<s>[INST] Why does $F(\sqrt{a+b+2\sqrt{ab}}) = F(\sqrt{a},\sqrt{b})$? Let $F$ be a field of characteristic $\neq 2$. Let $a \neq b \in F$, and $F(\sqrt{a},\sqrt{b})$ is of degree 4 over $F$. I've shown that $F(\sqrt{a}+\sqrt{b}) = F(\sqrt{a},\sqrt{b})$.
Observe that $x=\sqrt{a} + \sqrt{b}$ and $x = \sqrt{a+b+2\sqrt{ab}}$ both satisfy $p(x) = x^4 - 2(a+b)x^2 + (a-b)^2 = 0$. The problem is to conclude that $F(\sqrt{a+b+2\sqrt{ab}}) = F(\sqrt{a},\sqrt{b})$.
I can see that LHS $\cong$ RHS, but cannot see why they are literally the same field. For one thing, it's not clear to me that $\sqrt{a}, \sqrt{b} \in$ LHS.
 [/INST] As you and @D_S showed, $F(\sqrt{a+b+2\sqrt ab})=F(\sqrt a+\sqrt b)$ since they have the same minimal polynomial, so it suffices to show that $F(\sqrt a+\sqrt b)=F(\sqrt a, \sqrt b)$, which can be shown by showing that $\sqrt a \in F(\sqrt a+\sqrt b)$ (showing that $\sqrt b \in F(\sqrt a+\sqrt b)$ is very similar and showing that $\sqrt a+\sqrt b \in F(\sqrt a, \sqrt b)$ is trivial).
We start with $\sqrt a+\sqrt b$. The reciprocal of this element is $\frac{\sqrt a-\sqrt b}{a^2-b^2}$. We can multiply this by the scalar $a^2-b^2$ to get $\sqrt a-\sqrt b$ and then add this with our original $\sqrt a+\sqrt b$ to get $2\sqrt a$. Finally, divide by the scalar of $2$ to get $\sqrt a$.
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