$\mathbb Q$-basis of $\mathbb Q(\sqrt[3] 7, \sqrt[5] 3)$.

Question

Can someone explain how I can find such a basis ? I computed that the degree of $[\mathbb Q(\sqrt[3] 7, \sqrt[5] 3):\mathbb Q] = 15$. Does this help ?

Answer 1

Try first to find the degree of the extension over $\mathbb Q$. You know that $\mathbb Q(\sqrt[3]{7})$ and $\mathbb Q(\sqrt[5]{3})$ are subfields with minimal polynomials $x^3 - 7$ and $x^5-3$ which are both Eisenstein.

Therefore those subfields have degree $3$ and $5$ respectively and thus $3$ and $5$ divide $[\mathbb Q(\sqrt[3]7,\sqrt[5]3) : \mathbb Q]$, which means $15$ divides it. But you know that the set $\{ \sqrt[3]7^i \sqrt[5]3^j \, | \, 0 \le i \le 2, 0 \le j \le 4 \}$ spans $\mathbb Q(\sqrt[3]7, \sqrt[5]3)$ as a $\mathbb Q$ vector space. I am letting you fill in the blanks.

Hope that helps,