Let $y,z \in \mathbb{C}$ with $y^4 = 5$ and $z^6 = 15$. I want to show that $y \notin \mathbb{Q}(z)$.

So we have

$$ y = w_1 \cdot \sqrt[4]5 \;\;\;\;\;\;\; z = w_2 \cdot \sqrt[6]{15} $$

with $w_1$ and $w_2$ as 4th and 6th roots of unity respectively. With the cyclotomic polynomial we can find the primitive roots of unity, so that we have for example

$$ y = i^n \cdot \sqrt[4]5 \;\;\;\;\;\;\; z = ({\sqrt[3]{-1}})^m \cdot \sqrt[6]{15} $$

with $1 \leq n \leq 4$ and $1 \leq m \leq 6$.

Now, I'm not sure how to proceed. Can I just conclude that $i, -1, -i, 1$ (the 4th roots of unity) are independend from $({\sqrt[3]{-1}})^m$? I think I'm missing something here. Thank you for your help in advance.


$X^4-5$ and $X^6-15$ are irreducible (Eisenstein).

We have $[\mathbb{Q}(z):\mathbb{Q}]=6$ and $[\mathbb{Q}(y):\mathbb{Q}]=4$ suppose $y\in\mathbb{Q}(z)$, this implies that $[\mathbb{Q}(z):\mathbb{Q}]= [\mathbb{Q}(z):\mathbb{Q}(y)] [\mathbb{Q}(y):\mathbb{Q}]$ but $4$ does not divides 6.

  • $\begingroup$ Wow, thank you. $\endgroup$ – matt Jan 11 at 0:55

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