Find a polynomial with splitting field $\mathbb{Q}(\sqrt[5]{3},\zeta)$ over $\mathbb{Q}$, where $\zeta$ is primitive 15-th root of unity.

Question

I thought of $$f(x)=(x^5-3)(x^8-x^7+x^5-x^4+x^3-x+1)$$

since the second polynomial of the above product is the 15-th cyclotomic polynomial and therefore is irreducible over $\mathbb{Q}$ and all of its roots are powers of $\zeta$. But when I write down the roots of $f$ I think that the splitting field of $f$ is $$\mathbb{Q}(\sqrt[5]{3},\zeta,i)$$ and I cannot see why would $i \in \mathbb{Q}(\sqrt[5]{3},\zeta)$, so that my claim would hold.

Thank you for any help!

Answer 1

Your answer is correct. The roots of the cyclotomic polynomial are the $15$-th primitive roots which are all powers of $\zeta$. The roots of $x^5-3$ all have the form $\sqrt[5]{3}$ times a $5$-th root of unity. But all $5$-th roots of unity are powers of $\zeta$. So your polynomial indeed splits in $\mathbb{Q}(\sqrt[5]{3},\zeta)$, and since its roots generate the extension $\mathbb{Q}(\sqrt[5]{3},\zeta)/\mathbb{Q}$ we know this is a splitting field.