Let $p=n^2+1$ be prime, $a=\sqrt{p+\sqrt{p}},b=\sqrt{p-\sqrt{p}}$ and ${\rm Irr}\left(a,\mathbb{Q}\right)=x^4-2px^2+p\left(p-1\right).$
How do I check if $\;b\in \mathbb{Q}(a)$?
I tried by claiming that if $b\in \mathbb{Q}(a)$, then $b$ could be written as a linear format of $\mathbb{Q}(a)$, using its base, which is $\{1,a,a^2,a^3\}$.
That really got me nowhere or at least nowhere somewhat ''elegant''.
Is there another way?
Some of my guesses are using $b^{-1}$ or maybe $ab$, since we know a for sure belongs in $\mathbb{Q}(a)$.
One way is to show $b = a^3/n-a(p+1)/n$ by simplifying the expression on the right side of the equation. Another way is realize that the four roots of $f(x) = x^4-2px^2+p(p-1)$ are $\pm a$, $\pm b$. Combine this with the fact that the Galois group of $f(x)$ is cyclic of order 4 to deduce that $\mathbb{Q}(a)$ contains all four roots of $f(x)$, hence contains $b$. Note: Galois groups of polynomials of the form $x^4+ax^2+b$ are straightforward to determine; using Galois group of a biquadratic quartic for example.