Splitting field of a polynomial $f(x) =(x^2-3)(x^2-5)(x^5-1)$ over $\mathbb{Q}$.

Question

I was considering the splitting field E of the polynomial $f(x) =(x^2-3)(x^2-5)(x^5-1)$ over $\mathbb{Q}$.

I expected $E=\mathbb{Q}(\sqrt{5},\sqrt{3},\omega)$, where $\omega=e^{\frac{2\pi i}{5}}$.

But I saw a textbook that claimed $E=\mathbb{Q}(\sqrt{3},\omega)$; believing that $\pm\sqrt{5}\in \mathbb{Q}(\omega)$.

Can someone please show me how the latter answer is true if it is actually true? Thanks

Answer 1

Note that $$\begin{align}\frac{x^5-1}{x-1}&=x^4+x^3+x^2+x+1=x^2\,\left(t^2+t-1\right) \\&=x^2\,\left(t-\frac{-1+\sqrt{5}}{2}\right)\left(t-\frac{-1-\sqrt{5}}{2}\right)\,,\end{align}$$ where $t:=x+\dfrac{1}{x}$. Therefore, $\sqrt{5}$ is already in the field $\mathbb{Q}(\omega)$.