Decrease in algebraic degree on multiplying a real number with a root of unity

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: $$a, b = i$$.

• 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 :-( – Jyrki Lahtonen Oct 8 '18 at 16:18
• @IvanNeretin I believe $rw$ still has degree $4$. – nikhil_vyas Oct 8 '18 at 16:21
• @JyrkiLahtonen that would be interesting in its own right. – nikhil_vyas Oct 8 '18 at 16:22
• Not ruling out the possibility that the question I had in mind is the one where we collaborated :-( – Jyrki Lahtonen Oct 8 '18 at 16:24
• @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? – 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 $$d_1=4,\;d_2=2,\;d_3=2.
On the other hand, I don't see an easy way to make $$d_3<{1\over2}d_1$$. Maybe that's the lower bound you are after, though I can't be sure at the moment.
• Yeah $d_3 > cd_1$ for some constant $c$ would suffice for my application. – nikhil_vyas Oct 8 '18 at 20:59