Let $K/\mathbb{Q}$ be a Galois number field and let $K_{\infty}/K$ be its cyclotomic $\mathbb{Z}_p$-extension defined by $K_{\infty}=K\cdot \mathbb{Q}_{\infty}$, where $\mathbb{Q}_{\infty}$ is the unique (cyclotomic) $\mathbb{Z}_p$-extension of $\mathbb{Q}$.

Let $\tau \in \text{Gal}(K_{\infty}/K)$ be a topological generator for $\text{Gal}(K_{\infty}/K)$ and for each $g \in \text{Gal}(K/\mathbb{Q})$, we fix a lift $\tilde{g} \in \text{Gal}(K_{\infty}/\mathbb{Q})$ of $g$.

I would like to know under which conditions one has that $\tilde{g}\tau=\tau\tilde{g}$, for all $g \in \text{Gal}(K/\mathbb{Q})$. This seems to be true if one assumes that $K\cap \mathbb{Q}_{\infty}=\mathbb{Q}$, but is it possible to find a necessary and sufficient condition?

  • $\begingroup$ What do you mean by "cyclotomic $\Bbb Z_p$ extension? $\endgroup$ Dec 7, 2016 at 19:03
  • $\begingroup$ If $\zeta_{p^{n+1}}$ is a primitive $p^{n+1}$-th root of unity, then $\mathbb{Q}(\zeta_{p^{n+1}})$ has a unique subfield $\mathbb{Q}_n$ of order $p^n$ over $\mathbb{Q}$. Then $\mathbb{Q}_{\infty}$ is defined as the union of all such $\mathbb{Q}_n$. $\endgroup$ Dec 7, 2016 at 19:07

1 Answer 1


I'm not sure there's much one can say in general. The condition that $\Bbb Q_\infty$ and $K$ be linearly disjoint over $\Bbb Q$ gives a direct product representation which of course implies the commutator relation you want. However, if we take the fully general approach, we see

$$\text{Gal}(K_\infty/K)=\text{Gal}(\Bbb Q_\infty K/K)\cong\text{Gal}(\Bbb Q_\infty/\Bbb Q_\infty\cap K)$$

This of course is how we know $\text{Gal}(K_\infty / K)$ is isomorphic to a subgroup of $\widehat{\Bbb Z}$. But the usual s.e.s. is

$$1\to \text{Gal}(K_\infty/K)\to\text{Gal}(K_\infty/\Bbb Q)\to\text{Gal}(K/\Bbb Q)\to 1$$

and a lift of $g\in \text{Gal}(K/\Bbb Q)$ is just any element of $\stackrel{\sim}{g}\text{Gal}(K_\infty/K)$. And the condition any such element commutes with your $\tau$ is just that $\text{Gal}(K_\infty/K)\le Z( \text{Gal}(K_\infty/\Bbb Q))$ since all elements of $\text{Gal}(K_\infty/\Bbb Q)$ are in some coset of some $\stackrel{\sim}{g}$. And as such this condition is both necessary and sufficient.

This of course recovers the case you mention, as then

$$\text{Gal}(K_\infty/\Bbb Q)=\text{Gal}(K\Bbb Q_\infty/\Bbb Q)\cong\text{Gal}(K/\Bbb Q)\times \text{Gal}(\Bbb Q_\infty/\Bbb Q_\infty\cap K)$$

and the right factor is abelian.

  • $\begingroup$ Thank you very much for this answer. I am not 100% sure I get the last part where you say "This of course recovers the case you mention". Do you mean that the case $K \cap \mathbb{Q}_{\infty}=\mathbb{Q}$ satisfies $Gal(K_{\infty}/K)\leq Z(Gal(K_{\infty}/\mathbb{Q}))$, or something more, i.e. that the case I mention is also necessary for $Gal(K_{\infty}/K)\leq Z(Gal(K_{\infty}/\mathbb{Q}))$ to hold? $\endgroup$ Dec 8, 2016 at 20:59
  • $\begingroup$ @JakeHaider yes, I'm talking about the linearly disjoint case. It's not necessary, but if you have the linearly disjoint case it's clear that the given group is contained in the center because of the direct product representation $\endgroup$ Dec 8, 2016 at 21:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .