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Sorry if the title is not descriptive, I couldn't find a way to properly describe the problem.

In the proof of the AKS algorithm (link), the following is stated in Lemma 4.7:

"First note that since $h(X)$ is a factor of the cyclotomic polynomial $Q_r(X)$, $X$ is a primitive $r^{\text{th}}$ of unity in $F$."

I do not understand why $X$ is a primitive $r^{\text{th}}$ of unity in $F$. I can see that it is a root of unity in $F$, as $H(X)$ is a factor of $Q_r(X)$, which is in turn a factor of $X^r-1$. Hence $X^r \equiv 1 \pmod{p,H(X)}$. However, it is not clear to me why $X^k\not\equiv 1\pmod{p,H(X)}$ for all $0< k < r$.

Similar to the original AKS paper, Granville (link) states on the bottom of page 17 that

"(...), $\mathbb{F}$ contains $x$, an element of order $r$"

and on page 19, in the proof of Lemma 4.3:

"It can be shown that $x$ has order $r$ in $\mathbb{F}$ (...)"

Some background on notation:

  • $p$ is a prime. $r$ is an integer less than $p$.
  • $F$ and $\mathbb{F}$ refer to the finite field $(\mathbb{Z}/p\mathbb{Z})[X]/(h(X))$, where $h(X)\in (\mathbb{Z}/p\mathbb{Z})[X]$ is an irreducible factor of the cyclotomic polynomial $Q_r(X)$.
  • Granville uses lower capital '$x$' to denote the indeterminate in the polynomial ring.

Thanks in advance for any help.

  • By definition, the roots of the cyclotomic polynomial $Q_r$ are exactly the $r$-th primitive roots of unity. See https://en.wikipedia.org/wiki/Cyclotomic_polynomial – Dark Magician Oct 02 '23 at 15:50
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    If $p$ is a prime, and $\gcd(r,p)=1$ then the roots of $Q_r(x)$ are still roots of unity of order $r$. See for example here (the argument there became a bit kludgy - I have touched the theme many times in various guises, and that just seems to be the closest match). Observe that the gcd assumption is crucial. If $F$ is a field of characteristic $p$, it can never have roots of unity of order $p$, hence multiples of $p$ are also excluded. – Jyrki Lahtonen Oct 02 '23 at 16:24
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    A summary. Not sure it settles your questions. Feel free to comment. – Jyrki Lahtonen Oct 02 '23 at 16:55

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