MHB Prove: Q(ζp)=Q(ζip) | Cyclotomic Polynomials

[mathjam0990]

Let ζ_p be e^2πi/p. For an integer i, such that p does not divide i, prove Q(ζ_p) = Q(ζⁱ_p ).

I think this has something to do with both exponents of ζ_p (1 and i) being coprime to p, but I am not sure at all how to show the equality. If anyone could please help with an explanation, that would be great. Thank you.

 

[Opalg]

Let ζ_p be e^2πi/p. For an integer i, such that p does not divide i, prove Q(ζ_p) = Q(ζⁱ_p ).

I think this has something to do with both exponents of ζ_p (1 and i) being coprime to p, but I am not sure at all how to show the equality. If anyone could please help with an explanation, that would be great. Thank you.

Some caution is needed here, because $i$ is doing double duty, as an exponent and as the square root of $-1$. Just don't be tempted to write $\zeta_p^i$ as $(e^{2\pi i/p})^i.$

It should be clear that $\mathbb{Q}(\zeta_p^i) \subseteq \mathbb{Q}(\zeta_p)$ (because $\zeta_p^i \in \mathbb{Q}(\zeta_p)$). To prove the reverse inclusion you need to show that $\zeta_p \in \mathbb{Q}(\zeta_p^i)$. This will follow if you can show that $\zeta_p$ is a power of $\zeta_p^i$. That is where $i$ and $p$ being coprime will come in.

 

[mathjam0990]

To prove the reverse inclusion you need to show that $\zeta_p \in \mathbb{Q}(\zeta_p^i)$. This will follow if you can show that $\zeta_p$ is a power of $\zeta_p^i$. That is where $i$ and $p$ being coprime will come in.

Thank you for your response. I apologize, I should have mentioned that I knew the proof would be to show each is contained in the other. But I have no clue how to show $\zeta_p \in \mathbb{Q}(\zeta_p^i)$. Could you please explain how I would do that and I can respond via "reply with quote" if I happen to get lost? Thank you.

 

[Deveno]

The $p$-th roots of unity form a cyclic group of order $p$, under complex multiplication. The elements are, from your definition of $\zeta_p$:

$\{(\zeta_p)^k: k = 0,1,2,\dots,p-1\}$ (so $\zeta_p$ is a generator). The link between "$p$-th roots of unity" and an abstract cyclic group of order $p$ (or to be more specific: $\Bbb Z/p\Bbb Z$) is provided by the complex exponential (which lies beneath DeMoivre's formula, as well).

A (semi-) elementary result of group theory is that for a cyclic group $\langle a\rangle$, of order $n$, the order of $a^k$ is:

$\dfrac{n}{\gcd(k,n)}$

In particular, for $n = p$, if $\gcd(k,p) = 1$, then $a^k$ has order $p$ (I am using $k$ instead of your $i$ for the very reasons Opalg outlined in his first post).

 

[Opalg]

A possibly more simple-minded approach is to say that we want to find $r$ such that $(\zeta_p^k)^r = \zeta_p$. But $\zeta_p^k$ is a $p$th root of unity, so $(\zeta_p^p)^s = 1$ (for any integer $s$).

Therefore it would be good to find integers $r,s$ such that $kr = ps+1$, because then $(\zeta_p^k)^r = (\zeta_p^p)^s(\zeta_p)^1 = \zeta_p$. That ought to point you towards ideas connected with Euclid's algorithm.