show that $\mathbb{Q}(\sqrt[n]{2}) \neq \mathbb{Q}(\sqrt[n]{3})$ I would like to show that $\mathbb{Q}(\sqrt[n]{2}) \neq \mathbb{Q}(\sqrt[n]{3})$ for an even $n$.
I was hinted that I should use the following fact (which I already proved):
If $L/\mathbb{Q}$ finite field extension, and $A = L \cap \bar{\mathbb{Z}}$ (where $\bar{\mathbb{Z}}$ is the algebraic integer ring, or $O_L$) and $B \subseteq A$ sub ring with ${\rm Frac}(B) = L$ (${\rm Frac}(B)$ is the fraction field), then $n^2\cdot \Delta_{A/\mathbb{Q}}= \Delta_{B/\mathbb{Q}}$ where $\Delta_{A/\mathbb{Q}}$ is the discriminant of $A/\mathbb{Q}$ and $n \in \mathbb{Z}_{>0}$.
Thanks ahead.
 A: Look at the discriminant of the ring $\mathbb Z[\sqrt[n]{2}]$. By definition of discriminant this is $D_2=\prod_{i\neq j}(\sqrt[n]{2}\cdot \xi^i-\sqrt[n]{2}\cdot \xi^j)^2$, where $\xi$ is a primitive $n$-th root of unity. You can convince yourself very easily that this is $2^{n-1}\prod_{i\neq j}(\xi^i-\xi^j)^2$, and that $\prod_{i\neq j}(\xi^i-\xi^j)^2$ is an integer $d$, because it is an algebraic integer and it is fixed by $\text{Gal}(\overline{\mathbb Q}/\mathbb Q)$. In the very same way, the discriminant of $\mathbb Z[\sqrt[n]{3}]$ is $3^{n-1}\cdot d$. Now assume by contradiction that the two fields coincides. Then their ring of integers $\mathcal O$ coincide, and you can use the statement you quoted: since $\mathbb Z[\sqrt[n]{2}]\otimes \mathbb Q=\mathbb Q(\sqrt[n]{2})$, and the same for $\mathbb Z[\sqrt[n]{3}]$, then you must have that $\text{disc}(\mathcal O)=\frac{2^{n-1}\cdot d}{x^2}$ for some non-zero integer $x$, but it also needs to be $\frac{3^{n-1}\cdot d}{y^2}$ for some non-zero integer $y$. Now the fact that $n$ is even allows you to conclude immediately.
