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Let p be 1 mod 3 (separate question: work out 2 mod 3). What is the group structure of the abelian group $\mathbb{Q}_p ^* / \mathbb{Q}_p ^{*3}$?

$\mathbb{Q}_p ^*$ refers to the group of units in $\mathbb{Q}_p$, and $\mathbb{Q}_p ^{*3}$ is the group of units cubed.

I'm not really sure where to start with this. I believe the units in $\mathbb{Q}_p$ are just the non-zero elements. I can't seem to get a handle on $\mathbb{Q}_p ^{*3}$ though. Help and pointers would be much appreciated!

user76754
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    @vadim123, not quite sure what you're asking, but I'll do my best to answer. $\mathbb{Q}_p={ \alpha / \beta \colon \alpha, \beta\in\mathbb{Z}_p}$, so $\mathbb{Q}_p ^*$ should be the same, but with $\alpha _0 \neq 0$, so p doesn't divide $\alpha$. – user76754 May 09 '13 at 03:05
  • Thanks for pointing that out, hah. Any pointers for $\mathbb{Q}_p ^{*3}$? – user76754 May 09 '13 at 03:11
  • @vadim123, pardon me if I'm being thick, but I don't see how that question helps with this one. – user76754 May 09 '13 at 03:34
  • Dear user, This is essentially answered (in a more general setting) here: http://math.stackexchange.com/a/17767/221 Regards, – Matt E May 09 '13 at 03:38

1 Answers1

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The units in $\mathbb Q_p$ are the non-zero elements because $\mathbb Q_p$ is a field. To compute the structure of $(\mathbb Q_p^\times)^3$, we start by computing the structure of $\mathbb Q_p^\times$. Here I will assume $p > 2$ (and later $p > 3$) as these are particular cases.

We first may separate the elements of $\mathbb Q_p^\times$ by their valuation: each element $x \in \mathbb Q_p^\times$ may be written as $x = p^{v_p(x)} y$ with $y \in \mathbb Z_p^{\times}$, so that $$ \mathbb Q_p^\times = p^{\mathbb Z} \times \mathbb Z_p^\times.$$ Now let $y \in \mathbb Z_p^{\times}$; if $y \equiv y' \pmod{p}$ then $z = y/y' \equiv 1 \pmod{p}$, so that we may write $y = y' z$ with $z \in 1 + p \mathbb Z_p$ and $y'$ belongs to a set of representatives of all the non-zero classes (modulo $p$). Since $\mathbb F_p^\times$ is a cyclic group of order $p-1$ (this is where $p > 2$ is needed), such a set is the set of $p-1$-th roots of unity (which is also the image of $\mathbb F_p^\times$ by the Teichmüller lift), generated by a primitive root $\omega$: $$ \mathbb Z_p^\times = \omega^{\mathbb Z/(p-1)\mathbb Z} \times (1 + p \mathbb Z_p),$$ as a group product.

Finally, (this is again where $p > 2$ is needed), the logarithm map is a group isomorphism between $1 + p \mathbb Z_p$ and $p \mathbb Z_p$, so that if $u$ generates (additively and topologically) $\mathbb Z_p$, then $\exp (pu)$ generates $1 + p \mathbb Z_p$; equivalently, if $v_p(x-1) = 1$, then $x$ generates $1 + p \mathbb Z_p$. Combining all of these together, we find $$ \mathbb Q_p^\times = p^{\mathbb Z} \times \omega^{\mathbb Z/(p-1)\mathbb Z} \times x^{\mathbb Z_p}.$$ For example, for $p = 5$, we find $$ \mathbb Q_5^\times = 5^{\mathbb Z} \times \sqrt{-1}^{\mathbb Z/4\mathbb Z} \times (-4)^{\mathbb Z_5}.$$ In this group, we immediately find that the cubes are the elements of $$ (\mathbb{Q}_5^\times)^3 = 5^{\mathbb 3 Z} \times 1 \times (-4)^{\mathbb Z_5}$$ (since $3 \mathbb Z_5 = \mathbb Z_5$), which implies $Q_5^\times/(Q_5^\times)^3 \simeq 5^{\mathbb Z/3\mathbb Z}$ (i.e. it is a cyclic group of order $3$, generated by $5$).

Circonflexe
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  • How does $ \mathbb Q_5^\times = 5^{\mathbb Z} \times \sqrt{-1}^{\mathbb Z/4\mathbb Z} \times (-4)^{\mathbb Z_5}$ imply $ (\mathbb{Q}_5^\times)^3 = 5^{\mathbb 3 Z} \times 1 \times (-4)^{\mathbb Z_5}$ ? what about the term $\sqrt{-1}^{\mathbb{Z}/4 \mathbb{Z}}$ ? – MAS Jul 31 '19 at 10:34