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Question is to prove that

For any prime $p$ and any positive integer $m$ not divisible by $p$, there exists a primitive $m$-th root of unity if and only if $m$ divides $p-1$.

Let $\zeta=a_0+a_1p+a_2p^2+\cdots$ be a primitive $m$ th root of unity. We then have $$\sqrt[m]{1}=a_0+a_1p+a_2p^2+\cdots$$ Considering m th power and going modulo $p$ we see that $a_0^m\equiv 1\mod p$.

As $a^{p-1}\equiv 1\mod p$ for all non zero $a$, we can find $t<p-1$ such that $a_0^t\equiv 1\mod p$.

Suppose $(t,p-1)=1$ then we have $a_0\equiv 1\mod p$.

I have two questions to solve.

Does $a_0\equiv 1\mod p$ imply $\zeta=1$?

Does $a_0^t\equiv 1\mod p$ imply $\zeta^t=1$?

Positive answer for second question implies positive answer for first question.

If it is the case then i am done as i have $\zeta^t=1$, contradicting the condition that $\zeta$ is a primitive $m$ th root of unity.

The condition $a_0^t\equiv 1 \mod p$ with Hensel's lemma implies $\alpha^t=1$ for some $\alpha=a_0+b_1p+b_2p^2+\cdots$. This does not immediately say that $\zeta^t=1$.

Any hints are welcome.

In some sense, i am trying to prove/disprove that

Given $f(X)=X^n+c_1X^{n-1}+\cdots+c_n$ and $\alpha=a_0+a_1p+a_2p^2+\cdots$ and $\beta=a_0+b_1p+b_2p^2+\cdots$ we have $f(\alpha)=0$ iff $f(\beta)=0$

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  • $\begingroup$ $$x^m\equiv1\pmod p$$ Using [Discrete Logarithm][1] wrt to primitive root $g$, $m$ind$_gx\equiv0\pmod{p-1}$ $\implies$ ind$_gx=0\pmod{\dfrac{p-1}{(m,p-1)}}$ [1]: mathworld.wolfram.com/DiscreteLogarithm.html $\endgroup$ Commented Dec 26, 2016 at 6:12
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    $\begingroup$ @labbhattacharjee I do not understand your comment.. what is your conclusion? $\endgroup$
    – user87543
    Commented Dec 26, 2016 at 6:15
  • $\begingroup$ Do you want to show that there is a primitive $m$th root of unity in $\Bbb Z_p$ or in $\Bbb Q_p$? It looks like you're only working on the case of $\Bbb Z_p$. $\endgroup$
    – Stahl
    Commented Dec 26, 2016 at 7:00
  • $\begingroup$ @Stahl I forgot to say that primitive roots of unity are totally inside p adic integers as their valuation is zero... So, it is sufficient to look for roots of unity In p adic integers $\endgroup$
    – user87543
    Commented Dec 26, 2016 at 7:03
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    $\begingroup$ I'm not 100% sure how I'd remedy the argument you're making, since there are a few holes (you're not looking at existence at all, and you're not considering the case $(m,p - 1)\neq 1$ but $m > p - 1$) and it's not super clear how to go from $t$ to $m$ to me (why must $t < m$? what assumptions are you exactly making about $m$?). However, this, specifically theorem 3.1, should be helpful. Hensel's lemma is the key here. $\endgroup$
    – Stahl
    Commented Dec 26, 2016 at 7:24

1 Answer 1

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First, recall Hensel's lemma:

Theorem: Let $a\in\Bbb Z_p$. If $f(X)\in\Bbb Z_p[X]$, $f(a) = 0\pmod{p}$ and $f'(a)\not\equiv 0\pmod{p}$, then there exists a unique $\alpha\in\Bbb Z_p$ such that $f(\alpha) = 0$ and $\alpha\equiv a\pmod{p}$.

This might not be quite the version you're used to, but it's essentially the same, except we're starting with a $p$-adic number and polynomial with $p$-adic coefficients. Existence and uniqueness will both be important for us in the following.

Remark: This version of Hensel's lemma answers both of your questions in the affirmative. If you have $\zeta = a_0 + a_1 p + \dots$ with $\zeta^m = 1$ in $\Bbb Z_p$ ($p\not\mid m$), and $a_0 = 1$, then $\zeta = 1$, since $1$ is already an element of $\Bbb Z_p$ satisfying $X^m - 1 = 0$, and thus uniqueness implies $\zeta = 1$. The answer to your second question is again yes by uniqueness, because $\zeta^t$ still satisfies $X^m - 1 = 0$ (and so does $1$) and $\zeta^t\equiv 1\pmod{p}$.

Here is the way I'd probably make the argument.

Lemma: There exists a primitive $p-1$st root of unity $\zeta\in\Bbb Z_p^\times$.

Note that this proves the existence portion of your question, because if $m>0$ divides $p - 1$, then an appropriate power of $\zeta$ will be a primitive $m$th root of unity.

Proof: Consider the polynomial $X^{p - 1} - 1\in\Bbb Z_p[X]$, and the elements $1,2,\dots, p - 1\in\Bbb Z_p$. Each of these elements satisfies $a^{p - 1} - 1\equiv 0\pmod{p}$. Moreover, $-a^{p - 2}\not\equiv 0\pmod{p}$, so for each of the elements we considered we obtain an $\alpha\equiv a\pmod{p}$ with $\alpha^{p - 1} - 1 = 0$. So we have found $p - 1$ $p - 1$st roots of unity. To see that one is primitive, recall that a finite subgroup of the multiplicative group of a field is cyclic, and the elements in $\Bbb Q_p^\times$ satisfying $X^{p - 1} - 1 = 0$ form a subgroup. Those elements are precisely the $p - 1$ $\alpha$'s we found, because a degree $n$ polynomial over a field can have at most $n$ roots.

Now, we must show that if $m$ does not divide $p - 1$, there is not a primitive $m$th root of unity.

Lemma: Let $m > 0$ be a positive integer such that $p\not\mid m$. Suppose that $\alpha\in\Bbb Z_p^\times$ satisfies $\alpha^m = 1$. Then $\alpha$ satisfies $\alpha^n = 1$ for some $n\mid p - 1$.

Proof: (adapted from the notes in my above comment) Suppose $\zeta\in\Bbb Z_p^\times$ satisfies $\zeta^m = 1$. Then $\zeta\not\equiv 0\pmod{p}$, so there exists some $p - 1$st root of unity $\alpha$ with $\alpha\equiv\zeta\pmod{p}$ by our argument for the last lemma. Then $\alpha$ and $\zeta$ both satisfy $f(X) = X^{m(p - 1)} - 1 = 0$. (Note that $f'(\alpha),f'(\zeta)\not\equiv 0\pmod{p}$.) But now we have two elements of $\Bbb Z_p$ satisfying the same polynomial $f\in\Bbb Z_p[X]$ which are equivalent modulo $p$, so by uniqueness in Hensel's lemma, $\zeta = \alpha$. Thus, $\zeta$ is a $p-1$st root of unity, so it is a primitive $n$th root only for some $n$ dividing $p - 1$.

Note that we did not need to assume $\alpha$ is a primitive $m$th root for the last argument. The lemma holds for any $m$th root, and will thus hold for a primitive $m$th root.

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  • $\begingroup$ I was not thinking about the uniqueness portion of Hensel's lemma... Thanks for the answer... :) It is clear now $\endgroup$
    – user87543
    Commented Dec 26, 2016 at 17:40

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