# An automorphism of the field of $p$-adic numbers

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.

• What's an "algebraic" automorphism for you? Do you mean a field automorphism? – DonAntonio Jul 22 '13 at 12:51
• @DonAntonio An "algebraic" automorphism means that it may not be continuous. – Makoto Kato Jul 22 '13 at 13:06
• $\mathbb{Q}_p$ 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. – 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.

• Could you explain why $y \equiv 1 \pmod8$ has a $2$-adic square root? – Makoto Kato Jul 23 '13 at 22:00
• @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. – Jyrki Lahtonen Jul 24 '13 at 6:09
• @MakotoKato: see also this question. – Cantlog Aug 27 '13 at 22:40
• 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? – Germain Feb 18 '14 at 19:16
• 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 $3$. – 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.

• +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. – Jyrki Lahtonen Jul 24 '13 at 6:14
• 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.) – 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