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I would like to prove $ \Bbb Q_p^×/（\Bbb Q_p^×）^p$ is isomorphic to $\Bbb Z/p\Bbb Z×\Bbb Z/p\Bbb Z$ as a group.

I know $\Bbb Q_p^×$ is isomorphic to $\Bbb Z×\Bbb Z_p×\Bbb Z/（p-1）\Bbb Z$ using formal logarithm.

I tried to find group homomorphism $\Bbb Q_p^×$→$\Bbb Z/p\Bbb Z×\Bbb Z/p\Bbb Z$ of which kenerl is $（\Bbb Q_p^×）^p$, but I couldn't.

Thank you in advance.

Arturo Magidin
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Pont
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  • Thank you!, but is $G^p$ isomorphic to $\Bbb pZ×\Bbb pZ_p×\Bbb Z/（p-1）\Bbb Z$? – Pont Mar 12 '21 at 13:27
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    This is false when $p = 2$: the group $\mathbf Q_2^\times/(\mathbf Q_2^\times)^2$ has order $8$, not $4$. – KCd Mar 12 '21 at 20:07

1 Answers1

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Is it clear to you that (Hensel lemma) for $p$ odd: $$\Bbb{Q}_p^\times = p^\Bbb{Z}\times (1+p\Bbb{Z}_p)\times \langle \zeta_{p-1}\rangle$$

Then $(1+p\Bbb{Z}_p)^p=1+p^2\Bbb{Z}_p$ because it contains $(1+cp^k)^p\equiv 1+ c p^{k+1}\bmod p^{k+2}$ so any $1+p^2 a$ is of the form $\prod_{k\ge 1} (1+c_k p^k)^p$ with $c_k\in 0 \ldots p-1$.

Whence $$\Bbb{Q}_p^\times/(\Bbb{Q}_p^\times)^p\cong p^\Bbb{Z}/p^\Bbb{pZ}\times (1+p\Bbb{Z}_p)/(1+p^2\Bbb{Z}_p)\cong \Bbb{F}_p\times \Bbb{F}_p$$

For $p=2$ it is the same except that $(1+2\Bbb{Z}_2)^2=1+8\Bbb{Z}_2$ so $\Bbb{Q}_2^\times/(\Bbb{Q}_2^\times)^2\cong \Bbb{F}_2\times \Bbb{F}_2\times \Bbb{F}_2$.

reuns
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