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Let $U_n = 1 + p^n \mathbb{Z}_p = \{1+p^n x \mid x \in \mathbb{Z}_p\}$ for $n \in \mathbb{Z}_{\geq 1}$, where $U_0 = \mathbb{Z}_p^{*}$. I have two questions: I managed to show that $U_n$are subgroups of $\mathbb{Z}_p$. Can I immediately conclude that they are subgroups of $\mathbb{Z}_p^{*}$ ? Furthermore, I have no idea how to verify that

$\mathbb{Z}_p^{*} = \langle \omega \rangle \times U_1$, where $\omega$ is a primitive $(p-1)$th root of unity.

I would be very glad if someone could help. Any advice is greatly appreciated.

Crystal
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    $Z_p$ is local ring with maximal ideal $(p)$. So $U_n$ are always outside $(p)$. This shows $U_n$ are units of $Z_p$. To see $Z_p^\star$ isomorphic to that product. Consider $0\to U_1\to Z_p^\star\to (Z/p)^\star\to 0$. Now show there is a section $(Z/p)^\star\to Z_p^\star$.(Note that this is non-canonical splitting and elements of $Z/p^\star$ are coprime to $p$ once you have lifted to $Z$.) – user45765 Oct 21 '18 at 14:11
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    $U_n$ is a group under multiplication, while $\mathbb{Z}_p$ is a group under addition, so I'm not sure how you showed that $U_n$ is a subgroup of $\mathbb{Z}_p$... – Viktor Vaughn Oct 21 '18 at 15:23
  • At first $U_n$ is a subgroup of the multiplicative monoid $\mathbb{Z}_p^$, and $\mathbb{Z}_p^\times$ is the largest subgroup $ \subset \mathbb{Z}_p^$. @user45765 Do you use Hensel's lemma or a p-adic series to construct $\omega$ or what ? – reuns Oct 21 '18 at 22:10
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    @reuns $\omega$ is normally constructed by hensel via $x^{p-1}=1$ separable over $Z_p$ and admitting solutions. Then this lifts to $Z_p^\star$'s torsion part. p-adic series is using hensel without explicitly saying so. I do not know other ways to achieve this end. – user45765 Oct 21 '18 at 22:21
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    Thanks for all these comments that guided me in the right direction. Thanks to the comment of André 3000 I now see that my attempt to prove that $U_n$ are subgroups of $\mathbb{Z}_p$ is of course wrong. Nevertheless, I learn from my mistakes. – Crystal Oct 22 '18 at 22:09

2 Answers2

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@reuns

Here is a proof of $\mathbf Z^*_p = <\omega> \times U_1$ without Hensel's lemma (of course it's academic, because why shouldn't we use Hensel's lemma) ?

Algebraic lemma: Let $0\to A\to B\to C\to 0$ be a short exact sequence of finite abelian groups (in additive notation) s.t. the orders $a$ and $c$ of $A$ and $C$ resp. are coprime. Let $C'$ be the subgroup of $B$ consisting of all the elements killed by $c$. Then $B=A \times C'$, the projection $C' \to C$ is an isomorphism, and $C'$ is unique.

Proof. Everything follows from Bézout's theorem, which states the existence of $m, n \in \mathbf Z$ s.t. $ma+nc=1$. Since $(a,c)=1$, it is obvious that $A \cap C'=(0)$. Moreover, any $x \in B$ can be written $x=max + ncx$, which gives the direct product decomposition. If $B=A \times C''$, then clearly $C''$ is killed by $c$, so $C"$ is contained in $C'$, and finally $C'' = C'$. QED

Let us come back to the subgroups $U_n=1+p^n \mathbf Z_p$. Clearly $U_0/U_1 \cong \mathbf F^*_p$ and, for $n\ge 1$, the map $1+p^ny \to y$ mod $p$ induces an isomorphism $U_n/U_{n+1} \cong (\mathbf Z/p , +)$, and $U_1/U_n$ has order $p^{n-1}$ by induction. Applying the lemma to the exact sequence $1 \to U_1/U_n \to U_0/U_n \to \mathbf F^*_p \to 1$ for all $n\ge 1$, we get isomorphisms $U_0/U_n \cong U_0/U_{n-1}$ as well as induced isomorphisms $(U_n)' \cong (U_{n-1})'$. Taking the projective limit on $n$, we obtain $U_0=\varprojlim U_0/U_n$, hence the desired decomposition $\mathbf Z^*_p = <\omega> \times U_1$. NB: The uniqueness statement in the lemma was necessary to conclude about $\varprojlim (U_n)'$.

nguyen quang do
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  • Do you mean $U_1/U_n \cong U_0/U_n$ in the 4th line from bottom ? Indeed, I want know the order of $U_0/U_n$. – MAS Aug 11 '21 at 16:33
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A commenter has already explained why the $U_n$ are contained in $\mathbb{Z}_p$ using the fact the latter is a local ring. If you are familiar with the representation of elements $x$ of $\mathbb{Z}_p$ by infinite series $x=\sum_{n\geq 0}b_np^n$ with $0\leq b_n\leq p-1$, and multiplication in $\mathbb{Z}_p$ is given by the Cauchy product, from which it follows that $U_n$ is a subgroup.

Stefan Dawydiak
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