Minimal polynomial for $\mathbb{Q}(\sqrt{5}+\sqrt{3})$ over $\mathbb{Q}(\sqrt{10})$ I would like to find the minimal polynomial for $\mathbb{Q}(\sqrt{5}+\sqrt{3})$ over $\mathbb{Q}(\sqrt{10})$. Here is my thought process so far:
I know that the minimal polynomial for $\mathbb{Q}(\sqrt{5}+\sqrt{3})$ over $\mathbb{Q}$ is $x^4-16x^2+4$ (I think). I also know that $\mathbb{Q}(\sqrt{5}+\sqrt{3})=\mathbb{Q}(\sqrt{5},\sqrt{3})$ and that the only intermediate fields of the latter are $\mathbb{Q}(\sqrt{A})$ for $A=3,5,15$. Now $\sqrt{10}$ is not among any of these so the extension must be of degree 4. Then we can conclude (by uniqueness of the minimal polynomial), that $x^4-16x^2+4$ is the minimal polynomial of $\mathbb{Q}(\sqrt{5}+\sqrt{3})$ over $\mathbb{Q}(\sqrt{10})$.
Is my reasoning correct?
 A: You are correct that if $\sqrt 5 + \sqrt 3$ still has degree $4$ over $\mathbb Q(\sqrt {10})$ then the minimal polynomial will not change. However, I don't think you have very clearly justified it.
As you observe, $\sqrt {10}$ is not in  $\mathbb Q(\sqrt 3 + \sqrt 5)$ since otherwise $\mathbb Q(\sqrt {10})$ would equal one of those quadratic subfields and you know it does not. As a result, $x^2 - 10$ remains irreducible over $\mathbb Q(\sqrt 3 + \sqrt 5)$, so you get a tower of total degree $8$ by adding it in. Then you can reexamine this tower as $\mathbb Q(\sqrt 3 + \sqrt 5,\sqrt{10})/\mathbb Q(\sqrt {10})$ and see from the multiplicativity of degrees in towers that this degree is $4$, and that is what you want.
Maybe I am just being a little too picky, but there are cases where your logic would lead to a false conclusion, e.g. if you had a different degree $4$ extension (instead of $\mathbb Q(\sqrt{10})$ which was not in $\mathbb Q(\sqrt 3 + \sqrt 5)$ but overlapped in a quadratic extension. Then the degree would fall.
