Let $\mathbb{Q}_p$ be the field of $p$-adic numbers. Let $\alpha$ be a $p$-adic unit, i.e. an invertible element of the multiplicative monoid $\mathbb{Z}_p$. Consider the set $S = \{n \in \mathbb{Z}, n \gt 0\mid x^n = \alpha$ has a solution in $\mathbb{Q}_p\}$. Is $S$ an infinite set?
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Certainly not unless $|\alpha|_p=p^{nk}$ for some $k\in\mathbb Z$. – Alex Becker Jul 23 '13 at 02:47
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2@AlexBecker $\alpha$ is a $p$-adic unit, which means that $|\alpha|_p = 1$. Regards. – Makoto Kato Jul 23 '13 at 02:51
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Ah, missed that, sorry. – Alex Becker Jul 23 '13 at 03:04
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This is not my domain, but I am somewhat surprised that in the context of $\Bbb Q_p$, the term "unit" means neither (1) neutral element for multiplication, nor (2) invertible element (which would just be "nonzero" in $\Bbb Q_p$), but (3) element of absolute value$~1$. Long live ambiguity! By the way, if this is so, why not just talk about $p$-adic integers (where meaning (3) coincides with (2))? Clearly the $n$-th roots asked for are (unit) $p$-adic integers too. – Marc van Leeuwen Aug 02 '13 at 06:40
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@MarcvanLeeuwen It is customary in algebraic number theory that a unit of $\mathbb{Q}p$ means an invertible element of the multiplicative monoid of the ring $\mathbb{Z}_p$. In a ring theory, a unit of a ring $R$ means an invertible element of the multiplicative monoid of $R$. https://en.wikipedia.org/wiki/Unit(ring_theory) – Makoto Kato Aug 02 '13 at 07:04
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@MakotoKato: I knew the usual meaning of unit in ring theory; this is (2) of my comment. However the multipliciative moniod of $\Bbb Q_p$ is $\Bbb Q_p\setminus{0}$, as in any field, and all its elements are invertible. It is the implicit switch from $\Bbb Q_p$ to $\Bbb Z_p$ that surprises me. – Marc van Leeuwen Aug 02 '13 at 07:15
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@MarcvanLeeuwen It is also customary in algebraic number theory that a unit of an algebraic number field $K$ means an invertible element of the multiplicative monoid of the ring of algebraic integers in $K$. https://en.wikipedia.org/wiki/Dirichlet%27s_unit_theorem – Makoto Kato Aug 02 '13 at 07:24
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@MarcvanLeeuwen QUOTE:A $p$-adic number is called unit if it is not a multiple of a negative power of $p$ and its first digit is nonzero. p.5 A Tutorial on $p$-adic Arithmetic http://www.cs.ucsb.edu/~koc/docs/r09.pdf – Makoto Kato Aug 02 '13 at 07:30
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Of course, the term $p$-adic unit is abuse of terminology. In a correct terminology, it is an invertible element of the multiplicative monoid of the ring of $p$-adic integers or a unit of the ring of $p$-adic integers. But I rarely heard anyone said so. http://en.wikipedia.org/wiki/Abuse_of_notation – Makoto Kato Aug 02 '13 at 08:03
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By Hensel's lemma, if $p\nmid n$ and $x^n\equiv \alpha\mod{p}$ has a solution $x\in\left(\mathbb{Z}/p\mathbb{Z}\right)^\times$, then $x^n=\alpha$ has a solution $x\in\mathbb{Z}_p$. Now, $\left(\mathbb{Z}/p\mathbb{Z}\right)^\times$ is cyclic of order $p-1$, so for any $n$ with $(n,p-1)=1$, $x^n\equiv\alpha\mod{p}$ has a solution $x\in\left(\mathbb{Z}/p\mathbb{Z}\right)^\times$. It follows that $S$ contains all $n>0$ with $(n,p(p-1))=1$, so $S$ is infinite.
Minseon Shin
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Julian Rosen
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Nice! I was going to attempt an answer, but I misread the question and thought it was asking about the set $R$ consisting of all roots of $\alpha$. I think $R$ is finite iff $\alpha$ is a root of unity. I'm just not sure how to prove it without using the $p$-adic logarithm, a tool which I've glimpsed but not studied. Do you happen to know of a more elementary proof of this statement? – Chris Culter Jul 23 '13 at 04:38