Let $$f(X) \in \mathbb{Q} $$ such that $$f(X)= X^{15}-X^8-X^7+1=0$$ Determine splitting field over $\mathbb{Q}$ of $f(X)$
I know that i have to find roots of f but I have trouble.
$$\small f(X)=(x^{8}-1)(x^7-1) =(x^4-1)(x^4+1)(x^7-1) =(x^2-1)(x^2+1)(x^4+1)(x^6+x^5+x^4+x^3+x^2+x+1)(x-1) =(x-1)^2(x+1)(x^2+1)(x^4+1)(x^6+x^5+x^4+x^3+x^2+x+1)$$
So root are $\pm i, \pm 1, \pm \sqrt[4]{-1}$ and $\sqrt[7]{-1} $
So now the splitting field over $\mathbb {Q}$ is $\mathbb {Q}(i,\sqrt[4]{-1},\sqrt[7]{-1})$ ?
Hint.- A useful factorization of your polynomial is $$f(x)=(x^8-1)(x^7-1)$$ $$x^8-1=(x^2-1)(x^2+1)(x^2-\sqrt2 x+1)(x^2+\sqrt2 x+1)$$ $$x^7-1=(x-1)(x^6+x^5+x^4+x^3+x^2+x+1)$$ so the corresponding roots are $$\pm1,\pm i,\frac{\sqrt2(1\pm\sqrt i)}{2},\frac{\sqrt2(-1\pm\sqrt i)}{2}$$ and $$1,\cos\frac{2k\pi}{7}+i\sin\frac{2k\pi}{7}; k=1,2,3,4,5,6$$
from which your splitting field $K|\mathbb Q$ simplifying carefully, when elements are not $\mathbb Q$-linearly independent (for example i and -i and remember that $x^7-1$ defines an easy cyclotomic field of degree $6$ over $\mathbb Q$).