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After Hensel's Lemma there is the following proposition in my notes:

If $p$ is a prime and $m \in \mathbb{N}$ then there is a primitive $m^{th}$ root of unity in $\mathbb{Q}_p$ $\Leftrightarrow m \mid (p-1)$.

To prove this proposition we begin as follows:

$(\Rightarrow) $ Let $a$ be a primitive $m^{th}$ root of unity in $\mathbb{Q}_p$. That means that $m$ is the smallest natural number such that $$a^m \equiv 1 \pmod p$$ (correct?)

Can we suppose that $m$ and $p$ are coprime, and use Fermat's little theorem we have that $a^{p-1} \equiv 1 \pmod p$?

Can we conclude from this that $m \mid (p-1)$?

How do we prove the other direction? Perhaps using Hensel's Lemma?

Adam Hughes
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Mary Star
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    The other direction can be proven using Hensel (see Adam's answer). As professor Lubin commented here for all $a=1,2,3,\ldots,p-1$, the sequence $(a^{p^n})$ is Cauchy and converges towards a root of unity of order $|(p-1)$. Furthermore, those limits are distinct for they are distinct modulo $p$. This gives another constructive proof for the existence of those roots of unity. – Jyrki Lahtonen Jan 29 '15 at 22:53
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    Your first claim is incorrect for $p=2$, and also for odd $p$ I think it is not obvious, rather it would make up most of the proof for "$\Rightarrow$". One needs to show (at least) that the only roots of unity contained in $1+p\Bbb Z_p$ are $1$ for odd $p$ resp. $\pm 1$ for $p=2$. This is also missing from the accepted answer, see my comments there. – Torsten Schoeneberg Apr 05 '19 at 19:14

1 Answers1

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The easiest way to prove this is as follows:

You want to ask if the polynomial

$$x^m-1$$

has a root in $\Bbb Q_p$. We know if you pick one such primitive root, $\zeta$ then we can get all of the others by examining $1,\zeta,\zeta^2,\ldots, \zeta^{m-1}$.

Now, reducing this polynomial modulo $p$ we see that we want to know if there are $m^{th}$ roots of $1$ in $\Bbb F_p$, however we know by the classification of finite fields that the elements of $\Bbb F_p$ are exactly those that satisfy $x^p-x=0$. Now, since roots of unity are not $0$, we see all non-zero elements of $\Bbb F_p$ satisfy $x^{p-1}-1$. More completely we have the result from the structure of finite fields

If $F$ is a finite field of order $q$, then $F^\times$ is a cyclic group of order $q-1$.

From this we see that for each divisor $d|(q-1)$, we may take a multiplicative generator of $F^\times$, say we call it $g$, and note that $g^{(q-1)/d}$ is an element of order $d$ in the group, i.e. it is a $d^{th}$ root of $1$. And if $\zeta\in F$ is an $m^{th}$ root of one, then the multiplicative group $\langle \zeta\rangle\le F^\times$ must have order dividing the order of the group, i.e. $m|(q-1)$. So it is necessary and sufficient that $m|(q-1)$ for a finite field, $F$, of order $q$ to possess the $m^{th}$ roots of $1$.

Now, in our case the residue field is just $\Bbb F_p$, so we see it has all $m^{th}$ roots of $1$ for $m|(p-1)$ and no others. This implies immediately that if $m\not\mid (q-1)$, then $\Bbb Q_p$ does not possess $m^{th}$ roots of $1$, since if we had them in the bigger field, quotienting to the residue field would still keep them.

To see the other direction, we note that if $m|(p-1)$, in particular $m$ is coprime to $p$, hence since

$${d\over dx}(x^m-1)=mx^{m-1}$$

and $|m|_p=1$ so we satisfy the hypotheses of Hensel's lemma, and get a lifting back up to $\Bbb Q_p$. This completes the proof.

Adam Hughes
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  • Great!! Thank you very much!!! :-) – Mary Star Feb 14 '15 at 18:59
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    I think your argument about non-existence of other roots of unity is incomplete. Namely, you just use that the residue field is $\Bbb F_p$, but well, the (valuation ring of the) local field $K:=\Bbb Q_p (\mu_p)$ also has $\Bbb F_p$ as residue field, and $K$ obviously does contain more roots of unity. The issue is, I believe, in the last line of your argument for that direction: "quotienting to the residue field" would "keep" those roots, it just sends them to $1$. If by "keep" you mean it would respect their multiplicative order, well that exactly would need justification by more structure. – Torsten Schoeneberg Apr 05 '19 at 17:01
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    Compare my comment to the question, as well as the answer https://math.stackexchange.com/a/295693/96384, where at least it's noted that the case $p \vert m$ needs special treatment, and the answer https://math.stackexchange.com/a/49283/96384, which explicitly states the need to prove injectivity of the reduction map on roots of unity, and which before edited due to the comment by KCd made a wrong claim about that, similar to the one in the OP and here. – Torsten Schoeneberg Apr 05 '19 at 19:21
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    @TorstenSchoeneberg I'm not sure I see the substance to your objection--it probably is worth mentioning in passing, but it hardly seems to be as serious as you seem to think it is, far more basic than Hensel's lemma. If you are curious you simply use the Eisenstein criterion on $\Phi_p(1+x)$ to readily get irreducibility of the cyclotomic polynomial. – Adam Hughes Apr 05 '19 at 22:17
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    If you had mentioned it, even in passing, I would not be nagging, but now I still am. Let me put it this way: Of the four exercise steps in the second link above, your answer mentions exactly one (no. 3). We could agree that no. 1 is trivial. I am aware that no. 4 can be done with Eisenstein or more elementary methods, but maybe the OP is not. One still needs to fill in no. 2, injectivity of reduction on roots whose order is coprime to $p$. How do you show that? With Hensel's lemma maybe, which you do seem to find worth mentioning? – Torsten Schoeneberg Apr 06 '19 at 05:29
  • Surjectivity on finite sets is equivalent to a bijection when they have the same cardinality. I can agree that if we're being pedantic adding a note about eisenstein is fine, but even professional proofs don't spell out everything from basic set theory, fevers etiquette is to know your audience and know the context. I'm not certain where you get four exercises from, can you elaborate on that? All I see is one statement to be proven. – Adam Hughes Apr 11 '19 at 02:03
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    The four steps ("exercises") I was talking about are in this answer: https://math.stackexchange.com/a/49283/96384. I still do not see how you prove no. 2. Saying that the sets have the same cardinality is equivalent to the claim and not obvious either. – Torsten Schoeneberg Apr 11 '19 at 19:23