I want to show that for p-adic number field $\mathbb{Q}_{p}$, the maximal unramified extension is $\mathbb{Q}_{p}$ adjoin n-th roots of unity $\zeta_{n}$ for $(n, p)=1$.

It is obvious that if $(n,p) = 1$, $\mathbb{Q}_{p}(\zeta_{n})$ is unramified, so their compositum is also unramified. For the other direction, if I pick any unramified extension $T$ over $\mathbb{Q}_{p}$, I was only able to get its degree is $p^{m}$ for some $m$. I don't know whether this $T$ is separated or not (if not separated, we can not use fundamental identity $[L:K]=ef$ ), and even don't know whether $T$ is a finite extension. How should I prove $T$ is generated by some $\zeta_{i}$'s?

  • $\begingroup$ What's the definition of infinite unramified extension? $\endgroup$ – eduard Feb 13 '18 at 9:02
  • $\begingroup$ @eduard, an infinite extension is unramified over $k$ if every finite subextension is unramified over $k$. $\endgroup$ – Lubin Feb 13 '18 at 21:35

You make an incorrect statement about the degrees of unramified extensions of $\Bbb Q_p$. The truth is that there is one unramified extension of each degree.

When you think of the maximal unramified $\Bbb Q_p^{\text{un}}$ of $\Bbb Q_p$, you should think downstairs, of the algebraic closure $\Omega$ of $\Bbb F_p$, the prime field of characteristic $p$.

Just as you get $\Omega$ by adjoining the $(p^m-1)$-th roots of unity to $\Bbb F_p$, you also get $\Bbb Q_p^{\text{un}}$ by adjoining the $(p^m-1)$-th roots of unity. That’s Hensel’s Lemma.

For the converse, let $k$ be a finite unramified extension of $\Bbb Q_p$, say $[k:\Bbb Q_p]=m$. Then for the residue fields, you also have $[\kappa:\Bbb F_p]=m$, so the nonzero elements of $\kappa$ are the $(p^m-1)$-th roots of unity in characteristic $p$. Again Hensel says that the corresponding roots of unity are there in $k$.

I’ll leave it to you to show that if $\gcd(p,n)=1$, then there is $m$ such that $n|(p^m-1)$. In other words, you don’t need to adjoin all the $n$-th roots of unity for those $n$, you can adjoin the $(p^m-1)$-th roots of unity for all $m$.

  • $\begingroup$ Yes, I saw I made a mistake when thinking of the degree of finite extension of $\mathbb{F}_{p}$. Thanks a lot! $\endgroup$ – Zheng Xiao Feb 13 '18 at 22:59

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