I'm trying to prove that if $\mathbb Q(\alpha )/\mathbb Q$ and $\mathbb Q(\beta )/\mathbb Q$ have same degree, then $\mathbb Q(\alpha )$ and $\mathbb Q(\beta )$ are isomorphic. The natural map is of course $\alpha \mapsto \beta $ which is clearly a field isomorphism.
But what would happen in the case where for example instead of having $\mathbb Q(\alpha )$ we would have $\mathbb Q(\alpha ,\gamma )$ ? For example, how can I prove that $\mathbb Q(i,\sqrt 2)$ and $\mathbb Q(\sqrt[4]2)$ are isomorphic ? the problem is that $i\sqrt 2$ is an element of order $2$ where as $\sqrt[4]2$ has order $4$. Then I tried $i+\sqrt 2\mapsto \sqrt[4]2$, but I can't prove that it's an isomorphism. Any idea ?
Your statement is not true. Just take $\alpha =\sqrt 2$ and $\beta =\sqrt 3$. First of all $$\varphi :\sqrt 2\mapsto \sqrt 3,$$ is even not a ring homomorphism. Indeed, if it would be, then $$2=2\varphi (1)=\varphi (\sqrt 2^2)=\varphi (\sqrt 2)^2=\sqrt 3^2=3,$$ which is of course not correct. Moreover, suppose there is a homomorphism ring $\psi:\mathbb Q(\sqrt 2)\longrightarrow \mathbb Q(\sqrt 3)$, then it would be uniquely determinated by $\psi(\sqrt 2)$. So If there are $a,b\in\mathbb Q$ s.t. $$\psi(\sqrt 2)=a+b\sqrt 3,$$ then $$2=\psi(\sqrt 2)^2=a^2+b^2+2ab\sqrt 3,$$ and thus $ab=0$ and $a^2+b^2=2$. The only solutions are $(0,\pm\sqrt 2)$ and $(\pm\sqrt 2,0)$ that are not in $\mathbb Q^2$. Therefore, there is no such homomorphism.
Remark : However, the statement is true for finite field since there are unique up to isomorphism. I.e. if $K$ is a finite field, and $K(\alpha )/K$ and $K(\beta )/K$ have same degree extension, then $K(\alpha )\cong K(\beta )$.