Find the minimal polynomial of $\alpha=(1-\omega)/2$ over $\mathbb{Q}$ where $\omega$ is a primitive 8th root of unity

Question

What I though about this question was that $(1-\omega)/2 = 1/2 -\omega/2$ so $\mathbb{Q}[\alpha] = \mathbb{Q}[\omega]$ over $\mathbb{Q}$

Thus its minimal polynomial of $\alpha$ over $\mathbb{Q}$ is just the minimal polynomial of a primitive 8th root of unity..

Is it correct?

Thanks.

Answer 1

No, the minimal polynomials are different. What your argument implies is they have the same degree.

As $\omega^8=1$, you can deduce that the minimal polynomial of $\omega$ is $x^4+1=0$. Now by taking $\omega=1-2\alpha$ you see $(1-2\alpha)^4+1=0$. Then you should be able to determine the minimal polynomial of $\alpha$.