For roots of unity, the minimal polynomial is given by cyclotomic polynomials. Can we extend this to real algebraic multiples of roots of unity? For $r\omega_p$ I think there must be some way to combine the minimal polynomial of $r$ and minimal polynomial of $\omega_p$ to get the minimal polynomial of $r\omega_p$?

The conjugates of $r\omega_p$ will be the subset of $\{r'\omega : r'\ \text{is a conjugate of}\ r\ \text{and}\ \omega^p = 1\}$. But I could not find a counterexample or proof of whether it is exactly this set or not. We can get a polynomial with this set as roots by Resultant. But how do we check whether it is minimal or not? As @ancientmathematician and @Jyrki show this is not always the minimal set. How to find the minimal set?

Additional question: This proves that degree of $r\omega_p \leq$ degree of $r*\phi(p)$. Can we get an inequality for the other way? i.e. Can the degree of $r\omega_p$ be bounded below as a function of the degree of $r$?

  • $\begingroup$ I'm not sure it makes a difference to a general criterion (if one exists) of when this happens, but... are you primarily interested in the case where $p$ is a prime number? $\endgroup$ – Jyrki Lahtonen Apr 29 '17 at 5:22
  • $\begingroup$ @JyrkiLahtonen edited the question. $\endgroup$ – nikhil_vyas Apr 29 '17 at 5:23
  • $\begingroup$ No that would be interesting but I am looking for the solution the general case. $\endgroup$ – nikhil_vyas Apr 29 '17 at 5:23
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    $\begingroup$ Can you answer @JyrkiLahtonen 's question: is $p$ prime? For what it's worth,if $p$ is not prime examples like $p=8$, $r=\sqrt{2}$ show that some collapse is possible. $\endgroup$ – ancientmathematician Apr 29 '17 at 6:37
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    $\begingroup$ Just waking up. Surely the field of the $7$-th roots of unity contains an element $\zeta+\zeta^2+\zeta^4$ which is quadratic over $\mathbb{Q}$; if you take $r$ for " $\sqrt{b^2-4ac}$ " then $r\zeta$ will only have six conjugates and not the generic $12$. $\endgroup$ – ancientmathematician Apr 29 '17 at 6:48

If I understood it correctly you are asking whether the number of conjugates of $r\omega_p$ is always equal to $\phi(p)$ times the number of conjugates of $r$.

This is not necessarily the case. A simple example of that phenomenon is $r=\sqrt5$, $p=5$. This is because here $r\in K=\Bbb{Q}(\omega_5)$. More precisely, with $\omega_5=\cos(2\pi/5)+i\sin(2\pi/5)$ we have the well-known $$ \sqrt5=2(\omega_5+\omega_5^4)+1. $$ As $K$ is cyclic of degree four, the number $r\omega_5=\sqrt5\omega_5\in K$ has at most four conjugates. That is less than the predicted $2\cdot4=8$.

OTOH we have the following relatively general case, where we do get the predicted number of conjugates. Assume that the normal closure $L$ of $\Bbb{Q}(r)$ and the cyclotomic field $K_n=\Bbb{Q}(\omega_n)$ are linearly disjoint. Basically this means that if $L$ is the splitting field of the minimal polynomial of $r$, then we require that $L\cap K_n=\Bbb{Q}$ (two Galois extensions are linearly disjoint iff they intersect trivially).

Then the compositum of fields $F:=K_nL$ is also Galois over $\Bbb{Q}$, and the Galois group is a direct product of $Gal(K_n/\Bbb{Q})\simeq\Bbb{Z}_n^*$ and $Gal(L/\Bbb{Q})$. If $r'$ is any conjugate of $r$, and $\omega_n^k, \gcd(k,n)=1$ is any conjugate of $\omega_n$, then the direct product property implies that $r'\omega_n^k$ is a conjugate of $r\omega_n$.

  • $\begingroup$ I thought that whenever $Gal(L/\Bbb{Q})$ was not its own commutator subgroup it would follow (from Kronecker-Weber) that $L$ and $K_n$ are not linearly disjoint for some $n$. That is correct, but I also thought that this was sufficient to conclude that $r\omega_n$ would have a lower than predicted number of conjugates. This I'm no longer sure about, so I deleted that tangent. $\endgroup$ – Jyrki Lahtonen Apr 29 '17 at 12:56
  • $\begingroup$ Since the exact answer seems to be hard to characterise I have added an additional question. $\endgroup$ – nikhil_vyas May 4 '17 at 18:02

There need to be more hypotheses, as @Jyrki's clear answer shows.

In fact the easiest counterexample is, I suppose, to take $r=\omega_p$ itself!

However, you also ask how you could tell if the $\mathbb{Q}$ polynomial with roots $r' \omega$ splits. There is an algorithm to factorise any $\mathbb{Z}$-polynomial into irreducibles, so you could just use that. https://en.wikipedia.org/wiki/Factorization_of_polynomials

  • $\begingroup$ $r = \omega_p$ is not possible as $r$ is real while $\omega_p$ is a primitive root of unity. BTW since the exact answer seems to be hard to characterise I have added an additional question. $\endgroup$ – nikhil_vyas May 4 '17 at 18:02

Not in a straightforward way, since the root of an arbitrary real number will be (with probability one) transcendental. In particular, if $\alpha \in \mathbb{R}$ is transcendental, then we must also have (e.g.) $\sqrt{\alpha}$ transcendental; a reason being, the algebraic numbers are closed under multiplication, so if we had $\sqrt\alpha$ algebraic, then its square, $\alpha$, would be algebraic, too, which is a clear contradiction by assumption.

  • $\begingroup$ I meant $r$ to be an algebraic number, this is why I mentioned the minimal polynomial of $r$. I have updated the question. $\endgroup$ – nikhil_vyas Apr 29 '17 at 4:37

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