Is it true that $\mathbb{Q}(\sqrt{3}+\sqrt{5})= \mathbb{Q}(\sqrt{3+\sqrt{5}})$

Question

I am asked to show that the splitting field of $X^4-6X^2+4$ is $\mathbb{Q}(\sqrt{3}+\sqrt{5})$. By solving the quartic polynomial, I showed that the splitting field of this polynomial is in fact $\mathbb{Q}(\sqrt{3+\sqrt{5}})$. So my question is if these two fields are equal to each other and if so how.

Answer 1

To summarize the comments, let me mention that the usual exercise is to find the splitting field of $X^4-16X^2+4$. It is given by $\Bbb Q(\sqrt{3},\sqrt{5})=\Bbb Q(\sqrt{3}+\sqrt{5})$, see here:

Determine the minimal polynomial of $\sqrt 3+\sqrt 5$.

However, if really $X^4-6X^2+4$ was meant, then the splitting field is $\Bbb Q(\sqrt{3+\sqrt{5}})$. It is easy to see that $\sqrt{2}\in \Bbb Q(\sqrt{3+\sqrt{5}})$, but $\sqrt{2}$ is not contained in $\Bbb Q(\sqrt{3},\sqrt{5})$ - see here:

How to show that $\sqrt{2}$ is not in $\mathbb Q(\sqrt{3},\sqrt{5})$?

So the fields given in the title are not equal.