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I am looking for a non-trivial automorphism $\sigma$ of $\mathbb C_p$ such that $\sigma(\mathbb Q_p)\subset\mathbb Q_p$.

If $\mathbb C_p$ were spherically complete, then by Hahn-Banach theorem, that would be impossible. But $\mathbb C_p$ is not spherically complete, so we can not apply Hahn-Banach. And my problem is still opened....

Does anyone have a hint or an example of such automorphism? Thanks in advance

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    $\begingroup$ You are expecting an explicit formula? Good luck. $\endgroup$
    – KConrad
    Feb 7, 2015 at 3:42
  • $\begingroup$ No, existence would be already appreciated $\endgroup$
    – joaopa
    Feb 7, 2015 at 3:54
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    $\begingroup$ Oh, but then it's "easy" using Zorn's lemma: if $F$ is any field and $C$ is any algebraically closed field containing $F$ then any field automorphism of $F$ can be extended to a field automorphism of $C$. (Extensions of homomorphisms of fields to homomorphisms into algebraically closed fields are discussed in Lang's Algebra.) So any nontrivial ${\mathbf Q}_p$-automorphism of a finite Galois extension of $\mathbf Q_p$ extends to a nontrivial $\mathbf Q_p$-automorphism of $\mathbf C_p$. $\endgroup$
    – KConrad
    Feb 7, 2015 at 4:22
  • $\begingroup$ With your argument, there are uncountabily $\mathbb Q_p$ automorphisms of $\mathbb C_p$. Are there infinitely many $\mathbb Q_p$-automorphisms of $\mathbb C_p$ that are isometries of $\mathbb C_p$? $\endgroup$
    – joaopa
    Feb 7, 2015 at 4:31
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    $\begingroup$ Yes. Every element of ${\rm Gal}(\overline{\mathbf Q_p}/\mathbf Q_p)$ is an isometry on $\overline{\mathbf Q_p}$, which is a dense subset of $\mathbf C_p$, and they all extend by continuity to automorphisms of $\mathbf C_p$. The converse is true as well: every $\mathbf Q_p$-automorphism of $\mathbf C_p$ that is an isometry is the extension-by-continuity to $\mathbf C_p$ of an element of ${\rm Gal}(\overline{\mathbf Q_p}/\mathbf Q_p)$. $\endgroup$
    – KConrad
    Feb 7, 2015 at 5:48

2 Answers 2

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If you strengthen the condition $\sigma(\mathbf Q_p) \subset \mathbf Q_p$ to $\sigma$ being the identity on $\mathbf Q_p$, so $\sigma$ is a $\mathbf Q_p$-automorphism of $\mathbf C_p$, then a simple description is possible:

(1) every $\mathbf Q_p$-isomorphism between two finite extensions of $\mathbf Q_p$ extends (in many ways) to a field automorphism of $\overline{\mathbf Q_p}$, i.e., to an element of ${\rm Gal}(\overline{\mathbf Q_p}/\mathbf Q_p)$,

(2) every element of ${\rm Gal}(\overline{\mathbf Q_p}/\mathbf Q_p)$ extends uniquely to a continuous $\mathbf Q_p$-automorphism of $\mathbf C_p$,

(3) every continuous $\mathbf Q_p$-automorphism of $\mathbf C_p$ is an isometry of $\mathbf C_p$ and is the continuous extension of a unique element of ${\rm Gal}(\overline{\mathbf Q_p}/\mathbf Q_p)$.

Therefore any $\mathbf Q_p$-isomorphism between two finite extensions of $\mathbf Q_p$ extends (in many ways) to a continuous automorphism of $\mathbf C_p$, so in particular for any finite Galois extension $K/\mathbf Q_p$ every element of ${\rm Gal}(K/\mathbf Q_p)$ extends (in many ways) to a continuous $\mathbf Q_p$-automorphism of $\mathbf C_p$. Therefore starting with any non-identity automorphism of a finite Galois extension of $\mathbf Q_p$ we can lift it to a non-identity continuous $\mathbf Q_p$-automorphism of $\mathbf C_p$. For example, the conjugation on $\mathbf Q_p(\sqrt{p})$ extends (in many ways) to a continuous $\mathbf Q_p$-automorphism of $\mathbf C_p$. We only need the first two properties above. The point of the third property is to show which $\mathbf Q_p$-automorphisms of $\mathbf C_p$ come from ${\rm Gal}(\overline{\mathbf Q_p}/\mathbf Q_p)$: the automorphisms that are isometries (hence automatically continuous and thus automatically fixing $\mathbf Q_p$).

Proof of (1): This is a theorem from abstract algebra, having nothing to do with $p$-adic fields: every isomorphism between two fields extends (in many ways) to an isomorphism between their algebraic closures. The proof relies on Zorn's lemma, so it is not constructive. See Lang's Algebra, Theorem 2.8 of Chapter V. He proves any embedding $\sigma$ of a field $k$ into an algebraically closed field $L$ extends (by Zorn's lemma) to an embedding of any algebraic extension $E$ of $k$ into the same field $L$, and if $E$ is an algebraic closure of $k$ and $L$ is an algebraic closure of $\sigma(k)$ then he shows that embedding of $E$ into $L$ must be an isomorphism. Apply this theorem with $k$ being a finite extension of $\mathbf Q_p$ and $E = L = \overline{\mathbf Q_p}$.

Proof of (2): Let $\sigma \in {\rm Gal}(\overline{\mathbf Q_p}/\mathbf Q_p)$. Any $\mathbf Q_p$-automorphism of a finite extension of $\mathbf Q_p$ preserves the $p$-adic absolute value on that finite extension, so by viewing two elements of $\overline{\mathbf Q_p}$ inside a finite Galois extension of $\mathbf Q_p$ we get that $\sigma$ is an isometry on $\overline{\mathbf Q_p}$: $|\sigma(\alpha) - \sigma(\beta)|_p = |\alpha - \beta|_p$ for all $\alpha$ and $\beta$ in $\overline{\mathbf Q_p}$. Thus $\sigma$ is uniformly continuous on $\overline{\mathbf Q_p}$, so it extends in exactly one way to a continuous function on $\mathbf C_p$, namely by $\sigma(z) := \lim_{n \rightarrow \infty} \sigma(\alpha_n)$, where $\alpha_n$ is any sequence in $\overline{\mathbf Q_p}$ converging to $z$. It is straightforward to check for all $z$ and $w$ in $\mathbf C_p$ that $|\sigma(z) - \sigma(w)|_p = |z - w|_p$ and then that $\sigma$ is additive and multiplicative on $\mathbf C_p$. To show $\sigma$ on $\mathbf C_p$ is surjective, write $z\in \mathbf C_p$ as a limit of $\alpha_n \in \overline{\mathbf Q_p}$. Since $\sigma$ is an isometry the sequence $\sigma^{-1}(\alpha_n)$ in $\overline{\mathbf Q_p}$ is Cauchy and its limit $w$ in $\mathbf C_p$ satisfies $\sigma(w) = z$.

Proof of (3): Let $\sigma$ be a continuous $\mathbf Q_p$-automorphism of $\mathbf C_p$. Since $\sigma$ fixes each element of $\mathbf Q_p$, we have $\sigma(\overline{\mathbf Q_p}) \subset \overline{\mathbf Q_p}$. The field $\sigma(\overline{\mathbf Q_p})$ is algebraically closed and lies between $\mathbf Q_p$ and $\overline{\mathbf Q_p}$, so $\sigma(\overline{\mathbf Q_p}) = \overline{\mathbf Q_p}$. Thus $\sigma$ restricts to an element of ${\rm Gal}(\overline{\mathbf Q_p}/\mathbf Q_p)$, so $\sigma$ on $\mathbf C_p$ is the continuous extension of an element of ${\rm Gal}(\overline{\mathbf Q_p}/\mathbf Q_p)$, and it can be the continuous extension of at most one element of the Galois group since $\overline{\mathbf Q_p}$ is dense in $\mathbf C_p$. By the proof of (2), the continuous extension to $\mathbf C_p$ of an automorphism in ${\rm Gal}(\overline{\mathbf Q_p}/\mathbf Q_p)$ is an isometry on $\mathbf C_p$, so $\sigma$ is an isometry on $\mathbf C_p$.

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  • $\begingroup$ Using that the transcendence degree of $\mathbf{C}_p$ over $\overline{\mathbf{Q}_p}$ is $2^{\aleph_0}$, one can easily check that the cardinal of the automorphism group of the field $\mathbf{C}_p$ (or its subgroup of $\mathbf{Q}_p$-algebra, or $\overline{\mathbf{Q}_p}$-algebra) is not only uncountable, but has cardinal $2^{2^{\aleph_0}}$. $\endgroup$
    – YCor
    Mar 13, 2019 at 17:59
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  1. Transcendence degree ℂp/ℚp=20 which is enough to answer the question affirmatively.
  2. More explicitly: ℂp embeds into the field of p-adic Malcev-Neumann series = formal sums Σzipai with zi∈F̅p and {ai} well-ordered subset of ℚ. This gives an injection of the semi-direct product Gal(F̅p/Fp) X ℤ̂ ⊂ Aut(ℂp/ℚp).

  3. Reference: http://deepblue.lib.umich.edu/bitstream/handle/2027.42/26390/0000477.pdf?sequence=1

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