# Extending an automorphism while preserving commutativity

If I have a field $K$ and two automorphisms $\phi,\psi$ such that

1)$\phi$ is an automorphism of $K$ while $\psi$ is an automorphism of $\bar{K}$ the algebraic closure of $K$ and $\psi|K$ is also an automorphism

2)$\phi \circ \psi|_{K} =\psi|_{K} \circ \phi$.

Does it follow that we can extend $\phi$ to $\bar{K}$ such that it commutes with $\psi$.

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Let's first consider a group-theoretic analogue of your question:

Suppose that $G$ is a group, that $\psi$ is an element of $G$, and that $\phi$ is an element of $G^{ab} := G/[G,G]$. Can we lift $\phi$ to an element of $G$ so that it commutes with $\psi$?

In general the answer is no, since the centralizer of $\psi$ need not surject onto $G^{ab}$. E.g. if we take $G$ to the dihedral group of order $8$, and let $\psi$ be an element of order $4$. Then $G^{ab}$ is the Klein four group, while the centralizer of $\psi$ is just the group generated by $\psi$, so if we take $\phi$ to be any element of $G^{ab}$ not in the subgroup (cyclic of order two) generated by $\psi$, then there is no lift of $\phi$ to an element of $G$ which commutes with $\psi$.

Now to translate this into field theory, let $L/\mathbb Q$ be any Galois extension with Galois group $G = D_8$, and let $K$ be the subfield fixed by $[G,G]$. Then $G^{ab}$ acts as automorphims of $K$, but we cannot lift $\phi$ to an automorphism of $L$ which commutes with $\psi$ (as an automorphism of $L$). Consequently, we certainly can't lift $\phi$ to an automorphism of $\overline{\mathbb Q}$ which commutes with (any lift to $\overline{\mathbb Q}$ of) $\psi$.

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Well, I played around a bit with something I thought might make a counterexample, but I couldn't make it work (or rather, it always did work). Maybe something "more non-commutative" (?) than $\mathrm{Gal}(\sqrt[n]{a},\zeta_m)$ needs to be used; or maybe the statement is true. I'll record my efforts here as a CW answer in case it gives someone else an idea.

Let $K=\mathbb{Q}(\sqrt[3]{2},\zeta_{15})$, and let $\phi,\rho\in\mathrm{Gal}(K/\mathbb{Q})$ be the automorphisms defined by $$\phi(\sqrt[3]{2})=\zeta_{15}^5\sqrt[3]{2},\quad \phi(\zeta_{15})=\zeta_{15}$$ $$\rho(\sqrt[3]{2})=\sqrt[3]{2},\quad \rho(\zeta_{15})=\zeta_{15}^4$$ The automorphism $\rho$ certainly can be extended to $\overline{\mathbb{Q}}$; let $\psi\in\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ be any such extension. Then $\phi$ and $\psi|_K=\rho$ commute because $$\phi(\rho(\sqrt[3]{2}))=\phi(\sqrt[3]{2})=\zeta_{15}^5\sqrt[3]{2}=\zeta_{15}^{20}\sqrt[3]{2}=\rho(\zeta_{15}^5\sqrt[3]{2})=\rho(\phi(\sqrt[3]{2}))$$ $$\phi(\rho(\zeta_{15}))=\phi(\zeta_{15}^4)=\zeta_{15}^4=\rho(\zeta_{15})=\rho(\phi(\zeta_{15}))$$

Let $L=\mathbb{Q}(\sqrt[15]{2},\zeta_{15})$. Any extension $\sigma\in\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ of $\phi$ to an automorphism of $\overline{\mathbb{Q}}$ must restrict down to $L$ to do one of the following 5 things: $$\sigma|_L(\sqrt[15]{2})\in\{\zeta_{15}\sqrt[15]{2},\;\zeta_{15}^4\sqrt[15]{2},\;\zeta_{15}^7\sqrt[15]{2},\;\zeta_{15}^{10}\sqrt[15]{2},\;\zeta_{15}^{13}\sqrt[15]{2}\},\quad \sigma|_L(\zeta_{15})=\zeta_{15}$$ Now suppose we've taken our $\psi\in\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ such that $\psi|_L(\sqrt[15]{2})=\sqrt[15]{2}$ and $\psi|_L(\zeta_{15})=\zeta_{15}^4$ (e.g., take the automorphism of $L$ that does that, and lift it to $\overline{\mathbb{Q}}$). Then $\psi|_L$ and $\sigma|_L$ can never commute, because $$\sigma|_L(\psi|_L(\sqrt[15]{2}))=\sigma|_L(\sqrt[15]{2})=\zeta_{15}^{1,4,7,10,13}\sqrt[15]{2}\neq\zeta_{15}^{4,1,13,10,7}\sqrt[15]{2}=\psi|_L(\sigma|_L(\sqrt[15]{2}))$$ (except, of course, this is wrong; if $\sigma|_L(\sqrt[15]{2})=\zeta_{15}^{10}\sqrt[15]{2}$ then we get the same answer on both sides. This is also the case if we'd started with $\rho(\zeta_{15})=\zeta_{15}^{13}$, which is the only other possible choice for this particular setup.)

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Thanks for your efforts. I have been trying to to use similar ideas and I proved that If the question is true then it follows that any automorphism of $\mathbb{Q}(\xi)$ is continuous where $\xi$ is a primitive root of unity(using the inherited topology from $\mathbb{C}$) BUt I don't know weather this is true or not – omar Dec 20 '12 at 18:45