Let $\omega$ be a $\textbf{root of unity}$ with algebraic degree(degree of its minimal polynomial over $\mathbb{Q}$) $d_1$ over $\mathbb{Q}$ and $r$ be a $\textbf{real number}$ with algebraic degree $d_2$ over $\mathbb{Q}$. Can $r\omega$ have algebraic degree $= d_3$ strictly less than $d_1$ over $\mathbb{Q}$? Is there an explicit non-trivial lower bound for $d_3$ in terms of $d_1$ and $d_2$?

Note that if both multiplicands were real numbers then decrease in algebraic degree over $\mathbb{Q}$ is possible, example: $a, b = \sqrt{2}$. Similarly, if both multiplicands were roots of unity then decrease in algebraic degree over $\mathbb{Q}$ is possible, example: $a, b = i$.

  • $\begingroup$ I misread and thought you wanted $d_3<d_2$. I have a vague recollection of somebody settling that question here, but I couldn't find it :-( $\endgroup$ – Jyrki Lahtonen Oct 8 '18 at 16:18
  • $\begingroup$ @IvanNeretin I believe $rw$ still has degree $4$. $\endgroup$ – nikhil_vyas Oct 8 '18 at 16:21
  • $\begingroup$ @JyrkiLahtonen that would be interesting in its own right. $\endgroup$ – nikhil_vyas Oct 8 '18 at 16:22
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    $\begingroup$ Not ruling out the possibility that the question I had in mind is the one where we collaborated :-( $\endgroup$ – Jyrki Lahtonen Oct 8 '18 at 16:24
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    $\begingroup$ @nikhil_vyas I thought $\omega=\frac{1+i}{\sqrt2}$, so $r\omega=1+i$ has degree $2$ over $\mathbb Q$? Maybe I miss something again? $\endgroup$ – awllower Oct 8 '18 at 16:27

Well, $\omega=e^{2\pi i/8}={1+i\over\sqrt2}$ and $r=\sqrt2$ will do. This gives $d_1=4,\;d_2=2,\;d_3=2<d_1$.

On the other hand, I don't see an easy way to make $d_3<{1\over2}d_1$. Maybe that's the lower bound you are after, though I can't be sure at the moment.

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  • $\begingroup$ Yeah $d_3 > cd_1$ for some constant $c$ would suffice for my application. $\endgroup$ – nikhil_vyas Oct 8 '18 at 20:59

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