \forall q$) we get $$\forall q\in\Bbb Q\text{ the set }\Bbb Q \setminus \{q\} \text{ is open and dense.}$$

But $$\bigcap_{q\in\Bbb Q}\ \Bbb Q \setminus \{q\} = \varnothing \text{ is not dense}.$$

So $\Bbb Q$ is not complete as metric space w.r.t. $p$-adic absolute value.

$\Bbb Q$ is not complete w.r.t. $p$-adic norm

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