# Finite Unramified Galois Extension

I'm wondering whether finite unramified Galois extensions of p-adic field number fields (i.e. extensions of $$\mathbb{Q}_p$$ for some prime $$p$$) are cyclic? The absolute Galois group is isomorphic to the profinite completion of the integers. Hence I think that every finite unramified Galois extension is a finite quotient of the absolute Galois group (as the extension is then a subfield of the maximal unramified extension, which is exactly the one with Galois group the absolute Galois group). Then it should follow that its Galois group is a finite quotient of $$\hat{\mathbb{Z}}$$. Now, is it true that it must be $$\mathbb{Z}/{n \mathbb{Z}}$$? And is my argument above correct?

• Yes, the unramified extensions of such a field correspond to algebraic extensions of its residue class field. – Lord Shark the Unknown Jun 22 at 14:27
• @LordSharktheUnknown Thank you for the quick answer. So the residue class field would be $\mathbb{F}_q$ for $q$ a power of a prime. And then every finite algebraic extension is $\mathbb{F}_{q^m}$, which is a cyclic extension of $\mathbb{F}_q$ (generated by the Frobenius), is this correct? – M. Wang Jun 22 at 14:30
• Yes, and the corresponding extension of $p$-adic fields is got by adjoining the $(q^m-1)$-th roots of unity. – Lord Shark the Unknown Jun 22 at 14:32
• Thanks, is it true that every finite quotient of $\hat{\mathbb{Z}}$ must be ismorphic to $\mathbb{Z}/{n \mathbb{Z}}$ for an integer $n$? – M. Wang Jun 22 at 14:34
• Yes. Any finite index subgroup contains some $n \hat{\Bbb{Z}}$ so it reduces to the quotients of $\Bbb{Z/nZ}$. What you need to show is that $p$-adic completeness and Hensel lemma implies any finite unramified extension $K/\Bbb{Q}_p$ of degree $m$ contains a primitive $p^m-1$ root of unity (equivalently pick $a \in O_K$ of order $p^m-1$ modulo $\pi_K$ then $\lim_{l \to \infty} a^{p^{lm}} = \zeta_{p^m-1}$) – reuns Jun 22 at 16:57