Is an automorphism of the field $\mathbb{Q}_p$ of $p$-adic numbers the identity map? If yes, how can we prove it?

Note:We don't assume an automorphism of $\mathbb{Q}_p$ is continuous.

  • 3
    $\begingroup$ What's an "algebraic" automorphism for you? Do you mean a field automorphism? $\endgroup$ – DonAntonio Jul 22 '13 at 12:51
  • $\begingroup$ @DonAntonio An "algebraic" automorphism means that it may not be continuous. $\endgroup$ – Makoto Kato Jul 22 '13 at 13:06
  • 1
    $\begingroup$ $\mathbb{Q}_p$ is a topological field. An automorphism of $\mathbb{Q}_p$ as a topological field must be continuous. So just saying an automorphism of $\mathbb{Q}_p$ is ambiguous. $\endgroup$ – Makoto Kato Aug 2 '13 at 22:00

We use squares in a way that resembles the usual proof of the corresponding result about reals - we use squares to prove continuity of automorphisms.

Assume first that $p>2$ and that $\sigma$ is an automorphism of $\mathbb{Q}_p$. The key observation is that $1+px^2$ is a square in $\mathbb{Q}_p$ if and only if $x$ is in $\mathbb{Z}_p$. If $x$ is not a $p$-adic integer, then $\nu(1+px^2)$ is odd, so it cannot be a square. On the other hand (this is where we need $p>2$) by Hensel's lemma $1+px^2$ is a square, if $x$ is a $p$-adic integer.

If $1+px^2=y^2$, then clearly $1+p\sigma(x)^2=\sigma(y)^2$, so from the preceding paragraph we can deduce that $\sigma(x)\in\mathbb{Z}_p$ whenever $x$ is. But $\sigma(p)=p$, so we see that $\sigma^{-1}(p^k\mathbb{Z}_p)=p^k\mathbb{Z}_p$ for all natural numbers $k$. Thus $\sigma$ is continuous, and the claim follows from density of $\mathbb{Z}$ (they are all fixed points of $\sigma$) inside $\mathbb{Z}_p$.

If $p=2$ we need to make a small modification to the above argument. This time we see that $1+8x^2$ is a $2$-adic square, iff $x$ is a $2$-adic integer. This is because Hensel's lemma allows us to prove the existence of a $2$-adic square root of anything $\equiv 1\pmod8$. On the other hand if $x\notin\mathbb{Z}_2$, then either $\nu(1+8x^2)$ is odd (whenever $\nu(x)\le -2$) or $1+8x^2$ is an odd $2$-adic integer $\not\equiv1\pmod4$. It cannot be a square in either case. The rest of the argument works as above.

| cite | improve this answer | |
  • $\begingroup$ Could you explain why $y \equiv 1 \pmod8$ has a $2$-adic square root? $\endgroup$ – Makoto Kato Jul 23 '13 at 22:00
  • $\begingroup$ @Makoto: The short answer is that a sharpened version of Hensel lifting kicks in. We are recursively improving the approximate solution $y\approx y_1=1$ of the equation $y^2=1+8x^2$. The derivative has a positive exponential valuation, $\nu(2y_1)=1$, so the usual version doesn't work. But as the error $\nu(1+8x^2-y_1^2)\ge 3$ is small enough, this gives sufficient compensation, and the recursive steps in Hensel's lemma go through. Sorry, I don't have the time to go into details today. $\endgroup$ – Jyrki Lahtonen Jul 24 '13 at 6:09
  • 2
    $\begingroup$ @MakotoKato: see also this question. $\endgroup$ – Cantlog Aug 27 '13 at 22:40
  • $\begingroup$ How to prove that $1+p x^2$ is a square in $\mathbb Q_p$ iff $x \in \mathbb Z_p$? How are you defining $\nu (1+ px^2)$. Is it the $||_p$ as defined in Lang? $\endgroup$ – Germain Feb 18 '14 at 19:16
  • 4
    $\begingroup$ Maybe it’s late to be making a comment here, but I love this argument. I proved the triviality of the automorphism group of $\Bbb Q_p$ for myself by a far more elaborate argument, and this one really appeals to me. But it seems to me that talking about the cubehood of $1+p^2x^3$ should work just as well, and not need any special arguments, since it’s fine at $3$. $\endgroup$ – Lubin Feb 4 '16 at 5:31

The following proof is basically the same as Mr. Dietrich Burde's answer.

Let $\mathbb{Q}_p$ be the field of $p$-adic numbers. Let $\nu$ be the canonical additive valuation on $\mathbb{Q}_p$, i.e. $\nu(p) = 1$.

Let $\sigma$ be an automorphism of $\mathbb{Q}_p$. Since $\mathbb{Q}$ is dense in $\mathbb{Q}_p$ and $\sigma$ induces the identity map on $\mathbb{Q}$, it suffices to prove that $\sigma$ is continuous.

Let $U$ be the set of $p$-adic units. We first show that $\sigma(U) \subset U$. Let $\alpha \neq 0$ be a $p$-adic number. Suppose the set $S = \{n \in \mathbb{Z}, n \gt 0\mid x^n = \alpha$ has a solution in $\mathbb{Q}_p\}$ is infinite. If $x^n = \alpha$, then $n\nu(x) = \nu(\alpha)$. Hence if $\nu(\alpha) \neq 0$, then $\nu(\alpha)$ is divisible by infinite numbers of rational integers. This is absurd. Therefore $\nu(\alpha) = 0$, which means $\alpha \in U$.

Now suppose $\epsilon$ is a $p$-adic unit. The set $S = \{n \in \mathbb{Z}, n \gt 0\mid x^n = \epsilon$ has a solution in $\mathbb{Q}_p\}$is infinite as shown in here. Hence the similar set for $\sigma(\epsilon)$ is infinite. Therefore $\sigma(\epsilon) \in U$ by what we have shown above. This means $\sigma(U) \subset U$.

Let $\alpha \neq 0$ be an element of $\mathbb{Z}_p$. Let $n = \nu(\alpha)$. Then $\alpha = p^n \epsilon$, where $\epsilon \in U$. Since $\sigma(\alpha) = p^n\sigma(\epsilon)$, $\sigma(\alpha) \in \mathbb{Z}_p$. Hence $\sigma(\mathbb{Z}_p) \subset \mathbb{Z}_p$. Hence $\sigma(p^n\mathbb{Z}_p) \subset p^n\mathbb{Z}_p$ for every positive integer $n$. This means $\sigma$ is continuous.

| cite | improve this answer | |
  • $\begingroup$ +1 Thanks for the summary. I was on an autopilot typing my own answer, and didn't have the time to study Dietrich's link. $\endgroup$ – Jyrki Lahtonen Jul 24 '13 at 6:14
  • 5
    $\begingroup$ The same argument shows that if $K$ and $L$ are both local fields of characteristic 0 that admit a homomorphism from $K$ to $L$, then $K$ and $L$ must be extensions of ${\mathbf Q}_p$ for the same $p$ and this homomorphism must be continuous. (For example, this shows ${\mathbf Q}_p$ and ${\mathbf Q}_q$ are not isomorphic as abstract fields if $p \not= q$, although there are much easier proofs of that by counting roots of unity in the fields or looking at what integers are squares.) $\endgroup$ – KCd Aug 2 '13 at 6:06

The identity map is the only automorphism of the field of $p$-adic numbers, because of Schmidt's theorem implying that a field complete with respect to a discrete absolute value is not complete with respect to an absolute value, which is inequivalent to the original. For a proof, see here: http://www.math.utk.edu/~wagner/papers/padic.pdf

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.