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I have seen here how to show that $\mathbb Q$ is not complete with respect to the $p$-adic absolute value, where $p\geq5$. Is there a similar proof/idea for $p=2$ and $p=3$?

user26857
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User00000000
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    Please add what you already know about the completion $\mathbb Q_p$, because a good answer greatly depends on what one can already use at this point; if enough properties of $p$-adics are taken for granted, this statement is more or less trivial completely regardless of $p$. – Torsten Schoeneberg Apr 21 '20 at 15:30
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    No, there is no similar proof. The question that you quote uses the trick that repeating the operation $x\mapsto x^p$ is $p$-adically convergent, with limit a root of $X^p-X=\gamma_p(X)$ that’s congruent modulo $p\Bbb Z_p$ to the original $x$ you chose. The reason this doesn’t help you when $p=2$ or $3$ is that there, the only roots of $\gamma_p$ are already in $\Bbb Z$. So raising to the $p$-th power doesn’t help you and you’re thrown back onto Hensel or something similar. – Lubin Apr 22 '20 at 19:48
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    @Lubin The operation $x \mapsto x^p + p$ is $p$-adically convergent with limit a root of $X^p - X + p=0$ that's congruent modulo $p$ to the original $x$ you choose, but $X^2-X+2$ and $X^3-X+3$ are both irreducible over $\mathbf{Q}$. That is (arguably) a similar proof to the linked answer. – user760870 Apr 25 '20 at 15:17
  • A very pleasing observation, @user760870. If you give it as an answer, I’ll plus-one you. – Lubin Apr 25 '20 at 17:39

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Take a monic polynomial $Q(X) \in \mathbb Z[X]$ that does not have a rational root, but which has a root $a$ modulo $p$, and with $Q'(a) \not \equiv 0 \pmod p$. By Hensel's lemma, $Q(X)$ then has a root in $\mathbb Q_p$ (even in $\mathbb Z_p$), which is necessarily an irrational $p$-adic. Hence $\mathbb Q_p \neq \mathbb Q$, so that $\mathbb Q$ is not $p$-adically complete.

For $p \geq 5$, you can take $Q(X) = \frac{X^{p-1}-1}{X^2-1}$ and you recover more or less the construction in the linked answer.

For $p = 2$, you can take for example $Q(X) = X^3+3$.

For $p = 3$, you can take for example $Q(X) = X^2 + 2$.

Bart Michels
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