Minimal polynomials of real algebraic multiples of roots of unity.

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 $r$?

• 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? – Jyrki Lahtonen Apr 29 '17 at 5:22
• @JyrkiLahtonen edited the question. – nikhil_vyas Apr 29 '17 at 5:23
• No that would be interesting but I am looking for the solution the general case. – nikhil_vyas Apr 29 '17 at 5:23
• 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. – ancientmathematician Apr 29 '17 at 6:37
• 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 $12$. – 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 $r\omega_n$.

• 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. – Jyrki Lahtonen Apr 29 '17 at 12:56
• Since the exact answer seems to be hard to characterise I have added an additional question. – 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

• $r = \omega_p$ 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. – 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.

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