# If $y^4 = 5$ and $z^6 = 15$, then $y \notin \mathbb{Q}(z)$

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.