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Let $p$ be an odd prime, and let $E/\mathbb{Q}_p$ be a finite extension that contains a primitive $p$-th root of unity $\zeta_p$ but not a primitve $p^2$-th root of unity $\zeta_{p^2}$. Let $a,b \in E^*$ such that $[E : \mathbb{Q}_p(a,b,\zeta_p)] = p^m$ for some $m \geq 1$.

Is the Hilbert symbol $(a,b)_p$ calculated over $E$ necessarily equals $1$?

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No. Take $L/\mathbb{Q}_p$ the unique unramified extension of degree $p$.

Hence, the cyclic algebra $A'=(p,L/\mathbb{Q}_p,\sigma)$ is division (of degree $p$), where $\sigma$ is what you think. Since $\mathbb{Q}_p(\zeta_p)/\mathbb{Q}_p$ has degree $p-1$, which is prime to $p$, $A=A'\otimes_{\mathbb{Q}_p} \mathbb{Q}_p(\zeta_p)$ is still division of degree $p$. In fact, $A=(p,L(\zeta_p)/\mathbb{Q}(\zeta_p),\sigma')$, where $\sigma'$ is the canonical extension of $\sigma$.

The extension $L(\zeta_p)/\mathbb{Q}_(\zeta_p)$ is a cyclic Kummer extension of degree $p$, so you can write it down as $\mathbb{Q}_p(\zeta_p)(\sqrt[p]{d})/\mathbb{Q}_p(\zeta_p)$. Now take $a=p,b=d$, so that $\mathbb{Q}_p(a,b,\zeta_p)=\mathbb{Q}_p(\zeta_p)$.

By construction, $(a,b)_p$ is not $1$ over $\mathbb{Q}_p(\zeta_p).$

Now take $E/\mathbb{Q}_p(\zeta_p)$ any field extension of degree $p$ not containing $\zeta_{p^2}$ and which does not split $A$. Then $(a,b)_p$ is not $1$ over $E$.

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  • $\begingroup$ I am sorry for changing the question, but I wanted that my extension of degree $p^m$ will be Galois. Is it possible for your $E/\mathbb{Q}_p(\zeta_p)$ to be Galois? $\endgroup$ – Pablo May 14 '17 at 13:27
  • $\begingroup$ I would be surprised if you couldn't choose $E$ to be Galois over $\mathbb{Q}_p(\zeta_p)$. In fact, since the degree of $A$ is prime and the degree of $\mathbb{Q}_p(\zeta_p)$ divides the degree of $A$, $\mathbb{Q}_p(\zeta_p)$ splits $A$ if and only if $E$ is isomorphic to a subfield of $A$. Any extension $E/\mathbb{Q}_p(\zeta_p)$ to be Galois of degree $p$, not containing $A$, and not isomorphic to any subfield of $A$ would do the job. I'm pretty sure a suitable totally ramified Kummer extension of degree $p$ would work. $\endgroup$ – GreginGre May 15 '17 at 7:46
  • $\begingroup$ The reason I am insisting is because I think I have a proof in the Galois case, so I really want to know if your counterexample works in this situation. $\endgroup$ – Pablo May 15 '17 at 8:07

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