<s>[INST] How to show two field extensions are equal. How do I go about showing that $\mathbb{Q}(\sqrt{3}+\sqrt{5})=\mathbb{Q}(\sqrt{3},\sqrt{5})$?
Since $\mathbb{Q}(\sqrt{3}+\sqrt{5}) \subset\mathbb{Q}(\sqrt{3},\sqrt{5})$, we need to show the reverse inclusion. Does it suffice to prove that $\sqrt{2} \in \mathbb{Q}(\sqrt{3}+\sqrt{5})$?
Thank you
 [/INST] We have
$$\frac{(\sqrt{3} + \sqrt{5})^3 - 14(\sqrt{3} + \sqrt{5})}{4} = \frac{18\sqrt{3} + 14\sqrt{5} - 14\sqrt{3} -14 \sqrt{5}}{4} = \frac{4\sqrt{3}}{4} = \sqrt{3}$$
So $\sqrt{3} \in \mathbb{Q}(\sqrt{3} + \sqrt{5})$, and hence so is $\sqrt{5}$.

The general trick in these situations is to write one of the irrational numbers as a polynomial in their sum, with coefficients from $\mathbb{Q}$.
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